Physics Galaxy 2020-21: Optics & Modern Physics - Vol. 4 [4, 2 ed.] 9788183558938

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Physics Galaxy Volume IV

Optics & Modern Physics Ashish Arora Mentor & Founder

PHYSICSGALAXY.COM

World's largest en(yclopedia ofonline video lectures on High School Physics

G K Publications (P) Ltd

CL MEDIA (P) LTD. First Edition 2000 Revised Edition 2017

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My beloved wife

My Parents, Son, Daughter

Dedicated to

In his teaching career since 1992 Ashish Arora personally mentored more than 10000 IITians and students who reached global heights in various careerand profession chosen. It is his helping attitude toward studentswith which all his studentsrentember him in life for his contribution in their success and keep connections with him live. Below is the list of some of the successful students in International Olympiad personally taught by him. NAVNEET LOrWAL

International GQLD Medal in IPbO-2000 at LONDON, Also secured AIR-4 in IIT JEE 2000 PROUD FOR INDIA ; Navneet Loiwal was the first Indian Student who won first International GOLD Medal

for our country In International Physics Olympiad. DUNGRA RAM CHOUDHARY

HARSHIT CHOPRA KUNTAL LOYA LUV KUMAR

RAJHANS SAMDANI.

SHANTANU BHARDWAJ

SHALEEN HARLAlKA

AIR-1 in IIT JEE 2002

National Gold Medal in INPbO-2002 and got AIR-2 in IIT JEE-2002 A Girl Student got position AIR-8 in IIT JEE 2002 National Gold Medal in INPbO-2003 and got AIR-3 in IIT JEE-2003 National Gold Medal in INPbO-2003 and got AIR-5 in IIT JEE-2003 International SILVER Medal in IPbO-2002 at INDONESIA

International GOLD Medal in IPb0T2003 at CHINA and got AIR-46 in IIT JEE-2003

TARUN GUPTA

National GOLD Medal in INPbO-2005

APEKSHA KHANDELWAL

National GOLD Medal in INPbO-2005

ABHINAV SINHA

Honble Mension Award in APbO-2006 at KAZAKHSTAN

RAMAN SHARMA

International GOLD Medal in IPbO-2007 at IRAN and got AIR-20 in IIT JEE-2007

PRATYUSH PANDEY

International SILVER Medal in IPhO-2007 at IRAN and got AIR-85 in IIT JEE-2007

GARVIT JUNIWAL

International GOLD Medal in IPbO-2008 at VIETNAM and got AIR-10 in IIT JEE-2008

ANKIT PARASHAR

National GOLD Medal in INPbO-2008

HEMANT NOVAL ABHISHEK MITRUKA SARTHAK KALANI

National GOLD Medal in INPbO-2008 and got AIR-2S in IIT JE&-2008 National GOLD Medal in INPbO-2009 National GOLD Medal in INPbO-2009

ASTHA AGARWAL

International SILVER Medal in IJSO-2009 at AZERBAIJAN

RAHUL GURNANI

International SILVER Medal in IJSO-2009 at AZERBAIJAN

AYUSH SINGHAL

International SILVER Medal in IJSO-2009 at AZERBAIJAN

MEHUL KUMAR

ABHIROOP BHATNAGAR AYUSH SHARMA

International SILVER Medal in IPhO-2010 at CROATIA and got AIR-19 in IIT JEE-2010 National GOLD Medal in INPbO-2010

International Double GOLD Medal in IJSO-2010 at NIGERIA

AASTHA AGRAWAL

Honble Mension Award in APbO-2011 at ISRAEL and got AIR-93 in IIT JEE 2011

ABHISHEK BANSAL

National GOLD Medal in INPbO-2011

^ SAMYAK DAGA

National GOLD Medal in INPbO-2011

SHREY GOYAL

National GOLD Medal in INPbO-2012 and secured AIR-24 in IIT JEE 2012

RAHUL GURNANI

National GOLD Medal in INPbO-2012

JASPREET SINGH JHEETA

National GOLD Medal in INPbO-2012

DIVYANSHU MUND

SHESHANSH AGARWAL

National GOLD Medal in INPbO-2012 International SILVER Medal in IAO-2012 at KOREA

SWATI GUPTA

International SILVER Medal in IJSO-2012 at IRAN

PRATVUSH RAJPUT

' International SILVER Medal in IJSO-2012 at IRAN

SHESHANSH AGARWAL

International BRONZE Medal in IOAA-2013 at GREECE

SHESHANSH AGARWAL

International GOLD Medal in IOAA-2014 at ROMANIA

SHESHANSH AGARWAL

International SILVER Medal in IPbO-2015 at INDIA and secured AIR-58 in JEE(AdTanced)-20I5

VIDUSHI VARSHNEY

International SILVER Medal in IJSO-2015 to be held at SOUTH KOREA

AlvIAN BANSAL

AIR-1 in JEE Advanced 2016

KUNAL GOYAL

AIR-3 in JEE Advanced 2016

GOURAV DIDWANIA DIVYANSH GARG

AIR-9 in JEE Advanced 2016 International SILVER Medal in IPbO-2016 at SWITZERLAND

ABOUT THE AUTHOR

The complexities ofPhysics have given nightmares to many, but the homegrown genius ofJaipur, Ashish Arora has helped several students to live their dreams by decoding it. Newton Law ofGravitation and Faraday's Magnetic force ofattraction applyperfectlywell with this unassuming genius. APied Piper ofstudents, his webportal https;//www.physicsgalaxy.com. The world slargest encyclopedia ofvideo lectures on high school Physics possesses strong gravitational pull and magnetic attraction for students who want to make itbig inlife.

Ashish Arora, gifted with rare ability to train masterminds, has mentored over 10,000 IITians in his past 24 years ofteaching sojourn including lots ofstudents made it toTop 100 in nT-JEE/JEE(Advance)includingAIR-l andmanyinTop-lO. Apart from

that, he has also groomed hundreds ofstudents for cracking International Physics Olympiad. No wonder his student Navneet Loiwal brought laurel to the country bybecoming the first Indian to win aGold medal at the 2000 -International Physics Glvmniad in London (UK).

His special ability to simplify the toughest ofthe Physics theorems and applications rates him as one among the best Physics teachers in the world. With this, Arora simply defies the logic that perfection comes with age. Even at 18 when he started teaching Physics while pursuing engineering, he was as engaging as he is now. Experience, besides graying his hair, has just widened his horizon.

Now after encountering all tribes ofstudents - some brilliant and some not-so-intelligent -this celebrated teacher has embarked

upon anoble mission to make the entire galaxy ofPhysics inform ofhis webportal PHYSICSGALAXY.COM to serve and help global students in the subject. Today students from 221 countries are connected with this webportal. On any topic ofphysics students can post their queries in INTERACT tab ofthe webportal on which many global experts with Ashish Arora reply to several queries posted online by students.

Dedicated to global students ofmiddle and high school level, his website www.physicsgalaxy.com also has teaching sessions dubbed inAmerican accent and subtitles in 87 languages. For students in India preparing for JEE &NBET, his online courses will be available soon on PHYSICSGALAXY.COM.

FOREWORD

It has beenpleasurefor me to follow the progressEr.AshishArorahas made in teachingand professional career. In the last about two decades he has actively contributed in developing several new techniques for teaching & learning of Physics and driven important contribution to Science domain throughnurturingyoung students and budding scientists. Physics Galaxyis onesuch example ofnumerous efforts he has undertaken.

The 2nd edition of Physics Galaxyprovides a good coverage of various topics of Mechanics, Thermodynamics and Waves, Optics & Modem Physics and Electricity & Magnetism through dedicated volumes. It would be an important resource for students appearing in competitiveexaminationfor seeking admissionin engineering and medicalstreams. "E-version" of the book is also being launched to allow easy access to all.

The structure ofbook islogical and thepresentation is innovative. Importantly the book covers someofthe concepts onthe basis of realistic experiments and examples. The book has been written in an informal style to help students leara faster and more interactively with better diagrams and visual appeal of the content. Each chapter has variety of theoretical and numerical problems to test the knowledge acquired by students. The book also includes solution to all practice exercises with several new illustrations and problems for deeper leaming. 1am sure the book will widen the horizons of knowledge in Physics and will be found very useful by the students for developing in-depth understanding of the subject.

May 13, 2017 Prof. Sandeep Sancheti Ph. D. (U.K.). B.Tech. FIETE, MIEEE

President Manipal University Jaipur

PREFACE

For a science student, Physics is the most important subject, unlike to other subjects it requires logical reasoning and high imagination of brain. Without improvingthe level ofphysics it is very difficultto achievea goal in the present age of competitions. To score better, one does not require hard working at least in physics. It just requires a simple understanding and approach to think a physicalsituation.Actuallyphysicsis the surrounding of our everydaylife.All the six parts of generalphj^ics-Mechanics, Heat, Sound, Light, Electromagnetism and Modem Physics are the constituents of our surroundings. If you wish to make the concepts of physics strong, you should try to understand core concepts of physics in practical approach rather than theoretical. Whenever you try to solve a physics problem, first create a hypothetical approach rather than theoretical. Whenever you try to solve a physics problem, first create a hypothetical world in your imagination about the problem and try to think psychologically,

whatthe nextstepshould be, the bestanswer would begiven byyourbrainpsychology. Formaking physics strongin all respects and you should try to merge and understand all the concepts with the brain psychologically. The book PHYSICS GALAXY is designed in a totally different and fi"iendly approach to develop the physics concepts psychologically. The book is presented in four volumes, which covers almost all the core branches of general physics. First volume covers Mechanics. It is the most important part of physics. The things you will leam in this book will form a major foundation for understanding of other sections of physics as mechanics is used in all other branches of physics as a core fundamental. In this bookeverypart of mechanicsis explainedin a simpleand interactive experimental way. The bookis divided in seven major chapters, covering the complete kinematics and dynamics of bodieswith both translational and rotational motion. then gravitation and completefluid staticsand dynamics is coveredwith several applications.

The best way of understanding physics is the experimentsand this methodology I am using in my lectures and I found that it helps students a lot in concept visualization. In this book I have tried to translatethe things as I used in lectures. After every important section thereare several solved examples included withsimple andinteractive explanations. It mighthelpa student in a way that the student does not require to consult any thing with the teacher. Everything is self explanatory and in simple language.

Oneimportant factor in preparation ofph>^ics I wish to highlight thatmost ofthe student after reading thetheory ofa concept. start working out the numerical problems. This is not the efficient wayof developing concepts in brain. Toget the maximum benefit ofthebook students should readcarefully thewhole chapter at least three or four times with allthe illustrative examples andwithmorestress onsomeillustrative examples included in thechapter. Practice exercises included afterevery theorysection in each chapter is for the purpose of in-depth understanding of the applications of concepts covered. Illustrative examples are explaining some theoretical concept in theform ofanexample. After a thorough reading ofthechapter students can startthinking on discussion questionsand start working on numerical problems. Exercises given at the end of each chapter are for circulation of all the concepts in mind. There are two sections, first is the discussion questions, which aretheoretical andhelp inxmderstanding the concepts atrootlevel. Second section is ofconceptual MCQs which helpsin enhancing the theoretical thinkingofstudents and building logical skillsin the chapter. Third section of numerical MCQs helpsinthe developing scientific and analytical application of concepts. Fourth section ofadvance MCQs with one or more optionscorrecttypequestions is for developing advanceand comprehensive thoughts. Last sectionis the Unsolved Numerical Problemswhich includessome simpleproblems and some tough problemswhich require the building fundamentals of physicsfrombasics to advancelevelproblemswhich are usefulin preparationof NSEP, INPhOor IPhO.

In this second edition of the book I have included the solutions to all practiceexercises, conceptual, numerical and advance MCQs to support students who are dependent on theirselfstudy and not getting access to teachers for their preparation.

This book has taken a shape just because of motivational inspiration bymy'mother 20 years agowhen I just thought to write something for my students. She always motivated and was on my side whenever I thought to develop some new learning methodology for my students.

rdon't have words for mybest friend mywife Anuja for always being together with me to complete this book in the unique style and format.

I would like to pay my gratitude to Sh. Dayashankar Prajapati in assisting me to complete the task in Design Labs of PHYSICSGALAXY.COM and presenting the book in totallynew format ofsecond edition.

At last but the most important person, my father who has devoted his valuable time to finally present the book in such aformat and a simple language, thanks isa very small word for hisdedication inthis book.

In this second edition Ihave tried my best to make this book error free but owing to the nature ofwork, inadvertently, there is possibilityoferrors leftuntouched. Ishall be grateful to the readers, iftheypoint out me regarding errors and oblige me bygiving their valuable and constructive suggestions via emails for further improvement ofthe book.

Date:May,2017

Ashish Arora PHYSICSGALAXY.COM

B-80, Model Town, Malviya Nagar, Jaipur-302017 e-mails: [email protected] [email protected]

CONTENTS Atomic Physics

_

^

•

A

1.1 A Brief History to Atomic Physics 1.2 Thomson's Atomic Model

1.3 Ruthorford's Atomic Model

^

1.4 Bohr's Model ofanAtom

^

1.4.1 First Postulate

^

1.4.2 Second Postulate

1.4.3 ThirdPostulate

•

"

^

1.5 Properties ofElectron in Bohr's Atomic Model 1.5.1 Radius of nth Orbit in Bohr Model 1.5.2 Velocity ofElectron in nth Bohr's Orbit 1.5.3 Angular Velocity ofElectron in nth Bohr's Orbit 1.5.4 Frequency ofElectron in nth Bohr's Orbit 1.5.5 Time period ofElectron in nth Bohr's Orbit

^ ^ ^ ^ ^ ^

1.5.6. Current in nth Bohr's Orbit

^

1.5.7 Magnetic Induction at the Nucleus Due to nth Orbit 1.5.8 Magnetic Moment ofthe nth Bohr's Orbit

6 ^

1.5.9 Energy of Electron in nth Orbit

1.5.10 Energies ofDifferent Energy Level in Hydrogenic Atoms

7

1.6 Excitation and Jonization of an Atom 1.6.1 Frequency and Wavelength ofEmitted Radiation 1.6.2 Number ofLines Emitted During, de-excitation ofan Atom

^ 1' 12

1.7 The Hydrogen Spectrum

Chapter? ^

1.7.1 Spectral Series ofHydrogen Atom 1.8 Effect ofMass ofNucleus on Bohr Model 1.9 Use ofBohr Model to Define Hypothetical Atomic Energy Levels

12 21 25

1.10 Atomic Collisions

26

— — — —

DISCUSSION QUESTION CONCEPTUAL MCQs SINGLE OPTION CORRECT NUMERICAL MCQs SINGLE OPTIONS CORRECT ADVANCE MCQs WITH ONE OR MORE OPTIONS CORRECT

33 35 38 42

—

UNSOLVED NUMERICAL PROBLEMS FOR PREPARATION OF NSEP INPhO & IPhO

45

PhotoElectric Effect &MatterWaves

^

.

*^9-981

2.1 Electron Emission Processes 2.1.1 Thermionic Emission

2.1.2 Photoelectric Emission

2.1.3 Secondary Emission

30

2.1.4 Field Emission

30

2.2 Photoelectric Effect 2.2.1 Fundamental Laws of Photoelectric Effect 2.3 Experimental Study of Photo Electric Effect 2.3.1 Kinetic Energies of Electrons Reaching Anode 2.3.2 Reversed Potential Across Discharge Tube 2.3.3 Cut off Potential or StoppingPotential 2.3.4 Effect of Change in Frequency of Light on Stopping Potential

2.4 No. of Photon Emitted by Source Per second

31 51

35 36 57 57 58

62

2.5 Intensity of Light due to a Light Source 2.5.1 Photon Flux in a Light Beam 2.5.2 Photon Density in a Light Beam 2.6 Wave Particle Duality 2.6.1 Momentum of a Photon 2.7 De-Broglie's Hypothesis

2.7.1 Explanation of Bohr's Second Postulate 2.8 Radiation Pressure

2.8.1 Force Exerted by a Light Beam, on a Surface 2.8.2 Force Exerted on any Object in the Path of a LightBeam 2.8.3 Force Exerted by a Light Beam at Oblique Incidence

—

Chapters

62 63

63 70 70 70 71 72

72 72

7-3

2.8.4 Recoiling of an Atom Due to Electron Transition

75

2.8.5 Variation in Wavelength of Emitted Photon with State of Motion of an Atom

75

2.8.6 Variation in Wavelength of Photon During Reflection

75

DISCUSSION QUESTION

80

—

CONCEPTUAL MCQs SINGLE OPTION CORRECT

82

—

NUMERICAL MCQs SINGLE OPTIONS CORRECT

86

—

ADVANCEMCQs WITH ONE OR MORE OPTIONS CORRECT

92

—

UNSOLVED NUMERICAL PROBLEMSFOR PREPARATION OF NSEP, INPhO & IPhO

95

X-Rays 3.1 Introduction to X-Rays 3.1.1 Types of X-rays

3.2 Production Mechanism of X-rays 3.2.1 Continuous X-rays

3.2.2 Production of Continuous X-rays 3.2.3 Characteristic X-rays 3.2.4 Production of Characteristic X-rays 3.3 Moseley's Law

99-1161 100 100

100 100 100 102 102 103

3.4 Applications ofX-rays

103

—

DISCUSSION QUESTION

107

—

CONCEPTUAL MCQs SINGLE OPTION CORRECT

108

—

NUMERICAL MCQs SINGLE OPTIONS COI^CT

111

—

ADVANCE MCQs WITH ONE OR MORE OPTIONS CORRECT

113

—

UNSOLVED NUMERICAL PROBLEMS FOR PREPARATION OF NSEP. INPhO & IPhO

115

Nuclear Physics and Radioactivity 4.1 Composition and Structure of The Nucleus

117-1861 118

4.1.1 Size of a Nucleus

118

4.1.2 Strong Nuclear Force and Stability of Nucleus

118

4.2 Nuclear Binding Energy

119

4.2.1 Mass Energy Equivalence

124

4.2.2 Binding Energy Per Nucleon

124

4.2.3 Variation of Binding Energy per Nucleon with Mass Number

124

4.3 Radioactivity

127

4.3.1 Measurement of Radioactivity

127

4.3.2 Fundamental Laws of Radioactivity

128

4.3.3 Radioactive Decay Law

128

4.3.4 Half Life Time

129

4.3.5 Alternate form of Decay Equation in terms of Half Life Time

129

4.3.6 Mean Life Time

130

4.3.7 Calculation of Mean Life Time For a Radioactive Element

130

4.4 Radioactive Series

~

135

•

4.4.1 Radioactive Equilibrium

136

4.4.2 Simultaneous Decay Modes of a Radioactive Element 4.4.3 Accumulation of a Radioactive Element in Radioactive Series

136 136 140

4.5 Nuclear Reactions

4.5.1 Q-Value of Nuclear Reaction

•

•

•

'

'

,

141

141

4.6 Nuclear Fission

4.6.1 Fission of Uranium Isotopes and Chain Reaction

142

4.6.2 Liquid Drop Model

143

4.7 Nuclear Fusion

143

4.8 Properties of Radioactive Radiations

152

4.8.1 Alpha Decay

153

4.8.2 Beta Decay

153

4.8.3 Apparent Violation of Conservation Laws in b-decay

154

4.8.4 Pauli's Neutrino Hypothesis

155

4.8.5 Mass Defect Calculation For b-decay

155

4.8.6 Gamma Decay

156

—

DISCUSSION QUESTION

161

—

CONCEPTUAL MCQs SINGLE OPTION CORRECT

164

—

NUMERICAL MCQs SINGLE OPTIONS CORRECT

168.

—

ADVANCE MCQs WITH ONE OR MORE OPTIONS CORRECT

174

—

UNSOLVED NUMERICAL PROBLEMS FOR PREPARATION OF NSEP, INPhO & IPhO

177

187-3401

Geometrical Optics 5.1 Understanding a Light Ray and Light Beams

188

5.1.1 Different Types of Light Rays

188

5.1.2 Different Types of Light Beams

189

5.2 Reflection of Light

189

5.2.1 Regular or Specular Reflection

190

5.2.2 Irregular or Diffused Reflection

190

5.2.3 How we see an object in our surrounding 5.2.4 Laws of Reflection

, '

190

,

5.2.5 Vector Analysis of Laws of Reflection

5.3 Understanding Object and Image in Geometrical Optics

191

191 192

5.3.1 Object in Geometrical Optics

192

5.3.2 Image in Geometrical Optics

193

5.4 Reflection and Image formation by a Plane Mirror

5.5 Field of View for Image formed by a Plane Mirror

194 195

5.5.1 Field of View of an image

195

5.5.2 Field of View of a Mirror for an observer

195

5.6 Characteristics of Image formed by a Plane Mirror 5.6.1 Characteristic-} of Image formation by a Plane Mirror

196 196

5.6.2 Characteristic-2 of Image formation by a Plane Mirror

196

5.6.3 Characteristic-3 of Image formation by a Plane Mirror

197

5.6.4 Characteristic-4 of Image formation by a Plane Mirror

198

5.6.5 CharacleristicrS ofImage formation by a Plane Mirror

198

5.6.6 Characteristic-6 ofImage formation by a Plane Mirror

198

5.6.7 Characteristic-? ofImage formation by a Plane Mirror

198

5.6.8 Characteristics of Image formation by a Plane Mirror

199

5.6.9 Characteristic-9 ofImage formation by a Plane Mirror

200

5.6.10 Characteristic-10 of Image formation by a Plane Mirror

200

5.6.11 Characteristic-II of Image formation by a Plane Mirror

201

5.7 Understanding Shadow Formation

5.7.1 Umbra and Penumbra Regions 5.7.2 Antumbra Region

201 202 203

5.8 Spherical Mirrors

5.8.1 Standardterms related to Spherical Mirrors 5.8.2 Focal Length of a Spherical Mirror

5.8.3 Image Formation by a Spherical Mirror using Paraxial Rays 5.8.4 Standard Reflected Light Rays for Image Formation by Spherical Mirrors

5.8.5 Relation in focal length and Radius ofCurvature ofa Spherical Mirror 5.8.6 Image formation by Concave Mirrors 5.8.7 Image formation by Convex Mirrors

5.8.8 How an observer sees image ofan extended object in a spherical mirror 5.8.9 How image produced by a spherical mirror can be obtained on a screen 5.8.10 Sign Convention

5.9 Analysis of Image formation by Spherical Mirrors 5.9.1 Mirror Formula for Location of Image

5.9.2 Analyzing Nature ofImage Produced by a Spherical Mirror

5.9.3 Magnification Formula for Size and Orientation ofImage 5.9.4 Relation in Nature and Orientation of Image 5.9.5 Longitudinal Magnification of Image

5.9.6 Superficial Magnification by-a Spherical Mirror 5.9.7 Variation Curves ofImage Distance vj Object Distance 5.9.8 Effect ofMoving Object and Spherical Mirror on Image 5.9.9 Effect ofshifting Principal Axis ofa Mirror 5.9.10 Image formation ofdistant Objects by Spherical Mirrors

206

207 207 208 208

209 210 211 212 212 213 214 214 215 215 216 216

217 218 221 223 224 5.9.11 Concept ofReversibility ofLight 224 5.10 Refraction ofLight 226 5.10.1 Absolute Refractive Index ofa Medium 226 5.10.2 Relative Refractive Index ofa Medium 226 5.10.3 Laws ofRefraction 227 5.10.4 Vector form ofSnell's Law ofRefraction 227 5.10.5 Image Formation due to Refraction at a Plane Surface 228 5.10.6 An Object placed in a Denser Medium is seen from Air , 228 5.10.7 An Object placed in Air and seen from a Denser Medium 229 5.10.8 Shift ofimage due to Refraction ofLight by a GlassSlab 231 5.10.9 Shift due to Refraction ofLight by a Hollow thin walled Glass Box placed inside a Denser Medium 232 5.10.10 Lateral Displacement ofLight Ray by a Glass Slab 232 5.10.11 Lateral Displacement ofa Light Ray due to Refraction by Multiple Glass Slabs 5.10.12 Concept ofReflection by a Thick Mirror 5.11 Refraction ofLight by Spherical Surfaces 5.11.1 Analysis ofImage formation by Spherical Surfaces 5.11.2 Lateral Magnification ofImage by Refraction 5.11.3 Longitudinal Magnification of Image 5.11.4 Effect ofmotion ofObject or Refracting Surface on Image 5.12 Total Internal Reflection 5.12.1 Refraction ofLight Rays from a Source in a Denser Medium to Air 5.12.2 Cases ofGrazing Incidence ofLight on a Media Interface . 5.12.3 Refraction by a Transparent Medium of varying Refractive Index

233 233 239 240 243 243 243 247 248 249 249

5.12.4 Total Internal Reflection in a Medium of varying Refractive Index 5.12.5 Equation of Trajectory ofa Light Ray in a Medium of varying Refractive Index 5.13 Prism 5.13.1 Refraction ofLight through a Trihedral Prism • 5.13.2 Deviation Produced by a Small Angled Prism 5.13.3 Maximum Deviation ofLight Ray by a Prism 5.13.4 Condition of a Light Ray to pass through a Prism

250 250 254 255 256 256 257

5.14 Thin Lenses

5.14.1 Converging and Diverging Behaviour ofLenses 5.14.2 Primary and Secondary Focus ofa Lens 5.14.3 Standard Reflected Light Rays for Image Formation by Thin Lenses

264 264 265

5.14.4 Image Formation by Convex Lenses

5.14.5 Image formation by Concave Lenses 5.14.6 Focal length ofa thin lens 5.14.7 Focal length ofdifferent types ofstandard thin lenses 5.15 Analysis ofImage Formation by Thin Lenses 5.15.1 Lateral Magnification in Image Formation by a Thin Lens 5.15.2 Longitudinal Magnification by a Thin Lens

267 268 268 269 269 270

5.15.3 Variation Curves ofImage Distance vs Object Distance for a Thin Lens 5.15.4 Effect ofmotion ofObject and Lens on Image 5.16 Optical Power ofa Thin Lens ora Spherical Mirror 5.16.1 Combination of Thin Lenses 5.16.2 Combination of Thin Lenses and Mirrors 5.16.3 Deviation in a Light Ray due to Refraction through a Thin Lens 5.16.4 Combination of Two Thin Lenses at some Separation 5.16.5 Multiple images produced by a Lens made up ofdifferent materials 5.17 Lens andMirrors submerged in a Transparent Medium 5.18 Displacement Method Experiment to measure focal length ofa Convex Lens

270 271

278

5.19.3 Dispersive Power ofa Prism Material 5.19.4 Dispersion Analysis for a Small Angled Prism

279 280 283 283 284 285 285 286 286 291 292 292 293 293

5.19.5 Achromatic Prism Combination 5.19.6 Direct Vision Prism Combination

294 294

5.18.1 Condition offormation of Real Image by a Thin Convex Lens 5.18.2 Displacement Method Experiment 5.19 Dispersion ofLight 5.19.1 Dispersion of White Light through a Glass Slab 5.19.2 Dispersion of White Light through a Glass Prism

5.20 Optical Aberrations in Lenses and Mirrors 5.20.1 Spherical Aberrations

295 295

5.20.2 Methods to Reduce Spherical Aberrations

295

5.20.3 Chromatic Aberration in a Lens 5.20.4 Achromatic Combination of Lenses

296 297

5.21 Optical Instruments 5.21.1 The Human Eye

201 201

5.21.2 Camera

202

5.21.3 Angular Size ofObjects and Images 5.21.4 Simple Microscope 5.21.5 Magnification ofSimple Microscope 5.21.6 Compound Microscope 5.21.7 Magnifying Power of Compound Microscope 5.21.8 Tube Length of a Compound Microscope 5.21.9 Refracting Astronomical Telescope

202 202 303 203 304 304 204

— — — —

5.21.10 Magnifying Power of a Refracting Telescope 5.21.11 Tube Length of a Refracting Telescope 5.21.12 Reflecting Telescope 5.21.13 Terrestrial Telescope 5.21.14 Galilean Telescope DISCUSSION QUESTION CONCEPTUAL.MCQs SINGLE OPTION CORRECT NUMERICAL MCQs SINGLE OPTIONS CORRECT ADVANCE MCQs WITH ONE OR MORE OPTIONS CORRECT

305 305 306 206 306 311 312 319 325

—

UNSOLVED NUMERICAL PROBLEMS FOR PREPARATION OF NSEP, INPhO & IPhO

330

Wave Optics

^

6.1 Wave Theory 6.1.1 Dual Nature of Light 6.1.2 Wavefront of a Light Wave 6.1.3 Huygen's Wave Theory 6.2 Interference ofLight 6.2.1 Coherent Sources ofLight and Condition of Coherence 6.2.2 Theory ofInterference of Two Waves 6.2.3 Interference of two Coherent Waves of Same Amplitude 6.2.4 Intensity of Light at the Point ofInterference 6.2.5 Condition ofPath Difference for Interference 6.3 Young's Double Slit Experiment (YDSE) 6.3.1 Analysis of Interference Pattern in YDSE 6.3.2 Position of Bright and DarkFringes in YDSE Interference Pattern 6.3.3 Light Intensity on Screen in YDSE Setup 6.3.4 Fringe Width in YDSE Interference Pattern

'

6.4 Modifications in YDSE Setup • ' 6.4.1 Effect of Changing the direction of Incident Light in YDSE 6.4.2 Effect of Submerging YDSE Setup in a Transparent Medium 6.4.3 Path difference between two parallel due to a denser medium inpath of one beam 6.4.4 Effect ofPlacing a Thin Transparent Film infront of one of the slits in YDSE Setup 6.4.5 Concept of z-value in Interference Pattern of YDSE 6.4.6 Use of White Light in YDSE 6.4.7 Effect of Changing Slit Width in YDSE Setup 6.4.8 Fresnel's Biprism as a Limiting case of YDSE 6.4.9 Lloyd's Mirror as a limiting case of YDSE 6.4.10 Billet SplitLens as a limiting case of YDSE 6.4.11 Interference of Two Converging Coherent Parallel Beams of Light 6.5 Interference by Thin Films 6.5.1 Interference due to Thin Film in Reflected Light at Near Normal Incidence 6.5.2 Interference due to Thin Film in Transmitted Light a Near NormalIncidence 6.5.3 Interference due to a Thin LiquidFilm on Glass 6.5.4 Interference in Reflected Light by a Very Thin Film in Air 6.5.5 Interference in Reflected Lightfrom a Thin Film for Oblique Incidence

6.5.6 6.5.7 6.5.8 6.5.9

Interference in Transmitted Light from a Thin Film for Oblique Incidence Interference in Reflected Light due to Thin Wedge shaped Film Interference by an Air Wedge Shape of Interference Fringes in Reflected Light from different Air Wedges

6.5.10 Shape of Interference Fringes due to different types of Sources

341-408; .

^^2 342 342 343 343 344 344 345 345 345 349 349 350 359 351 353 353 354

354 354 355 352 357 35g 359 359 360 3gg 367 368 368 368 368 369 370 37j 371 371

6.6 Diffraction of Light 6.6.1 Explanation of Diffraction by Huygen's Wave Theory 6.6.2 Types of Diffraction of Light 6.6.3 Diffraction of Light by a Single Slit 6.6.4 Analysis of Diffraction of Light by a Single Slit 6.6.5 Diffraction Minima due to Single Slit 6.6.6 Diffraction Minima due to Single Slit

375 375 375 376 377 37g 37g

6.6.7 Observing Single Slit Diffraction Pattern on a Screen

379

6.6.8 Difference between Double Slit Interference and Single Slit Diffraction Patterns 6.6.9 Illumination Pattern due to Diffraction by a Single Slit 6.6.10 Diffraction by a Small Circular Aperture

379 380 3gO

i5.7 Polarization of Light 6.7.1 Representation of Unpolarized and Polarized Light

6.7.2 Circularly and Elliptically Polarized Light

3g2 383 3g4

6.8 'Methods of Polarizing an Ordinary Light

384

6.8.1 Polarization by Reflection

384

6.8.2 Brewster's Law

385

6.8.3 Polarization by Refraction

385

6.8.4 Polarization by Double Refraction

385

6.8.5 Polarization by Dichroism

386

6.8.6 Polarization by Scattering

386

6.8.7 Malus' Law

387

6.8.8 Intensity of Polarized light through a Polaroid (Polarizer)

387

6.8.9 Optical Activity of Substances

388

—

DISCUSSION QUESTION

391

—

CONCEPTUAL MCQs SINGLE OPTION CORRECT

392

—

NUMERICAL MCQs SINGLE OPTIONS CORRECT

395

—

ADVANCE MCQs WITH ONE OR MORE OPTIONS CORRECT

399

—

UNSOLVED NUMERICAL PROBLEMS FOR PREPARATION OF NSEP, INPhO & IPhO

401

answers & SOLUTIONS

Chapter 1

Atomic Physics

409 - 420

Chapter 2

Photo Electric Effect & Matter Waves

421 - 434

Chapter 3

X-Rays

435 - 440

Chapter4

Nuclear Physics and Radioactivity

441 - 454

Chapter 5

Geometrical Optics

455 - 504

Chapter 6

Wave Optics

505 - 524

1

AtoMic Physics FEW WORDS FOR STUDENTS

,

'

in your previous Classes'from several we have kiidie^d about existence ofatomsand nuclei However most ofus cannotcitemuch experimental evidencefrom them. In this chapter we will discuss

the experiments thatform the basis for our knowledge of atoms: i Nowiweexdmine manyfundamentalfach ofatomic structures and / ourreasonsfor believing inthem. Thus inthis chapter we aregoing 1 tpfranie^ ourgroundworkfor further discussion of atomicphysics ' in the following chapters.

CHAPTER CONTENTS

1.1

hB'riefHistory toAtomic Physics

-

'"1.6

.1.2 . • Thomson'sAtomic Model • >; r '1.3

Ruthorford'sAtomicModel

1.4.

Bohr's Modelofan Atom

1.5

•Excitation andlonization ofan Atom

1.7

TheHydrogen Spectrum

.

'Effect.ofMass ofNucleusonBohrModel

- 1.8

1.9

~

, Use ofBohrModH toDefine Hypothetical Atomic .

Properties ofElectron in Bohr's AtomicModel

•'

Energy Levels.

, .

,

1.10 ' Atomic Collisions -

^ii:i

COVER APPLICATION

•

Goldfol

Alpha Particles

Alpha partilces Source

An Atom

of Go d

. iLrad

Detector

Figure-(a)

i'.

•

'

. -'

Figure-(b)

Ruthorford's Alpha Scattering experiment based on which atomic model was defined by Ruthorford. Figure-(a) shows.the experimental setup and figure-(b) shows the atomic visualization of scattering of alpha particles.

'

"

• •:

Atomic Physics is that branch of physics in which we'll deal with the properties ofan atom. Thewhole Physics is based on ourphysical environment, surroundings andeverything inour surroundings is the multidimensional arrangement ofdifferent atoms.

.Afeniic Ph^}t%j

The actual observations are quite different. At rooiti temperature

or belownormalroom,hydrogenis vary stable,it neither emits radiation nor does the electron collapse into the nucleus. The

Hydrogen atom emits fixed wavelengths which areobserved in Hydrogen spectrum. Such observations could notbeexplained by classical concepts of physics.

Therewere somanyph)«icists whohad given theirexplanation

Lenard and Rutherford but no one was successful with his

These puzzles related to the Hydrogen atom andits spectrum was solved upto a greater extent byanatomic model which was presented byNiels Bohr, in 1913. Thiswas theageofModem

theorywhichcouldmakean atoma femiliarconcept.

Physics. "

Initiallythe wholeresearch was concentrated on the simplest atom-Hydrogen Atom andits spectrum. When a material body is heated, it emitselectromagnetic radiation. The radiationmay consists ofvarious components havingdifferent wavelengths. When these wavelengths are plotted on a calibratedscale then this plot is called the spectrumofthe material body.

Now we'll discuss this ^Bohr's Model ofatom" in detail with

If hydrogen gas enclosed in a sealed tube is heated to high temperatures, it emits radiation. If this radiation is passed through a prism, components of different wavelengths are deviated by different amounts and thus we get the hydrogen spectrum. When the'Hydrogen spectrum was taken then the mainfeatures ofthis spectrum are,somesharplydefined, discrete wavelengths exist in the emitted radiation. For example, the radiation of wavelength 6562 A was observed and then the radiation of4860 A was observed. Hydrogen atom do not emit

1.2 Thomson's Atomic Model

about the nature of an atom with their atomic models and

suggestions. Main physicists in this field were Thomson,

any radiation between 6562 Aand 4860 A. There were also so manylines ofelectromagnetic radiations above and below these

theoretical andanalytical expressions. Theatomic model of Bohr is based on three assumptions, these assumptions are called Postulates of the Bohr's model. Before proceeding to Bohr's Model First we'll discuss the basic concepts of Thomson's Model and Rutherford's experiment.

In 1898, J.J.Thomson suggested an atomic modelwhichbecame the most popular model of an atom in very early twentieth century. He suggested that the positive atom was spherical in

shape and the positive charges were uniformly distributed in thissphere of atomic dimensions orthe sphere was filled with positive charged matter ofuniform density andsufficient number of negative particles, electrons are embedded randomly throughoutthe volume just like seedsin watermelon to balance the positive charge.

two lines in visible region, ultraviolet, infi-a red region etc.

1.1 A Brief History to Atomic Physics In this age of classical Physics no one was presentwho could explain the theoryofhydrogen spectrum. Although Rutherford got someinteresting resultswith his a-scattering experiment

Uniformly distributed positive charge

and based on these observations, Rutherford proposed the model ofnuclear atom which remain accepted to a large extent

even today. According to this model all the positive chargeof the atom is concentrated at its centre called the nucleus of the atom. This nucleus contains almost all the mass of the atom. Outside this nucleus, there are electrons which move around it

Figure 1.1

at somesqjaration. This modelwasnot fullysatisfactorybecause it was not able to explain the contradictions,imposed by the classical theory of electromagnetics. According to Maxwell's electromagneticequation any acceleratedcharged particle must continuously emit electromagnetic radiation. The revolving electron should therefore, always emit radiations at all

Figure-1.1 shows an approximate-,picture of an atom what Thomson suggested.'Here the gray coloured sphere can be assumedas the positivelycharged uniformmatter and electrons (representedas small balls) are randomlyscatteredwithin the

temperatures and the radius of the circle should gradually

positive charge resembled the plum fi-uit seeds in a pudding

decrease and the electron should finally fall into the nucleus. But this is not the case actually exist.

hence this atomic model is sometimes called as "plum pudding

volume of this sphere. Here the electrons placed within the

moder.

lAtomic Physics _

1,3 Ruthorford's Atomic Model

In theRuthorford's planetarymodel, twobasicdifficulties exist.

In 1911, Earnest Rutherford performed acritical experiment that

electromagnetic radiation by an atom. Itisobserved that every

First is the emission of some characteristic frequencies of

showed that Thomson's model could not be correct. In this

atom emits some characteristic frequencies and no other

experiment a beam ofpositively charged alpha particles (helium

frequency isemitted. Ruthorford's model was not able toexplain

nuclei) was projected into a thin gold foil. It is observed that most of the alpha particlespassed through the foil as if it were emptyspace. But somesurprising results are also seen. Several

this phenomenon. Second difficulty was the revolution of electrons around the nucleus. These electrons undergo a centripetal acceleration. According to Maxwell's Theory of alpha particles are deflected from their original direction by electromagnetism, a centripetally accelerated, charge particle large angles.'Few alpha particles are observed to be reflected should continuously radiate electromagnetic waves of same back, reversing their direction oftravel as shown infigure-1.2. frequency asthat ofitsrevolution. Inthis model ifwe apply this classical theory it says that as electron radiates energy, the Viewing radius of its orbit steadily decreases and its frequency of screen

revolution continuously increases. This would lead to an ever-

increasing frequency ofemitted radiation andultimately atom will collapseas the electronplunges into the nucleus as shown in figure-1.4.

Source screen

Figure 1.2

IfThomson model isassumed to betruethatthe positive charge is spreaded uniformly in the volume of an atom then the alpha particle can never experience such a large repulsion due to Figure 1.4 which it will be deflected by such large angles as observed in the experiment. On the basis of this experiment Ruthorford 1.4 Bohr's Model of an Atom

presented a new atomic model.

Between 1913 and 1915 Niels Bohr developed a quantitative In this new atomic model

atomic model for the Hydrogen atom that could account for its spectrum. The model incorporated the nuclear model of the

it was assumed that the

positive charge in the atom was concentrated in a region that was small

/ /

relative to the size ofatom.

'

He

1

called ,

this

atom proposedbyRuthorfordon the basis of his experiments. We shall see that this model was successful in its ability to predictthe gross features of the spectrum emittedbyHydrogen atom. This model was developed specifically for Hydrogenic

\

concentration of positive ' \ charge, thenucleus ofthe

'

atoms. Hydrogenic atoms are those which consist ofa nucleus

\

f

j /

atom. Electrons belonging

interactions in an atom are not accounted in the Bohr's Model

JBOVi "to the atom were assumed'

that's why it'was valid only for one electron system or

to be moving in the large ^

volume of atom outside

with positive charge + Ze (Z = atomic number, e = charge of electron) and a singleelectron. Morecomplex electron-electron

hydrogenic atoms. Figure 1.3

the nucleus. To explain why these electrons were not pulled

The Bohr model is appropriate for one electron systems like H,

into the nucleus, Ruthorford said that electrons revolve around

He"^, Li"^^ etc. and itwas successful upto some extent inexplaining

the nucleus in orbits around the positively charged nucleus in the same manner as the planets orbitthe sun. Thecorresponding

the features of the spectrum emitted bysuch hydrogenic atoms. However this model is not giving a true picture of even these simple atoms. The true picture is fiillya quantum mechanical

atomic model can be approximately shown in figure- i .3.

Atomic physics

. affair which is different from Bohr model in several fundamental

1.4.2 Second Postulate

ways. Since Bohr model incorporates aspects ofsome classical

and somemodemphysics, it is nowcalledsemiclassical model. In the studyof atom, Bohr found that while revolving around Bohr has explained his atomic model in three steps called the nucleus the orbital angular momentum of the electron was . postulates ofBohr's atomic model. Lets discuss thesepostulates restrictedto onlycertain values, we saythat the orbitalangular momentum ofthe electron is quantized. He therefore took this one by one. as a second postulate of the model. The statement of second Postulate is, Bohr proposed that - ''During revolution around 1.4.1 First Postulate the nucleus, the orbital angular momentum ofelectronL could In this postulate Bohrincorporates andanalyze features ofthe not havejust any value, it can take uponlythosevalueswhich Ruthorford nuclear model ofatom. In this postulate it was taken are integral multiples of Plank's Constant divided by 2it that as the mass ofnucleus is so much greater then the mass of electron, nucleus was assumed to be at rest and electron

h .. I.e.

271

revolves around the nucleus in an orbit. The orbit of electron is

assumedto be circular for simplicity.Now the statement of first postulateis ''^During revolutionofelectron around the nucleus

Thus the angular momentumof electroncan be written as T ~

—

in circular orbit, the electric coulombianforce on electron is

balanced by the centrifugalforce acting on it in the rotating frame ofreference"

In

...(1.4)

Where«is a positiveinteger, knownas quantumnumber. In an orbit of radius r if an electron(mass m)revolvesat speedv, then its angular momentum can be given as L = mvr

...(1.5)

Now from equation-(1.4) and (1.5), we have for a revolving electron nh mvr =

In

...(1.6)

Equation-(1.6) is known as equation of second postulate of

Bohr model. Here the quantity ^h occurs so frequently in

Figure 1.5

If electron revolves with speed v in the orbit ofradius r. Then relative to rotating frame attached with electron, the centrifugal force acting on it is

modern physics that, for convenience, it is given its own designation fi, pronounced as "A-bar." In

mv

F r= cf

r

.

- 1.055 xlO-^'^J-s

...(1.7)

...(l.I) 1.4.3 Third Postulate

The coulombian force acting on electron due to charge of While revolution of an electron in an orbit its total energy is taken as sum of its kinetic and electric potential energy due to the interaction with nucleus. Potential energy of electron

nucleus (+ Ze) is F

electric

K{e){Ze) ^2

revolving in an orbit of radius r can be simply given as KZe'

...(1.2)

[/=-

K{e){Ze)

r~

Now according to first postulate from equation-( 1.1) & (1.2) we have

.

U=-

mv^

mv^ =

KZe^

KZe'

KZe'

...(1.8)

For kineticenergyofelectron,we assumethat relativisticspeeds

...(1.3)

Equation-(l .3) is called equation ofBohr's flrstpostulate.

are not involved so we can use the classical expression for kinetic energy. Thus kinetic energy of electron in an orbit revolving at speed v can be given as

K=\mv'

...(1.9)

iAtomic-^PKysics,' - '

' _5;:

Thus total energy ofelectron can begiven as KZe'

E = K+ U=

1.5 Properties of Electron in Bohr's Atomic Model ...(1.10)

Here we can see that while revolving inastable orbit, the energy of electron remains constant. From the purely classical viewpoint, during circular motion, aselectron isaccelo-ated, it should steadily loose energy by emitting electromagnetic

Now we'll discuss the basic properties ofan electron revolving in stable orbits. These stable orbits we call Bohr energy level. We have discussed thatthere aresome particular orbits inwhich electron can revolve around the nucleus for which first and

second postulates ofBohr model was satisfied. Thus only those orbits are stable for which the quantum number «= 1,2,3

radiations and it spiraled down into the nucleus and collapse Now for n^ orbit ifwe assume its radius is denoted by r and electron is revolving in this orbit with speed v^. We can represent

the atom.

Bohr in histhirdpostulate stated that''While revolving around

the nucleus inanorbit, itisinstable state, itdoes not emit any

all the physical parameters associated with the electron in n^'

orbit by using a subscript n with the symbol ofthe physical

parameter like r^, v^ etc. ener^ radiation during revolution. It emits energy radiation only when it makes a transition from higher energy level 1,5.1 Radius ofn^''Orbit in Bohr Model (upper orbit) to a lower energy level (lower orbit) and the energ;ofemittedradiation isequal tothe difference inenergies Radius ofelectron in Bohr's Orbit can be calculated using of electron in the two corresponding orbits in transition.'' thefirst two postulates ofthe Bohr's model, using equations(1.3) & (1.6) wehavefrom equation-(1.6)

Ifan electron makes atransition form ahigher orbit n.^ to alower orbit w, as shown in figure-1.69 (b). Then theelectron radiates

nh V

a single photon of energy

=

"

2nmr„

Substituting this value of inequation-(l .3), weget

A£ =-£„2-

=hv

2.2

Wh

r

...(1.12)

An^KZe^m r_ =

An^Ke^m

=0.529 x-|-A (a)

...(1.13)

(b)

Figure 1.6

Here

and E„^ are the total energies ofelectron inthe two

1.5.2 Velocity of Electron in

Bohr's Orbit

orbits «2 and n^ The emitted photon energy can beexpressed

By substituting the value of r^ from e9uation-(1.12) in

as hv where v is the frequency of radiated energyphoton. If be the wavelength ofphotonemittedthen the energyofemitted

equation-( 1.6) we can calculate the value ofv as

photon can also be given as

InKZe'

...(1.14)

V

nh

/^ =hv=^ K

...(1.11)

Similarlywhen energyis supplied to the atom by an external source thentheelectron will make atransition from lower energy level to a higher energylevel as shown in figure-1.6 (a). This process is called excitation of electron from lower to higher energylevel. In thisprocess the wayinwhich energyis supplied to the electron is very important because the behaviour of the

2nKe^ V. =

h

v^ =2.18 Xlo^x —m/s 1.5.3 Angular Velocityof Electron in

...(1.15)

Bohr's Orbit

Angular velocity of the electron in n"" orbit is given by

electron in the excitation depends on the process by which energy is supplied from an external source. This we'll discuss in

CO. =

detail in'later part of this chapter.

First we'll studythe basicproperties of an electron revolving around the nucleus of hydrogenic atoms.

CO.. =

...(1.16)

• .--

^

1.5.4 FrequencyofEIectroniii«*^Bohr'sOrbit

M^ =/^x7ir^

Frequency i.e. the number ofrevolution made by the electron j. th . per second in« orbit IS given as

M," "

- j 'Atorriic,..Ph^|ps1 •

r~5 n h

^^

^An^KZe^m

CO

^

y=

KZ e m

inthisequation-(1.21), for «= Ithe magnetic moment M, is also

"

known as"5o^rMjg/2ero«".

1.5.5 Time period of Electron in

Timeperiodof electron in

Bohr's Orbit

1.5.9 Energy ofElectron in

orbitis given by

Orbit

We've discussed that in orbit during revolution the total energyof electroncan be given as sum of kinetic and potential

rn =4-

energy of the electron as

r

nm

E =K+U

...(1.22)

An^K^Z^e^m Kinetic energy ofelectron in 1.5.6. Current in

orbit can be given as

Bohr's Orbit

...(1.23) Electrons revolve around the nucleus in the «

Bohr's Orbit

then due torevolution there iscurrent inthe orbit and according to the definition of current, the current in the

From equation offirst postulate ofBohr Model we have for

orbit willbe orbit

total coulombs passing through a point in one seconds, and in

an orbit an electron passes through a pointy^ times in one second so the current in the w'** orbit will be

=>

O 1 T


'•

E =-

In^K^Z^e^m :r-:

"

n 2f-'2'72 4_

„

=>

E=-Rch joules and iscalled as One Rydberg-Energy

^n~~

In K

Z

e

m

1Rydberg= 13.6eV=2.17x 10''^joules

The equation-(1.29) clearly shows that as the value of« j-«ubetween ^ increases, *1. the difference two consecutive energy 11 levels

decreases. It can be shown with the help of Figure-1.7, which shows the energy level diagram for a hydrogen atom.

2nr„ .

2.18xl0^x%

/« =

T7 s"

2x3.14x0.529x10"'° x"/2

. b Thusnuraberofrevolutionscompletedm 10 ®secondin« =2 ^^ . state are

A^=/^x 10"^

—^0 n = 6 —

«= 5

2.18x10^x10"^

A rr rPs-nsdpV z — 2 x^3 . 1 4 x 0 .529x10"'°

^

„=4

£=-0.8SeV

n =3

1.51 eV

„ =2

E—ZAeV

^

X ——

3 2 x 10^revolutions

# Illustrative Example 1.2

£ = -13.6 eV

« = l

Figure 1.7

What is the angular momentum ofan electron in Bohr's hydrogen atom whose energy is - 3.4 eV ?"

Now if we multiplythe numerator and denominatorofequation(1.28)by c/iwe get Solution X

ch

=>

-

Where R=

E^=-Kchx^eVn

^

X —^

Energy of electron m n Bohr orbit ofhydrogen atom is given " . , by,

n

,

...(1.30)

E=~^eV ,n

defined as Rydberg Constant and jjence

-34^ ^3.6

thevalue ofit isgiven as/?= 10967800 m"',which can betaken

"

approximatelyas lO'm"'.For«= 1andZ= 1theenergyisgiven

^

« =4

as

=>

«=2

•

_

Atomic Physics^

The angular momentum of an electron in

orbit is given as

# Illustrative Example 1.5

yih

L= -X-. Putting « = 2, we obtain

An electron in the ground state ofhydrogen atom is revolving

iTZ

in anti-clockwise direction in the circular orbit of radius R as

r^2A=:h. 2%

shown in figure-1.8. (i) Obtain an expression for the orbital magnetic dipolemoment

71 '

ofthe electron.

# Illustrative Example 1.3

The electron in a hydrogen atom makes a transition Wj wheren, and arethe principlequantum numbers ofthe two states. Assume the Bohr model to be Valid, the time perioil of the electron in the initial state is eight times that in the final

state. What are the possiblevalues of

and M2 ? Figure 1.8

Solution

The time period T of an electron in a Bohr orbit of principal quantum number n is n'h

Solution

Tccn^

(i) According to Bohr's second postulate

li T,

with the magnetic induction. Find the torque experienced by the orbiting electron.

3.3

Thus,

(ii) The atom is placedin a uniform magnetic inductionB such that the plane normal ofthe electron orbit makes an angle 3.0°

4

h

h

mvr= n — =

—

2n

As Tj = STj, the above relation gives'

2k

' (As for n = 1 first only) •

«2 j =>

-

h'

•

'

M] = 2^2-

...(1.31)

2nmr^

Weknow that the rate of flow of charge is current. Hence current

Thus the possible values of Wj and Uj are

in first orbit is

nj=2,«2=l; n] = 4, «2 = 2;

Now fi^om equation-(l .31) i = ev,= e

= 6, rt2~3; and so on # Illustrative Example 1.4 =>

How many times does the electron go round the first Bohr orbit ofhydrogen atom in b ?

1 =

2717*1

2717*1

27cr]

XVi

h

eh

2Kmr^

4^^^ m r 2

...(1.32)

Magnetic dipole moment,

Mj = i X Solution M,=

We know the fi^equencyof revolution of electron is

• eh

eh

An'^mr^

4Km

...(1.33)

(li) Torque on the orbiting electron in unifonn magnetic field is it is the number ofrevolutions in 1 second, it can be given as for

X = M

X B

n = 1 orbit of hydrogen atom as

/i =

T = Afflsin30°

2.18x10^ 2x3.14x0.529x10

,15 /j = 6.56 X10'= sec-

-10

-1 sec

eh x=

AKtn

B x - ^ =•

'2

ehB %Km

-

# IllustrativeExample 1.6

Practice Exercise 1.1

Determine the maximum wavelength that hydrogen in its ground state can absorb. What would be the next smaller wavelength that would work ? Solution

(i)

The innermost orbit ofthe hydrogen atom has a diameter

of 1.06A. Whatis the diameter of the tenthorbit? • [106 A]-

. . .

•

. '

(ii) Which energystate ofthe triplyionized beryllium (Be^ has the same electron orbital radius as that ofthe ground state

Maximum wavelength will correspond to the minimum energy ofhydrogen? Given 2 for beryllium =4 : transition ofanelectron. From ground state ofhydrogen atom-

the minimum energy transition is forn = Lton = 2, for which energyreleased willbe

-

[« = 2]

(iii)

•

^

InQ. (ii), what isthe ratio'ofthe enef^ state ofberyllitim

and that of hydrogen ? ri

'

[4]

• A£:,2 = (-3.4eV)-(-'13.6eV)

(iv) The orbital speed oftheelectron in the ground state of hydrogen isv. What will beitsorbital speed when it isexcited to

A5,2= 10.2eV

Thus 10:2 eV energy isabsorbed inthe form ofaphoton, ifX. be the energy state-3.4 eV ? its wavelength then

-

•: [^l

he

Y =10.2 X-1-6 xi(r'9 x=

^

..... -(v) Which energy state ofdoubly ionized lithium (Li^ has the same energy as that of the ground state of hydrogen ?

he

10.2x1.6 xlO~'^

Given Zfor lithium = 3. [« = 3]

,.(6,63x10'^'^)x(3x1Q:'^) 10.2x1.6x10:^^ ,

• A=12I8A

(vQ . ^In the Bohrmodel ofthe hydrogen atom, find theratioof,

•• (- .

the kinetic energy to Aetotal energy ofthe electron inaquantum ;

Fornext smaller wavelength the possibility is for an electron

transition from « ^ l;ton = 3,- for which the absorbed energy photon requiredds-

j

state«?

, . .

,,

[-1]

(vii)

. ^

,,

•

The total energyofthe electronin the first excitedstate

ofhydrogen is- 3.4 eV. What isthekinetic energy ofthe electron

ae,,=e,-e,

in this state?

A£i3 =(-:i.5kV)-(-13.6eV)

,

,

[+ 3.4 eV]

:'^A£,3 = 12.09ey

If X'be itswavelength then "it is given as

'•

1.6 Excitation and lonizatioh of an Atom'

he

• '-7^= 12.09 eV

'

•

(6.63xlO"^'')x(3xlO'*)' ^ ' ^ 12.09x1.6x10"'^ X' = 1027.5A

According to third postulate of Bohr model we've discussed when some energy is given to an electron of atom from an

external source it niay make a transition to theupper energy level.This phenomenon we call excitation ofelectronor atom and the upper energy level to which the electron is excited is

called excited state. To excite an electron toa higherstate energy

can besupplied to it in two ways. Here we'll discuss only the Web Reference at.ww.phvsicsgalaxv.com

Age Group- High School Physics j Age 17-19 Years

energy supply by an electromagnetic photon. Other method of .energy supplywe'll discusslater in this chapter.

Section-MODERN PHYSICS

Topic - Atomic Structure Module Number - r to 18

According to Plank's quantum theoryphoton is.defined as a packet of electromagnetic energy, which when absorbed bya

physical particle, its complete electromagnetic energy is

Atomic Physics

10

converted into the mechanical energy of particle or the particle

;£., = OeV

utilizes the energy of photon in the form of increment in its mechanical energy. When a photon is supplied to an atom and an electron absorbs this photon, then the electron gets excited to a higher energy level only if the photon energy is equal to the difference in energies ofthe twoenergy levels involved in

;£4 = -0.85 eV .£3 = -1.51 eV A£,3= 12.09eV

E

—2.6 eV (Hvpothctical level)

•£, =•-3.4 eV A£,2=102eV

=

eV

£, =- 13.6eV

the transition. Figure 1.9

For example say in hydrogen atom an electron is in ground

state (energy E^-~ 13.6 eV). Now it absorbs a photon and As discussed in previoussections, in an atom electroncan not makes atransition to«=3 state (Energy£3=-1.51 eV) then the takeup all energies. It can existonlyin someparticular energy levels which have energy given as- 13.6/«^e V. energy of incident photon must be equal to A£,3 = £3-£, M

,3 =(-l-51)-(-I3.6)eV

A£,3 = 12.09 eV Similarlyif wefind the difference in energiesof staten = 1 and w= 4, we get

^14 -^4 A£,4= (-0.85)-(-13.6)eV

A£,4= 12.75 eV Thuswhen a photon ofenergy 12.75 eVincidaitsonahydrogen

atom, the electron maybe excited to«=4 level ifit absorbs this photon. In the same fashion, the energy ofa photon required to excite the electron ofa hydrogenatom from ground state (w = 1)

to next higher level {n = 2) which we call first excited state is given as

Whena photon of energy 11 eV is absorbed by an electron in groundstate. The energy ofelectron becomes - 2.6eVor it will excite to a hypothetical energy level X somewhere between n = 2 and « = 3 as shown in figure-1.9,which is not permissible for an electron. Thus when in ground state electron can absorb

onlythosephotons whichhaveenergies equal to the difference in energies of the stable energy levels with groundstate. If a photon beam incidenton H-atoms having photon energynot equal to the difference of energy levels of H-atoms such as 11 eV, the beam willjust betransmittedwithoutany absorption by the H-atoms.

Thus to excite an electron fi-om lower energy level to higher

levels by photons, it is necessary that the photon must be of energy equal to the difference in energies of the two energy levels involved in the transition.

As we know that for higher energy levels, energy of electrons is less. When an electron is moved away from the nucleus to

00* energy level or at « = 00, the energy becomes 0 or the A£i2 =(-3.4)-(-13.6)eV

electron becomes fi-ee fi*om the attraction of nucleus or it is removed fi-om the atom. Infect when an electron is in an atom,

A£,2 = 10.2eV its total energy is negative

Here we have seen that in the ground state of a hydrogen atom electron can absorbs photons ofenergies 10.2eV, 12.09eVand 12.75eV to get excitedto « = 2,« = 3 and « = 4 levels.

E„ =~1Mz2

. This negative

sign shows that electron is under the influence of attractive

Now we'll see what will happen when a photon of energy equal

forces of nucleus. Whenenergyequal in magnitudeto the total energy of an electron in a particular energy level is given externally, its total energy becomes zero or we can say that energy, level or the electron is electron gets excited to

to 11 eV incident on this atom. From the above calculation of

removed fiom the atom and atom is said to be ionized.

energy differences of different energy levels we can say that if the electron in ground state absorbs this photon it will jump to a state somewhere between energy levels n = l and w = 3 as shown in figure-1.9. When electron in ground state absorbs a photon of 11 eV energy, its total energy becomes

£ = £1 + 11

We know that removal of electron fiom an atom is called

ionization. In other words, ionization is the excitation of an electron to « = 00 level.The energy required to ionize an atom is called ionization energy of atom for the particular energy level fiom which the electron is removed. In hydrogenic atoms, the

ionization energy for =>

£=-13.6 + lleV

=>

£=-2.6eV

A£.

state can be given as

= £„-£_

lAitoffiiciiPhysJcs A£"

11

13.6Z

=0-

This energy can.be converted toeVhy dividing this energy by theelectronic charge e,asifwavelength isgiven in the energy

2^ eV

n

in eF can be given as 13.62^

A£=|^(ineV)

eV

...(1.36)

Substituting the values of h, a and e weget

1.6.1 Frequencyand Wavelength ofEmitted Radiation

(6.63xl0-^'')x(3xl0^)

When an electron absorbs a monochromatic radiation from an external endrgy source then it makes a transition fromTa lower energy level to a higher level. But this state of the electron is not a stable one. Electron can remain in this excited state for a

verysmall internal at most of the order of 10"^ second.The time period for whichthis excited state ofthe electronexists is called

Xx(1.6xl0^l^)xl0"^° ^

AE=^^eV

...(1.37)

Hereinaboveequation-(1.37), lambdais in A units.

the life time ofthat excited state. After the life time ofthe excited

state the electron mustradiate energy and it willjumpto the Thisequation is themost important in numerical calculations, ground state. Sayif it wasin fifth energy level thenit maycome asitwill bevery frequently used. From equation-(1.34) &(1.36) totheground state byfollowing the path as5 ^ 3 ^ 1or it may we have follow 5->4->2-^Iorin somanyways, andin eachtransition it willemita photon ofenergyequaltothe energy difference of the two corresponding orbits according to Bohr's Third

1

1

«1 V"i

n-

eV

Postulate. 13.62-

Let us assume that the electron isinitially in «2 state and itwill

he

1 Wi

jump to a lower state «, then it will emita photon of energy

equal tothe energy difference ofthe two states «, and

1 n

eV

2J

as

v=-^=-RZ^

1

1

w,

n-.

...(1.38)

Where AE" isthe energy ofthe emitted photon. Nowsubstituting

the values ofE„^ ax\.&E„^ in above equation, we get AE = -

Here v is called wave number of the emitted radiation and is

+ 21.2 nth

'~t—2 ir2 3 ^ 2 ^ 1.Thusmaximum number of photons"emitted bya single electroh from state « are A^=«-l

It is clear that the energy of outer orbits is greater than the energy of inner orbits. When external radiation is given to the hydrogen atom then the electron in ground state jumps to a higher energy state and the atom is called now in excited state.

Anyexcited stateis an unstable stateand themaximumlifetime ofan excited state is ofthe orderof 10"® seconds, and afterthe lifetime of the excited state the electron jumps to the ground state again directly or indirectly by emitting one- or more electromagnetic radiations. It may have so many paths to come to ground state.

,• .,

,

The wavelength of the lines of every spectral series can be calculatedusing the formulagivenbyequation-( 1.37).. _

Five spectral sbies are observed in the Hydrogen Spectrum correspondingto the five energy levels of the Hydrogen atom and these five series, are named as on the names of their inventors. These series are

(1) Lyman Series (3) Paschen Series ...(1.41) '(5) Pfiind Series

1.7 The Hydrogen Spectrum

.,j

(2) Balmer Series (4) Breckett Series

These spectral series are shown in figure-1.11 n =

£=0

6.

« = 7-

h = 6-

£=-0.54 eV. n = 5-

Pfund Series

n = 4

£ = -0.85 eV

Brackett Scries -£ = - 1.51 eV

II = 3

Paschen Series £ = -3.4 eV

« = 2

Balmer Series £. = -1-3.6 eV

«= 1

Lyman Series Figure 1.11

When Hydrogen gas is discharged in a discharge tube (at High

(1) Lyman Series : The series consists of wavelengths of the

potential difference of the order of 10'^ volts), the Hydrogen

radiations which are emitted when electron jumps from a higher

iAjgnicfPh^ics

13

energy level to «= 1orbit. The wavelengths constituting this series lie in the Ultra Violet region ofthe electromagnetic spectrum,; ;

1

1

...(1.42)

, . •

For the first member of Lymanseries For Lyman Series v= R

•«i = l and

1

I

M

...(1.43)

4

n.=2,3,4.

Dividing equation-(l .42) byequation-(l .43), we get

First line of Lyman series is- the line corresponding to the transition «2 ~ 2 to «[ = 1, similarly second line ofthe Lyman

-^ = A X.

97 27

series is the line corresponding to the transition =3to Wj =1.

^ T - At,

5x6563

^

07 27 ^ 1

= 1215.37 A.

27 . (2) BalmerSeries : The series consists ofwavelengths pfthe radiations which are emitted when electronjumps from ahigher energy level to w= 2 orbit. The wavelengths consisting this # Illustrative Example 1.8 series liein thevisible region oftheelectromagnetic spectrum. Find the ratio ofionization energy ofBohr's hydrogen atom and (3) Paschen Series :-The series consists of wavelengths of hydrogen-like lithium atom. the.radiations which areemitted when electron jumps from a higher energy level to«= 3 orbit.-The wavelengths constituting Solution this series lieinthe Near Infra Red region ofthe electromagnetic Energy of an electron in ground state ofBohr's hydrogen-like spectrum.

atom is given by,

(4) Brackett Series : Theseries consists of wavelengths of the radiations which areemitted when electron jumps from a higher energy level to«=4 orbit. The wavelengths constituting this series lie in the Infra Red region of the electromagnetic

E=-

13.6Z-

eV

WhereZ=atomicnumber oftheatom.

spectrum.

I

The ionization energy of this atom is equal in magnitude to (5) Pfund Series ; The series consists of wavelengths of the radiations which areemitted when electron jumps from ahigher energy level to « = 5 orbit. Thewavelengths constituting this series lie in the Deep InfraRed region ofthe electromagnetic spectrum. >: -

energy ofground state =E^= 13.6 Z^.

(EJn

{Zuf

(EJm _(l

l3

Wecan findoutthe wavelengths corresponding to the firstline and the last line forremaining four spectralseries as mentioned in the case ofLyman Series.

a Illustrative Sample 1.9

Lets discuss the transition ofelectron between different orbits

Electronsofenergies 10.20eV and 12.09eV can causeradiation

indetail with thehelp ofsome examples.

to beemitted from hydrogen atoms; Calculate in eachcase, the principal quantum number of the orbit to which electron in the

# Illustrative Example 1.7-^

hydrogen atom is raised and the wavelength of the radiation emitted if it drops back to the ground state.

The wavelength of the first member of the Balmer series in

hydrogen spectrum is 6563 A. Find the wavelength of first

Solution

memberof Lyman series in the samespectrum. Solution

For the first member ofthe Balmer series

'

We know the orbital energy ofan electron revolving in

orbit

is given by

•

r 13.6 E=-^eV

.14-

Where n is the principal quantum number.

Solution

When

We know for transition w= 2 to « = 1, the energy released is

n=l E, =-13.6eV

« = 2£2=-3.4eV n = 3£3--1.51eV

=>

A£2, =(-3.4eV)-(-13.6 eV)

Here we can see that

The wavelength corresponding to this transition can be given 10.0eV = £2--^i and

as

\2.09ey = E^-E^

12431 X=

Thus by absorbinga radiation photon of 10.2 eV electronwill

10.2

= 1218.7 A

This radiations is in ultraviolet region.

make a transition to « = 2 state and by absorbing 12.09 eV

photon electron willmakea transition to « = 3 state. Nowafter # Illustrative Example 1.11 the life time ofexcited states, the electron in w= 2 and n = 3 will make transitions to lower states and ultimately come back to

ground state. In this process the possibilities of reverse transition are

n = 3

to

n= 2

n = 3

to

«= 1

« = 2

to

w= 1

In above three transitions the amount of energy released will be

A£32 = (-1.5IeV)-(-3.4eV)

Hydrogen atom in its ground state is excited by means of monochromatic radiation of wavelength 975 A. How many different linesare possible in theresulting spectrum? Calculate the longest wavelength amongst them. You may assume the ionization energyfor hydrogen atom as 13.6 eV. Solution

When an electron of hydrogen atom absorbs a wavelength

975 A. The energy in eV, it absorbs is given by •

= I.89eV

A£3, =(-1.51 eV)-C- I3.6eV) = 12.09 eV

A^2i =(-3.4eV)-(-13.6 eV)

=>

^

975

£ = 12.75 eV

As discussed earlier that 12.75 eV is the energy difference of n - l.and n-A. Thus electronwill make a transition from ground state to w= 4 orbit. From n = A when electron will start reverse

= 10.2eV

Thus wavelengths of radiations of corresponding transition

transitions and ultimately comesback to ground state then the number of possible transitions are

are

32

1.89

6577.2 A XT N= —^ =6,

^31

_ 12431 12.09 = 1028.2 A

,

_ 12431

These are also shown in figure-1.12 £.,=-0.85eV

n = 4-

A,.,i —

21

10.2

= 1218.7A

i

« = 3-

•£3 —1.51 eV •£2=-3.4 eV

n = 2-

# Illustrative Example 1.10 M= 1-

Determine the wavelength ofthe first Lyman line, the transition from«=2 to « = 1.In what region ofthe electromagneticspectrum does this line lie ?

£i=-13.6eV Ground State

Figure 1.12

omTc^ Physics

•Among these six transitions, the longest wavelength will be corresponding to the minimum energy which is released for transition «=4 to« = 3, theenergyreleased in thistransition is

(ii) Energy level diagram isshown infigure-1.13. n = 4 (-0.85 eV) « = 3 (-1.51 eV)

^^43 -^4 -^3

=> =>

n = 2 (-3.4eV)

- A£'43 = (-0.85eV)-(~1.51eV) A£43=0.66eV

n=l (-13.6eV)

The wavelength' correspondingto this energy is Figure 1.13 12431 0.66

A

X= 18834.8 A

Illustrative Example 1.12

A hydrogenatom in a statehaving a bindingenergyof 0.85 eV makes a transition toa statewith excitation energy 10:2 eV. (i) ^•Find the energyand wavelength of the photon emitted. (ii) Show the transition on an energy level diagram for hydrogen, indicating quantum numbers.

# Illustrative Example 1.13

Ultraviolet lightofwavelength 830A and 700A when allowed to fallon hydrogen atomin their ground stateis found to liberate

electrons with kinetic energy 1.8 e'V and 4.0 eV respectively. Find the value of Planck's constant. Solution

We known that

Solution

(i) Binding energy =

13.6

For radiation ofwavelengths

and X2 having energies 5, and

£*2 respectively, we get 0.85 eV =

he

13.6

«,=4

J Ej —£2 ~ he \X]

^„i=-0.85eV

13.6 + 10.2 =

c(X2~X,)

^

(4.0-I.8)xl.6xia"'^x700xI0^^''>c800xl0~'° (3xl0^)xl00xI0-'®

13.6

"2'

l_ X2

. , ^ (£;-£*2)^1^2

'Suppose the tfah'sitioh'takes'pldce to' stateTij'for which excitation energyis 10.2eV. In this casegroundstateenergy+ excitation energy= £"«2-

he

^.=17 "'"'^2= j-

=6.57 X10-3^ J-s.

Solving weget «2 ~ 2 and wehave

^„2=-3.4eV Energy ofphoton emitted=• = 3.4-0.85 = 2.55 eV

Now the wavelength emitted is

# Illustrative Example 1.14

Theionization energyofa hydrogen likeBohratom is4 rydberg. (i)Whatisthewavelength ofradiation emittedwhentheelectron

jumps from firstexcited statetothe ground state ? (ii) What is the radius offirst orbit for this atom ? Given that Bohr radius of

hydrogen atom = 5 x lO"" mand 1rydberg = 2.2x 10~^^ J.

. _ he Solution

1 =>

-12431 ,

X=4875A

(i) We know that ionization energy of an hydrogen likeatom is given as

^ " .AtomiG;Physicsi|

16

E,_^ =RchZ^

The average.kinetic energy is thus very small comparedto the energybetween the ground state and the'next higher energy state (13.6-3.4 =10.2 eV).-Any atomsin excitedstateemit light and eventually fall to the ground state. Once in the ground

j 12

«)2

=>. ' E,^^ =RchZ'

state, collisions with other atoms can transfer energy of 0.04 eV

Here it is given thatE^_^^ = 4 rydberg = 4 Rch joulethus,we on the average. A small fraction ofatoms can have much more have

Z^=4 =>

Z=2

Thus the wavelength emitted when e~makes a transition from w = 2 to « = 1 is

J

a discharge tube.

I

12

energy(inaccordance witii the distribution ofmolecular speeds), but evenkinetic energy that is 10 tim^ the averageis not nearly enough to excite atoms above the ground state. Thus, at room temperature,nearly all atomsare in the ground state.Atoms can be excited to upper states at very high temperatures or by passing current of high energy electrons through the gas, as in

2^

# Illustrative Example 1.16

^A =10967800x4x 4 • -4

The emission spectrum of hydrogen atoms has two has two

lines ofBalmer series with wavelength 4102 A and4861 A. To

=32903400 A

what series does a spectral line belong if its wave nmnber is equal to the difference of wave numbers of above two lines ?

X.=304A

What isthewavelength ofthisline?[Takei;= 1.097 x lO'm"'] (ii) Radius of

orbit of a hydrogenic atom having atomic

number Z is given as

Solution

r„ =0.529x-|-A

^"

According to given problem

Here Z=2 and n ~ 1 thus radius of first orbit is

^A] =EZ'^

r=0.529x ^A =>

r = 0.2645A

^A2 =RZ^

1

1

«1

«2

1

1

...(1.44)

# Illustrative Example 1.15

Estimate the average kinetic energy of hydrogen atoms (or molecules) at room temperature and use the result to explain whynearly all atomsare in the ground state at room temperature

•^A2 =RZ^

and

and hence emit no light.

2^ in'jf

1

1

1

Xi]

X2

1

RZ^

(ni'f 4

Solution

The wave numbers of above radiations are

Accordingto kinetic theorythe average kinetic energyofatoms or molecules in a gas is given by,

k=\kt

_

-1

, -

V, = ^

• Thus

V,

-

Vi =

1

and V2 = 7— 1.-

1

- Z="I X1.38 X10-23x300 ' =>

Z =6.2 X10-2'J.

V[ —V2 =i?Z^

or, in electron volt it is given as 6.2x10 K

—

4

Let the wave number ofthe third emitted line will be

-21

1.6x10"^^

(nz'f

=0.04eV. V

=

V, — Vn

...(1.45)

[Atomic Ph^ics ^ 17

AE==(AE)^-(AE)^ ...(1.46)

.(V)^ "2i

_5 AE=13.6x^[2/-Z,2j 36

Fromequation-(l.44),wehave 1

5.667 =I3.6xA[2 2 36 L

II'.

4

„2

(Given thatA5= 5.667 eV) ...(1.47)

=>

«|

• .•

4

Applying the law ofconservation ofmomentum, we have, ifP isthe momentum imparted totarget

4 (I.097xl0^)(l)2(4102xl0-'®) and

Solving for «2, we get «2 = 6. Similarly from equation-(].45), wehave

=4.

i

mgV =2P-mgV^

Here

= 2Z^ and

From so, the third line belongs to Brackett series becausethe transition is from level 6 to level 4.

•j =(1.097x10^X1)'

1.

1

L(4)' (6)'J =>

2Z^v =P-2Z,v,

and .

Thewavelength isgiven by equation-(1.46) so

= IZg asthey contain equal number of

•neutrons and protons. Hence

2ZgV = 2P-2ZgVi

=> 2Z^(v+Vi)-P •

•

...(1.48)

and 2Zg(v +Vi)-2P

.

...(1.49)

Dividing equation-(l .48) and(1.49), weget

^=2.62xl0-:6j^

If-=2 or Z^=2Z,

# Illustrative Example 1.17

...(1.50)

Substituting the value ofZ^ in equation-(1.47), we get Two hydrogen-lime atoms ^ and B are ofdifferent riiasses, and each atom contains equal number of protons and neutrons.

4V-V =3 or Z^=\

...(1.51)

The energy difference between the radiation corresponding to Fromequation-(l.50), first Balmerlines emittedby^ and B is 5.667 eV. When the

^.=2Z^=2 atoms Aand B, moving with thesame velocity, strikes a heavy target they rebound back with the same velocity. Inthis process Hence atoms Aand Bare the atom Bimparts twice the momentum tothe target than that respectively.

...(1.52)

(deuterium) and 2He'^ (helium)

A imparts. Identify the atoms ^ and 5.

# Illustrative Example 1,18 Solution

A hydrogeh-like atom' ofatomic number Zis in an excited state Theexcitation energies corresponding tothe first Balmer line ofquantum number 2n. Itcan emit amaximum energy photon of (« = 3 to « = 2) emitted by. "i is given by j

£„.^„=-(13.6eV)Z^

# Illustrative Example 1.19

A single electron orbits a stationary nucleus ofcharge + Ze, where Z is a constant and e is the magnitude of electronic

1_

...(1.54)

'2^"1

charge. Itrequires 47.2 eV to excite the electron from the second Bohr orbit to third Bohr orbit. Find

Ifw, =2n,the energyofthe emitted photon will be maximum for

(1)

transition W2 = 2 n to«, - 1. Thus

(if) The energy required toexcite the electron from the third to 1

E=-il3.6eV)Z'

The value of Z

the fourth Bohr orbit.

1

(iii) The wavelength ofelectromagnetic radiation required to

l")

remove the electron from first Bohr orbit to infinity.

^„..=-(13.6eV)Z2 Given that £

llfoA

(iv) The kinetic energy, potential energy and the angular

-1

momentum to the electron in first Bohr orbit.

An'

(v) Theradius ofthefirst Bohr orbit.

=204eV, weuse 1

204eV=-(13.6eV)Z'

...(1.55)

-1

An

(The horizontal energyofhydrogen atom =13.6 eV, Bohr radius =5.3 X10""m, velocityoflight=3 XlO^m/s, Planck's constant = 6.6xlO-^J.s). •

'

for thetransition ^2 = 2 n to«j = w, we have ✓

1

An^

Solution

1

n^J

Theenergy required to excite the electron from to W2 orbit revolving round thenucleus with charge + Zeisgiven by

Given that£2n_».„=40.8 eV, weuse

40.8eV=-(13.6 eV) Z^ f\An^

...(1.56)

=13.6Z2

n

J

eV

"I

«2

Dividing(1.55)by (1.56),we get

(i) According to the given problem, 47.2 eV energy isrequired

-1

204

' An^

40.8

r

1

An^

n^

toexcite theelectron from «, = 2 to «2 = 3 orbit. Hence . 472 = z2x 13.6

J 2?. 3.2

\'An' 5 =

2

1-4

Which gives 4«^ = 16 or« = 2.Using this value of«in either

(1.55) or (1.56) we getZ^ = 16 or Z= 4.

^

47.2x36 _ 25

13.6x5

Z=5.

(ii) Theenergy required toexcite theelectron from k, = 3 to«2 Theground state energy oftheatom corresponds to state « = 1. = 4 orbit is given by Putting n =1 and Z = 4 in equation-(1.53), the ground state energy of the atom is

J

^4-.£3 = 25X13.6 X

^2

a2

(A)^ £ =_(13.6eV)x •^=-217.6eV

, . '

(1)'

Since n = 2, the given excitedatom has

25x13.6x7 = ^ =16,53 eV

= 4.

Itwillemita photon ofminimum energy for a transition /72 = 4 to

(iii) Theenergy required to remove theelectron from the first

«, = 3.Using these values in(1.54), wehave

Bohr orbit to infinity (co) is given by

-£».=(-13.6eV)(4)^(^-^ = 10.58 eV.

£^-£, = 13.6xZ

1

L12 £^-£,=13.6x25eV

1

cc2.

|Momic'.Physics

^

'

Inorder to calculate the wavelength ofradiation, we use

different photons energies. Some ofthe emitted photons have energy2.7 eV, somehaveenergymoreand somehave lessthan

12431

^ 13.6x25 A

2.7eV

(i) Find the principal quantum number ofthe initiallyexcited

X=36.56lA

level 5.

(iv) Weknow for a given orbit for an electron

Total energy =Kinetic energy = ~ Potential energy For

, 19|

(ii) Find theionization energy for thegasatoms. (iii) Find the maximum and the minimum energies ofthe emitted photons.

« -1 Solution

Total energy in

orbit is

E=-

(i) Figure-1.14 shows the energy level Aand Bofhydrogen

13.6Z' eV

like atom. When light ofphoton energy 2.7eV isabsorbed, the electrons go to excited state. Now the atom emits six different

HereZ=5and«= l,thus =-I3.6x25eV

=>

.Ei=-340eV

Thus kinetic energyin first orbit is K,=+340eV and potential energy in first orbit is

photons such that sixdifferent transitions arepossible. This is only possible when the excited state corresponds to quantum number four. So the principal quantum number of state B must lie between 1 and 4. So either = 2 or = 3. Giventhat

the atoms ofthe gas make transition tohigher energy levels by absorbing monochromatic light ofphoton 2.7 eV. If =3,there will beno subsequent radiations with energy less than2.7 eV. Butradiations with energy less than2.7eVarepossible. Thisis possible when n^-l

'

•n = 4 •« = 3

4

Angular momentum ofelectron is mv, r, = -

-« = 2

h -•

6.63x10-34 ' '

2x3.14

'

A

^ Figure 1.14

mv,I '1 r, = 1.055X10-34

(v) Radius offirstBohrorbitis givenas

LetZbethecharge onthenucleus ofhydrogen likeatom, then

r, =0.529 x.^A =P

rj =0.529 x-iA

=>

ri =0.1058 A

# Illustrative Example 1.20

«=1

E=~Rch —r Now for

n =2

E,— and for

Rch

«=4 RchZ'

A gas ofidentical hydrogen like atoms has some atoms in the

"^4 = -

16

lowest (ground) energy level Aandsome atoms in a particular Thus energy released when electron makes a transition fi-om upper (excited) energy level B and there areno atoms in any « =4 ton = 2 is other energy level. The atoms of the gas make transition to higher energy level by absorbing monochromatic light ofphoton RchZ' F F -\ 2.7 eV. Subsequently, the atoms emit radiation of only six

l20' Practice Exercise 1.2

E,-E^^RchZ^ i_J_ 4 16

=>

(i)

Find theratio ofminimum to maximum wavelength of

radiation emitted byelectron in ground stateofBohr's hydrogen

:3 E^-E^ ='-^RchZ^

atom.

[3/4]

Further

2.7

RchZ^

(As £4-^2 =2.7 eVGiven)

(ii) Determine the wavelength of light emitted when a hydrogen atom makes a transition from « = 6 to « = 2 energy level according to the Bohr model.

^

MZ2 = 14.4eV

[4113.2 A]

(ii) Thus ionizationenergyof atom is •

'•

(iii)

.Ei^^ =MZ^=14.4eV

r. • ,

A double ionized Lithium atom is hydrogen-like with

atomic number 3:

(a) (iii) When electron makes atransition from «=4 then maximum energy is corresponding totransition from « = 4 to « = 1which

Findthe wavelength ofthe radiation required to excite

the electron in Li"^ from the first to the third Bohr orbit.

(Ionization energy ofhydrogen atom equals 13.6eV).

(b)

can be given as

Flow many spectral lines are observed in the emission

spectrum of the above excited system.

i.__L l2

AE_,^-Rch'Z^

4?

[114.26 A, 3]

(iv)

'-V6

'

'

An energy of68.0 eVisrequired toexcite a hydrogen like

atom from its second Bohr orbit to the third. The nuclear charge is Ze. Find the value ofZ, the kinetic energy ofthe electron in

the first Bohr orbit and the wavelength of the electromagnetic

If

radiation required toeject theelectron from thefirst Bohr orbit to infinity.

max

=tI 14.4 16

[Z = 6, 489.6 eV, 25.39 A.]

AE^ = 13.5eV

(v) The wavelength of the first line of Lyman series for hydrogen is identical tothat ofthe second line ofBalmer series

Similarly the minimum energyis corresponding to transition from « = 4 to w= 3 which can be given as AE_

I- A£43 =Rch Z^

AE^^ = RchZ'

9

1

1

16

A^min=i^^l4.4eV

for some hydrogen-like ion X. Calculate energies of the four levels of AT. Also find its ionization potential (Given : Ground state binding energy of hydrogen atom.13.6 eV).-. 54.4 eV, - 13.6 eV, - 6.04 eV, - 3.4 eV, 54.4 eV]

(vi) In hydrogen likeatom(atomic numberZ) is in a higher excited stateofquantum numbern. This excited atom can make a transition to first excited state by emitting photons of energies 10.20eVand 17.00eVrespectively. Alternativelythe atomfrom the same excited state can make a transition to the second excited

A£min=0.7eV

; Web Reference at www.Dhysicsgalaxv.com

j Age Group -High School Physics | Age 17-19 Years

•^iSection-MODERN Topic - Atomic Strticture .. Module Number-19 to 36

state by successively emitting two photons of energies 4.25 eV and 5.95 eVrespectively. Determinevaluesof« andZ, (ionization energy ofhydrogen atom = 13.6 eV). [« = 6. 2= 3]

(vii) A gas ofhydrogen like ions is prepared in such a way that ions are only in the ground state and the first excited state.

A monochromatic lightof wavelength 1218 A'is absorbed by the ions. The ions are lifted to higher excited states and emit radiation ofsix wavelength, some higher and some lower than

J-Atomic Physics21

theincidentwavelength.Findtheprincipal quantum number of Now the relative picture of atom will be same what we've all the excited state. Identify he nuclear charge on the ions, considered earlier as shown in figure-1.16 but electron mass is Calculate the values ofthe maximum and minimum wavelengths, replaced by its reduced mass.

-

[Z= 2, 4708.71 A, 243.74 A]

, •

rest

1.8 Effect ofMass ofNucleus on Bohr Model

+ Ze-

Upto this level we've discussed about the structure of a

hydrqgenic atom by considering nucleus at rest and electron

Figure 1.16

revolving around the nucleus. In fact we know that as no external

force is actingon a nucleus electron ,system, hencethe centre Now we can use all those relations which we've derived for of mass of the nucleus electron system must remains at rest. Bohr model just byreplacing by Such as the radius of Theoretically mass ofelectron isnegligible orsmall compared electron in orbit ofBohr's model is given as

to that of nucleu? and due to this we assume that centre of

mass of the atom is almost situated at nucleus that's why in Bohr atomic model it was assumed that in an atom, nucleus

remains at rest and electron revolves around it.But practically

..,(1.57)

r.=

4K^KZe^m.

But if we consider the motion of nucleus into account, the

the situation is a bit different. Actually centreofmassofnucleus. radius ofn* orbit will be given as

electron system is close to nucleus as it is heavy and to keep

centre of mass at rest, both electron and nucleus revolve around

their centre of mass like a double star system as shown in figure-1.15. If r is the distance of electron from nucleus, the

distances ofnucleus and electron from centre ofmass r, and can be given as

+mg) ...(1.58)

ATz^KZe^mgm^

^ mj-

r' = y

2

=

...(1.59)

Similarly ifwefind the speed ofelectron in

friMr

and

X

Bohrorbit, it can

be given by equation-1.14 as

Wx/+W„

InKZe' + Ze

V

=

nh

-Oe

...(1.60)

Here wecansee that in theabove expression ofspeed noterm

Nucleus

ofm^ (mass ofelectron) ispresent, hence itdoes not depend on electron mass, nochange will bethese in speed ofrevolution if we consider the motion ofnucleus into account. + Ze,^2

F,

-^HF,

Similarly we know the expression of energyof electron in nth orbit of Bohr model is given as

Figure 1.15

E=-

rt'h^

...(1.61)

Taking the motion of nucleus into account the expression of Now we can see that in the atom, nucleus and electron revolve

around their centre ofmass in concentric circles ofradii rj and r2to keep centre of mass at rest. In abovesystemwe can analyze

the motion of electron with respect to nucleus by assuming nucleus to be at rest and the mass of electron replaced by its

energy become E'=-

nW

reduced mass u , given as

...(1.62)

mxrm.

nji^+mg

E' =E

...(1.63)

Atomic

Thuswecansaythat the energyof electron willbeslightlyless

10967800 X

compared to what we've derived earlier. But for numerical

= 5483900 m-'.

calculations this small change can be neglected unless in a

given problem it is asked to consider the effect ofmotion of

X=

nucleus.

3x5483900

m

X=2430A

# Illustrative Example 1.21

# Illustrative Example 1.22

Calculate the separation between the particles of a system in

the ground state, the corresponding binding energy and wavelength of first line in Lyman series of such a system is positronium consisting of an electron and positron revolving round their common centre.

A p meson (charge - e, mass = 207 m, where misthemass ofelectron) can be captured bya proton to form a hydrogen like 'mesic' atom. Calculate the radius of the first Bohr orbit, the

bindingenergyand thewavelength ofthe firstlinein the Lyman series for such an atom. The mass of the proton is 1836 times

Solution

the mass ofthe electron. The radius of first Bohr orbit and the

The reduced massofthe s>^tem (electron and positron) isgiven

binding energyofhydrogen are 0.529 Aand 13.6 eV respectively. Takeii = 109678 cm-'.

by Solution

m.m

2

w + m

i

The reduced mass ofthe system is given by

Where m = mass ofelectron or positron.

(207m)(1836m) The radius offirst Bohr's orbit is given by

^ (207m +1836m) The radius offirst orbit is given by .

r, =

4n^Ke^(m/2)

Tj = 2 Xradius offirstBohr'sorbit ofhydrogen

47c^/:e^(lB6m)

r. =2x0.529 A r, =

r, =1.058A

1

Energy of first Bohr's orbit is given as

=

2K^K^z^e\m/2) ^1 =

0.529

...(1.64)

r, =0.002844A"

From Bohr's theory the ground state energy for hydrogen like atom with Z= 1 is given by

2n^K^e^p

E^=-jx13.6qV

=>

Xradius of first Bohr orbit of hydrogen atom

• h'

= - — Xground state energy ofhydrogen

=>

loo

2tr2_4, 2ji^.fi:^e''(186 m)

^1 =

h'

=-6.8 eV (Binding energy)

£,=-186 X13.6 eV The wavelength of Lyman series is given by £'i=-2530eV

j X

X=

= 7?.,

I

1

1^

Hence the binding energy is 2530 eV. The wavelength ofthe Lyman lines are given by

3R^ 2n^K^e'^li

1== p ±-JR m

1p «=2,3,4,...

...(1.65)

sAtomlc Phi^ics'

Where

23

= Rydberg constant for mesic atom.

(ii) • The radius of first Bohr orbit is given as h'

Forfirstline i =/fJ-L_ 1

r, =

'

An^Ke^m

Fr6mequation-(1.67), wehave ' %=

...(1.66)

3R..

2A96n'^Ke^m

n^=62A

Now, =>

n =25

=186i?

...

(iii) Rydberg constant for hydrogenic atoms is

.. =186x 10967800m-'

Substituting the value ofR^ in equation-(1.66), we get 3x186x109678

=>

'

m

ch-

=>

An^Ke^m

m

R —

z

ch^ Now when electron is replaced by p-meson, rydberg constant will change as

A,=653.6A. R —

# Illustrative Example 1.23

27i^/:^e\208 m) r—

ch^

R'=208 R=208 X10967800 m^'

A particle ofcharge equal to that ofan electron, - e, and mass 208 times the mass of electron (called a ji-meson) moves in a circular orbit around a nucleus ofcharge + 3e. (Take the mass of the nucleus to be infinite). Assuming that Bohr model of the atom is applicable to this system. (i) Derive an expression for the radius of the

Now the wavelength Xofthe radiation emitted when p-meson

makesa transition fromW2 = 3 to « = 1 is .

,

1 • 2

2

«2.

."l

Bohr orbit.

' 1 .1^ 3^

(ii) Find the value of n for which the radius of the orbit is approximately the same as that of the first Bohr orbit for the hydrogen atom.

^=8.R'

(iii) Find the wavelength of the radiation-emitted when the p-meson jumps from the third orbit to the first orbit. (Rydberg's constant.^ 10967800 m.) •,

A.=

8R' 1

8x208x10967800

Solution

m

-1

^=0.548 A

(i) We have the radius of

orbit of a hydrogen atom as # Illustrative Example 1.24

An^Kie^m

A pi-meson hydrogen atom in a bound state pf negatively

Ifelectron is replaced by a heavy particle of mass 20 times that of electron then the radius is given as 2j.2

.rfh r_ =

"

charged pion (denoted byji",

=273 mj and aproton. Estimate

the number of revolutions a 7C-meson make (averagely) in the ground state of the atom before it decay (mean life ofa 71-meson

= 10'® sec), (mass ofproton = 1.67 x,10"^'kg).

AK^K{3)e\20Sm), Solution

r„ =

n'h' 2A96n^Kze^m

...(1.67)

The frequency ofrevolution ofan electron.in hydrogen atom in first orbit is given by

Here we have not used reduced mass because it is given that mass ofnucleus is assumed to be infinite and here we take z=3.

fx-

AK^K^e'^m h'

Atomic Physics 1

24

Here in case of a t: meson the mass ofelectron is replaced by the

If effect of mass of nucleus is considered the new value of

reduced mass as

Rydberg constant can be given as

Wp(273 m) mp+273m - ,

2n^K^e^\i ch^

Thus the frequency ofrevolution of%meson will become

RM R

47C^i:e''m^(273 m) h\mp+n3m)

=

m + M

Percentage difference inthevalues ofR"and R^.is given as A/?

4x(3.14)^x9x10^ (1.6x10"'^)'^ x(1.67xl0-^')(273x9.1xl0~^')

i?j,-i;xlOO

R ~

R..

/, • (6.63x10"'^) -34^3

•^ =^x 100=^0.055% M

(1.67x10"^''+273x9.1x10"^')

=>

. K

y; = 1.77 XID'S sec"'

Thusin the lifetwo'ofn meson 10"^ sec, number ofrevolutions

# Illustrative Example 1.26

made by it is given as Calculate the difference between the ionization potentials of atomic hydrogen and atomic deuterium.

iV=/,xA/ 1.77 X ID'S X 10-8

=>

Solution

N= 1.77 X 10'® revolutions. '' J

'

The ionization energy for a hydrogenic atom can be given as

illustrative Example 1.25

energy required toexcite electron from Wj = I to «2 ~ given as'

Taking into account the motion of the nucleus of a hydrogen atom, find the expressions for the electron's binding energy in the ground state and for the Rydberg constant. How much (in percent) do the binding energy and the Rydberg constant, obtained without taking into account the motion ofthe nucleus, differ from the more accurate corresponding value of these

1

1

E = Rch 'I

E = Rch

quantities ?

•

1

1

1^

CX.2

E = Rch Joule

*

If motion of nucleus is considered then the value of Rydberg

Solution

constant can be given as

Ifmass ofnucleus is considered (not infinity) then the reduced mass of nucleus electron system can be taken as

R =

mM m + M

Here m is mass ofelectron and Mis that ofnucleus. The binding energy in ground state of hydrogen atom can now be given as

2K'K'e' ch^

Thus the ionization energy of hydrogen and deuterium atoms can be given as 2n'Kh'

£:=13.6x 41 eV

and

2n'K'e'

m

_

=>

^

mMi

.^m +Mf^

E = +

=>

mMi

m + M^

13.6M

(Here M^= 1840 m)

mMjy

\^m +M^ (HereMo=3680m)

The difference of E^^ & Ef^ is

E= ——rr eV m+M

We've hydrogen atom Rydberg constant is given as j . *

.2t^2

4.

2n^K^e^m .R =

ch'

h'

mM 1

mMi

rn + Mj^

m + Mff

Physics

. h'

. =>

3680

1840

3681

1841

on it in the rotating frame ofreference. There may be some

situation in which itis given that electron ofan atom is revolving under-the influence ofa new potential energy field given as

=5.88 X. 10^ J

U=f{r) [Non coulombian field] and using Bohr model, we are

=6.68xl0^^eV

required to develope theproperties ofelectron inthisnewatom.

# Illustrative Example 1.27

For this first we'll develope thefirst postulate equation for this new atom. Aselectron isorbiting inanew potential energy field

A muon isan unstable elementary particle whose mass is 207

centre of orbit given as

givenas U=f{r), it will experience an inwardforce towardthe

and whose charge iseither +eor-e. Anegative muon (p") can be captured by a nucleus to form a muonic atom. Suppose a

F=

imr)}

dU dr

...(1.70)

proton captures a negative muon (p"), find the radius of first

Bohr orbit and theionization energy ofthe atom. (Take mass of proton = 1836 where ismass ofelectron). Solution

Reduced mass of a system of two particles is given by F =

_ mM _ (207 m^){\ 836 m^) _ m+M " (207m,)+ (1836m^)

\du\

\dr\!

Radius corresponding to thereduced mass p is given by Figure 1.17

m.

•.=

ao =2.85x lO-i^m

(^186m,

Thus for a stable

Where Oq istheradius offirst Bohr orbit ofordinary hydrogen

orbit if electron ismoving with speed

in

the orbit ofradius r , we have .

atom. mv.

...(1.71)

fiio =5.29x lO-"m.

This equation-(1.71) will bethenew equation for first postulate and from Bohr model the second postulate is based on quantization ofangular momentum ofelectron, itsequation will

Ionization energy is given by =186£,

F' —

=>

remain same as

Fi=-2.53xl0^eV.

nh

Where £•, istheionization .energy ofordinary hydrogen atom.

1.9 Use ofBohr Modelto DefineHypothetical Atomic Energy Levels

2k

...(1.72)

Nowusing equation-(I.71) and (1.72), we can derive all the'

properties for electron motion like, radius of orbit, velocity ofelectron in orbit, angular velocity, fi-equency, time period, current, magnetic induction, magnetic moment and the total energy of energy levels for this hypothetical atom in the same

We'vealreadydiscussed that forhydrogenic atoms, Bohrmodel waywe'vederived thesefor properties for a general hydrogenic gives first and second postulates as •• • .i

atom. Lets discuss few examples to understand these concepts

..KZe^ -. mvl

in a better way. ...(1.68)

# Illustrative Example 1.28 and

nh mv

r

. " "

~

-T—

2%

...(1.69)

Here equation-(1.68) is given by balancing .the inward coulombian force on electron bythe outward centrifugal force

Suppose the potential energy between electron and proton at a

distance r is given by - ke^l3j^. Use Bohr's theory to obtain energy levels of such a hypothetical atom.

Atorfiic'PH^iCs'

1;;^ .^ # Illustrative Example 1.29

Solution

According to.given situation for an electron revolving in an «" orbitthe potential energy is given as ...(1.73)

U= .

dU F=

dr

ke' .

...(1.74)

Solution

In the given situation the centripetal force on electron in orbit is given by

' n

orbit electronrevolves at speed

mvl

Bohr's orbit and its energy levels.

3r-

The centripetal force on electron due tothis force isgiven as

If in

Suppose potential energy between electron and proton at separation r is given by U= klnr, where isa constant. For such a hypothetical hydrogen atom, calculate the radius of

then we have

ke^

If in

orbit speed of electron is v„ thenwehave mv^

mvl =

ke'

.(1.75)

From Bohr's second postulate, we have nh

According toBohr's qua.ntizatipn postulate, we have nh

(1.76)

- 2%

2%

From equation-(1.75) and(1.76), wehave -

_

^

'

nh

...(1.80)

...(1.81)

" 2tz4^

Energy ofelectron inn^level is

( nh Y ke^^

E=kE„^PE„

E„= ^mvl -klnr

Aiz^ke^m r„ -

' ^

Solvingequation-(1.79) and (1.80), weget

lTimr„

and

-...(1.79)

mvl - k •

,.(1.77)

21.2 n'-h

E^=--k\nr 3;.3

n'h and

,.(1.78)

ZTt^km^e^

Now energy-in

orbit is

„

I

2

ke

k

'e=-2

,,

nh

1-ln

C • 2t.2 ^ n h

\^Anmk j

1.10 Atomic Collisions

2

1 ke' , 2 (\ E= -7

6

^

E= -r -k\n j= 2 •271

2 7.2

ri, h ,

\ A% ke m

n3

In previous sections ofthe chapter we've discussed thatthere are two ways to excite an electron in an atom. One way, of supplying energytoanelectron isby electromagnetic photons, which we've already discussed. We've also discussed that an electron absorbs a photon only when the photon energy is

E

=

equal tothe difference in energies ofthe two energy levels of atom otherwisethe photon will not be absorbed.

iAtpfSe;'ph)«icsi

.271

The another waybywhich energy can be supplied to an elecfron ofanylattice in the colliding body. The energy lost can only be isbycollisions. To understand theenergy supply bycollisions we consider an example of a head on collision of a moving neutron with aStationaryhydrogen atom asshown infigure-l. 18.

absorbed by the atom involved in the collision and may get excited orionized bythis energy loss which take place ininelastic collision.

Here for mathematical analysis we can assume the masses of neutron and H-atom as same. neutron

H-atom

H neutron

rest

• H-atom neutron & excited H-atom

(Stationary) Figure 1.18

Figure 1.20

If in thiscase when perfectly electric collision takes place, we In ourcase ofcollision ofa neutron with a H-atom, theenergy know for equal masses here, neutron will come to rest and

hydrogen atom will move with thesame speed andkinetic energy

loss is

m\P- (half ofinitial energy ofneutron). This loss in

with which the neutron was moving initially as shown in figure-l.19 , .

energy can .be absorbed by H-atom only. We know that the minimumenergyrequiredtoexcitean H-atomis 10.2eV for« = 1

to n = 2. Thushydrogen atom canabsorb onlywhen thisenergy lossis equal to 10.2 eV. Ifthis energy lossin perfectly inelastic collision is morethen 10.2eV then H-atom mayabsorb 10.2eV energyforits excitation and restof the energywillremainin the collidingparticles(w & H-atom) as theirkineticenergyand the

H

collision will not be perfectly inelastic in this case. • neutron

at rest

moving. H-atom

For examplesaythe comingneutron which is going to collide headona stationaryH-atom hasinitial kinetic energy 24.5 eV. In this case if perfectly inelastic collision takes place, then the

Figure 1.19

Nowif weconsider the collision to beperfectly inelastic, then aftercollision both neutron andH-atom willmove together with speed Vj which can be given by law of conservation of

maximum energy loss can be given as

A£.»=y£,= i(24.5)eV

momentum as

=^.

mvQ = 2 mVj •I

V

...(1.82)

-v. = -

In this case the loss of energy can be given as the difference'in initial and final kinetic energies of the neutron and H-atom, given as -

. •

= 12-25 eV

From this energy H-atom can absorb either 10.2 eV or 12.09 eV

forits excitation fromk = 1to « = 2 or from « = 1to «= 3 energy level.

Thusin thiscasetheremaybethreepossibilities for,this collision.

u.

These are: AB^E.-E

'/

© It maybepossible thatthe collision isperfectly elastic and no energy is absorbed by H-atom as shown in figure-l.21 Before collision •

AE=

moving neutron with

AE=^mv'=^E,

K.E. j mv^ =24.5 eV

H-atom at rest

...(1.83)

Thus halfofthe initial kinetic energy will be lost in the collision.

After collision neutron

Oneimportant pointwhich weshouldunderstand carefully is, in collisions ofelementaryparticles and atoms, no energycan be lost as heat because here we can not consider the deformation

at rest

Figure 1.21

H-atom

i/wv2 =24.5 eV

Atomic^ Physics .i

28'

(ii) It maybepossiblethat H-atomwill absorb10.2eV energy during collision and both neutron and H-atomwill be moving withkinetic energy 24.5-10.2 = 14.3 eVaftercollision asshown in figure-1.22. In this casethe collision willbepartiallyelastic.

1 -.2:. Finalkineticenergy£^y= ywvJ+,y/«V2 =12.41 eV. ,

In this case the final speeds of neutron and H-atom can.be and energy. Herebymomentumconservation law, we have

Before collision---•

-Q--

7

"

WV = mv, +WV,

moving neutron

H-atom

'

with

at rest

.

K.E.i,„v2.24.5eV

^ "

After collision

•

V|

*•

^

'

V=V,+V,

-

/-i o

^

moving neulron with

fnu~2tnv

H-atom at rest

v = u/2

K.E. j ntv' =24.5 eV

' _

...(1.88)

The energy ofexcitation AE is given by n

After collision

T|

Vj •

>

A£=

1

o

1

,•

." .o •-

y m{ul2)

from n = 1 to /»= 3

-- . Figure 1.23

^

-

•

.

f • 2

AE =. T4'mu

•

-• " -

,...(1.89)

i^fomic= [physics

•29

Theminimum.exGitation energy for a H-atom is for transition = 1to = 2 which corresponds to an energy of 10.2 eV.

Let Q be the energylost in the reaction. Then

65 eV =y mv^ +y {4m)v^+Q From equation-(1.89)

^ X(1.0078 X1.66 X "• •'

'^

10.2 X(1.6 Xl0-'9)

=>

.

zrc T7

1

2 mu^

mv} ^

65eV=Tr/«v,-h——

m'=6.24x lOWs. 65eV=

# Illustrative Example 1.31

A neutron ofkinetic energy 65eVcollides inelastically with a singly ionized helium atom at rest. It is scattered at an angle of 90° with respectto its original direction-

=>

,

.

(ii) If the atoms gets de-excited subsequently by emitting

65eV =

+-^(65 eV)-i-g

...(1.93)

Now ionization energy of He atom is

(i) Find the allowed values of the energyof the neutrons and that ofthe atom after the collision.

+Q

/£H^+ = 13.6x4 = 54.4eV Thus energy ofelectron in first orbit ofHe atom

./£He+=-54.4eV.

.radiation, find the fi-equencies ofthe emitted radiation. •

[Given ; Mass of Heatbm= 4 x (massof neutron) lonization energyofHatom= 13.6eV].

Similarly, the energyof electron in second, third and fourth orbits would be—13.6 eV,-6 eVand-3.4eV respectively. The energy lost must have been used in excitingthe atom.

Solution

First possibility: Let0= 54.4-13.6 = 40.8 eV

(i) The situation'is shown in figure-1.24.

•|mv,^ =65x j

= "I" -40.8

= 8 eV nearly

So,

1 ~mvi2--i(8eV)=6.4eV

This will be the energy ofrecoil electron.

Neutron

From equation-(l .92),

Atom

=>

Figurel.24-

-ymw^+ymx 16v2^

Applying the law of conservation of momentum, we have •• ..

. , ww = 4w

cos, 0

(1.90)

WV] =4 wv^sin 0

...(1.91)

=4 X~(4m)v2^ 65 eV+6.4 eV = 4 KE ofatom =>

From these equations, we get

= 17.85 eV

Second possibility: Let2=54.4-6 = 48.4eV.

m-4v^cos0 and

KE of atom = ~

Calculating in the same ways as above, we have energy of

V, =4v^sin0

recoil neutron = 0.28 eV

u^-\-v^-\6 Vjv [sin''0 + cos'^ 0]

energyofatom= 16.32 eV

'

= 16

•proceeding as above, let- ' u

+v.

16

...(1.92)

2 = 54.4-3.4=51 eV

=

\ lA.::;Atomic'Phys:rcsj

.30

Inthiscase, theenergy ofrecoil electron=- 2.25 eV(negative).

velocity. The atom subsequently returns to its ground state

This is meaningless.

with emission ofradiation ofwavelength 1.216 x 10"' m. Find the velocity of the scattered electron.

Hence

2=40.8eV,48.4eV. Solution

(ii) Itisobvious from first partthattheatom isexcited tothird state or second state. The frequencies of emission can be calculated as follows:

The energylostbythe electron in exciting the hydrogen atom equals the energycorresponding to 1.216 X10"^ m,

n = 3

h = 2

1

u

6.63xl0"^''x3.0xl0^ 1.216x10

"n=l

-7

Total 3 Lines

=>

?.= 16.36 xlO-'9j

Figure 1.25

Figure-1.25 shows the number of emitted frequencies. These

Now, the initial energy ofelectron is20 eV =32 x 10~'^ J.Hence

are

the kinetic energy ofthe scattered electron is

(i)

A]

=RZ''

1

£:=32x i0-'9j_ 16.36 xio-'^J

1

=>

=(1.097x l0^)x4x

£= 15.64 X10"'^J

The velocity v ofthe scattered electron is given by

= A] =(1.097x 10^)x4x Ax(3xlo8) 4 V, =9.85x1015 Hz 1

(ii)

f 2E^ v =

1/2

I w^

1

-19

1/2

9.11x10'^' v=I.86xl0^ms-'.

=(1.097xl0^)x4x A A2

2x15.64x10

# Illustrative Example 1.33

y

According to the classical physics, an electron in periodic

V2=

=(1.097 X10')x4 XI X(3 X10^) V2 = 11.7x1015 Hz

(iiO

^3

J 22

32

motion will emit electromagnetic radiation with the same

frequency as that of its revolution. Compute this value for hydrogen atom in quantum state. Under what conditions does Bohr's quantum theory permit emission of such photons due to transitions between adjoining orbits ? Discuss the result obtained.

'=(1.097xl0')x4x ^

V3 ^2^. =(1.097 X10') X4X^ X(3 X10^ = 1.827 xlO'5 Hz

# Illustrative Example 1.32

An electron ofenergy 20 eV collides with a hydrogen atom in the ground state. As a result of the collision,the atom is excited to a higher energystateand the electronis scatteredwith reduced

Solution

The orbital frequency of y

S5

orbit is given by

electron speed - •'

•

"

orbital circumference

"

I Time^ x-—X2m/7 So 2ji „2 g ^2 me

rA

1

v„ =

"

47tGo

,

iAtomic,Physics

-31

From thelaws of electromagnetic the'ory, the frequency ofthe The energies radiation emitted bythiselectron will also bev^. According to Bohr's theory, the frequency of the radiation for the adjoining orbits is given by -E„-\

1

me

h

1

12431

12431 ^2=-^eV=40.9eV

and v„ =

and E2 of the twoemitted photons in eKare

Thus total energy

(.7-1)'

£• = .5,+ ^2= 11+ 40.9= 52.3 eV. me

{2n-\)

me

(2/t-l)

Let Mbe the principal quantum number of the excited state. Using equation-(l .94) we have for the transition from m= mto M = l.

v„ =

A'dd 2rp-{n~\f

£:=_{54.4eV)

A comparison of this expression with the classical expression above show that the difference in their predictions will be large

1_

But £•=52.3 eV. Therefore

for small«, i.e., for«=2, \lr?= 1/8 and(2m- l)/2n^(M-1)^ =3/8,

52.3 eV = 54.4 eVx

so that the frequency given by quantum theory is 3 time that calculated from classical theory. However, for very large n, the quantum expression radius to that obtained from classical theory because for m

J 1^

oo 2m-1

2m

2m^(m-1)^

2}?-d

Which gives, m^ = 25 orm= 5.

1

Consequently, the predictions of Bohr's theory agrees with that of the classicaltheory in the limit of verylarge quantum numbers. This correspondence is called as *^Bohr's Correspondence Principle".

The energy ofthe incident electron = 100 eV (given). The energy

supplied to He"*" ion = 52.3 eV. Therefore, the energy of the electron left after the collision = 100-52.3 = 47.7 eV.

r WebReference at www.phvsicsgalaxv.com

# Illustrative Example 1.34

! Age Group - High School Physics j Age 17-19 Years A100 eVelectron collides witha stationary helium ion(He"^ in

f Section-MODERN PHYSICS

its ground state and excites to a higher level.After the collision,

!• Topic r Atomic Structure

He"^ ions emits two photons in succession with wavelength

! Module Number - 37 to 48

1085 A and304A.Findtheprincipal quantum number ofthe excited state. Also calculate the energy ofthe electron after the collision. Given/i = 6.63 x 10"^''Js. Solution

The energy oftheelectron inthem'^ state ofHe"^ ion ofatprnic number Z is given by

£„=-(13.6eV),^

Practice Exercise 1.3

(i) Show that for large values ofprincipal quantum number, the frequency of an electron rotating in adjacent energy levels of hydrogen atom and the radiated frequency for a transition between these levels all approach the same value. (11)

Determine the separation ofthe first line ofthe Balmer

series in a spectrum of ordinary hydrogen and tritium (mass

M

number 3).TakeRydberg's constant/? = 10967800 m"'

For He^ion, Z= 2. Therefore

[2.387 A]

(13.6eV)x(2)' £_ = -

77 54.4 E_=—eV

...(1.94)

(ill) Aphoton ofenergy 5.4852 eV liberates an electron from the Li atom initially at rest. The emitted electron moves at right angles to the direction in which photon moves. Find the speed

Atomic Physics

32

iV^, =6.02x 10^Wr'andm^ =9.1 10"^'kg.

momentum and energy to obtaina relation between the minimum distance s ofthe particle from the nucleus in terms of Z, Z, v and b. Show that for ^ = 0, 5 reduces to the distance of closest

[14.2 m/s, 0 = 88.9°]

approach Tq given by

and the direction in which the Li^"^ ion will move. lonization

potential of Li atom = 5.3918 V.Atomic weight of Li = 6.94 g,

1

(iv) A neutron moving with speed v strikes a hydrogen atom in ground state moving towards it with the same speed. Find the minimum speed ofthe neutrons for which inelastic collision may take place. Take mass of neutrons and that of hydrogen

/•„ =

I

2I 4neo

471:

2ZZ'e^ =0

mv

•/'2

ZZ'e

isl.67xlrtg. [3.13 X 10'^ m/s]

(vii)

(v) A uniform magnetic field B exists in a region. An electron projected perpendicular to the field goes in a circle. Assuming Bohr's quantization rule for angular momentum, calculate (a) the smallest possible radius of the electron (b) the radius of the nth orbit and (c) the maximum possible speed of the electron. [(a)

2neB

nh

heB

\ 2neB '

V2nm^ ^

(vO A particle of mass m, atomic number Z, initial speed v and impact parameter b is scattered by a heavy nucleus of atomic number Z. Use the principle of conservation of angular

A small particle ofmass m moves in such a way that the

potential energy U=- j

where 6is a constant and r is

the distance of the particle from the origin (Nucleus). Assuming Bohr model ofquantization ofangular momentum and circular

orbits, show that radius of the >7th allowed orbit is proportional

toVw .

,

'

.

"

Advance IllustrationsVideos at www.phvsicsgalaxv.com Age Group - Advance Illustrations Section - Modern Physics Topic - Atomic and Nuclear Physics Illustrations - 54 In-depth IllustrationsVideos

^Atomic Physics 33

Discussion Question Q1 -1 Balraer series was observed and analyzed before the other Q1 -10 Galaxies tend to be strong emitters ofLyman-aphotons series. Can you suggest a reason for such an order ? (from the «=2to«= 1transition in atomic hydrogen). But the intergalactic medium—^the very thin gas between the galaxies— Q1 -2 You are examining the spectrum ofaparticular gas that is tends to absorb Lyman-a photons. What can you infer from excited in a discharge tube. You are viewing the discharge these observations about the temperature in these two through a transparent box that contains the same gas. Under environments ? Explain.

what conditions would you expect to see dark lines in the spectrum?

Q1 -11 The total energy ofthe hydrogen atom is negative. What significance does this have ?

Q1-3 The first excited energy ofa He^ ion is the same as the ground state energy of hydrogen. Is it always true that one of Q1-12 Findoutthewavelength ofthefirst lineoftheHe+ ion the energies ofany hydrogen like ion will be same as the ground ina spectral series whose frequency width isAv = 3.3 x 10^^ s~''. state energy of a hydrogen atom ?

Q1 -4 An atom isinits excited state. Does the probability ofits

Q1 -13 Very accurate measurements ofthe wavelengths oflight

emitted bya hydrogen atom indicate thatall wavelengths are coming to ground state depend on whether the radiation is slightly longer than expected from the Bohr theory. How might already present or not ? If yes, does it also depend on the the conservation ofmomentum help explain this ?[Hint: Photons wavelengthof the radiation present ? carry momentum andenergy, both ofwhich must beconserved.] Q1-5 At room temperature, most of the atoms of atomic hydrogen contain electrons that are in the groundstateor « = 1

Q1 -14 Inthe Bohr model for thehydrogen atom, the closer the electron is to be nucleus, thesmaller isthetotal energy ofthe

energy level. A tube is filled with atomic hydrogen. atom. Is thisalsotrue in the quantum mechanical picture of the Electromagnetic radiation with a continuous spectrum of hydrogen atom ? Justifyyour answer. wavelengths, including those in the Lyman, Balmer, andPaschen series, enters one end of this tube and leaves the other end.

Theexisting radiation is found to contain absorption lines. To which one(or more) ofthe series dothe wavelengths ofthese absorption lines correspond ? Assume that once an electron

Q1-15 The materials (phosphors) that coat the inside of a

fluorescent lamp convert ultraviolet radiation (fi-om the mercuryvapor discharge inside the tube) into visible light. Could one alsomake a phosphor thatconverts visible lighttoultraviolet ?

absorbs a photon and jumps to a higher energylevel, it does not absorb yet another photon and Jump to an even higher

Ejq3lain.

energy level. Explain your answer.

Q1 -16 Consider the line spectrum emitted from agas discharge

tube such as a neon sign or a sodium-vapor or mercury-vapor Q1-6 When electromagnetic radiation is passed through, a lamp. Itisfound thatwhen the pressure ofthevapor isincreased, sample ofhydrogen gasat room temperature, absorption lines the spectrum lines spread out over a larger range of areobserved in Lyman series only. Explain. wavelengths, that is, areless monochromatic. Why ?

Q1 -7 The difference in the frequencies ofseries limit ofLyman Q1 -17 Which wavelengths will be emitted byasample ofatomic series and Balmer series is equal to the frequency ofthe first hydrogen gas (in ground state) ifelectrons ofenergy 12.2 eV line ofthe Lyman series. Explain.

collide with the atoms ofthe gas ?

Q1-8 When an electron goes from the valence band to the

Q1-18 As a body is heated to a very high temperature and

conduction band in silicon, its energy is increased by 1.1 eV.

becomes self-luminous, the apparent color of the emitted radiation shifts from red to yellow and finally to blue as the

The average energyexchanged in a thermal collision is ofthe order of A:/" which is only 0.026 eV at room temperature. Howis a thermal collision able to take some of the electrons from the

temperature increases. Why the color shifts ? What other changes in the character of the radiation occur ?

valence band to the conduction band ?

Q1-9 Doesit takemoreenergyto ionize(free) the electron ofa hydrogen atom that is in an excited statethan onein the ground state ? Explain.

Q1-19 Explain whythe Bohr theory is applicable onlyto the hydrogen atom and to hydrogen-like atoms, such as singly ionized helium, doublyionized lithium, and other one-electron systems.

.34

^

^c• „

Q1 -20 What are the most significant differences between the Bohr model ofthe hydrogen atom and the Schrodinger analysis ofthat atom ? What are the similarities ?

Q1-24 The numerical value of ionization energy in eVequals

theionizationpotential in volts. Does the equality hold ifthese quantities are measured in someotherunits ?

Q1-21 Howmany wavelengths are emitted byatomic hydrogen Q1-25 When the outermost electron inan atom isin an excited in visible range (380 nm- 780 nm) ? In the range 50 nm to .state, the atom ismore easily ionized than when theoutermost 100nm ?

electron is in the ground state. Why ?

Q1-22 Stars appear to have distinct colors. Some stars look red, someyellow, and othersblue.What is a possibleexplanation

Q1-26 Elements in the gaseous state emit line spectra with well-defined wavelengths. But hot solid bodies always emit a continuous spectrum, that is, a continuous smear of wavelengths. Can you account for this difference ?

for this?

Q1-23 What will be the energycorresponding to the first excited state of a hydrogen atom if the potential energy ofthe atom is taken to be 10 eV when the electron is widely separated from

the proton ?Can we still write

=E^/n^ ? ~

iAtomic; Physics

35,

ConceptualMCQsSingle Option Correct 1-1 According to Bohr's theory of the hydrogen atom, the 1-8 Inthefollowing figure-! .26 the energy levels ofhydrogen total energy ofthehydrogen atom with itselectron revolving in atom havebeen shown alongwith sometransitions marked A, the «th stationary orbit is : B, C,Z)and£.Thetransitionsy4,£andCrespectivelyrepresent: (A) (B) (C) (D)

Proportional to « Proportional to Inversely proportional to « Inversely proportional to

OeV

« = 5

•-.544eV

n = 4

1-2 According to Bohr's theory of the hydrogen atom, the radii ofstationary electron orbits arerelated totheprincipal

-0.85 eV

B

« = 2

cc l/«2

(B)

(Q r cc«

cc 1/k

•D

A

quantum number « as :

(A)

E

•C

n = 3

n=\

--1.5eV

• - 3.4 eV

—13.6 eV

p) r Figure 1.26

1-3 The wavelengths involved in the spectrum of deuterium (j D) are slightly different from that of hydrogen spectrum,

(A) The first member of Lymanseries,third member ofBalmer

because'

series and second member of Paschen series

(A) (B) (Q p)

.

.

Size ofthe two nuclei are different Nuclear forces are different in the two cases Masses ofthe two nuclei are different Attraction between the electron and the nucleus is different

in the two cases

P) The ionisationpotential ofhydrogen, second member of Balmer series and third member of Paschen series

(Q The series limit ofLyman series, second member ofBalmar series and second member of Paschen series

.

''

P) The series limit of Lyman series, third member of Balmer series and second member of Paschen series

1-4 The energy of the electron of hydrogen orbiting in a stationary orbit of radius is proportional to :

(A)

.

(Q

1-9 The electron in a hydrogen atom makes a transition from an excited state to the ground state. Which of the following

P) l/r„

statements is true ?

(D) l/^„

(A) Its kinetic energy increases and its potential and total energies decreases

1-5 The shortestwavelengthof the spectrumfor transition of an electron to« = 4 energylevelofa hydrogen likeatom(atomic number = 2) is the same as the shortest wavelength of the Balmer series ofhydrogen atom. The value ofZis:

(A) 2

P) 3 . " " -•

(Q 4

P) 6

1-6 Which of the following series in the spectrum of the hydrogen atom lies in the visibleregion of the electromagnetic spectrum ? (A) Paschen series (Q Lyman series

P) Balmer series p) Brackett series

1-7 The angular momentum of an electron in an orbit is quantized because it is a necessary condition for the compatibilitywith: (A) The wave nature of electron p) Particle nature ofelectron (Q Paulli's exclusion behaviour p) None of these

p) Its kinetic energy decreases, potential energy increases and its total energy remains the same

(Q Its kinetic and total energies decrease and its potential energy increases

P) Its kinetic,potential and total energies decrease 1-10 In the Bohr's model ofhydrogen atom, the ratio of the kinetic energy to the totalenergy oftheelectron in quantum state is:

(A)-l ' (Q -2

.

'

•

• p) +1 p).2 ^

1-11 According to Bohr's theory of the hydrogen atom, the speed ofthe electron in a stableorbitisrelatedto theprincipal quantum number n as (C is a constant) :

(A) (Q v^=Cxn

p) v„=C/h p) v„ = Cx«2

1-12 In the Bohr model of a hydrogen atom, the centripetal forceis furnished by the coulombattraction betweenthe proton and the electron. If is the radius ofthe ground state orbit, m

Atornic-Physics

36%

is the mass and e is the charge on the electron and Bq is the vacuum permittivity, the speed ofthe electron is : (A) 0

(B)

(Q

(D)

1-18 The difference in angular momentum associated with electron in two successive orbits of hydrogen atom is :

(A)f (C)f

^AnSQaQm

1-19 If radiation ofall wavelengthsfromultraviolet to infrared 1-13 If elementswith principal quantumnumber n > 4 werenot allowed in nature, the number of possible elements would be : (A) 60 (B) 32 (Q 4 (D) 64 1-14 ^Bohr's atomic model gained acceptance above all other models because it:

(A) Is based on quantum hypothesis (B) Explained the constitution of atom (Q Assumed continuous radiation of energy by orbiting

is passed through hydrogen gasat room temperature absorption lines will be observed in the

(A) Lymanseries (Q Both (A) and (B)

(B) Balmerseries p) Neither (A) or (B)

1-20 Which ofthe following force is responsible for a-particle scattering ? (A) Gravitational (B) Nuclear (Q Coulomb p) Magnetic

electrons

1-21 A Hydrogen atom and Li"*^ ion are both in the second

(D) Explained hydrogen spectrum

excited state. If£^and aretheir respective angular momenta, andE^ andE^^ their respective energies, then:

1-15 Pauli's exclusion principle states that'no &vo electrons in an atom can have identical yalues for: (A) One of the four quantum nurnbers ^ (B) Two ofthe four quantum numbers (Q Three ofthe fouf quantum numbers (D) All four quantum numbers

(A)

and

P) (Q

= 1^1 and = 1^1 and

p)

and

1-16 Energy levels A, B and C ofa certain atom correspond to

increasing values ofenergy i.e. E^E^>E2

1-23 The wavelength ofradiation emitted due to transition of electron from energy levelE to zero is equal to X. The wavelength

ofradiation (X.j) emitted when electron jumpsfrom energy level 3E -^to zero will be:

h

Figure 1.27

(A)

=

+^

(C) X,] + X2 + X3 = 0

(D)

+A.'

1-17 When white light (violet to red) is passed through hydrogen gas at room temperature, absorption lines will be observed in the

(A) Lyman series (Q Both (A) and (B)

(B) Balmer series (D) Neither (A) or (B)

Figure 1.28

(A) fx

(B) fx

(Q

(D)

!;Atomic_Physics

,

1-24 A neutron collies head-on with a stationary hydrogen atom inground state. Which ofthe following statements is/are correct ?

1-28 Hydrogen H, deuterium D, singly ionized helium He"^ and doubly ionized lithium Li"^all have one electron around the

(A) Ifkinetic energy ofthe neutron isless than 13.6 eV, collision

nucleus. Consider n= 2and w= 1transition. The wavelengths ofthe emitted radiations areXj, and X^ respectively. Then

must be elastic

approximately:

(B) Ifkinetic energy oftheneutron isless than 13.6 eV, collision maybe inelastic

(Q Inelastic collision maytakeplace on when initial kinetic energy ofneutron is greater than 13.6 eV

(D) Perfectly inelastic collision cannottake place

(A) X^ =1X^=242X^=2>42X^ (B) X^ =X^ =2X^ =2,X^ XQ X, = X,=4X^ = 9X, . P) 4X^ =2X^=2X^ = X^

1-25 An electron inhydrogen atom after absorbing- an energy 1-29 Which ofthe following curves may represent the energy photon jumps from energy state to n^. Then it returns to ofelectron inhydrogen atom asa function ofprincipal quantum ground state after emitting six different wavelengths inemission

number n:

spectrum.'The raergy ofemitted photons iseither equal to, less

than or graterth^ the-absorbed photons: Then wj,and (A) = =3 (B) 02 =5,«j-3

(C) ^2 =4,"1 =2

(D) «2=4,ni-i

1 -26 Mark correct statements:

(A) Bohr'stheoryis applicable to hydrogen alonebecause its nucleus is most light

(B) Binding energy ofelectron (inground state) of isgreater than that of jHin ground state (Q Allthe linesofBalmerseries lie in visible spectrum p) None of these

1-27 Figure-1.29 represents transitions ofelectrons from higher tolower state ofa hydrogen atom. Which transition represents the line ofBalmer series:

(C) 0 -E

« = 4'

n = 3

1-30 Ahydrogen atom inground state absorbs 12.1 eV energy. Theorbital angular momentum ofelectron is increased by:

m= 2

(B) f (9 i

«=I

P) zero

Figure 1.29

(A) 1 (C) 3

(B) 2 P) All 1,2, and 3

1-31 Magneticmomentdueto the motionofthe electron in energy state of hydrogen atom is proportional to : p) (A) n (C) p) ,,3

Atomic Physics-

NumericalMCQs Single Optioris Correct 1-1 Theenergy required to excite a hydrogen atom from n = 1 1-8 The different lines in the Lyman series have their to « = 2 energy state 10.2 eV. What is the wavelength of the wavelengths laying between: P) 900 Ato 1200 A radiationemitted bythe atom when it goesback to its ground (A) Zero to infinite

(C) 1000 Ato 1500 A

state ?

(A) 1024 A (Q 1218A

(B) 1122 A P) 1324A

p) SOOAtolOOOA

1-9 The orbitalelectron of the hydrogenatomjumps fromthe

ground state to a higher energy state and itsorbital velocity is

1-2 ConsiderBohr'stheoryfor hydrogen atom.The magnitude reduced to one third ofits initial value. Ifthe radius ofthe orbit ofangular momentum, orbit radius andfrequency oftheelectron in the ground state is r, then what is the radius of the new orbit ? in energystatein a hydrogen atom areL, r &/respectively. (A) 2r P) 3r Find out the value of 'x', if the product f r L is directly (Q 4r P) 9r proportional to if: (A) 0 (C) 2

(B) 1 P) 3

.

1-10 If we assume that penetrating power of any

radiation/particle is inversely proportional to its de-Broglie

1-3 For the first member ofBalmer series ofhydrogen spectrum,

the wavelength is X. What is the wavelength of the second member?

(A) 30^

(B) YgX

(Q 1?^

(D) fx

wavelength of the particle then :

(A) a proton andan a-particleafter getting accelerated through samepotential difference willhave equal penetrating power. P) penetrating power ofa-particlewillbe greater thanthat of proton which havebeen accelerated bysame potential difference. (Q proton's penetrating power will be less than penetrating power of an electron which has beenaccelerated bythe same potential difference.

p) penetrating powers can not be compared as all these are particles having no wavelength or wave nature.

1*4 In a new system of units the fundamental quantities are

planks constant Qi), speedof light (c) and time (7). Then the dimensions of Rydberg'sconstant will be :

(A) (Q

excited states ofhydrogen will be: (A) 1:8 P) 8:27

P) h^c-^r^ P) /z"' c P

c® r'

1-11 According to Bohr's theory the ratio of time taken by electronto completeone revolutionin first excitedand second

(Q 8^:272

P) 4:9

1-5 In a hypothetical atom, if transition from « = 4 to w= 3 produces visible light then the possible transition to obtain 1-12 The area of the electron orbit for the ground state of hydrogenatom is^. What will be the area of the electronorbit infrared radiation is: (A) « = 5 to « = 3 (Q « = 3 to w= 1

corresponding to the first excited state ? (A) 4 A P) 8A

P) rt= 4 to « = 2 p) none of these

(Q 16A

1-6 If first excitation potential of a hydrogen like-atom is Velectron volt, then the ionization energy ofthis atom will be: (A) V electron volt W

AV

1-7 lonisatiorienergy for hydrogen atom in the ground state is

E. Whatisthe ionisation energy ofLi"^ atomin the.2"'' excited state ?

•.

(A) E

P) 3£ -

hydrogen is

(G) n = 6

electron volt

P) cannot be calculated by given information

(C) 6E

1-13 Ifthe wave-number of a spectral line ofBrackett series of 9

times the Rydbergconstant. What is the state

from which the transition has taken place ? (A) « = 4 p) « = 5

P) — electron volt (Q

P)'32A

p) 9:E

.

"•

p) n = l

Ionization potential ofhydrog^ atom is 13.6 V. Hydrogen atom in the ground state is excited by monochromatic radiation ofphotons of energy-12.09 eV. The number of spectral lines emitted by the hydrogen atom, according to Bohr's theory,will be:,

(A)Pne;.

'•(^;';i[iree

' / p) Two •

P)-^dur

„" •'

LAtomic Physics

3£f5

1-15 An electronjumps from the first excited stateto the ground 1-22 In a hydrogen atom following the Bohr's postulates the stage ofhydrogen atom..What will be the percentage change in product of linear momentum and angular momentum is the speed ofelectron ?

(A) 25% (Q 100%

proportional to/T* where '«'is the orbit number. Then 'xMs:

•

(B) 50% (D) 200%

(A) 0 (Q -2

(B) 2 (D) 1

electron from the neutral helium atom. The energy(in eV) required

1-23 The velocity ofan electron insecond orbit oftenlyionized sodium atom (atomic number Z= 11) is v. Thevelocity ofan

to remove both the electron from a neutral helium atom is:

electron in its fifth orbit will be:

(A) 382 (Q 51.8

(A) V

(B) fv

(C) |v

22 (D) fv

1-16 An energyof 24,6 eV is required to remove one of the

(B) 492 (D) 79.0

1-17 A neutron beam, in which each neutron has same kinetic

energy, is passed through a sample of hydrogen likegas (but not hydrogen) in ground state and at rest. Due to collision of neutrons with the ions of the gas, ions are excited and then they emit photons. Six spectral lines are obtained in which one

ofthe linesis ofwavelength (6200/51) nm. Which gasis this ? (A) H (Q

(B) He^ (P) Br3

1-18 In previous questicm whatis theminimum possible value of kinetic energy of the neutrons for this to be possible. The mass of neutron and proton can be assumedto be nearly same. Use

=12400 eVA.

(A) 31.875eV (C) 127.5 eV

(B) 63.75eV (D) 182.5 eV

1-24 A positronium consists of an electron and a positron revolving about their common centre of mass. Calculate the

separation between the electron and positron in theirfirstexcited state:

(A) 0.529A (Q 2.116A .

(B) 1.058A p) 4.232A

1-25 A positronium consists of an electron and a positron revolving about their common centre of mass. Calculate the

kinetic energy of the electron in ground state: (A) 1.51 eV (B) 3.4ey (Q 6.8eV (p) 13.6eV

1-26 A hydrogen atom is in an excited state of principle quantum number n. It emits a photon of wavelength X when

1-19 In Millikan's oil dropexperiment, a chargedoil dropof mass 3.2 X10"^'* kg is held stationary between two parallel

returns to the ground state. The value ofn is :

plates 6mm apart by applying apotential difference of1200 V (A) ^XRCkR-l) between them. How many excess electrons does the oil drop carry? Takeg = 10ms~^: (A) 7 (Q 9

(B) 8 (D) 10

(Q

(B)

XR

(>J?-1) XR

(D)

XR-\

1-27 Theratio ofmagnitude ofenergies ofelectron inhydrogen

1-20 In a hydrogen like atomthe energyrequiredto excitethe

atom in first to second excited states is :

electron from 2"*^ to S"' orbit is 47.2 eV. What is the atomic

(A) 1:4 (Q 9:4

number ofthe atom ?

(A)-2

.

(B) 4:9 P) 4:1

....

(B).3

1-28 Monochromatic rafriatipn ofwavelength Xisincident on a hydrogen sample in ground state. Hydrogen atoms absorba .fractionoflight and subsequentlyemit radiations ofsix difierent 1-21 Twohydrogenatoms are in excitedstate with electronsin wavelengths. Find the wavelength X: w= 2 state. First one is moving towards left and emits a photon.' (Q 4

of energy

(P) 5

-

towards right. Second one is moving towards

right with same speed and emits aphoton ofenergy £"2 towards right. Taking recoil of nixcleus.into account during_emission process : . . / . . •

(A).£j>£2 (C) £j--£'2'

(A)-975 A (C) 2248A

.

(B) 1218A P) 4316A

1-29 Out ofthe following transitions, the frequency ofemitted photon will be maximurn for.:

(B) ./ • •(A)V«^'5t6n'^.3-.. • r;. p) n = 6tOff = 2 (D) information insufficient - : ;(Q.'«^23p«-i';.p

Atomic Physics ;

40:

1-30 Imagine a neutral particle of same mass m as electron revolving around a proton of mass only under newton's gravitational force. Assuming Bohr's quantum condition, the radius ofelectron orbit is given by: (A)

atom equal to: 1

7

(B)

3Z^R

(B)

nm^GMp

16

(Q

2 7-2 GMpN^h

(C)

The longest wavelength of photon that will be emitted has wavelengthX(given in terms of Rydberg constant^? ofhydrogen

nhGM„

P)

4n^m^

^ 4Km

1-31 Determine the ratioofperimeters in 2"^ and 3"^ Bohrorbit in He^ atom:

P)

3Z^R

3Z^i?

1-37 Oneofthe linesin the emission spectrum ofLi2"*" hasthe same wavelength as that of the 2"^^ line of Balmer series in hydrogen spectrum. The electronic transition corresponding to this line is:

(A) f (Q I

(A) n = 4-^/7 = 2 (C)« = 8^n = 4

.(B) ig

(D)f

P)n = 8^rt = 2 P) n=12-^« = 6

1-38 In Bohr's theory the potential energy of an electron at a Kr'

1-32 The photon radiated from hydrogen corresponding to

position is

2"^^ lineofLyman series isabsorbed bya hydrogen likeatom 'A" in 2"^ excited state. Asa result thehydrogen likeatom 'AT makes

quantized energy of the electron in nth orbit is:

a transition to

orbit. Then,

(A) X= He+, n = 4 (Q Ar-He^« = 6

(B) AT- Li++,« = 6 (D) A'=Li^,«-9

1-33 An aparticlewith a kinetic energyof2.1 eV makes a head on collision with a hydrogen atom moving*towards it with a kinetic energy of8.4 eV.The collision : (A) must be perfectly elastic (B) may be perfectly inelastic (Q maybe inelastic P) must be perfectly inelastic

(A)

nh f K 2n\m

(q n{^]

(where A!" is a positive constant); then the

,1/2

(B)

P)

2jcl^m J nh (

1/2

2iz{k)

1-39 Iffirst and second frequencies in transition to

orbital

are relatedbythe relation Vj = kv^, then the first frequency in the transition to second orbital will not be equal to :

(A) V. (i-l)

P) (1-A)V2

1-34 A hydrogen atom is initially at rest and free to move is in the secondexcitedstate. It comesto ground state by emitting a

(Q V2-Vj

P)

photon, then the momentum of hydrogen atom will be approximately: (in kg-m/s)

1-40 The ratio of de-Broglie wave length of a photon and an electron ofmass 'w' having the same kinetic energy £ is: (Speed oflight = c)

(A) 12.1x0-27 (Q 3 X10-2^

p 6.45x10-27 P) 1.5 X0-27

1-35 An gas ofH-atoms in excited state

absorbs a photon

(A)

2mc

P)

mc

\

E

of some energy and jump in higher energy state n^. Then it returns to ground state after emitting sbcdifferent wavelengths in emission spectrum. The energy of emitted photon is equal,

(Q

2mc

mc

P)

~Y

less or greater thanthe energy ofabsorbed photon thenWj and will be:

(A) Kj =5, «2 = 3 =4,«2~3 (Q

P) Wj=5, «2~2 P) =4, M2~2

1-36 Imagine an-atom made of a nucleus ofcharge (Ze) and a hypothetical particle of same mass but double the charge of the electron. Apply the Bohr atom' model and consider all possible transitions of this hypothetical particle to the ground state.

1-41 A monochromatic radiation of wavelength Xis incident on a sample containing He"*". As a result the Helium sample starts radiating. A part of this radiation is allowed to pass through a sample of atomic hydrogen gas in ground state. It is noticed that the hydrogen sample has started emitting electrons whose

maximum Kinetic Energy is37.4 eV. (he = 12400 eVA)Then Xis: (A) 275 A P) 243 A (Q 656A P) 386A

jAtomic Physics

1-42 Ofthefollowing transitions in hydrogen atom, theone which gives an absorption line ofhighest frequency is : (A) n = \ton = 2 (Q « = 2to/7=l

(B)« = 3tort = 8 (D)« = 8to« = 3

1-46 An orbital electron in the ground state ofhydrogen has an angular momentum L,, and an orbital electron in the first

orbit in the ground state oflithium (double ionised positively) has an angular momentum Zj-Then; (A)Zj=Z2 (B)Z, =3Z2 1-43 An electron ofthe kinetic energy lOeV collides with a (C) ^2 =31, (p) L^ =9L^

hydrogen atom in 1st excited state. Assuming loss of kinetic

energy inthe collision tobe quantized which ofthe following statements is INCORRECT.

(A) Thecollision maybeperfectly inelastic (B) The collision may be inelastic (Q The collisionmay be elastic

1-47 The ratio ofthe maximum wavelaigth ofthe Lyman series in hydrogen spectrum to the maximum wavelength in the Paschen series is:

(A)

p) The collision must be inelastic

1-44 Ifin the first orbit ofahydrogen atom the total energyof

3 105

(Q ~

(B)

108

theelectron.is-21.76,x lQ-'%then itselectric potential energy will be:

'

(A) -43.52 X10-19 J (C) -10.88x10-19;

" r '

(B) -21.76x10-19; p) -13.6x10-19;

1-45 In thefigure sixlines ofemission spectrum are shown. Which ofthem will be absent in theabsorbtion spectrum.

4 1

2

5

-X

(Q 4,5,6

Suppose Zj, £2and£3areminimum energies required sothat the atoms H, He% Li"*"*" can achieve their first excited states respectively, then:

(A) £,=£2 =£3 (Q E^£3 p) £,=£2 =£3

1-49 The radius offirst Bohr orbit ofhydrogen atom is0.53 A. Then the radius of firstBohr-orbit of mesonic atom (negative meson hasmass 207times thatofelectron butsame charge) is : (A) 2.85X lO-i^m (B) l.ObxlO-i^m (Q 0.53xlo-iOm

3

Figure 1.30

(A) 1,2,3

1-48 Consider atoms H, He^, Lif^ in tfieif'^buny states.

(P) 7.0xio-i2m

Atomic-Physics

42

AdvanceMCQs with One or More Options Correct 1-1 The ground state and first excited state energies of 1-7 Which ofthe following transitions inHe"*" ion will give rise hydrogen atom are-13.6 eVand-3.4 eVrespectively. Ifpotential toa spectral line which has the same wavelength assome spectral line in the hydrogen atom ? energy in ground state is taken to be zero. Then : (A) potential energyin the firstexcited statewould be20.4eV (A) K= 4tow = 2 P)« = 6ton = 2 total energyin the first excitedstate wouldbe 23.8 eV (Q « = 6 to « = 3 p) n = 8 to « = 4 (Q kineticenergy in the first excitedstate wouldbe 3.4 eV P) total energyin the ground state wouldbe 13.6eV 1-8 In the Bohr model of the hydrogen atom, let R, Vand E represent the radius ofthe orbit, speed ofthe electron andthe 1-2 An electron is excited from a lower energy state to a higher magnitude of total energy of the electron respectively. Which energy state in a hydrogen atom. Which of the following of the following quantities are proportional to the quantum quantity/quantities decreases/decrease in the excitation ? number «? (A) Potentialenergy (B) Angular speed (A) VR (^) RE (Q Kinetic energy

P) Angular momentum

1-3 An electronin a hydrogenatom makes a transition

{D)|

(Q f

where nj and ^2 ^re principle quantum numbers of the two states. Assume the Bohr model to be valid. The time period of the electron in the initial state is eight times that in the final

1-9 In an electron transition inside a hydrogen atom, angular momentum ofelectron may change by

state. The possible values of and «2are • (A)«j-4,«2 = 2 P) /7^ = 8,W2 = 2

(A) h

(Q nj=8,«2~l

P) «i=6,«2^2

1-4 An electron in hydrogen atom first jumps from second

P) -7C ^ h

(Q

2k

(D) A 471:

excited state to first excited state and then from first excited

state to ground state. Let the ratio of wavelength, momentum and energy of photons emitted in these two cases be, a, b and c respectively. Then:

(A)c=i

P) fl= 9/4

(C) 6 = 5/27

P) c=5/27

1-10 A beam of ultraviolet light of all wavelength passes

through hydrogengas at roomtemperature, in thex-direction. Assume that all photons emitted due to electron transitions insidethe gas emergein the y-direction. Let^ and B denotethe lights emerging from the gas in the x-and y-directions respectively:

1-5 The magnitude of energy, the magnitude of linear

(A) Someof the incident wavelengths will be absentin A P) Only those wavelengths will be present in B which are

momentum and orbital radius ofan electron in a hydrogen atom

absent in A

corresponding to the quantum number n are E, P and r respectively. Thenaccording to Bohr's theoryofhydrogen atom :

(Q B will contain some visible light P) B willcontainsomeinfraredlight

(A) EPr is proportional to —

1-11 Whenever a hydrogenatom emits a photon in the Balmer

P) P/E'is proportional to « (Q Er is constant for all orbits P) Pr is proportional to n

series:

(A) It may emit another photon in the Balmer series p) It miist emit another photon in the Lyman series (Q The secondphoton, if emitted, will have a wavelength of

1-6 The wavelengths and frequencies of photons in transitions 1,2 and 3 for hydrogen

about 122 nm

(b) It may emit a second photon, butthewavelength ofthis photon cannot be predicted

likeatom areX.,, Xj, >13, vj, V2 and V3 respectively. Then : (A) V3 = Vj+V2 v,v-

P) v,=

(Q ^ ~

V777777777777777777777777Z,

Figure I.OO

V1+V2

^

X.iX,2

1-12 Which of the following statements about hydrogen spectrum is/are correct ? (A) All the lines ofLyman series lie in ultraviolet region P) All the lines ofBalmer series lie in visible region

(C) All the lines of Paschen series lie in infrared region P) none of these

[Atomic Physfcs

43:

1-13 Aneutron collides head-on with a stationary hydrogen

1-18 Suppose the potential energy between electron and

atomin groundstate. Which ofthe following statements is/are correct ?

proton at adistance r is given by-^^-.

(A) ifkinetic energy oftheneutron isless than 13.6 eV, collision

2r^ Using Bohr's theory choose the correct statements :

must be elastic

(B) ifkinetic energyoftheneutron is less than 13.6 eV, collision

(A) Energy in the nth orbit is proportionalto (B) Energyin the nth orbitis proportional to

maybe inelastic

(C) Energy is proportional to rrp- {m :mass ofelectron)

(Q inelastic collision maytakeplaceonlywheninitialkinetic p) Energy isproportional to

(m: mass ofelectron)

energy ofneutron is greater than 13.6 eV

(D) perfectly inelasticcollision cannottakeplace

1-14 If, in hydrogen atom, radius of frequency ofrevolution ofelectron in

1-19 An electron in an hydrogen atom has total energy of

Bohr orbit is r^, orbit is^ and area

enclosed by w''* orbit is A^, time period ofelectron is

then

which of the following graphs is/are correct ?

Mi. (A)

1-20 When Z is doubled in an atom, which of the following statements are consistent with Bohr's theory ? (A) Energy of a state is double

(B)

logn log

(Q

-3.4 eV. Choose the correct statement (s): (A) The kinetic energy ofthe electron in that orbit is 3.4 eV P) The potential energy ofthe electronin that orbit is- 6.8 eV (Q Angular momentum ofthe electron in that orbit is h/n p) Angular momentum ofthe electron for that orbit is 2h/%

IT

P) Radius of an orbit is doubled (C) Velocity ofelectrons in an orbit is doubled p) Radius of an orbit is halved

1-21 The electron in a hydrogen atom jumps back from an excited state to ground state, by emitting a photon of 16 ^ wavelength , whereis Rydberg's constant. In place

(D)

15 K

logn

ofemittingonephoton, the electron couldcomebackto ground state by

1-15 Mark correct statement(s): (A) Bohr's theory is applicable to hydrogen alone because its nucleus is very light

(A) Emitting 3photons ofwavelengths

and ^3 such that

15R 16

^2 ^3 (B) Binding energy ofelectron (in ground state) ofjrf is greater (B) Emitting 2 .photons of wavelength A,,and than that ofjH' in ground state

such that

(Q all the lines ofBalmer series lie in visiblespectrum

J_

1

(D) none of these

A,

X2 ~ 16

1-16 A photon of energy 10.5 eV isallowed to interact with a hydrogen atom in its ground state. Then :

(C) Emitting 2 photons of wavelength X^ and X^ such that

15R

16

(A) the photon is completely absorbed bythe H-atom (B) the photon cannot excite the H-atom and comes out with energy 10.5 eV (Q thephoton transfers 10.2eV energyto H-atomexcitingit to

(D) Emitting 3photons ofwavelength Aj, Aj andA3 such that 16

first excited state

P) none of these

1-22 The photon radiated from a hydrogen atom corresponding

1-17 When a hydrogen atom is excited from ground state to first excited state:

(A) Its kinetic energy increases by 10.2 eV (B) Its kinetic energy decreases by 10.2 eV (Q, Its'potential energy increases by 20.4 eV

(D) Itsangular momentum increases by1.05 x lO'^^ J-s

to 2"^^ line ofLyman series isabsorbed by a hydrogen like atom 'JT in 2"^ excited state. Asa result thehydrogen likeatom 'X" makes a transition tow'^orbit. Then predict 'X and the quantum number:

(A) He"',«= 4 (Q X=He\n = 6

(B) X=Li^,n = 6 (D) X=Li^,n^9

Atomic Physics ;

44

1-23 A particular hydrogen like atom has its ground state (A) Ifkinetic energy oftheneutron islessthan20.4eVcollision must be elastic binding energy 122.4 eV. It is in ground state. Then p) Ifkineticenergyoftheneutronislessthan 20.4eVcollision (A) Its atomic number is 3 maybe inelastic (B) An electron of90eV can excite it (Q An electron ofkineticenergynearly91.8eVcan bebrought (Q Inelastic collision may be take place only when initial kinetic energy ofneutron is greater than 20.4 eV to almost rest by this atom P) An electron of kinetic energy 2.6 eV may emerge from the P) Perfectly inelastic collision can not take place atom when electron ofkinetic energy 125 eV collides with this 1-28 The figure above shows an energy level diagram for the atom hydrogen atom. Several transitions are marked as I, II, III, Momentum ratio ofphotons is . The diagram is only indicative and not to scale. 1-24 If radiations ofallowed wavelengths from ultraviolet to infrared are passed through hydrogen gas at room temperature, absorption lines will be observed in the nTT (A) Lyman series (B) Balmer series c 3 t.--

(C) Both (A) and ( B ) ;

-

:

-

. O-rg'. •

"

II-:-

.-rv:

•:y

:v.i

37

1-25 In the hydrogen atom, ifthe reference level of potential energy is assumed to be zero at the ground state level. Choose the incorrect statement.

(A) The total energy ofthe shell increases with increase in the valueof«

Figure 1.00

(B) The total energy ofthe shell decrease with increase in the value of n

(A) The transition in which a Balmer series photon absorbed

(Q The difference in total energy of any two shells remains

is vi:

the same

(B) The wavelength oftheradiation involved in transitionll is

P) The total energy at the ground state becomes 13.6eV

486 nm.

1-26 Choose the correct statement(s) for hydrogen and deuterium atoms (considering the motion ofnucleus) (A) The radius offirst Bohr orbit ofdeuterium is less than that of hydrogen P) The speed ofelectron in first Balmer line of deuterium is more than that of hydrogen (Q The wavelength offirst Balmer line ofdeuterium is more than that of hydrogen P) The angular momentum ofelectron in the first Bohr orbit ofdeuterium is more than that ofhydrogen

1-27 A neutron collides head-on with a stationary hydrogen atom in ground state. Which of the following statements are correct (Assume that the hydrogen atom and neutron has same mass)

(Q IV transition will occur when a hydrogen atom is irradiated with radiation ofwavelength 103nm. p) IV transition will emit the longest wavelength line in the visible portion of the hydrogen spectrum.

1-29 A hydrogen atomis in the 4^^^ excited state, then: (A) the maximum number ofemitted photons will be 10 p) the maximum number ofemitted photoris will be 6 (Q it can emit three photons in ultraviolet region P) if an infrared photon is generated, then a visible photon may follow this infrared photon 1-30 An electron with kinetic energy £" eV collides with a hydrogen atom in the ground state. The collision is observed to be elastic for :

(A) 0

=4.9-4.5 = 0.4eV

=1.5eV

# Illustrative Example 2.4

IfE be the energy of each ejectedphoto electron momentum of electrons is

P= ^2mE

Light ofwavelength 1800 Aejects photoelectrons from aplate ofa metal whose workfunction is 2 eV. If a uniform magnetic

field of5 XlO'^tesla isapplied parallel toplate, what would be Weknowthat changeof momentum is impulse. Herethewhole the radius of the path followed byelectrons ejected normally momentum of electron is gained when it is ejected out thus

from theplatewithmaximum energy.

impulse on surface is given as Solution

J= ijlmE Energy of incident photons in eFis given as

Substituting the values, we get maximumimpulse 12431

J= V2x9.1xI0"^'x0.4xl.6xl0"'^

1800

Aswork function ofmetal is2 eV, themaximum kinetic energyof ^

y = 3.45X10"^^ kg m/sec.

ejected electrons is

# Illustrative Example 2.3 In an experiment tungsten cathode which has a threshold 2300 A

=>

^nm = 6.9-2eV '^n^=4.9eV

isirradiated byultraviolet lightofwavelength 1800 A.Calculate (i) Maximumenergyof emittedphotoelectron and

If

be the speed of fastest electrons then we have

(ii) Work flmction for tungsten.

-4.9 X1.6x10-19joule

(Mention both the results in electron-volts) Given Planck's constant./; = 6.6 x 10"^"* joule-sec,

2x4.9x1.6x10-^®

leV= 1.6 X10~'^joule andvelocity oflightc = 3 x lO^m/sec

9.1x10"^'

=>

Vjj^ = 1.31 XlO^m/s

Solution

The work function of tungsten cathode is

When an electron with this speedentersa uniform magnetic field normallyit follows a circular path whose radius can be given by

,

he

(i)=T— ^th

,

mv r =

qB

[As qvB=-^]

I243I

9.1x10"^'xl.31xl0^ r =

1.6x10-^^x5x10"^ =>

(i)=5.4eV r=0.149m

Photo Electric Effect & Matter Waves 5

,54

(iii) The energy of incident radiation is 1.89 eV thus the

# Illustrative Example 2.5

corresponding wavelength is The radiation emitted, when an electron jumps from « = 3 to n = l orbit is a hydrogenatom, falls on a metal to producephoto

X=

12431 1.89

•A

electrons. The electrons from the metal surface with maximum

=>

kinetic energy are made to move perpendicularto a magnetic field of(1/320) T in aradius of 10"^ m.Find(i)thekinetic energy

X=6.577.2 A.

# Illustrative Example 2.6

ofelectrons (ii) work function ofmetal and (iii) wavelength of radiation.

Photoelectrons are emitted when 4000 A radiation is incident on a surface of work function 1.9 eV. These photoelectrons

Solution

passthrough a region containing a-particles. A maximum energy electroncombinewith an a-particle to form a He"*" ion, emitting a single photon in this process. He"^ ions thus formed are in

(i) The energy ofelectron in the atom is given by

„

energy level ofhydrogen

their fourth excited state. Find the energies (in eV) of the photons,lying in the 2 to 4 eVrange, that are likelyto be emitted

13.6 ,,

E__ = -—r- eV

during andafterthe combination. Take/j= 4.14 x 10"'^eVs.

For w= 3, we have „

£3=

13.6

Solution

. ..

^=-1.51 eV

(3)'

The energy ofthe incident photon is

and for « = 2, we have

£ =}j^= hp. = 124M Y

£2=-^=-3.4eV '

Now,

X

=>

(2)2

AE32 = £3-£2=-1-51 + 3.4

4000

£. = 3.1 eV

From Einstein's photoelectric equation, the maximum kinetic energy of the emitted electrons is

-1.89eV = /iv

Now these radiations incident on a metal surface. The photo electrons emitted from the metal surface move perpendicularly in magnetic field B. Let v be the velocity of photo electrons in a

KEITiuX =/A'-d)=3.1eV-1.9eV=1.2eV '

It is given that

(electron with £^3^) +He ^ He"'" +photon(in 4*^ excited state).

circle ofradius r. Then,

The fourth excitedstatecorresponds to w= 5. ForHe"^ ion Z=2. Now,we know that the energyof the electron in the state for a hydrogen like ion is

.2 mv

= qvB-

mv = qBr

£„=-(13.6eV)4-

...(2.6)

n

/7 = 5 X10~^kg-m/s

Thus,

The kinetic energy of photo electrons is

^

p^ 2m

£, =-(13.6eV) x

'

.

(2)2

(5)'

=-2.18eV

The energy of the emitted photon in the above combination ICXIU

I

2x(9.1x.10'^31^

reaction is

J-

25x10"^®

2x(9.IxlO~^')x(1.6xlO-'^)

^ = ^n,ax + (-^5) eV

£^»0.86eV (ii) According'to Einstein's photo-electric equation

/iv =(l)+jmvJ^

=>

£=1.2eV + 2.18eV = 3.38eV

This energy is within the range 2 to 4 eV. After the recombination reaction, the electron may undergo

transitions from a higher level to a lower level, thus emitting

photons. Using Equation-(2.6), the energies in the lo^r electronic levels ofHe"'' ions are

=>

1.89= (t)+0.86

=>

(t)=1.03eV

'

(-13.6eV)> (j)), photoelectrons are ejected and move toward anode with negative polarity.

Intensity/^ >/, Frequency v (same for both radiation)

Incident light v>v.,

^0

Cathode

0

Anode

Reverse voltage Figure 2.10

This voltage V^, we can calculate from equation-(2.9) by

substituting v^= 0hence 1

2

yWVmax -eVQ=0 V

Figure 2.9

jr I 2 eVQ=^mv^^

Now theelectrons which areejected with very low kinetic energy are attracted back tothecathode because ofitspositive polarity. Those electrons which have high kinetic energies will rush

1

toward, anode and may constitute the current in circuit.

^0=

In this case the fastest electron ejected from cathode will be

^0=

retarded during its journey to anode. As the maximum kinetic 1

2

..(2.10)

/?V-(j)

..(2.11)

We cansee onemore thinginfigure-2.10 thatthegraphs plotted

7

energyjust after emission at cathode is j ifpotential for two different intensities /j and1^, Vq issame. Current inboth difference across the discharge tube is Kthen the speed v^with the cases incut offatsame reverse potential V^. The reason for this is equation-(2.10) and (2.11). If is clearthat the valueof

which electrons will reach anode can be given as I

2

•-eV=

•/

...(2.9)

Thus alltheelectrons which are reaching anode will have speed less then or equal to . Remaining electrons which have relatively low kinetic energy will either be attracted to cathode justafter ejection orwillreturn during theirjourneyfrom cathode

depends onlyon the maximum kinetic energy of the ejected electrons which depends only onfrequency oflightandnoton intensity of light. Thus in above two graphs as frequency of incident light is same, the value of

potential difference

is also same. This reverse

at which the fastest photoelectron is

stopped and current in the circuit becomes zero is called cut off

potential or stopping potential.

Photo Eiectric Effect & Matter Waves

58

2.3.4 Effect of Change in Frequency of Light on Stopping Potential

If we repeat the experiment by increasing the frequency of incident light with number of incident photons constant, the variation graph ofcurrent with voltage will beplotted as shown in figure-2.11.

Figure 2.12

Frequency (vj > v,)

Hereequation-(2.14) can be writtenas

=eFo= /j(v-vJ

...(2.16)

This equation-(2.16) is called Einstein'sPhotoElectricEffect equation which gives a direct relationship between themaximum kinetic energy stopping potentialfrequency of incident light and the threshold frequency.

Figure 2.11

# Illustrative Example 2.7

This graph is plotted fortwo incident light beams of different

frequency Vj and Vj and having same photon flux. As the Find the frequency oflight which ejectselectronsfrom a metal numberofejected photoelectrons are same in the twocases of incident light here we can see that the pinch offvoltage ^01 as

surfecefully stoppedbya retarding potential of3 V. The photo

well as saturation current

Find the work function for this metal.

are same. But as in the two cases

electric effect begins inthis metal atfrequency of6 x 10'*^ sec"'.

thekinetic energy offastest electron are different asfrequencies are different, the stopping potential for the two cases will be Solution different. In graphIIas frequency ofincident lightis more, the maximumkineticenergyofphotoelectron willalsobehigh and The threshold frequencyfor the given metal surfaceis to stop it high value of stopping potential is needed. These Vjj^ =6xlO"'Hz here

and

can be given as

Thus the work function for metal surface is

...(2.12)

=> and

V •^02 =

hV2

...(2.13)

In general for a given metal withwork function cj), if Vq is the stopping potential for an incident light offrequency v then we

.

(j)=3.978x 10-19 j

As stopping potential for the ejected electrons is 3 V, the maximum kinetic energy ofejected electrons will be

have

KE=1>qW

eV^-hv-^ =>

eVQ =.hv-hv^

(t)-6.63xlO"3tx6xlo"'

...(2.14)

...(2.15)

KETn3X =3x l.6x 10-19J

SJ_ =4.8X10-"J According to photo electric effect equation, we have hv = hv,,In +

max

Equation-(2.15) shows that stopping potential is linearly proportional to the frequency v ofincidentlight.The variation => frequency ofincident light is (j)-Iof stopping potential with frequency v can be shown in figure-2.12.

jpfvoto-Elcctrtc-Effect &Matter Waves

59

3.978x10"'^+4.8x10"'^ 6.63x10"^^

V =

According to photo electric effect equation we have

v = 1.32x1015Hz

^max-^-^l'

# Illustrative Example 2.8

.

^S_=2.85-1.24eV

^

^max = l-61eV

Electrons with maximum kinetic energy 3eV are ejected from a metal surface byultraviolet radiation of wavelength 1500 A. The stopping potential for these ejected electrons can be given Determine the work function of the metal, the threshold wavelength of metal and the stopping potential difference required to stop the emission of electrons.

Solution

as

y _ ^^max

^

„

Vf,=

1.6leV

,

,

= 1.61 volts.

Energy of incident photon in e Fis „

# Illustrative Example 2,10

12431

Determine the Planck's constant h if photoelectrons emitted =>

E:=8.29ev

from a surface of a certain metal by light of frequency

According to photo electriceffect equation, we have E = ^^KE T

=>

^-=E-KEmax

"

^=8.29-3eV

=>

(j)=5.29eV

Solution

From photo electric effect equation, we have

Threshold wavelength for the metal surface corresponding to work function 5.29 eV is given as .

Here

/iVj =(f) + eFQj

...(2.17)

and

hv^ = isf + 2eV^

...(2.18)

Subtracting equation-(2.17) from equation-(2.18), weget

_ 12431 ,

5.29 ^

/i(v2-Vi) = e(Fo2-Fo,)

=2349.9 A

Stopping potential for the ejected electrons can be given as ^max

and those ejected by light of frequency 4.6 x lo'^ Hz by a reverse potential of 16.5 eV.

max

=>

2.2 X10'^ Hz are fiilly retarded by a reverse potential of6.6 V

3eV

(n2~^oi)(l-6xlO-^^) h =

(V2-V,)

= 3 volt h =

# Illustrative Example 2,9

(16.5-6.6)(l.6xl0-^^) (4.6-2.2)xl0'^

/z = 6.6x 10-34j-s.

Calculate the velocity ofa photo-electron, ifthe work function ofthe targetmaterial is 1.24eV and the wavelength ofincident

light is4360 A. What retarding potential isnecessary tostop the emission of the electrons ? Solution

Energy of incident photons in e Fon metal surface is „

^

'

12431

.e=2.85eV

•

# Illustrative Example 2,11

When a surface isirradiated with lightofwavelength 4950A, a photo currentappears whichvanishes if a retarding potential greater than 0.6 volt is applied across the photo tube. When a different source of light is used, it is found that the critical retarding potential is changed to 1.1 volt. Find the work function

ofthe emittingsurfaceand the wavelength ofsecondsource. If the photo electrons (after emission from the surface) are subjected to a magnetic fieldof 10tesla, what changeswill be observed in the above two retarding potentials.

Photo Electric Effe^& Matter.Waves j

60

Solution

Solution

In first casethe energy of incident photon in eV is

(a) Let the work function of the surface be If v be the frequency ofthelightfalling onthe surface, thenaccording to Einstein's photoelectric equation, the maximum kinetic energy

12431 •

4950

eV

KE

=2.51 eV

tTlHX

ofemitted electron is given by

The maximum kinetic energyof ejectedelectrons is We know that, = 0.6eV

Thus work function ofmetal surface is given as

=>

\^^ere

T7

^

(j)=2.51-0.6 eV (l)=1.91eV

=cut-offpotential.

^

In second case the maximum kinetic energy ofejected electrons

A

eK„=^-(i> »

eX

AV = V

Now,

^^0

will become

^02

-V

^01 he

^niax2^^^02

eX-,

£2 = 1.91 + l.leV

EVo=0 e

eXi

e

J X2

Thus the incident energy of photons can be given as

=>

e

e

1_ X^j

X^-X2 X.]X.2

...(2.19)

(b) Fromequation-(2.19),wehave

Thus the wavelength of incident photons in second case will

he _ AF"o(A.iX.2)

be

c

^ =>

3.01 ^

he

X-4129.9A

—X2

-10(1.3 - 0.9)[(4000 X10"'") X(4500x 10"*'®)]

500x10

-10

When magnetic field is present there will be no effect on the

stopping potentials asmagnetic force can notchange thekinetic

— =1.44xlO-^V/m e

energy of ejected electrons. We use.

# Illustrative Example 2.12

e

cell be reduced from Xj to Xj A, then what will be the change in =>

(b) Light is incident on the cathode of a photocell and the stopping voltages are measured ,for light of two different wavelengths. From the data given below, determine the work function of the metal of the cathode in eV and the value of the universal constant hde.

Wavelength (A)

Stopping voltage (volt)

4000

1.3

4500

• 0.9

eX e

^ - h£--v

(a) If the wavelength of the light incident on a photoelectric the cut-off potential ?

^0

eX

-6

® 4000x10 -10

-1.3

(j)=2.3eV

# Illustrative Example 2.13

A lowintensity ultraviolet lightofwavelength 2271Airradiates a photocell made ofmolybdenum metal. Ifthe stopping potential is 1.3 V, find the work fimction ofthe metal. Will the photocell work ifit is irradiatedby a high intensity red light ofwavelength 6328 A?

fPhoto'E(6cyjc Effectac MatFer Waves 61

Solution

Theenergy in

f-.,

incident photons is Hence, 2271

=>

Which gives T'=2T.

£'=5.47eV

As stopping potential for ejected electrons is1.3 V, the maximum kinetic energyofejectedelectronswill be

It is given that peak emission at temperature T occurs at a

wavelength = 9000 A. If is the wavelength for peak emission at temperature T', then from Wien's displacement law\,T= constant, we have

^

^ _ = 1.3ev

Nowfrom photoelectric effect equation, we have E = ^+KE * max

^'„ =^„|!-=9000Ak^=3000A. According to the problem, the kinetic energy of the emitted

photo-electrons = excitation energyfor transition «=2to«=3,

=>

i>=E~KE T max

=>

(i)=5.47-1.3eV

which is

^nw=^32={13.6eV)[^-i

(l)=4.17eV

Energy in eV for photons ofred light ofwavelength 6328 A is 6328

,

^™ =13.6'

if)=4.14-1.89eV

=>

(})=2.25eV

from the surface. After the increase oftemperature the peak radiation from the black body caused photoemission. To bring

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these photoelectrons to rest, a potential equivalent to he

Age Group-High School Physics | Age 17-19 Years

excitation energy between the « = 2 and « = 3 Bohr levels of hydrogen atom isrequired. Findthe workfunction of the metal.

.Section-MODERN PHYSICS Topic - Photoelectric Effect Module Number-14 to 18

'




-

£

due to wavelength

4144A =(4144xl0-i0m)

=>

plate

2.425x10'^

4x(3.14)x(2)2 A7=4.82xl0'6m-2s-'

The maximum kinetic energy of the photo-electrons emitted

from theplate having work ftinction ij) = 1.17 eVis given by

=>

'^max-2.58-1.17

^

^max = l-41eV

Themaximum velocity ofphoto-electrons ejected isgiven as

(1.2xl0"^)(4144xl0-'^) (6.63xl0"^^)(3xl0^) Kj =0.5x1012

Number ofphotons

y'«vL=1.41eV 2x1.41x1.6x10"'^ 9.1x10"^'

dueto wavelength 4972 A 10(2.4xl0"^)(4972xl0~'") (6.63xl0"^^)(3xl0^)

=>

=>

llktX

= 7.036 X 10^ m/s.

Theradius ofthe circle traversed byphoto electron inmagnetic field B is given by

=

Total photonsA7=«j +«2="0.5 x 10^2 +0.575 x 10'2 =:>

joule

~_Ek

"" {hcl-k)~ he Number of photons

4.125x10

Number ofphotons striking per squaremeter persecond on the

Energy incident on surfecefor each wavelength in 2 seconds.

The number of photons of wavelength X, are

^

=

/•= 2.424x lO'Vs.

X(1(^ m2)

£ = (1.2 X10-7) X(2)-2.4 X10-'joule

—

£7-1.075x 1012

r =

=>

mv_(9.1xlQ-^^)(7.b36xlQ^) (1.6xl0"'^)x(10"'')

mv

,

[Asg-vP-——]

j-=40.0 x-l0"^metre=4.0cm.

# Illustrative Example 2.23

a Illustrative Example 2.24

A smallplateofa metal (workfimction =1.17 eV)is placedat a distance of 2 m from a monochromatic light source of

A40 Wultraviolet-light source ofwavelength 2480 A illuminate a magnesium {Mg) surface placed 2 m away. Determine the number ofphotons emitted from the source per second and the numberincidenton unit areaof the Mg surfece per second. The photo-electric work function for Mg is 3.68 eV. Calculatethe

wavelength 4800 Aand power 1.0 watt. The light falls normally on the plate. Find the number of photons striking the metal plate per squaremeter per second. If a constantmagnetic field

ofstrength 10^ tesla is applied parallel to the metal surface, findthe radiusofthelargest circular pathfollowed bythe emitted photo electrons.

kinetic energy of the fastest electrons ejectedfrom the surface. Determine themaximum wavelength forwhich thephotoelectric effect can be observed with Mg surface.

Solution

Solution

Energy of incident photon ineVis

The energy of incident photon in eFis given by

r,_ 12431

-

4800

zr

12431

^=1480"^^

£=2.58eV

=>

£ = 5.01 eF

^

£-2.58 X 1.6 X 10-'9j

=>

£ = 5.01x1.6x10-19;

=>

£=4.125x10-'^

=>

£ = 8xl0-'9j

Photo Electric

68

The number of photons emitted per second bythe source can be given as

MatterJ/Vayes .

Energy required toeject photo-electron isgiven by (2.22eV)(L60x 10-'^J/eV) =3.55x Now exposure time

3.55xlO"'^J

40 N= 8x10

-19

t =

= 5x lO'^s-i

0.1 W

47tm

The number of photons incidenton Mg surface per unit area

x(47ix10"^V^)

/ = 355s.

per second 5x10 N' =

19

4Tcr^

5x10

19

4x3.14x(2)^

7y' = 1.04x lO'^s-^m-^

12431

(b) Incident photon energy \neV\sE= ^5^ ^^ =>

£:=4.9eV

£ = 4.9x1.6x10-'^

The K.E. of the fastest electron is given by ^

KE max =E-^T

^

2

KE^ = 5.01 eV- 3.68 eV= 1.33eV TTIdX

£=7.84xl0-'9j

(c) Atthe cathode (area 4x10"^ m^), the photon flux is

Now the threshold wavelength for Mg surfece can be given as N=

for its work function 3.68 eKis

X"h=^A 3.68

'0.1 W^ 4xlQ^m^ ,47c m^J 7.84x10"'^ J

V=4.06 XlO'^photons/s

X^ =33'1SA

With an efficiencyof 5%, the photo-current is

1=0.05Ne- (0.05) (4.06 x 10'^) (1.60 x 10"'^C)

# Illustrative Example 2.25

7=3.25 X10-«A=32.5 nA

A mercury arc lamp provides 0.10 W of UVradiation at a

wavelength ofX= 2537 A (allother wavelengths having been absorbed by filters). The cathode of photoelectric device (a photo-tube) consists of potassium and has an effective area of

(d) The stopping potential for ejected electrons can be given as

_ ;iv-(t)o _ 4.86 eV-2.22 eV 0

4 cm^. The anode is located at a distance of 1 m from radiation

source. The work function forpotassiumis (t)^ = 2.22 eV.

=>

e

e

Ko=2.64V

(a) According to classical theory, the radiation from the arc spreads outuniformly in space as spherical wave. Whattimeof # Illustrative Example 2.26 exposureto the radiation should be required for a potassium atom(radius 2 A) in the anodeto accumulate sufficient energy Amonochromatic point source Sradiating wavelength 6000 A, withpower 2 watt, an aperture A of diameter 0.1 m and a large to eject a photo-electron ? screen SOare placed as shown in figure-2.17. Aphotoemissive (b) What is the energy of a single photon from the source ? detector D ofsurface area0.5 cm^ is placedat the centreofthe (c) What is the flux of photons (number per second) at the screen (see the figure)." The efficiency of the detector for the cathode ? To what saturation current does this flux correspond photoelectron generation per incident photon is 0.9. if the photo-conversion efficiency is 5% (i.e., if each photon (a) Calculate the photon flux at the centre of the screen and has a probability of 0.05 of ejecting an electron). (d) What is the cut off potential

the photocurrent in the detector. SC

? A

-1,

Solution S.

(a) The UV energy flux at a distance ofone metre =

0.1

'.J

D

1 \

wattper square metre.

Cross-sectional area of atom A = tzP'

=>

\ /

.4= 7c(2x 10~'®m)^ = 4;rx lO-20m^

6 m

Figure 2.17

•»

; Phdto Efectric Effect

Matter Waves

69

(b) If a concave lens L of focal length 0.6 m is inserted in the aperture as shown, find the new values of photon flux and photocurrent. Assume uniform average transmission of 80%

Thus the concave lens forms its image at a distance//2 i.e., 0.3 metre on the left of the lens. This we get from lens formula

1 = 1_1

from the lens.

/

(c) Ifthe work function ofthephotoemissive surface is 1 eV, calculate the value, of the stopping potential in the two cases (without and with the lens in the aperture).

V

u

1 = 1+1 = ^ + ^ V

f

u

-0.6

-0.6

1

0.3

v=-0.3m Solution

Now all the photons are focussed at a distance 0.3 m from the lens or 5.7 m from the detector. The detector transmits 80% of

(a) Energy of Photon is

^ he 6.63x10"^^ x(3xl0^)

photons. So the number of photons reaching the detector 80

n'=

~ ~ 6000x10^'® £ = 3.315 X10-'>ule

Solid angle subtended by detector Solid anglesubtended by aperture

Number ofphotons emitted per second from the source are

80

Power n

=

n

=

n'= TrtTT Xphoton flux x

^

(0.5x10"*/5.7)^

Tlx (0.05)2z 7(0.3)27

Substitute the value of photon flux and obtain the value of

2

=>

100

Energy of each photon 3.315x10

Xphoton flux

-19

« = 6.033 x lO'^photorisec,-l

Here the solid angle subtended by detector is less than that subtended by aperture on the source. This shows that all the photons reaching the detector are allowed by aperture. In this

Photon flux

(bp =

^P =

=>

,—

area of detector

0.5x10

-4

per m^ per second

({)p =2.956 X10'Vm^ sec.

way the number of photons reaching the detector per m^ per Thus photon current = 0.9 x Photon flux x e

second is given by Photon flux

(bp =

7 6.033x10

18

4x(3.14)x(6.0)^ =>

=>

/= 0.9XPhoton flux X1.6x 10"'^Amp.

=>

7=0.0213 pA.

m -2c-l

(j)p = 1.33 XlO'^photons/m^-sec

(c) The stopping potential is independent ofphoto-current. In both the cases, according to Einstein's photo-electric equation, this is given by

Photo current = ne

E= W+eV^ 1= [9.0 Xphoton flux x area ofdetector] e

=>

/=[0.9 XphotonfluxX(0.5 x 10^)] x 1.6x 10"'^ Amp.

=>

/=0.096 pA.

=>

3.315 X 10"'^ = leK+eK 3.315x10

-19

1.6x10"*^

(b) The source is at the focus of concave lens.

eV=\eV+eV.

2MeV=\eV+eV^ K^=2.06-l K= 1.06 volt.

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Age Group - High School Physics | Age 17-19 Years

0.3 m

Section - MODERN PHYSICS r«--0.6 m--W 5.7 m-

Topic -Wave Particle Duality Module-Number - 3 to 11

Figure 2.18

Photo Electric Effect &-Matter Waves •

70

Practice Exercise 2.3

(vii)

When the sun is directly overhead, the surface of the

earth receives 1.4 x 10^ W/m^ ofsunlight. Assume that the light

(i) Calculate the number ofphotons emitted per second by ismonochromatic with average wavelength 5000 Aand that no a 10 watt sodium vapour lamp. Assume that 60% of the light is absorbed in between the sun and the earth's surface. consumed energy is converted into light. Wavelength of the The distance between the sun and the earth is 1.5 x 10" m. (a) Calculate the number ofthe photons falling per second sodium light = 5900 A. on each quare meter of earth's surfacedirectlybelowthe sun. [1.7 X 10"] (b) How many photons are there in each cubic meter near

(ii)

the earth's surface at any instant ?

Abulb lamp emits light ofmean wavelength of4500 A (c)

The lamp is rated at 150 watt and 8% of the energy appears as emitted light. Howmany photons are emitted bythe lamp per

How manyphotons does the sun emits per second ?

[3.5 X lO^', 1.2 X 10'3, 9.9 X 10''^]

second.

[27.17 X 10"!

2.6 Wave Particle Duality

(iii)

The true nature of light is very difficult to asses. In 1801Young's double slit experiment it was shown that light exhibit fundamentalpropertieslike a wavediffraction and interference. After several years Maxwell discovered that light was a wave

0.05 kg of ice at - 20°C is to be converted into steam at

100°C. The ice is first heated with a420Wattsheaterfor5 minutes.

After this, itisheated with aninfi^a-red lamp ofX, = 10,000 Afor 23 minutes and 20 seconds at an efficiencyof50%. Find the rate

at which the photons are striking the ice when heated with infi:a-redlamp.

Specific heat of iceat-20''C = 500cal/kg

Specific heatofwater= 10^ cal/kg

of oscillatingelectricand magnetic fields, an electromagnetic wave. On the other hand Plank's equation theory, photoelectric effect and campton effects indicates that light behaves like a particle, a photon, withbothenergyand momentum. Thus light shows wave particle duality.

Latent heat offusion ofice = 8 x 10"* cal/kg

Latentheatofvaporization ofwater= 5.42x 10^ cal/kg.

2.6.1 Momentum of a Photon

[2 X 102%ec-']

According to relativistic theorythe total relativistic energy£" of a particle is related to the magnitudep of the momentum and

(iv)

Alight source, emitting three wavelengths 5000 A, 6000 A

mass m

and 7000A hasa totalpower of 10"^ Watt and a beam diameter of 0.002 m. The power density is distributed equally amongst the three wavelengths.The beam shines normally on a metallic surface of area 10"^ m^ and having a work function of 1.9 eV. Assuming that each photon liberates an electron, calculate the charge emitted per unit area in one second.

as

£:2^^2^2 + ^2^4 For a photon, its mass is zero, thus we have E=pc E p = — ^ c

[93.76 coul/mVsec]

For a photon its energy can be given as

(v) A monochromatic light sourceofintensity5 mW emits. 8 X 10^^ photons per second. This light ejects photoelectrons

Thus photon momentum is given as

fi-om a metal surface. The stopping potential for this setup is 2.0 eV, calculate the work fiinction ofthe metal. [1.9 eV]

(vi) A potassium surface is placed 75 cmawayffoma lOOW bulb. It is found that the energy radiated by the bulb is 5% of the input power. Consider each potassium atom as a circular disc of diameter 1 A & determine the time required for each atom to absorb an amount ofenergy equal its work function of 2.0 eV.What is the answer ifthe atom is assumed to be spherical. [57.6 s]

...(2.35)

Pphoton

^

-

Q

/'photon^ X

...(2.36)

This relation in equation-(2.36) gives the momentum ofeach photon of an electromagnetic wave having a wavelength X.

2.7 De-Broglie's Hypothesis In year 1923, as a graduate student Louis de-Broglie made a surprising suggestion which later confirmed in 1927 by different experiments. By the year 1921 it was all accepted that light

|£hoto Efefric Effect &Matter Waves behaves like a particle, a stream ofphotons andthewavelength

nh

of light is related to the magnitude of its momentum as x=

h

...(2.37)

/'photon

Using the above accepted fact, Broglie started thinking the reverse ofit. He suggestedthat since light waves could exhibit particle like behaviour, particles ofmatter should exhibit wave

mvr =

...(2.42)

2n

Thus if a wavelength is associated with the electron, the quantization ofatomic orbitals leads toconsider integral number ofwavelengths that fit intoeachorbit Forexample figure-2.19 shows the different orbits in which for different values of n

electron wave isforming astationarywave. Figure-2.19(a) shows likebehaviour. De-Broglie proposed thatall moving mattCT has an orbit inwhich thestationary wave isformed fortheprinciple

.a wavelength associated with it, just as a wave does and this wavelength is related to the magnitude of its momentum by an equation of the same form as given in equation-(2.37) as

quantum number « = 3. Similarlyfigure-2.19(b) shows an orbit correspondingto « = 4 and figure-2.19(c) shows an orbitwhich is larger then that of«=3 but smaller then that of« = 4 in which

we can see that in this orbit wavelength is not fit properly thus X=—-—

...(2.38)

/'particle

The wavelength associated with a particleis calledits de-Broglie wavelength. The de-Broglie hypothesis implies that the wave particle duality has a universal and symmetrical character'i.e. waveshave particle properties and moving particles have wave properties. It was also assumed that as the speed ofa particle increases its wave character also increases and the wavelength associated with the particle may have a significant practical values. For particles in nature moving very slowly, the wave character associated with the particle will not have much significance.

standing waves can not be formed for this orbit hence the orbit is not stable.

orbit

fern = 3

(a)

2.7.1 Explanation ofBohr's Second Postulate

The de-Broglie hypothesis brought Bohr's second postulate in a new light. It was stated that the magnitude of angular momentum ofthe orbiting electron in a hydrogenic atom was an integral multiple ofh/2n. Given as ...(2.39)

orbit

fern = 4

Asin anorbit speed ofelectron is oftheorder of10^ m/sitmay have a significant wave character associated with it. As we know in an atom stable orbits are those in which electron does

not radiate any electromagnetic radiation. Thus ifwe describe electron as a wave, its stable orbits in an atom are those that

(b)

satisfy the conditions for a stationary wave so that the complete wave energy is sustained in that orbit only.Thus for stationary waves to exist in an orbit the circumference ofthe orbit must be

an integral multiple ofthe wavelength, which can be given for an orbit of radius r

...(2.40)

orbit

For electron, its de-Broglie wavelength \ can be given as for 3 < n < 4

^

mv

...(2.41)

Now from equation-(2.41) & (2.42) we have (c) 2w = n — mv

Figure 2.19

Pholo''ae^trid iffedt: rM^er VVav^j

si

a_ = 0.7

2.8 Radiation Pressure

a.-0.3

We've discussed that a,photon carries mornentum, given as .

, P

...(2.43)

Pwatt

When a photon beam incidents on a surface,andabsorbed, the

Figure 2.21

total momentum can^ied by the beam is also transferred to that surface hence a force is exerted on the surface. Ifthe surface on

whichthe^beam incidentis reflectingthen dueto the reflection in.photon beam-the change in momentum,of photons will be twice the value compared to the.case when it was absorbed. Thus 'force exerted on surface will also get almost doubled. Now we'll discuss different cases ofincidence ofa light beam

In'this case 70% of the incident photons are reflected back and 30% are absorbed by the'body. Thus the photon which is absorbedwill impart a momentum

r,' OJFX

X

''he •

X

to

h'

X .—^

he ' ' X

. ,..(2.48)

F=

...(2.44)

he

2h 1, 0.3P?.

I.IF

2.8.1 Force Exerted by a Light Beam on a Surface

Figure-2.20 shows a black body ofmass m placed on a smooth surfeceon which a light beam of crosssectional area S incident. The beam is produced by a torch of power P watt. If X is the wavelength oflight produced by torch, the number ofphotons emitted per second are

'2>fi

the body. Thus net force acting on body can be given as F=

FX

'

to the bodyapd the photon.

which is reflected will impart the change in momentum

on different surfaces.

N=

h

If in any of the case we wish to find pressure on the surface on which light beam is incident.In aboverelationswe can use.light

intensity

% i^st^ad ofpower Pofthe light source;

2.8.2 Force Exerted on any Object in tiie Path of a Light Beam

Figure-2.22 shows a big lamp of power F watt which produces a uniform parallel beam oflight ofcross sectional area S. Thus the intensity ofthis light beam will be '

Black Body

/= -^ w/m^•

--...(2.49)-

P watt

• r """M

area = S

1

1

Figure 2.20

We know that momentum in each photon is

h

Power = P watt

Area =

P=x-

'

Area = 5

....(2.45)

(a)

As all the photons incident on the black body will be absorbed by it, here the total momentum absorbed by the body per second or force exerted on the body is F=

FX he

P

...(2.46)

c

In above case ifthe surface ofbody is perfectly reflecting like a Power = P watt

mirror then the force exerted on body will become Area = nR? Area = S

...(2.47) ,

c

Figure 2.22

Similarly consider another case as shown in figure-2.21. Now

the surface ofbodyon which light beam is incident is having a jf jj, reflection coefficient

0.7 and absorption coefficient a^=0.3.

path ofthis beam a black body sphere is placed as

shown infigure-2.22(a). Inthiscase only those photons will be

,Ph^o EJectric Effect &Matter Waves 73''

incident on the sphere which pass through the cross sectional areaA=idf which isthe projection ofsphere on across-sectional plane. Thus the power incident on sphere is

Inthe path ofauniform light beam oflarge cross-sectional area

P- = IA=I%R^

...(2.50)

Thus forceexertedon sphere will be

C

C

Illustrative Example 2.27

and intensity I, a solid sphere ofradius R which is perfectly reflecting isplaced. Find the force exerted on this sphere due to the light beam.

...(2.51)

'

Solution

Similarlyasshown infigure-2.22(b) a black body cone isplaced in the path ofthe lightbeam. Here alsothe projection of cone along a cross-sectional plane of beam is^ =71/?^ hence the force

Light Intensity /

exerted on cone due to the beam will remain same as F =

InR-

...(2.52)

2.8.3 Force Exerted bya LightBeamat Oblique Incidence / dF%mQ

As shown in figure-2.23 whena lightbeamincidenton a mirror

at an angle 0 to normal, it will be reflected at same angle. If power of lightbeam isP watt, themomentum ofphotons inthe beam per second will be ...(2.53)

Figure 2.24

To find theforce onsphere, we consider asmall elemental strip ofangular width onitssurface atanangle 0from its horizontal diameter as shown. TheareadSof this stripon the surface of sphere is

Ap sin 9

1I

dS = 2nR sin0 • RdQ

.*..(155)

Nowas shown in figure-2.24, dA is the projection of the slant strip areadSalong the cross-sectional plane of thelightbeam

Reflecting Surface

and it is given as

dA = dSeos 0

...(2.56)

Ap cos 9

-f-.

Here thepower oflightincident on thisstrip is

>F

Apcos9

'

dP=ldA

Thus the momentum of photons per second incident on this strip are ,

dP

IdA

dp= — = ~

...(2.57)

Here wecan see thatthese photons areincident atan angle 0to

v.

tip sin 9

the normal

ofthis strip and as sphere isperfectly reflecting.

These are reflected at the same angle 0 to Pwatt

as shown in

figure-2.24. Figure 2.23

Here as shown in figure-2.23 there will be no change in the component of this momentum along the mirror due to reflection

butalong normal momentum oflightbeam is changed byItsp cos0. Thus force exerted on themirror isalongitsnormaland is given by IP

F= 2 Ap cos 0 = — cos c

...(2.54)

Herethe changein momentum ofphotons is alongthe normal andthus force exerted on this stripalongthe normal is dF = 2dp cos0 ^ —— cos0 Thus net force on sphere will be given as

F=Ji^FcosG ^pi dA_^2

F=

cos

0

... (2.58)

Photo "Electric Eff^t &Matter VVaves ]

74

Now to find the force on the cone, we consider an elemental

strip ofwidth dx on the lateral surface ofcone ata distance x

F= J—(27ii! sin 0cos0• cos

from the vertex O ofconeas shown in figure-2.26. If the radius of thestripis r its surface areais

0

2^

iLE^— [cos^ BsinBt/G c

...(2.60)

dS = 2Tir • dx

J 0

41 nR'

Herebysimilartriangles in conewehave

cos 0

F=

^ IizR'

F=

R

^ [1-0]

^1r^+h^

=>

InR'

... (2.59)

F=

r = X sin 0

Thus area of strip can be given as

We can see that equation-(2.58) is exactly same as equation-

(2.59). Thus for asphere placed in the path ofa light beam, force exerted on sphere isindependent from thenature ofthesurface ofsphere. Butthis happens onlyfor a sphere.

...(2.61)

dS = 27a: sin 0 •

IfdA betheprojection ofslant strip ofarea dSalong thecrosssectional planeof the lightbeam, it can be given as sin 0

-••(2.62)

Lets discuss the same for a cone with reflecting surface in the

IfdPisthe power oflight beam incident on the strip then

path ofa light beam, illustrated innext example.

dP=IdA

Illustrative Example 2.28

...(2.63)

Now themomentum persecond ofthelightphotons dpincident Figure-2.25 shows acone ofradius Rand heightwith perfectly on the strip is reflecting lateral surface, isplaced inthe path ofalight beam of _ dP IdA ...(2.64) intensity/. Find the force exerted on this cone.

^

c

c

Here also we can see that from figure-2.26 the momentum of

photons is changing only along normal due to reflection of these photons, which will exert a force on cone along normal

Intensity = / watt/m^

direction. Ifthis force is dF, it can be given as ...(2.65)

dF=2dPsmQ

Thus net force on cone can be given as

Figure 2.25

sin 0 Solution

=>

F=J'2(^sin^0 F

sin^ 0

rfHicosiS

|^£/5sin^0

v!/ oFsm0

F

•27ad!xsin 0

4/7tsin 0 F=

Figure 2.26

iR-

^xdx 0

iphoto Electric Effect & Matter Waves

75

2.8.5 Variation in Wavelength ofEmitted Photon with State of +H'

F

r =

Motion of an Atom

4/71 sm^ . 4I c

As we've discussed that when a photon is emitted by transition

ofan electron in an atom,the atomrecoils and therecoil velocity can be obtainedbymomentumconservation. In previoussection we've discussed that the emitted photon wavelength is given

R

4{r^+h^)

by Rydberg's formula as 1

[As sin 0 =

21 kR-

4{R^+H^) ...(2.66)

F=

1

...(2.70)

R

Equation-(2.66). gives the force exerted by light beam on a perfectly reflecting cone in its path. 2.8.4 Recoiling of an Atom Due to Electron Transition We know that when an e~ in a hydrogenic atom makes a transition from a higher energy level to a lower energy level

«p thedifference ofenergies ofthe two energy level isreleased in the form ofan electromagnetic radiation photon. As we've discussed that every photon has a momentum associated with it, we can say that when a photon is emitted from an atom, it

This equation-(2.70) or Rydbergformula is valid only ifall the energy released by electron during transition from energy level

«2 to«j will transform into thephoton energy. Butin ourcase if all the energy released by electron is carried by the photon then no energy is left with which the atom an recoil. If it does not recoil, the law of momentum conservation will violet. Thus in

previous section equation-(2.69) only gives an approximate value ofrecoil velocity. Practically we can say that the energy released by electron during transition is shared between the emitted photon and the kinetic energy ofrecoiled atom. IfX'is the wavelength ofemitted photon and is the recoil velocity of atom, we can write the energy equation as

must recoil in opposite direction as shown in figure-2.27 Rch2^

1

1

he

r

+ ~mvi

...(2.71)

Using momentum conservation we can write

|r =Mv„ H-atom

Photon

Figure 2.27

Here the wavelength of emitted photon can be given as 1

1

...(2.67)

The momentum ofthis emitted photon can be given as

p=-^=hRZ:^

1

1

...(2.68)

If the after emission ofphoton the atom recoils with speed v^, using momentum conservation principle, we can write

...(2.72)

Solving equation-(2.71) and (2.72), we can calculate the exact value ofrecoil velocity and the wavelength ofemitted photon X'. Note that the value of will be slightly more compared to that calculated in previous section by equation-(2.67). As the difference between the two is very small, for numerical applications students can use equation-(2.69) and (2.70) for calculations of emitted wavelength and recoil velocity if very high accuracy is not needed in the application. 2.8.6 Variation in Wavelength of Photon During Reflection We've discussed that when light incident on a body a force is exerted due to the momentum associated with the light photons. Consider the following example. A torch of power P in front of a body ofmass Mis lit just for a time At so that torch will emit a pulse of energy AE which is given as AE=PAt

...(2.73)

Ifwavelength light is X, the momentum in this pulse is hRZ'^ v„ =

M

h _0

...(2.69) AP =

AE

(.YD

h X

AE

...(2.74)

Photo Electric,Effect^^ Matter Waves ;

76-

As the total energy ofsystem isalso constant, the equation for A£

energyconservation can be given as ...(2.78)

Power = P watt

Ap

Now solving equation-(2.77) and (2.78) we can get the value of speed ofblock as well as the wavelength X' ofreflected light pulse. This process gives accurate results but for numerical problems, itis relatively lengthyprocedure. Students are advised to directly use equation-(2.75) or assume that the reflected wavelength is approximately same because the difference is very small which can be neglected for numerical problems.

(a)

Power = Pwatt

(b)

Figure 2.28

# Illustrative Example 2.29

Asshown in figure-2.28 we can see thatwhen theenergy pulse is reflected from the body if its surface is perfectly reflecting. Due to this the body will gain a velocity v as shown in

With whatvelocitymustan electron travel sothatitsmomentum

is equal to that ofphoton with awavelength of?. =5200 A.

figure-2.28(b). Solution

Hereif we assumethat the wavelength ofreflected and incident

light are same, we can say that themomentum ofreflected pulse is same in magnitude but direction is opposite. Thus by

Momentum of photon 6.626x10

momentum conservation we can write

^

Ap = Mv - Ap

kg.m/s.

Where v is the velocity of electron.

... (2.75)

Using the above equation we can find the speed of theblock after reflecting the pulse. But this result will only be an approximate value. Ifwe carefully look into thesituation once again, aquestion arises, ifwhole ofthe energy ofpulse isreflected

back, how the block has gained the kinetic energy

5200x10""^

Momentum ofelectron = wv = 9.1 x 10"^'v

2 tsp=Mv =Mv

X

-34

Equating the two momenta, wehave 9.1 X10-^'v=6.626x l(r3''/5200x 10-10 >

v = 1400m/s.

# Illustrative Example 2.30

MvP-. Hydrogen gas in theatomic state is excited toan energy level

Thus it is sure that when the incident energy pulse is reflected

from the block, someof its energyis transmitted toblock'as its

such that the electrostaticpotentialenergyof H-atombecomes - 1.7 eV. Now the photo-electric plate having work function

kinetic energy and remaining is reflected as energy pulse of (|) = 2.3 eV is exposed to the emission spectra of this gas. energy A£". Now the way ofaccurate calculations for the speed Assuming allthetransitions tobepossible, find theminimum de-Brbglie wavelength of ejected photo-electrons. ofblock is given here. The number of photons in the incidentpulse are

Solution

AE

N=

Given that electrostatic potential energy of H-atom is

i'Yx)

PE=-\JqV N=

AEX he

... (2.76) . We know that kinetic energy is given as PE

Here Xis the wavelength ofincident light.

=-^=0.85eV

If?.' is the wavelength ofreflected light, the equation for Thus total energy momentum conservation can be written as

N^=Mv~N-^

p=_I.7 + 0.85=-0.85eV

•••(2-77)

13.6 £

= —

= -0.85 eV

[Photo^ Electric Eff^ &Matter Waves 2_ 13-6

n'- =

0.85

77

= 16

and

n=A

_

Aa 2 • •

Hence the atom isexcited to state n=4.The maximum energy is emitted when electrons will make a transition fi-om w= 4 to « = I for which energy emitted is

# Illustrative Example 2.32

Find the ratio ofde-Broglie wavelength ofproton and a-particle

AE=- 0.85-(-13.6) = 12.75eV

Now this photon energy when incident on ametal plate having work function 2.3 eV, the kinetic energy offastest electron ejected

which have been accelerated through same potential difference. Solution

can be given as

Kinetic energy gained by a charge q after being accelerated

^max =^-'1'= 12.75-2.3 = 10.45 eV

through a potential difference Fvolt,

Theminimum de-Broglie wavelength is given by .j

_

h

qV= ^m^P•

h

(/')max " pm(K.E.)^

V=J^ m

=>

6.63x10-^^^

V2x(9.].1x10-3')(10.45x1.6x10-'9) =>

mv= ^2mqV

Now we have, de-Broglie wavelength is given as h

^^ = 3.8xlO-iOn]-3.8A

_

h

-jlmqV

# Illustrative Example 2.31

fripqpVp Ana-particle anda proton arefired through thesame magnetic fields which is perpendicular to their velocity vectors. The a-

Substituting V^=V

particle and the proton move such that radius of curvature of

4x2

their pathissame. Findtheratio oftheirde-Broglie wavelengths. Solution

# Illustrative Example 2.33

Magnetic force experienced bya charged particle ina magnetic field is given by,

canform a standing wave between theatoms arranged in a one

Eg=q'^y.B =qvBsmQ In our case,

Hence,

= qvB

[As

mv

Bqv = ^

=>

Assume that the de-Broglie wave associated with an electron

mv = qBr

0 = 90°]

dimensional array with nodes at each of the atomic sites. It is found that one such standing wave is formed if the distanced

between theatoms ofthearray is2 A. Asimilar standing wave is again formed if d is increase to 2.5 A but not for any intermediate value of d. Find he energy of the electrons in electron volt and the least value ofd for which the standing wave of the type described above can forms.

The de-Brogliewavelength

X= ~mv

^ qBr

Solution

From the given .situation it is clear that stationary waves are ^a-panicle 5i'-proton

Since,

'

^=1

formed for two successive values of2 A and 2.5 A thus we can directly say that

y=0.5A ;^=2x0.5 = 1A

Photo Electric Effect & Matter Waves

78

Thus the energy of electrons can be given as E =

he

®

(6.63xl0-^'^)(3xI0^)

E-

(5x1.6x10-")

2m

?.= 2.48625 X10"^ m= 2486A

(h/xy E =

E =

2m

(6.63x10"^'^)^ (10"'®)^x2x9.1xl0-^^

h deBroglie

h

p

£• = 2.41 X lO-i'^J 2.41x10 £ =

1.6x10

6.63x10 deBroglie

-17

-19

•eV

-34

-^2x(9.1xl0-^^)x(2xl.6xl0-'^)

Wie=8.6877A

£• = 150.6 eV

Now we use

Forformation of stationary waves between two atomic sitesthe X

minimum separation must be that is0.5 A. # Illustrative Example 2.34

X

2486

^deBroglie

8.6877

=286.18

(c) After time t, thepotential Vofthe sphere is given by

In a photo electric effect set-up, a pointsource light ofpower 3.2 X 10"^ W emits monoenergetic photons of energy 5.0 eV.

47teo yr

The source is located at a distance of0.8 m from the centre ofa

where q = charge on the sphere.

stationary metallic sphere ofwork function 3.0eV andofradius 8.0 X10"^ m. The efficiency of photo-electron emissionis one

We also use stopping potential as

for every 10^ incident photons. Assume that the sphere is

^ -^photon "*1*

located and initially neutral, and that photo-electrons are instantly swept away after emission.

When potential ofsphere becomes 2volt, photo electric emission (a) Calculate the number of photo-electrons emitted per stops as maximum kinetic energy ofpTiotoelectrons is2eV. second.

(b) Find the ratio of the wavelength of incident light to the DeBroglie wavelength ofthe fastest photoelectrons emitted. (d) (c) It is observed that the photoelectrons emission stops at a

(9xl0^)«(1.6xl0-^^) ^

certain time t after the light source is switchedon. Why ? (d) Evaluatethe time t.

8x10"^ Solving we get

Solution

n = l.llxi07

(a) The energy reachingthe sphereis givenby E'=

P

2

. XTtr^ =

Now, we use

Pr^

« 10^

t = —r

l.llxio'

TTlcUv

'

P)

mdx

is doubled butn remains the same

2-28 The frequency and the photon flux of a beam of light falling onthesurface ofphotoelectric material are increased by 2-34 A metal surfaceis illuminatedbya light ofgiven intensity a factor oftwo; This will and frequency to cause photoemission. If the intensity of (A) Increase themaximum kinetic energy ofthephotoelectrons, illumination is reduced to one fourth ofits original value keeping as wellas photoelectric currentbya factor of two frequency ofradtion constant, then the maximum kinetic energy (B) Increase the maximum kinetic energy ofthe photo electrons of the emitted photoelectrons would become : andwould increase thephoto electric current bya factor oftwo (A) Unchanged P) l/16thoforiginalvalue (Q Increase the maximum kinetic energy ofthephoto electrons (Q Twice the original value p) Four times the original value by a factor oftwo and will have no effect on the magnitude of thephoto electrons by a factor oftwo and will have noeffect on 2-35 A point source causes photoelectric effect from a small the magnitude ofthe photoelectric current produced P) Noproduce any effect onthekinetic energy oftheemitted metal plate. Which ofthe following curves may represent the electrons but will increasethe photo electric currentbya factor of two.

2-29 Let

saturation photocurrent as a function of the distance between

the source and the metal?

be the maximum kinetic energy ofphotoelectrons

emitted by light of wavelength X., and

corresponding to

wavelength IfA.j = 2X^ then : (A) 2K^=K^ P) K^=2K^ P) ^,> 2^:2 (Q K^.): (A)

nh iTzXm

(B)

Xm

Xm Xm

Innh

(Q

21!L

P) increase themaximum kinetic energy ofthephoto-electrons bya factor greater thantwoandwould increase thephotoelectric saturation current by a factor of two.

(Q increase the maximum kinetic energy ofthephotoelectrons by a factor greater than two and will have no effect on the magnitude ofthe photoelectric saturation current produced. P) increase themaximum kinetic energy oftheemitted photo electrons by a factor of two but will have no effect.on the saturation photoelectric current.

P)

2-11 Minimumlight intensitythat can beperceived bynormal humaneye is about 10"'® Wm-2. What is theminimum number

2-5 No photoelectrons are emitted from a metal if the wavelength oflightexceeds 6000 A.Thework function ofthe ofphotons ofwavelength 660 nm that must enter the pupil in one second,for one to see the object? Area of cross-section of metal is approximately equal to: thepupil is 10"^ m2? (A) 3.315xl(r6j P) 3.315 xl0-'9j

(Q 2.07x10-'^ J

p) 2.07 X10-22 J

(A) 3.318 xlP (Q 3.318 xlO^

P) 1.453x103 p) 1.453x105

2-6 5% of the energy supplied to a lamp is radiated as a visible

light. How many quanta of light are emitted per second by 2-12 The average wavelength of de-Broglie wave associated 100wattlamp.Assumethe average wavelength of visiblelight with a thermal neutron ofmass m at absolute temperature T is given by (here k is the Boltzmann constant):

as555nm?

(A) 0.75x10'^ (C) 2.16 X10^^

P) 1.39x10'^ P) 2.83 X10'^

h

(A)

•JmkT h

2-7 All electrons ejected from a surface by incident light of

(Q

yl3mkT

_

P)

h

•^2mkT h

P)

lylmkT

wavelength 2000A canbestopped before travelling 1m in the direction ofuniform electric field of4 N/C. The work function of the surface is:

(A) 4eV (Q 2eV

P) 6.2eV P) 2.2 eV

.

2-13 What percentage increase in wavelength leads to 75% loss of photon energy in a photon-electron collision? (A) 200% P) 100% (Q 67.7% P) 300%

[photo Elecfric Effect &Matter Waves

2-14 Aparallel beam oflight ofintensity7and cross section

2-19 The work function ofasubstance is 4.0 eV. The longest

area S is incident on a plate at normal incidence. The

wavelength oflightthatcancause photoelectron emission fi-om this substance is approximately: isVand the work function ofthe plate is(p (hv > 3) & 2.36 eV (4 -> 3)]

2-36 At a given instantthereare 25%undecayed radioactive

nuclei in a sample. After 10 second thenumber ofundecayed nuclei reduces to 12.5%. Calculate: (0 Mean life ofthe nuclei

is one for every 10^ incident photons. Assume that the sphere (ii) The time in which the number of undecayed nuclei will

is isolated and initially neutral, and that photo electrons are instantly swept away after emission.

further reduce to 6.25% ofthe reduced number. Ans. [(i) 14.43 s, (ii) 40 s]

Photo Electric Effect & Matte^Vyay^_^

: 98___

2-37 When a beam of10.6 eV photons ofintensity 2.0 W/m^ 2-41 Irradiating themetal surface successively by radiations fells on aplatinum surfece ofarea 1.0 10"^ m^ and work function ofwavelength 3000 Aand 5400 A, itisfound that the maximum 5.6 eV, 0.53% ofthe incident photons eject photoelectrons. Find velocities ofelectrons areintheratio2:1. Findtheworkfiinction the number of photoelectrons emitted per second and their ofthe metal surface, meV.

minimum energies (in eF)-Take 1eV= 1.6 x10"'^ J. Ans. [6.25 >= lO", 0]

2-38 In a photoelectric effect experiment, photons of energy 5 eV are incident on the photocathode of work function 3 eV.

Ans. [1690]

2-42 A filter transmits onlytheradiation ofwavelength greater than 4400 A. Radiation from a hydrogen discharge tube goes

through such a filter and isincident on ametal ofwork function

= lO'^ m"^ s~\ saturation current of 2.0 eV. Find the stopping potential which can stop the

For photon intensity

4.0 pA is obtained. Sketch the variation of photocurrent against the anode voltage in the figure below for photon

photoelectrons. Ans. [0.55 volts]

intensity (curve A) and= 2 x lO'^ m"^ s"' (curveB). 2-43 The peak emission from a black body at a certain

temperature occurs at awavelength of9000 A. On increasing the temperature, the total radiation emitted isincreased 81 times. Atthe initial temperature when the peak radiation from theblack body is incident on a metal surface, it does not cause any Ans. [ -6^-2

0

2

4

6

V/yol. k)

2-39 In a photoelectric setup, the radiations from theBalmer series ofhydrogen atom are incident on a metal surface ofwork

photoemission from the surfece. After the increase oftemperature the peak radiation from the black body caused photoemission. To bring these photoelectrons torest, a potential equivalent to theexcitation energy between the«=2 and «= 3 Bohr levels of hydrogen atom isrequired. Find the work function ofthe metal. Ans. [2.25 eV]

function 2 eV. Thewavelength ofincident radiations liesbetween 450 nm to 700 nm. Find the maximum kinetic energy of

2-44 The radiation emitted when an electron jumps from n = 3

photoelectron emitted. (Given hc/e = 1242 eV-nm).

ton = 2 orbit of hydrogen atom falls on a metal to produce

photoelectrons. The electrons emitted from the metal surface 2-40 Monochromatic radiations ofX= 640.2 nm from a neon

withmaximum kinetic energy aremade to moveperpendicular

. . lamp irradiates photosensitive material made of cesium on toa magnetic field of 1 T ina radius of10 m. Find tungsten. Thestopping potential is measured as 0.54 V. The source is replaced by an iron source and its 427.2 nm line (a) the kinetic energy of the electrons, irradiates thesame photocell. Predict thedifference in stopping (b) theworkfunctionofthemetaland

potential. (Round offto singledigit) Ans. [1 eVJ

(c) the wavelength ofthe radiation. • (Planck's constanth = 6.62 x Ans. [(a) 0.86 eV, (b) 1.03 eV, (c) 6570 A]

Js)

3

X-Rays FEW WORDS FOR STUDENTS

We have already discussed about the energy levels in atoms. In

this chapter we will study about the utilization of the energy releasedfrom atomic transitions in heavy atoms in theform of X-rays. Now we will study from the discovery ofX-rays to the fundamental utility along with the different production

Metal

target

___ .HighV-sr- voltage

X rays

source

mechanism ofX-rays, Vacuum

Electrons Heated filament

CHAPTER CONTENTS

3.1

Introduction toX-Rays

3.2

Production Mechanism ofX-rays

3.3

Moseley'sLaw

3.4

Applications ofX-rays

Filament

voltage

COVER APPLICATION

Figure-(a)

Figure-(b)

Figure-(a) shows a commercial X-Ray machine used in various diagnostic labs in which X-rays are used^to take photographs of

organs of human body. Figure-{b) is a typical X-Ray image of human hands taken on a commercial X-Ray machine.

,X:RaysJ ;ioo

3.1 Introduction to X-Rays

(d) The wave nature ofX-rays makes them useful tool for the study ofthe structure ofcrystals &molecules. This was later

The discovery of X-rays was accidental by Wilhelm K.

discovered.

Roentgen in Germany. X-rays were discovered while investigating the discharge ofelectricity through rarefied gases, the same phenomenon and equipment through which cathode rays were first noticed. At the time of researches going on cathode rays, X-rays were also being produced but Roentgen

3.1.1 Types of X-rays

According to thequality ofX-rays, these can be divided intwo groups, these are

was the first person tonotice them when hewas investigating

(I) Soft X-rays: These are X-rays with low penetrating power.

the characteristics of cathode rays.

The wavelength range ofthese X-rays are from 10 Aand above hence their frequency is small and theyhave smaller energy.

As has been found when cathode rays (fast moving electrons) These are produced atcomparatively low potential difference

are braked bya material, their kinetic energy isconverted into electromagnetic radiations. These are known as X-rays. These are electromagnetic radiation with very short wavelengths extending from 0.01 Ato100 A. Since inthe visible region, the shortest wavelength are ofviolet rays which are visible tothe

and high pressure.

(II) HardX-rays: These are X-rays having high penetrating power. The wavelength range ofthese X-rays are ofthe order of 1A so they have high frequencies and hence high energies. eye have wavelength ofabout 4000 A, X-rays are invisible to These are produced at comparatively low pressure and high eye.

potential difference.

X-rays are produced in X-ray tube consist ofa glass or metal 3.2 Production Mechanism of X-rays envelope enclosing the assembly of a cathode, an anode (generally referred as target) spaced a certain distance apart

On thebasis ofproduction mechanism. X-rays are classified in

and connected to a source of extremely high tension (EHT)

two broad categories

supply. The cathode acts as a source ofelectrons and the anode

1.

Continuous X-Rays

as a source of X-rays. The field set-up between the cathode

2.

Characteristic X-Rays

and the anode accelerates the electrons toenergies oforder lO'^ tolO^eV.

Nowwe'll discuss about the production mechanism of X-rays in detail.

The invisible X-rays are detected by observing their effects. Among other things X-rays produce a strong photochemical action which blackens photographic plates. They are also

3.2.1 Continuous X-rays

capable of ionizing gases and causing fluorescence in Continuous X-rays, as their name implies, have a continuous phosphors. For measurement purposes, these are used inmainly spectral distribution. They are produced when electrons for their applications oftheir photochemical and ionizing effects.

accelerated in a vacuum strike a target and lose kinetic energy

In ionization chambers, the intensityofX-rays is determined by

in passing through the strong electric field ofthe target nuclei,

Roentgen discovered that the X-rays has following

3.2.2 Production of Continuous X-rays

characteristics.

Inthegeneral X-ray production mechanism there isa discharge tube (generally called Coolidge tube) operated onhigh potential

measuring the saturation current due to the ionization of the andresulting in a continuous X-ray spectrum. Asaccording to gas enclosed in the chamber, because the saturation current is the classical physics any accelerated charged particle emits electromagnetic radiation ofcontinuous spectral distribution. proportional to theintensity ofX-rays.

(a) X-rays are generated whenever high energy cathode rays difference. The cathode used in the tube is the source ofcathode strike solid materials. Generally the greater the density of rays i.e. fast moving electrons, and theanode used in thetube impacted material the moreX-rays areproduced. is thetargetmaterial whichis the source ofX-rays. TheX-rays

(b) Matter is more or less transparent to X-rays. Wood and

areproduced bythe bombardment ofanode material bycathode rays, so there will be a lot of heat generation in the anode

flesh areverytransparent bone andmetals lessso,which makes

material. The anode material should have high melting point as

the use of X-rays in medicine so useful.

(c) X-rays are unaffected (undeviated) byelectric and magnetic

duetohighenergetic electrons, a hightemperature isgenerated. For this purpose target is selected of high atomic mass. The coolidge tube assembly for the production ofX-rays is shown

fields, hence these are uncharged.

infigure-3.1.

'X-Rays

101

eV= ~ mv^

...(3.1)

2eV

...(3.2)

v = m

When this electron comes out fi-om the atom ofanode material

the speed of this electron will be veryless as compared to its initial speed. Thus the difference of kinetic energy of this electron is emitted in the form of an X-rays photon fi-om the

X-rays *

anode atom. Figure 3.1

The mechanism for the production of continuous X-rays can be explained on the basis of figure-3.2. Figure showsan atom of the anode material of high atomic weight with its electron configuration. In the Coolidge tube an electron is projected towards the anodewith an accelerating voltageV. Sothe kinetic energy of the projectile electron will be eV. As shown in the figure when the projectile electrons enters into the extremely high electricfield ofthe nucleus ofthe atom of the anode material, it experiences a strong electric force towards the nucleus ofthe atom of the anode material, it experiencesa strong electricforce towards the nucleus of the atom and dueto this strong attraction the velocityof this electron when it emerges fi"om the atom will be highly reduced and negligibly compared with the initial velocity ofthe projectile electron. This electron in the influence of the highly positive nuclei experiences a very high acceleration

and according to the classical theoryeveryaccelerated charged particle emits the electromagnetic radiations, so this electron will also emit electromagneticradiations, these electromagnetic radiations are called X-rays. According to the law of conservation of energy, the energy of these electromagnetic radiation will be equal to the decreasein the kinetic energy of the projectile electron. This amount can be calculated. E = hv

X-ray photon

Those electronswhich pass through the atom verycloseto the nucleus, will be more accelerated and the photon energy corresponding to these electrons will be more as compared to thoseelectrons whichpassthrough the atom at relatively large distance from nucleus. Themaximum energyofX-photon will be corresponding to that electron which looses almost all of its energy during passing through the atom. The photon correspondingto this electronwill have the shortestwavelength among all the photons radiated by other electrons. Ifthis shortest wavelength is then we have

AE= y1 ,

•)

=

he

12431 X

Xc = ^F7 ~ —77—A eV V

This is theminimumwavelength of X-rays emittedfrom an X-ray tubewhich we call short wavecut off.Thus fi-om equation-(3.3) we can seethat the maximum energyor minimum wavelength of X-rays emitted depends only on the potential difference applied across the discharge tube.

Thus we can obtain X-rays inany range X^ toco byapplying an appropriate voltage across discharge tube which will fix X^ and other photons emitted firom tube will have wavelength more

thanX^ andranging uptooo. Thats why these X-rays arecalled continuous X-rays. The basic intensity per unit wavelength versus wavelength plot (wavelength spectrum) of continuous X-rays is shown in figure-3.3.

Projectile Electron

Figure 3.2

If the initial velocity ofthe projectile electron is v, then as the velocity is gained due to the accelerating voltage V, we have

...(3.3)

Wavelength Figure 3.3

__

i

As we can see in the graph that the intensityof emittedX-rays Ifcavity is created inK-shell thenit maybefilled bytheelectron willbemaximum (maximumnumberofphotons) fora particular ofeither of L, M, N ... shells. If cavityis filledbyan electron of valueof wavelengthat a particular acceleratingvoltageacross L-shellthen the energyreleasedwill be equalto the difference thedischarge tube. Ata particular voltage theintensity ofX-rays inenergies oftheK-shell and L-shell. This iscalled AT^-line and can be variedbychangingthe currentin the circuitbecausethe if the cavity is filled byan electron of M-shell then the line is

intensity of X-rays (number ofphotons) is proportional to the'

called Kp-line. Similarly K^, Kg,... lines may also exist. This

number of electrons attacking the anode. The broad continuous spectrum beyond the peak intensity is referred as

series of lineris called K-series of characteristic X-rays.

"Bremsstrahlung".

Ifcavity iscreated in L-shell thenaccording tothe transition of

3.2.3 Characteristic X-rays

this is called L-series. There may also be M-series as shown in

electrons from higher orbits there will be L^, L^,...lines, and

figure-3.5(a). Figure-3.5(b) shows thewavelength spectrum for TheseX-raysare called characteristicX-rays becausethey are the characteristicX-rayswhen cavityis created in K-shell. characteristicofthe elementused as target anode. Characteristic X-rays has a line spectral distribution unlike to continuous X-rays. The wavelength spectrum of these X-rays is also a ° r \ L-series continuous spectrum but this spectrum is crossed over by distinct spectrallines. The frequencies corresponding to these lines are the characteristic ofthe material of the target i.e. anode material.

3.2.4 Production of Characteristic X-rays

Production of these X-rays can be explained with help of the figure-3.4. These X-rays are produced when the projectile

•-

/ M-series

electron towards the anode collides with an internal electron of

the atom of the target material, and then secondary emission will happen.Now this willcreate a cavityin the inner shellof the target material, and then secondaryemissionwill happen. Now this will create a cavity. There are so many ways in which the

(a)

cavitymay be filled. The cavity can be filled by an electron of higherorbitwhichmakethetransition fromitsorbitto the lower orbit in which cavity is created. When an electron make such a form of electromagnetic radiation, this energy is the characteristic line for the X-ray spectrum.

Projectile Electron

Bremsstrahlung

(b) Figure 3.5

Projectile Electron

Thus the characteristic lines of the characteristic X-rays

depends only on the target material and not on the accelerating voltage. One more thing we can see here that the characteristic Orbital ^ Electron

Figure 3.4

X-rays are emitted only when the projectile electron make a collision with another bound electron ofthe atom ofthe target material. When cathode rays pass through the target then the

X-Rays

103

probabilityof collision of a projectileelectron to collide with the bound electron is very less. Majority of the projectile electron to collide with thebound electron is veryless. Majority

person is allowed to stand between a fluorescent screen and the X-ray unit. The deep shadow of bones is formed on the

of the projectile electrons will pass through the anode without making the collisionand they produce continuous X-rays. So always both type ofX-rays, continuous and characteristic will be emitted. Only a single type can not be emitted.

Instead of the fluorescent screen, a photograph known as radiograph can alsobe. taken. X-rays are alsousedto diagnose diseases in the lungs, kidneys, intestinesand other parts of the

fluorescent screen and the fracture of the bone can be detected.

body.

(2) Radio-therapy : X-rays are used to destroy malignant tumorsand to cure skindiseased. Longexposure ofX-rays kills the germs in the body and hard X-rays are used to destroy

3.3 Moseley's Law

Moseley found that the wavelengths of characteristic X-rays depend in a well defined manner oh the atomic numbers of the tumors very deep inside the body. elementthat emitX-rays (thetargetelement). Thisdependence, is known as Moseley law, may be written as (3) Industry: Xyrays are used to detect defect in radiovalves, tennis balls, rubber tyres and the presence of pearls in oysters. -Jv =a{Z-G) ...(3.4) They are also used to test the uniformity of the insulating materials and the quality of oil paintings. Where a is known as screening constant so the term (Z - a) here is effective atomic number for the electron which takes

part in the transition which results in characteristic X-rays and a is Moseley constant. It can be seen that for

line the electron make a transition

from L-shell to K-shell, and during this transition this electron move in the electric field ofthe nucleus and the one electron of

the K-shell, so the screening on the transition electron is only due to the remaining one electron of the K-shell so we can take the effectiveatomic number for this transition as (Z-1) so the frequency emitted for the line can be given as

V^-fl(Z-l) The wavelength ofthe characteristic X-rays can be calculated by the relation

(4) Engineering : X-rays are used to detect cracks in structures and blow holes in metals. They are used to test the quality of weldings, moulds and metal coatings. They also help in detecting any crack in the body of the aeroplane and motor cars.

(5) Detection departments : X-rays are commonly used to detect the smuggling of precious metals at the custom posts and to detect the explosives and other contraband goods like opium in sealed parcels and in leather cases. They are also used in mints, where coins are made and every person has to pass before an X-ray unit after the day's work is over. (6) Research : X-rays are used in research to study the structure of crystals, arrangements of atoms and molecules in matter and their behaviour on different materials.

{=R(Z-of

1

1

For AT^-line we use a = 1 and = 2 and wavelength of the AT^-line can be given as

...(3.5)

= 1, thus the

# Illustrative Example 3.1

An X-rays tube operates at 20 kV.Find the maximum speed of the electrons striking the anticathode, given the charge of

electron = 1.6x 10^'^coulombandmassofelectron =9 x 10~^'kg. Solution

=>

v=|-fe(Z-l)^

Thus for ^^-linea =

, where 7? istheRydberg constant.

When an electron ofcharge e is accelerated through a potential difference V, it acquires energy eV. If m be the mass of the electron and v the maximum speed of electron, then

3.4 Applications of X-rays (1) Surgery: X-rays can pass through blood and not through bones. They are used to detect the fracture of bones, diseased organs and foreign bodies and growth in the human body. A

2eV m

X-Rays

1104

Dividing (3.6) and (3.7) we get

Substituting the given vales, we get

2x(1.6xl0"^^)x20,000 9x10

-31

= 8.4 X10^m/sec.

>^1-415A = 0.708A.

# Illustrative Example 3.2

(a) AnX-ray tubeproduces a continuous spectrum ofradiation with itsshort-wavelength endat0.45 A.What isthemaximum energy ofa photon in the radiation ? (b) From your answer to (a), guesswhat order of acceleratingvoltage (for electrons) is required in such a tube ?

# Illustrative Example 3.4

If the short series limit ofthe Balmer series for hydrogen is 3644 A, findthe atomicnumber of the elementwhich giveX-ray

wavelengths down to lA. Identify the element. Solution .

Solution

Ifthe short series limit of the Balmer series is corresponding to transition « = coto w= 2 which is given by

(a) Short wavelength is given as

Maximum photo energy is given as he

^max ^max

Ti

'^min

12431 ^„.x=W=27624.44eV

The shortest wavelength corresponds to « = co to « = I. Therefore X is given as

^^-27.624keV

J

(b) The minimum accelerating voltagefor electronsis 1

e

i.e.

l_

1

ofthe order of30 kV.

# Illustrative Example 3.3

The wavelength of the characteristics X-ray

3644

line emitted

from zinc (Z= 30)is 1.415 A. Find the wavelength ofthe

= 911

Z-1 =30.2

line

emitted from molybdenum (Z = 42).

Z=31.2:::3].

Solution

Thus the atomic number ofthe element is 31 which is gallium.

According to Moseley's law, the frequency for X series is given by

# Illustrative Example 3.5

A material whoseX absorption edge is 0.2 A is irradiated by X-rays ofwavelength 0.15 A.Find themaximum energy ofthe

v oc (Z- 1)^

=>

f=c(Z-l)^

photoelectrons that are emitted from the X shell. ...(3.6)

Where ^ is a constant. Let V be the wavelength of emitted from molybdenum, then

line

Solution The binding energy for X shell in eV is ho

...(3.7)

12431 0.2

eV=62.155KeV

!X-Rays 105

Theenergy oftheincident photon in eVis

# Illustrative Example 3.8

he _ 12431

X

0.15

82.873 KeV

The wavelength of

X-rays produced by a X-ray tube is

Therefore, the maximum energy ofthe photoelectrons emitted

0.76 A. What isthe atomic number ofthe anode material ofthe

from the ^ shell is

tube ?

=£--^^=82.873 - 62.155 KeV Solution

^

-^max=20.718 KeV

X-rays are produced when an electron makes a transition # Illustrative Example 3.6

from w=2to «=1to fill avacancy in K-shell. The wavelength of X-raylines is givenby

Calculate the wavelength ofthe emitted characteristic X-ray from atungsten (Z= 74) target when an electron drops from aM

1

={z-iy •Ka

shell to a vacancy in the K shell.

j__j_ l2 22

1

^Ka

Solution

4

(Z-l)2 =

Tungsten is a multielectron atom. Due tothe shielding ofthe nuclear charge bythe negative charge ofthe inner core electrons, each electron issubject toan effective nuclear charge which

3RX Ka

(Z-\f =

3x(1.097xl0^)x(0.76xl0-^®)

is different for different shells. = 1599.25 "

For an electron in the K shell (a = 1) thus effective nuclear

(Z-1)2 ^1600

charge is given as

Z- 1 = 40

2:eff=(Z-a);

Z=41.

# Illustrative Example 3.9

Here aselectron drops from Mshell {n =3)toX" shell (« = 1), the radiated emission wecall X-ray andfrom Mosleys law the The X-absorption edgeof an unknownelementis 0.171 A:

wavelength emitted ofXp X-ray is given as l2

^Kp

32

10967800 X(74-1)2

(a) Identifythe element.

(b) Find the average wavelengths ofthe

and

lines.

(c) Ifa 100 eVelectron strike thetargetofthiselement, whatis the minimum wavelaigthoftheX-rayemitted ?

''/tp

=0.192 A

Solution

# Illustrative Example 3.7

From Moseleys law, the wavelength oiK series ofX-rays is given bytaking a = 1inmodified inrydberg's formula given as

Apotential difference of20kV is applied aX-raytube. Findthe minimumwavelength of X-raysgenerated. Solution

Y=E.{Z- 1)2 ^1—Lj for£:iineswhere,« =2,3,4,... (a) For ^T-absorption edge, weput« = qo, inabove expression

The X-rays produced under an accelerating potential Vwill

gives

have varying wavelengths with the minimum due to the entire

energy ofthe accelerated electron being lostin a singlecollision with thetarget atoms. Here theshortest wavelength generated of X-rays can be given by X =

12431 V

z=

A

= 0.6215A

=>

1

(0.171xl0~'°)(1.097xl0^)

Z=74.

The element is Tungsten.

+ 1

X-Raya

[106; Practice Exercise 3.1

(b) Forline: 1

(I)

= R{14-\f

^^=0.228 A

[12.431 keV, 3 x lO'® Hz, 6.63 x 10'^"* kg-m/s]

For Spline:

(II) What potential difference should be applied across an X-ray tube toget X-ray ofwavelength not less than 0.10 nm ?

=R{i4-\y

Whatisthe maximum energy ofa photon ofthisX-ray injoule? [12.431 kV, 1.99 Xi0-'5 J]

X.^=0.192A For

Find the energy, the frequency and the momentum ofan

X-rayphoton ofwavelength O.lOnm.

'Ka

line:

J—=R{14-\f

1—-r

^Ky

Xf^^=o.mA (c) The shortest wavelength corresponding toanelectron with

(III) Find the maximum potential difference which may be applied across an X-ray tube with tungsten target without emitting any characteristic X orL X-ray. Theenergy levels of thetungsten atom with an electron knocked outareas follows. Cell containingvacancy

K

L

M

Energy inkeV

69.5

11.3 2.3

[Less than 11.3 kV]

kinetic energy 100 eV is given by , _ ^

_ 12431 X

E

=>

100

^)

A free atom of ironemits

atom = 9.3 x 10"^° kg.

'X,=12431A

[3.9 X 10''" eV]

# Illustrative Example 3.10

(v)

Iron emits

X-ray of energy 3.69 keV. Calculate the

timestaken by an iron

The

X-rays of energy 6.4keV.

Calculatethe recoilkineticenergyof the atom.Mass ofan iron

X-ray emission line oftungsten occurs atX. = 0.21 A.

photon and a calcium

photon to

cross through a distance of 3 km.

What is the energydifference between K and L levels in this [10 gs by both]

atom?

(vl)

Solution

The wavelength of

X-ray of tungsten is 21.3 pm. It

takes 11.3 keV to knock out an electron from the L shell of a

We know that

line is the radiation emitted when an electron

tungsten atom. What should be the minimum accelerating voltage across an X-ray tube having tungsten target which

from I-shell{n = 2)makes a transition toA-shell («= 1) tofill a allows productionof vacancy in it. Thus the released photon will have an energy equal totheenergy difference ofZ.-shell and A^shell, which is [69.5 kV]

X-ray ?

given by he

12431

''Ka

0.21

AE =

(vii) •eV

A£=59.195KeV

Theenergy ofa silver atom with a vacancy inK shell is

25.31 keV, inLshell is3.56keVandinMshellis 0.530keVhigher than the energyofthe atomwith novacancy. Findthe frequency

of

Ap and

X-rays ofsilver.

[5.25 X 10'® Hz, 5.98 x lo'® Hz, 7.32 x lo" Hz] Web Reference at www.Dhvsicsgalaxv.com

AgeGroup - High School Physics 1Age 17-19 Years

Advance Illustrations Videos atwww.Dhvsicsgalaxv.com

Section-MODERN PHYSICS

Age Group -Advance lilustrations

Topic-X-Rays

Section - Modern Ph^^ics

Module Number - 1 to 13

Topic - Atomicand Nuclear Physics Illustrations - 54 In-depth Illustrations Videos

iX-RaysI 107

Discussion Question Q3-1 In aCoolidge tube, electrons strike the target and stop Q3-16 Howare x-raysproduced ?Explain the origin ofthe line inside it. Does the target get more and more negatively charged spectra and the continuous spectra. What limits the minimum as time passes? •

size ofX-ray wavelengths?

Q3-2 Can X-rays beused for photoelectric effect?

Q3-17 When asufficient number ofvisible light photons strike a piece ofphotographic film, the film becomes exposed. An X-rayphoton ismore energetic than avisible light photon. Yet,

Q3-3 X-ray and visible light travel atthe same speed invacuum. Do theytravelat the samespeedin glass?

Q3-4 Characteristic X-rays may beused toidentify the element from which theyare coming. Can continuous X-rays be used

most photographic films arenotexposed bytheX-ray machines used at airport security checkpoints. Explain what these

observations imply about the number ofphotons emitted by the X-ray machines.

for this purpose?

Q3-18 The drawing shows the X-ray spectra produced by an Q3-5 Is it possible that in a Coolidge tube characteristic X-raysare emittedbut not

Q3-6 Can than

X-rays?

X-ray ofone material have shorter wavelength

X-ray ofanother?

X-ray tube when thetube isoperated at two different potential differences. Explain why the characteristic lines occur at the

same wavelength inthetwo spectra, while thecutoffwavelength Xq shifts tothe right when a smaller voltage is used tooperate the tube.

Q3-7 Cana hydrogen atom emitcharacteristic X-ray?

1 Higher voltage

Q3-8 Why isexposure to X-ray injurious tohealth but exposure to visible lightis not, whenbothare electromagnetic waves?

Q3-9 When a Coolidge tube is operated for some time it

Bremsstrahlung

becomes hot. Where does the heat come from ? Wavelength

Q3-10 Can X-raysbe polarized ?

Q3-11 In terms of biological damage, ionization does more damage when you standin front of a very weak (low power) beam ofX-rayradiation than in front ofa stronger beam ofred light. How does the photon concept explain this paradoxical

1

Lower voltage

X -

.

situation?

Q3-12 Whyshould a radiologist be extremely cautions about

Wavelength

X-ray doses when treating pregnant women ? Figure 3.6

Q3-13 Does theconcept ofphoton energy shed any light (no pun intended)on the question of whyX-rays are so muchmore penetrating than visible light ? Explain.

Q3-14 What isthebasic distinction between x-ray energy levels

Q3-19 In the production of X-rays, it is possible to create Bremsstrahlung X-rays without producing the characteristic

X-rays. Explain how thiscanbeaccomplished byadjusting the electric potential difference used to operate the X-ray tube.

and ordinary energy levels ?

Q3-20 The short wavelength side of X-ray spectra ends Q3-15 Can x-raysbe emittedbyhydrogen?

abruptly ata cutoffwavelength X.^. Does this cutoffwavelength depend on the target material used in the X-ray tube > Give your reasoning.

X-Rays

ri,08

ConceptualMCQs Single Option Correct 3-1 X-rays are produced when an element ofhigh atomic weight is bombarded by high energy:

(A) Protons (Q Neutrons

(B) Electrons P) Photons

3-8 With which characteristic of the target does the Moslems law relates the frequencyof X-rays ?

(A) Density (Q Atomic number

p) Atomic weight p) Interatomic space

3-2 Which of the following principle is involved in the generation of X-rays ?

3-9 X-rays areproduced in anX-ray tube operating at a given accelerating voltage. Thewavelength ofthe continuous X-rays

(A) Conversion ofkinetic energy into potential energy

has values from:

(B) Conversion of mass into energy (C) Conversion ofelectric energyintoradiant energy p) Conversion of electric energyinto em waves

(A) Otooo

3-3 In X-raytube when the accelerating voltage Vis halved, thedifference between thewavelengths ofK^ line andminimum wavelength of continuous X-ray spectrum : (A) Remains constant (B) Becomesmore than two times (C) Becomeshalf

(B) \.

mm

to 00 where

mm

> 0

(Q Otok

3-10 The wavelength of A-rays for lead isotopes Pb^^^ Pb^^^, Pb^*^ are Xj, are X3 respectively.'Then ; (A)

P) X,>X2>X3

•(C) Xjv P) X'>X;v'^ andhardX-rays (Q Very had X-rays andlow-frequency y-rays (D) Soft X-rays and y-rays

3-7 Choose the correct statement:

(A) Energy of an atom with K shell electron knocked out is more than the energy of an atom with L shell electron knocked out.

P) Energy of an atom with L shell electron knocked out is more than energy ofan atom with K shell electron knocked out.

(Q Energy of

photon is the sum ofthe energies of an L

3-2 Let Xp and denote the wavelengths ofthe X-rays of and photon. the K^, Xp and lines in the characteristic X-rays for ametal: P) Total energy emitted in from ofX^ photons is more than (A)

(B)

the total energy emitted in from of Xp photon from an X-ray setup.

(Q

X,

3-3 Two electrons starting from rest are accelerated byequal potential difference:

(A) Theywillhavesamekineticenergy (B) They will have same linear momentum

(C) Theywill havesamedeBroglie wave length (D) Theywill produce X-rays ofsame minimum wave length when they strike different targets.

3'8 Ina Coolidge tube experiment, theminimum wavelength of the continuous X-rayspectrum is equal to 66.3 pm, then (A) electrons accelerate through a potential difference ofabout 12.75kV in the Coolidgetube

P) electrons accelerate through a potential difference ofabout 18.75'kV in the Coolidge tube (Q de-Broglie wavelength of the electrons reaching the anticathode is ofthe order of 10 mm

p) de-Brgolie wavelength of the electrons reaching the anticathode is 0.01 A

3-4 When an electron moving at a high speed strikesa metal surface, which of the following are possible? 3-9 The potential difference applied to an X-ray tube is

(A) The entire energy ohheelectron maybe converted into an

X-ray photon.

Anyfraction oftheenergy oftheelectron maybeconverted into an X-ray photon

(Q The entire energy of the electron may get converted to

increased. As a result, in the emitted radiation (A) the intensity increases

p) the minimum wavelength increases (C) the intensity decreases P) the minimum wavelength decreases

heat

p) The electron may undergo elastic collision with the metal

3-10 A beam of electrons striking a copper target produces

surface

3-5 Mark correctstatement(s):

X-rays.Its spectrum is as shown. Keeping'the voltage same if the copper targetis replaced with a different metal,the cut-off wavelength and characteristic lines of the newspectrum will

(A) circumference of orbit of an electron in Bohr's model is

change in comparision with old as:

equal to an integer multiple of deBroglie wavelength of the electron

P) Kinetic energy of electron increases with increase of principle quantum number

(Q when anX-rayphoton isemitted, onlyenergy conservation law is satisfied

P) none of these

Figure 3.11

3-6 In anX-ray tube thevoltage applied is 20KV. The energy

(A) Cut-off wavelength may remain unchanged while

required to remove an electron from L shell is 19.9 KeV. In the

characteristic lines maybe different.

X-rays emittedbythe tube: (takehe = 12420eVA) (A) Minimumwavelength will be 62.1pm p) Energy of the charactersticX-rays will be equalto or less

p) Both cut-offwavelength andcharacteristic lines mayremain unchanged.

than 19.9 KeV

different.

(Q LjjX-raymaybe emitted P) L^X-raywillhaveenergy 19.9KeV

P) Cut-off wavelength will be different while characteristic lines may remain unchanged.

(C) Both cut-offwavelength and characteristic lines may be

X-Bays j

[114 3-11 Which of the following statements is/are correct for an X-ray tube ?

(A) on increasing potential difference between filament and

(B) The minimumwavelength increases (C) The intensity remains unchanged . P) The minimumwavelength decreases

target, photon flux of X-raysincreases

(B) on increasing potential difference between filament and 3-15 X-ray incident on a mato-ial: target, frequency of X-rays increases

(C) onincreasing filament current, cutoffwavelength increases (D) onincreasing filament current, intensity ofX-rays increases

(A) P) (Q P)

Exerts a force on it Transfer energy to it Transfersmomentum to it Transfers impulse to it

3-12 The intensity of X-rays firom a coollidge tube is plotted

against wavelength Xas shown inthefigure-3.12. Which ofthe

3-16 Regarding X-ray spectrum, which of the following

following statements is/are correct:

statements is are correct:

(A) The characteristic X-ray spectrum is emitted due to excitation of inner electrons of atom

P) Wavelength ofcharacteristic spectrum depend on potential difference across the tube

(Q Wavelength of continuous spectrum is dependent on the potential difference across tube P) None of these 3-17 Which ofthe following statements is/are false ? Figure 3.12

(A) On inaeea ngthez (atomic number) oftargetX,^ decreases (B) On increasing the accelerating voltage of tube - X^ increases

(Q On increasing the power of cathode, /q increases P) On increasing the power of cathode, decreases 3-13 For a given material, the energy and wavelength of characteristic X-ray satisfy:

(A) EiKJ>EiK^)>E{K;) (C) X(KJ>XiK^)>HK^)

(B) E{MJ>EiLJ>E(K^) P) XiMJ>X{LJ>X(KJ

3-14 The potential difference applied to an X-ray tube is increased. As a result, in the emitted radiation : (A) The intensity increases

(A) The energyofphotoelectrons emitted fi-om a givenmetal by soft X-rays have less energy than those emitted by hard X-rays

p) To increase the intensity of X-rays, the filament current should be increased

'

(C) The characteristic X-rays have continuous range of wavelengths

P) X-rays werenamed so becauseof their mysterious nature at the time of discovery

3-18 Which of the following statements is/are true ?

(A) The wavelength of soft X-rays is less than that of hard X-rays.

P) X-rays are produced duringacceleration ofelectrons (C) The anticathodeis a metal of lowatomicweight P) The wavelength ofthe characteristic X-rays depends upon the nature ofthe metal of the target

•X-Rays 115

UnsolvedNumericalProblemsforPreparation ofNSEP, INPhO &IPhO Fordetailedpreparation oflNPhOandIPhO students can refer advancestudy material on www.physicsgalaxy.com 3-1 The X-ray coming from a Coolidge tube has a cutoff 3-10 The electric current in an X-ray tube (from the target to wavelength of80pm. Find thekinetic energy ofthe electrons the filament) operating at 40 kV is 10mA.Assume that on an hitting the target.

average, 1% ofthe total kinetic energy oftheelectrons hitting

Ans. [15.5 keV]

the target are converted into X-ray. (a) What is the total power

3-2 Ifthe operating potential in an X-ray tube is increased by

target every second ?

emitted as X-rays and (b) how much heat is produced in the

1% bywhat percentage does thecutoffwavelength decrease ? Ans. [Approximately 1%]

Ans. [(a) 4 W (b) 396 J]

3-3 The distance between the cathode (filament) and the target 3-11 The X-rays ofaluminium (Z- 13) and zinc (Z= 30) in an X-ray tube is 1.5 m. Ifthe cutoffwavelength is30 pm, find have wavelengths 887 pm and 146 pm respectively. Use theelectric field between thecathode and thetarget. Ans. [27.7 kV/m]

Moseleys law Vv =a(Z- b) to find the wavelength ofthe X-ray of iron (Z= 26).

Ans. [198 pm]

3-4 The short-wavelength limit shifts by 26 pm when the operating voltage in an X-ray tube is increased to 1.5 times the

original value. Whatwasthe original approximate value ofthe operating voltage ? in KV. Ans. [15.9 kV]

3-5 The electron beam in acolour TV is accelerated through

3-12 Acertain element emits X^X-ray ofenergy 3.69 keV. Use thedata from the previous problem to identify theelement. Ans. [Calcium]

^

certain elements are given below. Draw

32 kV and then strikes the screen. What is the wavelength of aMoseley-type plot ofVv versus Zfor

themostenergeticX-rayphoton? Ans. [38.8 pm]

radiation

p Ca Mn Zn Br '

Energy (keV) 0.858 2.14 4.02 6.51

9.57 133

3-6 When 40 kV is applied across an X-ray tube, X-ray is 3-14 Use Moseley's law with = 1tofind the frequency ofthe obtained with a maximum frequency of9.7 x 10'^Hz. Calculate he value of Planck constant from these date.

X^ X-ray ofLa (Z = 57) ifthe frequency ofthe X^ X-ray of Cu (Z = 57) is known to be 1.88 x 10Hz.

Ans. [4.12 X 10"'^ eV-s] Ans. [7.52 x 10'» Hz]

3-7 The Xp X-ray ofargon has awavelength of0.36 nm. The

minimum energy needed toionize an argon atom is 16eV. Find 3-15 The

and Xp X-rays ofmolybdenum have wavelengths

the energy needed to knock out an electron from the X shell of

0.71 Aand 0.63 Arespectively. Find thewavelength ofZ-^ X-ray

an argon atom.

ofmolybdenum.

Ans. (3.47 keV]

Ans. [5.64 A]

3-8 An X-ray tube operates at 40 kV. Suppose the electron 3-16 The wavelengths ofX„ and L X-rays ofamaterial are converts 70% of,ts energy,ntoaphoton at each collision. Find 21.3 pm and 141 pm respectively. Find the wavelength ofX„

the lowest three wavelength emitted from the tube. Neglect the x-ray ofthe material energy imparted to the atom with which the electron collides.

Ans. [44.3 pm, 148 pm, 493 pm]

P'"]

3-9 TheKaX-rays ofmolybdenum haswavelength 71 pm. If

3-17 Heat at therate of200 W is produced in an X-ray tube

the energy of a molybdenum atom with a X electron knocked out is 23.32 keV, what will bethe energyofthis atom when anL

operating at 20 kV. Find the current in the circuit. Assume that

only a small fraction of the kinetic energy of electrons is

electron is knock out ? In eZ.

converted into X-rays.

Ans. [5.82 keV]

Ans. [10 mA]

[116

^

,

3-18 Fill the blanks in each ofthe following statements:

X-Rays ]

by ahydrogen like element is 0.32 A. Calculatethe wavelenth of

Xp line emitted bythe same element. (a) In an X-ray tube, electrons accelerated through apotential difference of 15000 V strike a copper target. Findthe speed of Ans. [0.27 A] the emittedX-rays insidethe tube.

(b) Inthe Bohr model ofthe hydrogen atom, find the ratio of 3-21 Apotential difference of20 KV isapplied across an X-ray thekinetic energy tothe total energy oftheelectron inaspecific quantum state. Ans. [(b) - 1]

tube. Theminimum wave length ofX-rays generated is Ans. [41]

3-22 CharacteristicX-rays offrequency4.2 xlO^^Hzareemitted

3-19 Suppose a monochromatic X-ray beam ofwavelength from a metal due to transition from L - to K-shell. Find the 10 pmissent through aYoung's double slitand theinterference atomic number ofthemetalusingMoseley's law. Take Rydberg pattern isobserved onaphotographic plate placed 40cmaway constanti?= 1.1 x lO'm"'. from the slit. What should be the separationbetween the slits so that the successivemaxima on the screen are separated by a

Ans. [42]

distance of 0.1 mm?

3-23 An X-raytube isoperated at20kV and the current through

Ans. [4 X 10"' m]

the tubeis 0.5 mA. Find the total energyfalling on the target per second as the kinetic energy of the electrons in J.

3-20 ThewavelengthofthecharacteristicX-rayX^lineemitted

Ans. [lO]

4

Nuclear Physics andRadioactivity FEW WORDS FOR STUDENTS

In preceding chapters we have discussed about

structure ofan atom and different experiments involving properties of atomic structure. Now. we consider the composition and properties of the atomic nucleus. In this chapter we will discuss about the forces holding the nuclear

Alpha partides

Gold film

•Alpha partides -

Gold film

•

• —•

• . •

•

t

•

/ _ /

7

matter within the nucleus, the nucleus and the nuclear structure. We next examine the

"• e_ •

• —

»

~"Nudeus

(W

condition of nuclear stability and the natural

radioactive processes ofalpha, beta and gamma decay. Finally we considernuclear reactions and the energy involved in these reactions.

CHAPTER CONTENTS

4.1

CompositionandStructure ofThe Nucleus

4.5

Nuclear Reactions

4.6

Nuclear Fission

4.2

-Nuclear Binding Energy

4.3

Radioactivity

4.7

Nuclear Fusion

4.4

Radioactive Series

4.8

Properties ofRadioactive Radiations

COVER APPLICATION

Flgur,e-(a)

Flgure-(b)

Nuclear reactors are today major source of power across the globe. Flgure-(a) "shows a specific nuclear power plant in which power generation is done by nuclear reactors. Figure-(b) shows the internal block diagram of a nuclear reactor.

Nuclear Physics and Radioactivity .

118

After Rutherford's discovery of the nucleus in 1911, scientists

(2) Isobars : Elements having same mass numbers and

slowly realized that the nucleus must be made of smaller different atomic numbers are called isobars. Thus isobars have constituent parts. Earlier it was proposed that the nuclei of same A, but different Z. atoms heavier then hydrogen consisted of hydrogen nuclei, called protons, and electrons. The protons would provide the (3) Isotones : The elements which have same number of mass andpositive charge andelectrons which were presumed neutrons are called isotones. Thus isotones have same {A-Z). to be in the nucleus, would neutralize some of the charge. This theorycouldsatisfactorily account fora nucleusofatomicmass A and atomic number Zby assuming that the nucleus contained

(4) Isodiaphers: Those elements which havesame difference in neutrons and protons are called isodiaphers. Thus in

Aprotons andA-Zelectrons. The proton-electron model ofthe isodiaphers {A-2Z) issameandthis value {A-22) iscalled isotopic nucleusgavethe correctcharge and approximately the correct number of the element. mass for any nucleus. However this proton electron model of nucleushad manydifficulties. It wasnot ableto explainspin of 4.1.1 Size of a Nucleus particles emitted during decayprocesses from thenucleus. This

this model. For such reasons, Ruthorford suggested in 1920 that the nucleus should contain a neutral particle ofsame mass

The Ruthorford scattering experiment provided the first estimates of nuclear sizes. At that time a variety ofexperiments have been performed to calculate the nuclear dimensions. It was found that the volume of a nucleus is directly proportional

model does not conserve angular momentum in this

phenomenon. This factalonewasa strongreason fordiscarding

as that ofprotonbutnot ableto provepractically. Later in 1932

to the number of nucleons in it. Thus it can be said that the

James Chedwick demonstrate the existence of such a neutral

densityof nucleons is approximately same in the interiorsofall

particle in his experiments. This particle was given the name

nuclei.

neutron.

4

Now we'll discuss the ftindamental properties ofnucleus having protons and neutrons are the major constituents.

,

,

Ifanucleus is ofradius R, its volume is y %R., thus here R is directlyproportionalto the number of nucleonsor mass member A ofthe element. Hence we have

4.1 Composition and Structure, of The Nucleus Because protons and neutrons both appear in the nucleus, they are collectively called nucleons. According to today's accepted proton-neutron model of the nucleus, the number of protons in a nucleus is called atomicnumber of the elementZ, and the number of protons plus the number of neutrons is called mass number ofthe element/4. Some timesy4 is also called

nucleon number. A shorthand notation is often used to specify Z and A along with the chemical symbol for the element. It is generallywritten as

R^ozA or

RozA^^^

or

R= R^ 1/3

Here

...(4.1)

is named fermi constant and its value is given as = 1.2x10"'^-1.2 fin.

...(4.2)

As nuclei do not have sharp boundaries, the value of may also have slight deviations from its value given by equation-(4.2).

2X or

4.1.2 Strong Nuclear Force and Stability of Nucleus Similarly in nuclear reactions the symbol used for a proton is

}p or [h . Aneutron is denoted by qii. In caseofelectron we

We've discussed that inside a nucleus in a very small volume

use _] e (It has 0 mass and its charge is - 1). Someparticular

nucleons are bounded together. The positively charged protons inside the nucleus repel one another with a very strong electrostatic force. It is surprising that due to such a large repulsive force what keeps the nucleus from flying apart ? It is

names are given to different group of elements having some similarity in nuclear structure. These are (1) Isotopes : Elements those nuclei have same number of protonsbut differentnumber ofneutrons, are known as isotopes. Thus isotopes have same Z, but different X. Isotopes of same elements have same chemical properties because they have the same number and arrangement ofelectrons.

clear that there must be some kind of attractive force which

hold the nucleons together as many atoms contain stable nuclei. The gravitational force of attraction between nucleons is two weak to counteract the repulsive electric force, so we can say

that a different type of force must hold the nucleus together.

[Nuclear Physics and Radioactivity

119

This force is strong nuclear force and is one of the four fundamental forces ofnature;

We can see that in the graph shown in figure-4.1 almost all points representing stable nuclei fall above the straight line N=Z, reflecting that the number of neutrons are greaterthen

The mostimportant feature of the strong nuclearforce is it is independent of electric charge. At a givenseparation between

number of protons.

nucleons, approximately some nuclear force ofattraction exist

In veryheavynuclei as numberofprotons are large,therecomes a point when balance of repulsive and attractive forces can not

between twoprotons, twoneutrons or between a proton and a neutron. The range of action of the strong nuclear force is extremely short. When two nucleons are very close at separation of the order of about 10"'^ m, the nuclear force of attraction between the two is very large and almost zero for large separations. For electric force we already know that the range of action ofelectricforce is verylarge.The electricforce

be achieved by an increased number of neutrons. For heavy nuclei, the size is also large and due to the limited range of strong nuclear force, extra neutrons in the nucleus cannot

balance the long range electric repulsion of extra protons. In naturethe stable nucleus withthe largestnumberofprotons is

Bismuth ^gj^Bi which contains 83 protons and 126 neutrons.

between two charges is very large at close separation and All nuclei above bismuth in nature are unstable due to decreases to zero gradually as separation increases to very unbalanced internal force and spontaneously break apart or large values.

The veryshort range ofstrongnuclear force playsan important role in stability of nucleus. For a nucleus to be stable, the electrostatic repulsion between protons must be balanced by the attraction between nucleons due to strong nuclear force.

rearrange the internal structure of nucleus. This phenomenon of spontaneous disintegration or rearrangement in internal" structure of nucleus is called radioactivity. This phenomenon wasfirst discovered byHenriBecquerel in ]896.In laterpart of this chapter we'll discuss radioactivity in detail.

4.2 Nuclear Binding Energy Inside a nucleus one protonsrepel all other protons. Since the electrostaticforcehas a longrange of action. Buta proton or a neutron due to strong nuclear force attracts only its nearest neighbours, thus in. a large sized nucleus to counter balance

the repulsive force more number of neutrons are required to maintain the stability ofnucleus.

We've discussed that in a stable nucleus, because of strong nuclear forceofattraction, the nucleonsare held tightlytogether in a small volume.As systemis stable we can relativelysay that

the totalpotential energyofsystem is negative and to separate all the nucleus from eachother someenergymustbe supplied tobreakthenucleus. Themorestable thenucleus is,thegreater is the energyneededto break it apart. This required energy, we call binding energy ofthe nucleus.

Binding energy

Nucleus

(smaller mass) Separated nucleons (greater mass) Figure 4.2

The origin ofbinding energycan beeasilyexplainedif we look into the masses of different elements in table-4.1. In nuclear

Stable nuclei

physics, mass ofatoms and nuclei are often measured in atomic mass unit (amu). 20

40

60

Proton number Z

80

We can see clearly fi-om the data given in table-4.1 that the mass of each element atom is less then the sura of masses of its

Figure 4.1

constituent particles.

Nuclear Physics and Radloacttvity j

;i2o

Table 4.1

Element

z 0 1

2

3.

Neutron

Hydrogen

Helium

Lithium

Symbol

A

Atomic Mass, amu

z

1.008 665

14.

1

n

H

He

Li

1

1.007 825

2

2.014 102

3

3.016 050

3

3.016 029

4

4.002 603

6

6.018 891

6

6.015 123

7

7.016 004

8

8.022 487

7

7.016 930

15.

16.

*17. 4.

Beryllium

Be

•

6.

7.

8.

9.

Boron

Carbon

Nitrogen

Oxygen

Fluorine

B

C

N

F

.

10.

11.

12.

13.

Neon

Sodium

Magnesium

Aluminum

Ne

Na

Mg

A1

Sulfur

Chlorine

Atomic Mass, amu

Si

28

27.976 928

29

28.976 496

30

29.973 772

30

29.978 310

31

30.973 763

32

31.972 072

P

S

CI

33

32.971 459

34

33.967 868

35

34.969 032

36

35.967 079

35

34.968 853

36

35.968 307

37

.

36.965 903

Argon

Ar

36

35.967 546

10

10.012 938

37

36.966 776

11

11.009 305

38

37.962 732

12 ~

12.014 353

39

38.964 315

40

39.962 383

10

10.016 858

11

11.011 433

39

38.963 708

12

12.000 000

40

39.963 999

13

13.003 355

41

40.961 825

14

14.003 242

15

15.010 599

40

39.962 591

41

40.962 278

19.

20.

Potassium

Calcium

K

Ca

12

12.018 613

42

41.958 622

13

13.005 739

43

42.958 770

14

14.003 074

44

43.955 485

15

15.000 109

45

44.956 189

16

16.006 099

46

45.953 689

17

17.008 449

47

46.954 543

48

47.952 532

14

0

Phosphorus

A

10.013 535 18.

5.

Silicon

Symbol

9.012 182

9 10

Element

14.008 597

15

15.003 065

16

15.994 915

17

16.999 131

18

17.999 159

19

19.003 576

17

17.002 095

18

18.000 937

19

18.998 403

20

19.999 982

21

20.999 949

21.

Scandium

Sc

45

44.955 914

22.

Titanium

Ti

46

45.952 633

23.

Vanadium

V

47

46.951 765

48

47.947 947

49

48.947 871

50

49.944 786

48

47.952 257

50

49.947 161

51

50.943 962

18

18.005 710

48

47.954 033

19

19.001 880

50

49.946 046

20

19.992 439

52

51.940 510

21

20.993 845

53

52.940 651

22

21.991 384

54

53.938 882

23

22.994 466

24

23.993 613

54

53.940 360

55

54.938 046

22

21.994 435

24.

25.

26.

chromium

Manganese

Iron

Cr

Mn

Fe

54

53.939 612

56

55.934 939

57

56.935 396

23

22.989 770

24

23.990 963

23

22.994 127

58

57.933 278

24

23.985 045

59

58.934 878

58

57.935 755

59

58.933 198

60

59.933 820

25

24.985 839

26

25.982 595

27

26.981 541

27.

Cobalt

Co

Nuclear Physics and Radioactiwty z 28.

29.

30.

31.

32.

33.

34.

Element Nickel

Copper

Zinc

Symbol

A

Ni

58

57.935 347

96

60

59.930 789

97

96.906 018

6 1

60.931 059

98

97.905 405

100

99.907 473

Cu

Zn

Gallium

Germanium

Arsenic

Selenium

Ga

Ge

As

Se

-

35.

36.

Bromine

Krypton

Br

Kr

1

37.

38.

Rubidium

Strontium

121

Rb

Sr

Atomic Mass, amu

62

61.928 346

64

63.927 968

63

62.929 599

64

63.929 766

65

64.927 792

64

63.929 145

65

64.929 244

66

65.926 035

67

66.927 129

z

Element

Symbol

A

Atomic Mass, amu 95.904 675

43.

Technetium

Tc

99

98.906 252

44.

Ruthenium

Ru

96

95.907 596

68

67.924-846

70

69.925 325

45.

Rhodium

Rh

69

--^68.925'581

46.

Palladium

Pd

71

70.924 701

70 72

98

97.905 287

99

98.905 937

100

99.904 217

101

100.905 581

102

101.904 347

104

103.905 422

103

102.905 503

102

101.905 609

104

103.904 026

69.924 250

105

104.905 075

71.922 080

106

105.903 475

73

72.923 464

108

107.903 894

74

73.921 179

110

109.905 169

76

75.921 403

107

106.905 095

74

73.923 930

75

74.921 596

47.

Silver

Ag

•

48.

107.905 956

109

108.904 754

74

73.922 477

106

105.906 461

76

75.919 207

108

107.904 186

77

76.919 908

110

109.903 007

78

77.917 304

111

110.904 182

80

79.916 520

112

111.902 761

82

81.916 709

113

112.904 401

79

78.918 336

80

79.918 528

81

80.916 290

78

77.920 397

80

79.916 375

81

80.916 578

82

81.913 483

83

82.914 134

84

83.911 506

49.

50.

Cadmium

108

Indium

Tin

Cd

In

Sn

114

113.903 361

116

115.904 758

113

112.904 056

115

114.903 875

112

11 1.904 823

114

113.902 781

115

114.903 344

116

115.901 743

117

116.902 954

86

85.910 614

118

117.901 607

85

84.911 800

119

118.903 310

87

86.909 184

120

119.902 199

84

83.913 428

86

85.909 273

87

86.908 890

88

87.905 625

39.

Yttrium

Y

89

88.905 856

40.

Zirconium

Zr

90

89.904 708

52.

Antimony

Tellerium

Sb

Te

121.903 440

124

123.905 271

121

120.903 824

123

122.904 222

120

119.904 021

122

121.903 055

123

122.904 278

124

123.902 278

93.906 319

125

124.904 435

95.908 272

126

125.903 310

127

126.905 222

91

90.905 644

92

91.905 039

94 96

51.

122

41.

Niobium

Nb

93

42.

Molybdenum

Mo

92

91.906 809

94

93.905 809

95

94.905 838

92.906 378

53.

Iodine

1

128

127.904 464

130

129.906 229

127

126.904 477

131

130.906 119

Nuclear Physics and Radioactivity i

:12Z

z 54.

55.

56.

57.

58.

Element

Symbol

Xenon

Cesium

Barium

Lanthanum

Cerium

Xe

Cs Ba

La

Ce

A

Atomic Mass, amu

Holmium

Ho

165

68.

Erbium

Er

162

161.928 787

164

163.929 211

166

165.930 305

125.904 281

128

127.903 531

129

128.904 780

130

129.903 509

131

130.905 076

132

131.904 148

134

133.905 395

136

135.907 219

69.

Thulium

133

132.905 433

70.

Ytterbium

129.906 277

131.905 042

134

133.904 490

135

134.905 668

136

135.904 556

137

136.905 816

138

137.905 236

138

137.907 114

139

138.906 355

136

135.907 14

138

137.905 996

140

141.909 249

59.

Praseodymium

Pr

141

140.907 657

60.

Neodymium

Nd

142

141.907 731

143

142.909 823

144

143.910 096

62.

63.

Promethium Samarium

Europium

P m Sm

Eu

Gadolinium

Gd

167.932 383

170

169.935 476

Tm

169

168.934 225

Yb

168

167.933 908

170

169.934 774

171

170.936 338

172

, 171.936 393

173

172.938 222

176

173.938 873

71.

72.

Lutetium

Hafnium

Lu

Hf

74.

Tantalum

Tungsten

Ta

W

•

175.942 576

175

174.940 785

176

175.942 694

174

172.940 065

176

175.941 420

177

176.943 233

178

177.943 710

179

178.945 827

180

179.946 561

180

179.947 489

181

180.948 014

180

179.946 727 181.948 225

145

144.912 582

182

146

145.913 126

183

182.950 245

148

147.916 901

184

183.950 953

150

149.920 900

186

185.954 377

147

146.915 148

-

75.

Rhenium

Re

185

184.952 977

187

186.955 765

184

183.952 514

144

143.912 009

147

146.914 907

148

147.914 832

186

185.953 852

149

148.917 193

187

186.955 762

150

149.917 285

188

187.955 850

152

151.919 741

189

188.958 156

154

153.922 218

190

189.958 455

151

150.919 860

192

191.961 487

153 64.

166.932 061

168

174'

73.

61.

164.930 332

167

139.905 442

142

Atomic Mass, ai

67.

123.906 12

126

132

A

Element

124

130

Symbol

z

76.

77.

152

151.919 803

154

153.920 876

155

154.922 629

156

155.922 130

157

156.923 967

158

157.924 111

160

159.927 061

65.

Terbium

Tb

159

158.925 350

66.

Dysprosium

Dy =

156

155.924 287

158

157.924 412

160

159.925 203

161

160.926 939

162

161.926 805

163

162.928 737

164

163.929 183

78.

Osmium

Iridium

Platinum

Os

Ir

Pt

191

190.960 603

193

192.962 942

190

189.959 937

193

191.961 049

194

193.962 679

195

194.964 785

196

195.964 947

198

197.967 879

79.

Gold

Au

197

196.966 560

80.

Mercury

Hg

196

195.965 812

198

197.966 760

199

198.968 269

.

200

199.968 316

201

200.970 293

202

201.970 632

204

203.973 481

iNiiclear Physics and Radioactivity

Z

Element

Symbol

123

A

Atomic Mass, amu

Z

Element

Symbol

A

Atomic Mass, arau ~

81.

,82.

83."

Thallium

Lead

Bismuth

TI

,Pb

Bi

203

202.972 336

205

204.974 410

204

Polonium

Po

91.

Protactinium

Pa

233

92.

Uranium

U

232

232.037 168

233

233.039 629 234.040 947

203.973 037 205.974 455

234

207

206.975 885

235

235.043 925

208

207.976 641

238

238.050 786

210

209.984 178

214

213.999 764

237

237.048 169

239

239.052 932

209

93.

Neptunium

Np

——

233.040 244

206

212

84.

7

209.980 388 211.991 267

210

209.982 876

214

213.995 191

216

2i6.001 790

218

218.008 930

85.

Astatine

At

218

218.008 607

86.

Radon

Rn

220

220.001 401

222

222.017 401

87.

Francium

Fr

223

223.019 73

88.

Radium

Ra

226

226.025 406

89.

Actinium

Ac

227

227.027 751

90.

Thorium

Th

228

228.028 750

230

230.033 131

232

232.038 054

233

233.041 580

94.

Plutonium

Pu

239

239.052 158

240

240.053 809

95.

Americium

Am

243

243.061 374

96.

Curium

Cm

247.

247.070 349

97.

Berkelum

Bk

247

247.070 300

98.

Californium

Cf

251

251.079 581

99.

Einsteinium

Es

252

252.082 82

100.

Fermium

Fm

257

257.095 103

101.

Mendelevium

Md

258

258.098 57

Nobelium

No

259

259.100 941

103.

Lawrencium

Lr

260

260.105 36

104.

Rutherfordium

Rf

261

261.108 69

105.

Hahnium

Ha

262

262.113 84

• 102.

For examplewe discuss for 2He atom. Wecompare the mass of

Einstein mass energy relationship some amount of mass from

atom to that of its constituents. To calculate the masses of the

independent nucleons is converted into energy and released

components, we can either add the masses of two protons, two

when nucleons bounded with each other to form a stable

neutrons and two electrons or we can add the masses of two

nucleus. Thisis the energywhatholdthe nucleons together in

H-atoms and two neutrons. Using the values from table-4.1, we

a nucleus, we call ''BindingEnergy' of nucleus. So when this energy is supplied to a nucleus, it splits into its constituent

have

particles.

2m^+2m^=2 (1.007825)+2 (1.008665) =4.0329804 •

...(4.3)

Nowwe can see that in table mass 2He atom is 4.0026034

For the above reaction given in equation-(4.4), ifwe calculate the difference in masses of nucleons and that of nucleonsX, then it is given as

which is less then the value given in equation-(4.3), the sum of masses ofconstituentparticles. Similarthing can be verifiedfor all the elements, also the reason for this can be explained by

This difference in masses of independent nucleons and mass

equation-(4.4) which is a basic nuclear reaction for formation of

ofnucleus is called''Mass Defect' ofthe nuclear reaction. Using

Aw= Zmp+ (A-Z)m

...(4.5)

a nuclei ^X. In this reaction we can see that when Z-protons mass defect Aw we can find the energy released in a nuclear and (A-Z) neutrons fuse together to form a nucleus 2X, some

reaction, such as the binding energy ofabove nucleus X can be

amount of energy must be released as nucleus is more stable

given as

form ofnucleus

AEg^ =Awc^ ZWp+ {A-Z)m

Energy

...(4.6)

...(4.4)

If we think fromwhere this energy come,we can simply say by

In thesimilarwaywecanfindthebindingenergyforanynucleus in nature for a known composition.

Nuclear Physics and Radioactivity :

ii24

4.2.1 Mass Energy Equivalence

AEv

AEy^

Ay

We'vediscussed in previous section that using massdefect of a nuclear reactionwe can find the energyreleasedin the nuclear

process. We know generally nuclear masses aregiven in atomic mass unit where

and these values aresuch that {BE^^


AE=-0.092MeV

[9.435 MeV]

Because B is negative for this reaction, hence ^Be is unstable against decay to two alpha particles.

(vil) Find the binding energy of^gFe. Atomic mass of^^Fe is 55.934939 amu. Mass of a proton is 1.007825 amu and that of

;. Web'Referenceat www.physicsgalaxv.com

I Age Group - High School Physics [ Age 17-19 Years

neutron = 1.008665 amu. [496.95 MeV]

: Section-MODERNPHYSICS

. Topic- NuclearStructure &Radioactivity Module Number - 1 to 16

4.3 Radioactivity As we've discussed that inside a nucleus electrostatic attraction

Practice Exercise 4.1

is counterbalanced by short range strong nuclear forces and nucleus becomes stable. Despite the forces are balanced, many

(!) In a thermo-nuclear reaction 1.00 x 10 ^kghydrogen is nuclides are unstable because of nuclear size or the limited range of nuclear forcesor dueto slight imbalancein small sized converted into 0.993 x 10"^ kghelium. (a)

Calculate the energy released in joule.

(b)

If the efficiency of the generator be 5%, calculate the

energy in kilowatt hours. [(a) 63.0 X lO'O J, (b) 8.75 kWh]

(ii)

nuclides, these nuclidesspontaneously disintegrateinto other nuclide. This phenomenon of spontaneous disintegration we callradioactivity. In further section ofthe chapterwe'll discuss the aspects of radioactivity by which unstable nuclide disintegrate to achieve stability.

Find the binding energy and the binding energy per 4.3.1 MeasurementofRadioactivity

nucleon ofthe nucleus of gO. Given, atomic mass of^gO (w) =

15.994915 amu, mass ofproton (wp—1.007825 amu, mass ofa Radioactivity ofan element is measured in terms of"activity" neutron (m„)=1.008665amuandlamu=931.5MeV. Theactivityofasampleofanyradioactivenuelideistherateat [127.62 MeV, 7.976 MeV]

which the nuclei ofits constituent atoms disintegrate. If are

Nuclear Physics arvd Radioactiyityi

1128

thenumberofnucleipresentin a radioactive sampleat an instant Thus due to randomness, approximately the amount of disintegrations perunitvolume persecond (called disintegration then activity of this sample is given as

density) remains constant in thewhole volume ofthesubstance. dN

...(4.8) 43.3 Radioactive Decay Law

Here

dN

is negative as with time always no. of elements

This lawrelatesthe activityof a substance withthe number of decreases due to disintegration, due to negative sign, is active or undecayed atoms present in a group of radioactive always takenpositive. The activity of a substance is measured atoms at an instant oftime. This law is stated as in termsof"dps" or disintegrations per second. The SIunit of "TTie activity ofa radioactive element at anyinstantisdirectly activity is named after Bequeral, defined as 1bequeral= 1 Bq=\ disintegration/sec

proportional to the number ofundecayed active atoms {parent atoms) present at that instant.''^

Generallyactivities of radioactive samples in nature are very highthat's whybequeral isa verysmall unitfornormal practice, Letus consider that at r= 0,thereareA'q parent atoms arethere moreoften MBq or GBq are used.Forthe sametraditionalunits in a substance and after a time ?, A^-atoms are left undecayed. curie (C/) and rutherford{Ru) are alsoused. These are defined Thisimplies that in theduration from t = Q\.ot = t,N^-Natoms are decayed to their daughter element. If in further time from as t-t\.ot = t + dt,dN more atoms will decay then at time r = r,we can say that the activity of the element is given as

1 Ci = 3.7 X lO^'dis/sec and

1 Ru = 10^dis/sec

dN

Roughly curie was originally defined as activity of 1 gm of

^Ra. Similarly itwas observed that 1kg ofordinarypotassium

...(4.9)

A =

Now according to Radioactive Decay Law, we have

hasan activity of about 1 raCi (10~^ Ci) because in ordinary potassium smallproportion ofradioisotope

is alsopresent.

dN_ dt

xN

43.2 Fundamental Laws of Radioactivity ^ .

dN A =

Onthebasisofexperiments performed byRuthorford and Soddy

dt

= XN

...(4.10)

Here Xis the proportionality constant, we call decay constant for the decayprocess. The value of decay constant differs for we summarize as fundamental laws ofradioactivity. These are different elements. From equation-(4.10) we can see that ifX, is high the elementwillhave high value ofactivityand ifXisless, (1) Radioactivity is purely a nuclear process, it is not the activity will be relativity less. Thus we can say that the concerned in any manner with the extranuclear part of atom. decayconstantfor a radioactive elementgives.a relativecriteria (2) As radioactivity is a nuclear process, it is independent of its stability ifthe value of Xfor an element is more, it is more fromany chemicalpropertyof the element.As we'vediscussed active or relatively less stableand iffor an elementXis less, it is that radioactive property of an element is only the process more stable.From equation-4.10, we can also write

some conclusions were made for behaviour of radioactive

elementsand the properties ofRadioactivity, these conclusions

concerned with nucleus of the element. It does not effect the

electronic configuration of the element. If this element takes partin a chemical reaction theproduct formed willalsohavethe radioactive property in the same fraction by which the radioactive atom is present in the molecule of product.

(3) Radioactivity is a randomprocess, its studyis onlypossible by laws of probability mathematically. In a group of several radioactive atoms, which one will disintegrate first is just a matter of chance.

dN dt X=

Thus decayconstant of a processcan be given as "activityper atom" (as given in equation-(4.11)). This shows that for a given radioactive element the activity per atom always remains

constant where as we've already discussed that with time the overall activity of a substance decreases with time as number of parent elements continuously decreases with time.

(4) As radioactivity is a random process, the disintegration densitythroughout the volumeofa radioactive element remains Now from equation-(4.10), we can write constant. If an elementX decaysto a daughter nuclide Ythen in a given volume of element, all portions of volume will have same ratio of number of atoms of Y to that ofX

...(4.11)

N

-dr=~^

...(4.12)

•Nuclear Physics and Radioactivity

"12^1

Herenegative sign shows that ~ , the rate at which the active Insome nuclear power plants amajor problem isdisposal ofthe

elements are disintegrating is negative or number of active elements are decreasing with time. Now we have from equation-(4.12)

radioactive wastes since some ofthenuclide present in waste have long halflives.

For a radioactive element in a sample if at / = 0, No nuclei are present ofactive parent element and during observation after dN

//

, ,

^dt

t= ^ are leftthen thisduration 7canbetaken as halflifeof

Now we integrate this expression within time limits from /=0, to this element. From radioactive decay equation wehave t = t,WQ have

N=NQe \r

N

...(4.15)

t

If-I

Here a.t t-T,N= ~

Xdt

thus wehave inabove equation

N,

[inivE =-Xt

In I^I=-XT

\nN-\nN^ =-'kt In I

\n(2)=XT

Xt

^ In (2) _ Q.693 N=N^e-^'

...(4.16)

...(4.13)

Here equation-(4.13) gives thenumber ofactive parent atoms N 4.3.5 Alternate formofDecay Equationin terms ofHalfLife present attimet inthemixture. Thisequation, wecallradioactive Time decay equation.

If one radioactive element X decays to a daughter nucleus Y with a decay constant Xthen for theprocess nuclear reaction, is

From equation-(4.13) we can have

written as

U[=XN^e-'^' A c =A cO

.-X/

X

^

Y

...(4.14)

Ifinitially nuclei ofelement Xare present then after time t, Here A^^ = XN^is the initial activity of substance at r = 0. number of nuclei present in the sample are given by decay equation-(4.14) is another form ofradioactive decay equation. equation given as This equation can be used to find activity of a radioactive

N=N^e-^

substance at any time instant.

...(4.17)

If werearrange the equation,we have 4.3.4 HalfLife Time

In ^ =-XT

In previous section we've discussed that during decay of a radioactive sample, the amount of radionuclide fall off

exponentially with time. Every radioactive sample has a

^0

We know that half life of substance is defined as

y,^In(2)

characteristichalflife. Halflifetime is defined as the time duration

in which halfof the total number ofnuclei will decay or left undecayed. Say for example at any instant we look into the quantity of a sample of radioactive element, it is observed that

X or

we can write

In (2) ...(4.19)

after every3 hour the number of undecayed parent element reduces to halfthus accordingly the half life of the nuclide is 3 hour. Some half lives are only a millionth of a second for highly active elements and some less active elements have half life in billions ofyears.

...(4.18)

Now from equation-(4.18) we have In

jV_

ln(2)

Nr.

T

Nuclear Physics and Radioactivityj

:i3o

time t is dN. The sun in the number ofthe average is really an

ln|^l=ln(2r"-

integration ofthe quantity t ^iVbetween

and 0particles; the

denominator is the sumofthe particles, or the integration ofdN over all the particles:

Taking antilog on both sides, we get idN

— =(2r"^

A^o

^^

...(4.22)

Mean or average life = —

N=N^{2)-IIT

I

...(4.20)

dN

^^0

Equation-(4.20) is an alternate form ofdecay equation useful pQj. fornumerical applications.

integration in the numerator, we use the activity ofthe

substance to find dN, given as

4.3.6 Mean Life Time at

Fora given radioactive substance, some nuclei disintegrate in dN=- XN dt thebeginning andsome willdisintegrate aftera longtime. Thus wecansaythat somenuclei haveveryshortlifeandsomehave We make the substitution for dN in the numerator ofequationverylargelifetimefordisintegration. Fora radioactive element (4.22).For the averagelife,we then have mean life is defined as

-1 tXN dt

Mean life time

T

>"

=

Mean or average life =

Sum of lives of all nuclei in a sample Total number of nuclides in sample

r=| m X

...(4.23)

_ N,

From radioactive decay equation we have

N-Nff-'^'

...(4.21)

The above numerical value is comes out to be reciprocal of

Thus equation-(4.23) becomes CO

decay constant oftheelement. This cannotbedirectly calculated

dt

but we can calculate the value ofmean life time for a radioactive

element using basic probabilitylaws as explained in the next

Mean life -

N,

...(4.24)

section.

This integration is done by parts. The result comes out as 4.3.7 Calculation of Mean Life Time For a Radioactive Element

The half-life ofa sampleofradioactive nucleiis thetime forhalf ofthe sample to disintegrate. The average or mean life of the

sample isdifferent. Some oftheatoms in thesample exist much longer than others before decaying. To determine themean life, consider an analogy. Imagine a collection of 10people withthe following death statistics; 3 die at age 60 y, 2 at age 70 y, 4 at 75 y, and 1 at a venerable 90 y. The average age is found by multiplying thenumber dying ateachage, summing theresults, and dividing the sum by the total sample size : Average age =

3(60y) + 2(70y) + 4(75y) + l(90y)

...(4.25)

Mean life =

Thus the numerical value ofmean life is the reciprocal ofthe

disintegration constant for the respective radioactive element. # Illustrative Example 4.7

The halflife ofradon is 3.8 days.Afterhow many dayswill only one twentieth of radon sample be left over ? Solution

10

•ny

The average lifetimeofan initial sample ofradioactive atoms is found in a similar way. LetA^be the number ofatoms that still

We know that

Here

=0.693/7

7=3.8 day

exit at time t. Between t and t + dt, we lose a few ofthese hearty

atoms: dN of them decay. Thus the number of atoms that live a

%=

0.693 3.8

= 0.182 per day

jNuciear Physics and Radioactivity

131

Ifinitially atr= 0, the number ofatoms present be N^, then the number of atoms N leftaftera time t is given by

log

'21100

N

20 =>

50

T=

Nn

100

log2

99

^ r=50x

e^=20

logio(2) Iog,o(100)-Iog2(99)

Taking log on both sides, we get 2-1.9956

2.303logio 20 2.303 logio 20 0.182

0.301

. T=50

X/ = ln{20)-2.3031ogio20

T= 3420 sec = 57 min.

16.45 days.

# Illustrative Example 4.8

One gm of a radioactive material having a half life period of 2 years is kept in store for a duration of4 years. Calculate how much of the material remains unchanged ?

§ Illustrative Example 4.10

1 g of a radioactive substance disintegrates at the rate of

3.7 XlO"^ disintegrations per second. The atomic mass ofthe substance is 226. Calculate its mean life. Solution

Given that activity of substance is Solution

^^=3.7xi0"0dps According to radioactive decay law we have

The number of atoms in 1 gm ofsubstance are

N=Nq2-"'^ N=

Here we have T=2 yrs. and t = 4 yrs.

Thus we have

N=Nq2

=>

1x6.023x10^^ 226

N=2.66 X10^' atoms

If X is the decay constant of the substance, we know that

KJ 1 iV=^ = ^gm

Activity

A-XN

Thus after 4 years 0.25 gm of the material will be left.

^ N

a Illustrative.Example 4.9

1gm of radioactive substance takes 50 sec to lose 1 centigram. Find its halflife period.

ao

.

=>

X=

=>

3.7x10'

^ s'

2.66x10^'

A.= 1.39 X 10-11 s"^

Thus mean life of the radioactive substance is Solution X

Given that after 50 second the amount remaining is 0.99 gm as out of 1 gm, 1 centigram is lost. Now fi-om radioactive decay T

equation, we have

=

I ^-11

1.39x10"

N=N^2-*''^ 0.99 = (1)2-50/r

r_ = 7.194xl0i°s

Nuclear Physics and Radioactivity!

!:132

No. ofatoms in 2.5 mg ofsubstance are

Illustrative Sample 4.11 There is a stream ofneutrons with akineticenergyofO.0327 eV. If the half life of neutron is 700 seconds, what fraction of

neutrons will decay before they travel a distance of 10 km.

(mass oftheneutrons= 1.6758 x 10"^'kg).

2.5x10"^ x6.023x102^

^ =>

230

Ar=6.54 XlO^^atonK

We know that activity is given as • A=XN

Solution

or

decay constant is

Given that kinetic energy ofneutrons is

1= ^ ^ N

=0.0327 X(1.6 X10"*^ J

8.4 X= .2-

6.54x10

2x0.0327x(1.6xl0"^^) 1.675x10

^

v^= 625xl0'^

=>

v = 2500m/s

18

X=1.28xl0-'®s-^

-27

Thus half life of substance is

ln(2) T=

Time to travel a distance of 10 km

0.693 T=

lO'^m t =

2500m/s

,-18

1.28x10" = 4s

7=5.41 xio'^

After 4 second number of neutrons left can be given as

# Illustrative Example 4.13

N=Nq2-"

When n= yt no. ofhalf-lives. Here nJL

= 2

4

1

-1/175 _

= 0.996

after two days ?

A^=0.996N„

Thus fraction of neutrons decayed is Nq~N

In an experiment on two radioactive isotopes of an element (whichdonot decayintoeachother), theirmass ratio at a given instant was found to be 3. The rapidly decaying isotopes has larger mass and an activityof 1.0 p. curie initially.The halflives of the two isotopes are known to be 12 hours and 16 hours. What would be the activity of each isotope and their mass ratio

Solution

0.004Aro

f=

= 0.004

Let the two isotopes are A and B such as their halflives are

r^ = 12hr. a Illustrative Example 4.12 and

An experiment is done to determine that halflife of a radioactive substance that emits one beta particle of each decay process.

7B = 16hr. m.

Initiallytheir mass ratio is given — =3

[As

Measurements show that an average of 8.4 beta particles are

emitted each second by 2.5 milligram of the substance. The atomic weight of the substance is 230. Find the half life of the substance.

Given that initially activity ofisotope^ is 1 pCi

^^=lpa Activity ofisotope A after 2 days will be given as

Solution

Given that activity of the substance is A = 8.4 dps

A =^ 16'

> W5]

tNuclear. Physics and Radioactivrty

133

Aftert = 5 hrs, the activity presentin the solution is

iLlC/

=>

- 0.0625

=>

^=3.7xl0''x2-'^3dps

For the second isotope B it atomic mass assumed to be same

the initialactivity canbe given as

Given that in 1cm^ ofblood sample is

dps. Iftotal value of

blood is Vthen total activity is lii{2)

ln(2)

4)^x12

16

3xln-

296

3.7x10"

60

,1/3

v=

=>

"^OB

After two days activity of second isotope willbe

3.7xl0"x60 296x2'^^

cm

K=5952cm^ = 5.9521tr.

# Illustrative Example 4.15 In an ore containinguranium,the ratio of

A)B

Aqj 32

A

32

to ^®^Pb nucleiis

3.Calculate the age oftheore, assuming that allthelead present in the ore is the final stableproduct of

Takethehalf-lifeof

to be4.5 X10^ years. Solution •

^5 = 0.03125 pCi

No. of atoms of isotopes A andB left after two days are

Presentlythe ratio of

to ^®^Pb nuclei is 3

N.

N;,=NA2r

N

Ni=N,(2r' ^'a

1

=3 Pb

Ifwe consider no. ofuranium atoms are

3 Then no. of lead atoms

# Illustrative Example 4.14

Given thatall lead atoms arethedecayproduct ofuranium thus

Asmall quantityofsolution containing Na^radionuclide (half in the beginning at ?= 0, no. of uranium atoms can be taken as life15hours) ofactivity 1.0microcurie isinjected intotheblood ofa person.A sample ofthe bloodof volume 1 cm^ taken after 5 hours shows an activity of 296 disintegrations per minute.

N,u=AN,

If areoforeis t thenfi^om radioactive decay equation, we have

Determine thetotalvolume ofblood in thebody oftheperson. Assume that the radioactive solution mixes uniformly in the

N^=N,^{2r'''

blood of the person (1 curie = 3.7 x 10'® disintegration per second). w/4.5 X10®_ 4

(2)-

Solution

Giventhat initial activityis

/ = 4.5xl09

^ = ipa

=>

^ = 3.7xl0'®xi0^dps

=>

^ = 3.7x lO'^dps

yrs

logio(2) ^

_ 4.5x10^x0.125 0.301

r= 1.868 xloVs.

Nuclear Physics an^ Radioactiyityj

UZ4 _ _ ;

After time t no. of atoms of equation as

# Illustrative Example 4.16

can be calculated by decay

N,,=N,,,{2r"-

In the chemical analysis of a rock, the mass ratio of two radioactive isotopes isfound tobe 100:1. Themean lives ofthe

two isotopes are 4 X10^ years and 2x 10^ years respectively. If

N 14

N,014

it is assumed that, at thetime offormation ofthe rock, the atoms

An '012

An •"'012

ofthe two isotopes were in equal proportion, estimate the age

hit

(2)

[As Aq,2 remain unchanged]

of the rock. The ratio of the atomic weights of the two isotopes

1.1 X10-'2 = 1.2 X10-^2(2)-''^

is 1.02:1.

1.2

Solution

Given that the presentmass ratio of the twoisotopes is m,

^=100

Iogio(2)

m2

The ratio of no. of atoms can be given as N,

m =— X^ m2

N2

= lOQx

0.0377 t =

1

ICQ

1.02

1.02

0.301

xSVaOyr

x5730yr

= 719.1 yr.

From radioactive decay equation, we have

and

N^ =N^^e-'"''

.-.(4.26)

j Web Reference at www.phvsicsgalaxv.com

^2=^20^"'^''

...(4.27)

Age Group - High School Physics | Age 17-19 Years j Section-MODERN PHYSICS

Dividing equation-(4.26)by equation-(4.27) 1

^ Topic - Nuclear Structure &Radioactivity ; Module Number-16 to 30

1

1

In

A

Practice Exercise 4.2

N,

t =

1

1

2.3031og

(!) The half lifeperiodof radium is 1590 years. Afterhow manyyearswill one gram of the pure element. (a) be reduced to one centigram, (b) loseone centigram.

100 1.02 yrs

t =

I

1

[10564.78 years, 23.06 years]

.2x10^ 4x10^ J = 1.833 xlO'Vs-

(ii) The disintegration rate ofa certainradioactive sampleat any instant is 4750 disintegrations per minute. Five minutes latertheratebecomes 2700disintegrations per minute. Calculate

# Illustrative Example 4.17

A bone suspected to have originated during the period of Ashokathe Great, was found in Bihar. Accelerator techniques

the halflife ofthe sample. [6.135 minutes]

14.

gave its

C

— ratio as 1.1 x 10

Is the bone old enough to

C

have belongedto that period ? (Takeinitial ratio of

= 1.2X10"^^ andhalflifeof"'C = 5730 years). Solution

Given that initial ratio of'''C to iVn '014 N,012

is

= 1.2x10"'^

with

(iii)

Calculate the activity of one gm sample of jgSrwhose

halflife period is 28.8 years. [5.106 X 10'^ dps]

(iv) The halflife ofa cobaltradio-isotope is 5.3 years.What strength will a milli-curie source of the isotope have after a period ofone year. [0.87 mCi]

[j^clear-.Physics and Radioactiy^^^ (v)

Aisample contains 10"^ kgeach oftwo substances Aand

that the heavy nuclides having mass number A = An, where n is

5 with half lives.4 seconds and 8 seconds respectively. Their an integer, can decay into one another in descendingorder of atomic masses are in the ratio of 1:2. Find the amounts ofA and

mass number constituting a radioactive series. Thus there can

B after an interval of 16 seconds.

befour radioactive series having mass number specified by4w,

[6.25 X 10"^ kg, 2.5 X 10-3

,

...

(vQ At a given instant there are 25% undecayed radio-active nuclei in a sample. After 10secondsthe number of undecayed nuclei reduces to 12.5%. Calculate (a) mean-life ofthe nuclei, and (b) the time in which the number ofundecayed nuclei will further reduce to 6.25% ofthe reduced number.

An + 1, 4« + 2 and An + 3. Table-4.2 lists the four radioactive

series. Amongall series,ttiehalflifeofNeptuniumis soshortas compared to the age of solar system that the members ofthis

series are not found on Earth. This series is produced in laboratory by bombarding other heavy nuclei with neutrons. From table we can see that all three natural radioactive series

terminate with the end product lead (Pb)and the artificial series terminatesin Bismiith(Bi)

[(a) 14.43 s, (b) 10 s] Table 4.2

(vil) 1gram ofcesium-137 ClJCs) decays by p-emission with a half-life of 30 years. What is (a) the resulting isotope? (b) the numberofatoms leftaftet 5years? (c)Ifinitialactivity ofsample is 1 mCi then what is the activity ofcesium after 5 years ? [(a) "^Ba, (b) 3.913 x IQ^', (c) 0.89 mCi.]

Mass Numbers

Series

-Parent

An

Thorium

•^^Th

An + \

Neptunium

AiJ+2 An + 3

Uranium Actinium

^jNp-

Half - life, y Stable End Product ' 1.39x10'" 2.25x10"-

238,, 92U

.4.51x10"

235,,

7.07x10*

92'-'

^Pb

(viil) The^riormal activity ofa livingmatter containing carbon is found-to^be 15 decays per minute per gram of carbon. An archaeologicalspecimen gives6 decaysper minute per gram of carbon. If the half-life of carbon is 5730 years, estimate the approximate age ofthe specimen. [7575.40 years]

Figure-4.4showsthe uranium series{A = An '+2 series)which starts from the parent nucleus and terminates in All the intermediateelementsofthe seriesare shownin figure. We can see here that some of the radioactive nuclei have

alternate decay modes, decaying by either alpha emission or

betaemission like^'®Po, ^'''Bi, ^'^i etc.

(ix) -• A radioactive isotopeA'hasa halflifeof3 second. Initially a given sample ofthis isotope contains 8000 atoms. Calculate (a) its decay constant, (b) the time when 1000 atoms of the

isotopeXremain in the sample, and (c) thenumberofdecayper second in he sampleat/ = fj. 1(a) 0.23i'j-', (b) 9 s; (c) 2318-'] -

4.4 Radioactive Series

In the discovery of radioactivity we know it was formed that many of radioactive elements are relatively heavy. It was seen that sometime when one nucleus decays into another, the resulting daughter nucleus is also unstable or radioactive and it subsequently decays into another daughter nucleus until a stable element appears. This series ofall daughter nuclides we call radioactive series and the end product of the series is a stable isbtope. There are four radioactive series discovered and majority of radioactive elements can be considered to be members of one

a decay \

ofthese radioactive seri^. Out ofthese four, three are naturally found radioactive series and one is artificial laboratory made series. The reason that there are exactly four series follows from

p decay

thefact that when aradioactive elenient decays bya-decay, the mass number ofnucleus reduces by 4. Due to this we can say

Figure 4.4

[136^^ _

Nuclear Physics and Radioactivity

4.4.1 Radioactive Equilibrium

Similarlyfora radioactive elementwithdecayconstantXwhich decays by both a and p-decay is given that the probability for

In a radioactive series ifwe talk about an intermediate element,

an a emission, isP, and that for p-emission is

it is produced due to the decay of its previous element and it decaysto the next element of the series. In the equation-{4.28)

constant of the element can be split for individual decaymodes, like in this casethe decayconstantsfor a and P decays separately

shown an intermediate element J decays to K with a decay

constant X.j &K decyas to L withdecay constant J

K

N,

N.

^

L

then thedecay

can be given as

...(4.28) and

Ifat an instant, iVj nuclei ofJ arepresent, itsdisintegration rate 4.4.3 Accumulation ofa Radioactive Element inRadioactive can be given as

1

Series ...(4.29)

As J decays to K, the above relation in equation-(4.29) also gives the formation rate of nuclei of K. If at this instant nuclei ofK are present its decay rate can be given as

In a radioactive series, we've discussed that each element decays into its daughter nuclei until a stable element appears. Consider a radioactive series shown below A,

A.

^3

...(4.30)

To analyze mathematically the above series, we assume initially If at some instant the production rate and decay rate of the element becomes equal then the amount ofi^ appears to be a constant as the number of nuclei ofATproduced per second are equal to the number of nuclei oiK disintegrating per second.

at /• = 0,Nq atoms ofparent elements, arepresent which decays to the element ^2 ^ decay constant Xj, thus after time t, number ofundecayed nuclei ofA, present at a timeinstantt can be given be decay law as

This situation for the intermediate element AT is called radioactive

equilibrium. We can also state that this equilibrium is a dynamic equilibrium in which the amount ofK element appears to be a constant along with the process ofits continuous formation by decaying element /and its continuous disintegration to element L. Thus for an element condition of radioactive equilibrium is

N^=N^e-^i'

:..(4.32)

Dueto disintegration ofA^, nuclei ofA^ are formed and these startdecaying with a decay constant X2 to another element Ay Let at an instant t, undecayed nucleiofA^ are presentthen the decay rate ofA^ at this instant can be given as

Rate offormation = Rate ofdisintegration Decay rate of Here fore element K to be in radioactive equilibrium, we have

=

...(4.31)

A~= X.N^

...(4.33)

Dueto disintegration of.4j, A2 isproduced thustheproduction rate ofnuclei ofA^ will bethedecay rateofnuclei of.4pthus production rate of.^2 instant canbe given as

4.4.2 Simultaneous Decay Modes of a Radioactive Element

Production rate of A^ = X, We know that due to radioactive disintegration a radio nuclide transformed into its daughter nucleus. Depending on the nuclear structure and its unstability a parent nucleus may undergo either a or p emission. Some times a parent nucleus may undergo both types of emission with the probabilities of a-decay or P-decay. The amount ofdaughter nuclide produced by a and p-decay will be in the probability ratio of a and p decays.

...(4.34)

Now in a further time dt^ if 0^2 nuclei of element A2 are accumulated thenthe accumulation rateofnuclei ofelementy42 can be given as

dN2 dt

= X,W,-X2W2

dN.

...(4.35)

Now we analyze the decay ofsuch elements which disintegrates with two or more decay modes simultaneously. If an element decays to different daughter nuclei with different decay

solving gives the numberofnucleiofelement yfjas a function

constants A,p

oftime t. On solving equation-(4.35) we get

... for each decay modethen the effective

Equation-(4.35) is a simple linear differential equation which on

decay constant of the parent nuclei can be given as

XiWo N

2

= ——

(X1-X2)

(g-V-e^^i')

...(4.36)

iNuclear Physics and Radioactivity' 137

Here we can see that in the begging as

= 0, due to #IllustrativeExample4.19

disintegration of^p-^2 being formed and as the amount of^, is decreased and that of increases and that of

is increased, decay rate of

The mean lives ofa radioactive substance are 1620 years and

decreases. After a time when both 405 years for a-emission and p-emission respectively. Find out

decay rate becomes equal, the element ^^2 will be said to be in the time during which three fourth ofasample will decay ifitis

radioactive equilibrium and later the amount of A^ start decaying both by a-emission and P-emission simultaneously.

decreasing with time. Thus it is the state of radioactive

equilibrium when the yield ofthe radionuclide A^ is maximum. Lets take some examples to understand the above phenomenon better.

# Illustrative Example 4.18

Solution

The decay constants for a and p emissions are 1/1620 and 1/405 per yearrespectively.

In this case effective decay constant for both decays

A radionuclide with half life 14.3 days is produced in a reactor at a constant rate ^ = 2.7 x 10^ per second. How soon

simultaneously is

after the beginning ofproduction ofradionuclide will its activity be equal to^ = 1x 10^ disintegration/sec

1

Solution

Let t be time in which the given sample decays three fourth. Therefore, the fraction of sample undecayed in timet is 1/4.

In the reactor just after production of radio nuclide, it starts

Hence

decaying. The accumulation rate of the radio nuclide can be given as

• N=N^J4 Nowfi-om decayequation

.

A.

dN

q-XN JV

=^dt

I

r_^

r

J q-XN

J

dt

t=

ln(4)

= 1.386x324=449yr

# Illustrative Example 4.20

q-XN = t

Lead ^°^Pb is found in acertain uranium ore due to disintegration

_ ^ -X/ q-XN=qe

of uranium. What is the age of uranium ore if it now contains

0.8 gm of^°^Pb for each gram of

given the halflife of

uranium=4.5 x 10^ years.

When activity

Solution

1 x 10^ dps then 14.3

206 grams of Pb is producedby238 grams of

, (2.7

' In(2) "'"U.V 14.3

,

Hence 0.8 gramsof Pbwillproduced by

{21

log,„(2) '"Sioll? 14 3

^ X0.8 =0.92427 grams of^^^U Now fi-om decay equation we use w = Wg

/'= 9.55 days.

1 = 1.92427

^

Nuclear Physics and

•138,

Here

1

•e T= halflifeperiod andN=number ofmolecules,

Whe

= 1 + 0.92427 = 1.92427

m,

Further,

= ^-(0-693 ;/4.5 X10^)

mass of the sample x Avogadro's number

1.92427

N=

0.693?

4.5x10^

mass number

= In (1.92427)

Let t|ie total mass ofuranium mixture be M.

= 2.303 log,0 (1.92427)

Thep, mass of^92^,

4.5x100.693

... —rx^M 0-006 ,, Mi=

[2.303 log,0 (1.92427)]

100

mass of^9|U,

=4.25 xlOV

0.71

^2=

# Illustrative Example 4.21

Find the half-lifeofuranium, given that3.32 x 10 ^gmofradium isfound pergmofuranium inold minerals. The atomic weight

and

mass of^92^, 99.284 ,.

ofuranium and radium are 238 and 226 and half-life ofradium is

1600years(AvogadroNumberis6.023 x 10^^/gm-atom). Again,

0.006 ,. 6.03x10^^ Arj= ^^Mx234

100

Solution

Invery old minerals, theamount ofanelement isconstant this implies that theelement exist in radioactive equilibrium thus

0.693

and

N,

Nr

^

and

Mx

6.03x10

23

238

100

R

235

100

7.1x10^ 99.284

myAR

234

0.693 ..0.71M,,6.03x10^^ X 'X •

R.=

'u

and

0.693 ..99.284M..6.03x10"

Ri—

^ 4.5x10^"

100

Thus activities are in the ratio

^ ^

1x226

0.006

X 1600yr

0.71

234x(2.5xl0^) ' 235x(7.1xl0^)

3.32x10""^ x238

0.01026:0.000426

T,,=4.7xl0V

99.284

# Illustrative Example 4.22

Asample ofuranium is a mixture ofthree isotopes

• 238x(4.5xl0^)

'^IpJ

and^92U present in the ratio of0.006%, 0.71% and 99.284%

respectively. The half-lives ofthese isotopes are 2.5 x 10^ years, 7.1 X10^ years and 4.5 x 10^ years respectively. Calculate the contribution to activity (in %) of each isotope in this sample.

: 0.00927

Totil activity = 0.019956 %A

ctivities,-1^^x100=51.41% 0.000426

Solution

Activity

23

0.71., 6.03x10^^

XuNy and

6.03x10

'"2.5x10^'' 100 „

here we can use

0.006M

0.019956 0.00927 0.019956

X 100=2.13%

X 100=46.45%

238

[Nuclear Physics and Radioactivity

139

# Illustrative Example 4,23

-ipt-N)E^

A radionuclide with half life T is produced in a reactor at a constantratep nuclei per second. Duringeach decay, energy

=ptE.-

is released. If production of radionuclide is started at r= 0,

pEqT In-

(1-e-^O

# Illustrative Example 4.24

calculate

Nuclei of a radioactive element A are being produced at a

(a) rate of release of energy as a function of time

constant rate a. The element has a decay constant X. At time

(b) total energy released upto time t

t = 0, there are

nuclei ofthe element.

(a) Calculate the number N ofnuclei ofA at time t

Solution

(b) If a = 2Nq X, calculate the number ofnuclei of^ after one As productionrate of radionuclideisp, the accumulation rate of these nuclide in the reactor can be given as

half-lifeofA, and also the limiting valueofiVas t -> oo Solution

[Where

X= (a)

=

dN

p-XN = di

Rate of decay = - X N, and rate of formation = a, thus accumulation rate ofelement is dN dt

'p-lN'

dN = t

P

a-XN

.

=-dt

Integrating this expression, we get

p~XN

=>

= a-XN

XN=p(l-e-'^')

In

.(4.37)

Thus after time / the activity ofradionuclide in the reactor is

a-XN = -Xt

a-XNo a-XN

a-TJ^Q

A^ =XN=p{l-e-'^')

a-XN^(a-XNQ)e-^'

Given that during each decay an energy Eq is released, thus rate ofenergy release or power ofreactor P at time / is

N= Y [a-(a-XNQ)e

P^A.^E. In. =pEf^(l~e^)' [Where A.--^]

(b) \fa = 2Nr,X

N=j-[2N^X-{2NqX-N^ X)e-^']

Upto time /, number of undecayed nuclei can be given by equation-(4.37) as

At the time ofhalf-life,

In a time /, total number of nuclei produced are pt. Thus upto time /, number ofnuclei decayed are

N^=pt-N Thus total energy released upto time / is

Et=N^>^E^

T=(p.693/X) So,

N=N^,[2-e-^-^^']=^

Limiting value ofN (as /

oo)

N=N,[2-e^]=2N,.

Nuclear Physics and Radioactiyity;

!140

(v)

; Wjeb Reference afwww.phvsicsgalaxv.com

^'Co decays to ^^Fe by P+emission. The resulting ^^Fe

is in its excited state and comes to the ground state by emitting

? Age: Group - High School Physics | Age 17-19 Years I;Section, -IvlODERN PHYSICS ' Topic -i^udear Structure&Radioactivity

7-rays. Thehalf life of p^ decay is 270 days and that of the

y-emission is 10"® sec. Asample of^^Co gives 5x10^gamma raysper second. Howmuchtimewillelapsebefore the emission

rate ofgamma rays drops to2.5 x 10^ per second ?

Mbdule Number - 31 to 37

[270 days]

Practice Exercise 4.3

(vO

Aradionuclide Ajgoes through thetransformation chain

(1) A radio nuclide with decay constant >,[ transforms A, >Aj >Aj (stable) with respective decay constants X, into a radionuclide with decay constant 7^2' Assuming that and Xj- Assuming that at the initial moment thepreparation at the initial moment the preparation contained only the radio contained only the radionuclide A^ equal in quantity to Ajq nuclide Aj. Consider initially therewere nuclei ofA[ were nuclei, find the equation describing accumulation of the stable there at/=0, find:

isotope Aj.

(a) The equation describing accumulation of radio nuclide d2wifiitime. (b)

The time interval after whichthe activityof radio nuclide

^2 reaches itsmaximum value. 4.5 Nuclear Reactions [(a) m

=

j Kn

A-) ~Ai

(11) Bi can disintegrate either by emitting an a-particle or by emitting a p-particle.

(a) Write the two equations showing the products of the decays. (b) The probabilities ofdisintegration by a- and P-decays are

intheratio 7/13. Theoverall half-life of^^^Bi isone hour. If 1g ofpure ^'^Bi istaken at 12.00 noon, what will bethe composition ofthis sample at 1 p.m. the same day ? T1 + a, ^11 Bi ^ 212 Bi ^ 2Upo + e- + V, (h) 0.50

When two nuclei came close to each other, nuclear reaction can occur that results in a new nuclei being formed. Generally it is

very difficult to.bring nuclei very close as they are positively charged and repulsion between them keeps them beyond the range where they can interact unless they are moving very fast toward each other to decrease the distance of their closest

approach to start the nuclear reaction. In Sun nuclear reactions are very frequent as the temperature is millions ofkelvin, ftie nuclei have sufficient speed to bring then very close to each other and the energy released in reaction maintains the temperature there.

g - Bi, 0.175 g-TI, 0.325 g - Po]

(iii) A^^®Bi radionuclide decays via the chain ^'®Bi ^

Most of the nuclear reactions occur in two stages or steps. In

first step an incident particle strikes a target nucleus and the two combine to form a new nucleus called compound nucleus,

206p|^ (stable), where the decay constants are

which have atomic number and mass number equal to the sum

X, = 1.6 X10~^s""\X.2=5.8 X10"®s"'. Calculatea&p activities

of target nucleus and the striking particle. The compound nucleus is generallyfohned in excitedstate becauseof the initial kinetic energy of the striking particle. It is observed .that the

^"^0 A.2

of the ^'^Bi preparation of mass 1.00 mg a month after its manufacture.

W =A^oX, exp. (-X0 = 0.72 Xio"part/s,^„=

compound nuclei have very short life time oftheorder of10"'^ seconds. In second step these compound nuclei may decay in one ca: more different ways dq5ending on their excitation energy.

. (iv) A human body excretes (removes by waste discharge, sweating etc.) certain materials by a law similar to radioactivity. If technetium is injected in some form in a human body, the

body excretes halfthe amount in 24 hours. A patient is given an

Lets discuss an example when an a-particle incident on a N nucleus with some kinetic energy. The corresponding nuclear reaction is given in equation-(4.38).

injection containing ^^Tc. This isotope is radioactive with a + ^H

->'lF

->^|0 + 1H

...(4.38)

half-life of 6 hours. The activity from the body just after the injection is 6 pCi. How much time will elapse before the activity falls to 3 pCi ?

Here fluorine nucleus

[4.8 hours]

some excess energy which automatically lead it to eject a particle

F, formed in an excited state. This has

IKTudear Physics and Radioactivity

141 j

like in this case, a proton (|h ). As the excited nucleus lasts There aretwo major categories ofnuclear reactions, we'll discuss only for a short time, it iscommonly omitted from theequation

in detail. These are

ofnuclear reaction thus above reaction can also be written as

(i) Nuclear fission

N

^ 'bO + Ih

+

(iQ Nuclear fusion

...(4.39)

lets discuss these in detail.

Nuclear reaction such asthose written inabove have a general form

4.6 Nuclear Fission A + a

...(4.40)

-> B + b

Figure-4.5 shows the variation ofBinding Energy per nucleon

Here uppercaseletters represent the nuclei and the lowercase letter represent the particles. The shorthand form of above

with mass number of differentelement. Wehave discussedthat

the middle weight elements are more stable-then the heavy weight elements. Thus if we can break a large nucleus into smaller nuclei energy must be released. Ifthe released energy is ...(4.41)

reaction is written as

A(a,b)B

morethen that required,to split the nucleusthen the difference

For the example we've taken above equation-(4.39) can be energy can be converted into useful forms. written as

',^N(a,p) 'lo

Thephenomenon ofsplittinga heaviernuclei intotwo or more lesser weightfragments is knownas nuclear fission.

...(4.42)

4.5.1 Q-Value of Nuclear Reaction

For example in the figure say

0-value ofa nuclear reaction is defined as the difference in the rest mass energies of particle and nuclei before reaction and

energy required is

Binding energyper nucleon

those ofproducts formed after reaction. Forexample inanuclear

is a heavy element with then to break it into nucleons

C/j = 150£^

reaction.

...(4.45)

Ifnuclei splits into two fragments Yand Zaccording toreaction A + B

givenin equation-(4.46)

...(4.43)

C + D

The 2-value of above reaction is defined as

150

'X

Q=im^ +m^-m^-mj)(?

56 t:

75. 33

...(4.46)

and Z is

a reaction g is a negative quantity then for reaction to take place energy must be supplied to the system.

26

^

90

...(4.44) Then the amount of energy released in formation of nuclei Y

Ifin areaction energy isreleased, ^ isa positive quantity. Ifin

31-B

> 60y ^

U2 =60Ej.+ 90E^

...(4.47)

209R,

4tt j|/w59

19

/Ll2c

^

I

50

92"

- 3^Li

• -3 eo

.5 CO 1

Nucleon number A

Figure 4.5

Nuclear Physics and Radioactivity]]

142

Where Eyand

are the binding energy per nucleon ofnuclei

YandZ respectively. Thus theenergy released in thefission of nucleusXper. fission Af" can be given as

i,n +

^ flu* ^ "{Ba +

+3l,n ...(4.50)

The above reaction is one out of many possible reactions like

AE=t/2-17i

!in + f|U

=60£y+90.E^-150£^

...(4.48)

'SXe +

+2!,n ...(4.51)

But it is observedthat the maximum yield of the reaction is for

The above energy A£' or g-value of this reaction can also be equation-(4.50). Some reaction may produce as many as 5 neutrons but the average number of neutrons produced per calculatedbymass defectAw of this reaction as fission is 2.5.

AE = Anic^

Fission processes have the possibility of some practical uses because wehavepositive ^-values ofthereaction. Thepositive The fission fi-agments formed byfission ofa heavy nuclei areof 2-value occurs because the heavy element break up into two unequal size becausethe heavy nuclei have a greater neutron fragments that aremore tightly bound andtherefore have less toproton ratio as compared to lightnuclei, thusthefragments total mass energy. ...(4.49)

= {nix-my-

will have more neutrons. Hence to achieve stability generally

out of twofission fragments, one is a middle weightfragment Aheavy nucleus undergo fission when it hasenough excitation and other is a relatively heavy element because of excess energy so that the compound nucleus oscillate violently. For neutrons in the heavy nucleus. To reduce this excess, two or example (which isabout 99.3% innatural uranium) undergo three neutrons are emittedbythe fragments as soonas theyare fission reaction onlybyfast neutronswhosekinetic energyare formed and subsequent beta decays followed by y-emission more then 1 MeV. Whereas few nuclei like are able to split adjust their nip ratio to stable values. into twonucleibyjust absorbing neutron. For suchcases very slowneutrons are required to start fission reaction as in fast neutrons the contact time ofneutron with the nuclei is very less

Fission

to start fission with such nuclei.

fragment

Another important aspect of nuclearfission is the amount of large magnitude ofenergy released. Forexample during fission

Neutron

A. f I'X" Neutron

of each uranium nuclei about200 MeV energyis released. In all fission reactions majorfraction of the energyreleasedis in the formofkinetic energyofthe fragments. Roughlyabout80 to 85

percent ofthe energy released is carried bythe fragments of fission as their kineticenergy. About2-3 percentis in the form ofkinetic energy of neutrons. About 2-5 percentis in the form ofemitted y-rayphotons duringfission andthe remaining10-15

Fission

fragment

Q Neutron

percent ofthe total energy is the form of subsequent betaand gamma decays ofthe fission fragments.

Figure 4.6

4.6.1 Fissionof Uranium Isotopes and Chain Reaction In 1939 Otto Hahn and his colleagues found that a uranium nucleus, after absorbing a neutron, splits into two fragments each with a mass smaller then the original nucleus. Figure shows

a fission reaction in which a uranium "I Unucleus issplit into

barium 'jgBaand krypton ^gKrnuclei. The reaction starts when a ^9! U nucleus isabsorbs aslow moving neutron (kinetic energy - 0.04 eV or less)wecall thermal neutron,transforms

into a compound nucleus, ^9! U. This compound nucleus have

avery short life time and disintegrates quickly into 'sgBa and 3g Kr and three neutrons. The reaction can bewritten as

In each fission we've discussed that an average of about 2-5 neutrons are released. Without these neutrons practically the utilization of the fission energy becomes difficult. If we place

the fissioning nuclei quantity in a properamount it is possible to obtain a self sustained fission reaction. The neutrons from the fission can cause other fission. Each fission releasing more

neutrons and energy as shown in figure-4.7. Such a process is called a chain reaction. If such a process occurs in an uncontrolled fashion, it results in a very big explosion. If the some of neutrons from each fission are stqjped by some external mechanism to limit the uncontrolled chain reaction then the

kinetic energy(now limited) offragments can be extracted and utilized to do useful work. This process is carried out in nuclear fueled power plants.

[Nuclear Physics and Radioactiwty

143

The neutrons lose their kinetic energy by collisions with moderator and theyslowdown initial theirkineticenergy is of the other of thermal neutrons to start further fission events.

Detailed explanation ofnuclearreactoris not given here as in earlier classes you have studied about a nuclear reactor in detail. -» Lost

Weadvise you to refer the same once again.

neutron

Lost ---- neutron

4.6.2 Liquid Drop Model Lost neutron

Nuclear fission of uranium can be understood on the basis of

liquid drop model ofnucleus. When a neutron strikes the nuclei, it is captured by the nucleus and it is transformed into

neutron

the excited which due to excess energy oscillates in a varietyofways. Figure-4.8(a) shows the stages of oscillations ofthe asa liquid drop. The drop inturn becomes a spheroid,

a sphere, an oblate spheroid, a sphere and a prolate spheroid First

Second

Third

Fourth

generation

generation

generation

generation

neutrons

neutrons

neutrons

again and so on.

Figure 4.7

The major problemin abovecasesof energyproductionis, the products ofthe fission process are extremely radioactive. These are in majorityy-emitters with large lives, whichis a wizardfor

Time

(a)

health.

Most ofthe nuclear reactors use uranium as their nuclear fuel.

As we've discussed that in natural uranium only 0.7% is rest is In chain reaction plays an important role. Thusbefore usinginnuclear reactor naturaluranium isprocessed and enriched to about 3.5% of the isotope

Some students may think why^^^U do not participate inachain

(b) Figure 4.8

Oscillating stages of compound nucleus

Theinertia ofthe moving liquid molecules causes the drop to overshoot sphericity and go to the opposite extreme of

reaction. The answer is simple.As we've discussedthat to start

distortion. It is observed the nuclei exhibit surface tension and

fission in

we need very fast neutrons having kinetic

these can vibrate like a liquid drop when in excited state. For

energies more than about 1 MeV. Other wise when struck by a neutron, it captures the neutron to become and

small excitation it is possible to oscillate like this as distortion in nucleus is recovered back by its surface tension. But when

according to equation-(4.50) it decays to byemitting two excitation is large and distortion in nucleus is large,once the p particlesand immediately fissions whenstruckbya slow excitation energyof nucleus is given off in the form ofy-ray moving thermal neutron. The reactions photon, the surface tension due to shortrange strongnuclear forces is not able to bring back the nucleus into its spheical 4. 238tt 0n + 92 U '9IU + Y ...(4.52a) shape and the nucleus splits into two parts as shown in 239

92 239

U -

93 Np

239

93

Np + _oe + Y

> 'I'Pu + -0' _'ne +

...(4.52b) ...(4.52c)

In a nuclear reactor, the probability that ofa neutron will cause

to fission is more for a slow moving neutrons. Thus there is less probability that the fast neutrons produced in fission will cause other fission to occur and chain reaction will start. To

slow down these neutrons (to initiate more fission events) in a nuclear reactor moderator is used which may be ordinary water

carbon or heavywater(D2O).

figure-4.8(b). 4.7 Nuclear Fusion

If we again look at the figure-4.9 which shows variation of binding energy per nucleon ofelements with mass number, we

can seethat for heavyweightelementsaveragebinding energy per nucleon is about 7.8 MeV and for middle weight stable elements it is about 8.6 MeV. Thus we can say that the average energy released per nucleon by fission is the difference

between these two values about 0.9 MeV per nucleon.

Nuclear Physics

il44

Radioactivity]

For another example consider a general fusion reaction, given as

2

+ 3^

»

Here 2 nuclei ofan element Fission

+ AE(fusionenergy)

and3 of^B fuse together toform

a single middle weight nuclei

Fusion

...(4.54)

If we know the values of

binding energy pernucleon for theelements A, B and CasE^,

and E^then the amount ofreleased energy can be given as 100

A£' = 38£^-10£^-18£^

150

Nucleon number A

...(4.55)

Equation-(4.55) is an alternative wayto calculatefusion energy ifthe bindingenergypernucleon forall the elements of reaction

Figure 4.9

are known. If masses of individual nuclei are known AE can

Now if we take a look on the left portion of this curves, the

steepness is relativelyhigh or wecan saythat the differencein bindingenergypernucleon between verylightnuclei andmiddle weight nuclei are much more compared to heavyand middle weight elements. Thus if two verylowweight nuclei are fused together to produce a middle weight nuclei having greater binding energy, the energy released per nucleon will be very high compared to energyreleased pernucleon in a fission event. This process is callednuclearfusion. For example lookat the fusion reaction given in equation-(4.53) •H

+ tn

> ;He

+

In a fusion reaction, reaction will start only when two nuclei are

brought sufficientlycloseto each other so that the short range nuclear force can pull then together. We know at large

separations coulomb force dominates and due to strong electrostatic repulsionbetween positivenuclei, it is very difficult to bring nuclei so close that strong nuclear force dominates and start the fusion reaction. This distance at which fusion

starts is called coulomb potential energy barrier and it is ofthe 0"

...(4.53)

Here two deuterium isotope fuse together and releases

approximately 4 MeV energywhichwe can easilycalculateby using mass defect of the reaction. This energy is about the same energy per nucleon released in a fission event. But the most important thing is most of the fusion reaction products such as 2He here are stable and non radioactive. Proton

also be calculated by mass defect of the reaction.

Neutron

2

Tritium jH

Deutenum jH nucleus

nucleus

(contains two neutrons)

(contains two neutrons)

order of 10"^'' m. At such close distance nuclei will approach each other only when they have large kinetic energies which is possibleonly at high temperatures of the order of hundreds or thousands ofmillion kelvin.

Such reactions which takes places only at such high temperatures are called thermonuclear reactions. Under such conditions all the atoms are completely ionized in gaseous form. Such a high temperature gas of positive and negative charged particles is called plasma. The main problem in nuclear reactions is to confine plasma for a long enough time so that collisions among the ions can lead to fusion. There are three methods of confining the plasma named, magnetic confinement, inertial confinement andZ-pinch which is a slight modificationof inertial confinement. Detailed analysis of these methods is beyond the scope of this book.

Fusion

# Illustrative Example 4.25 On disintegration of one atom of the amount of energy obtained is 200 MeV.The power obtained in a reactor is 1000 kW. How many atoms are disintegrated per second in the reactor ? What is the decay in mass per hour ? Solution

Neutronqh

Helium oHe

nucleus

Figure 4.10

Power produced in the reactor is =>

E = 1000kW=1000x lO^W

[Nuclear Physics and Radioactivity =>

liii

P=10^J/s

The number ofatoms of^^^U consumed in 1000 hoursare

A^, = 10'^x{l000x3600) =36xl0^

10^

P =

1.6x10"'^

eV/s

Now 1 gm-atoms of

mass of^^^U consumed in 1000 hours ofoperation are

= 6.25xl0'SMeV/s

Asineach disintegration 200 MeV energy isreleased, number

w-

of atoms disintegratedper secondare N=

6.25x10'^

has 6 x 10^^ atoms. Therefore, the

,

=>

=3.125 xlo-S-'

36x10^^ ^

6x10^^

x235 = 1410gm

®

m = 1.41 kg

# Illustrative Example 4.27

Energyreleased per second is 10^ J

• The fission type of war head of some guided missiles is

Energy released per hour is

estimated to be equivalent to 30000 tons of TNT. If

E= 10^x60x601

3.5 X10® joules ofenergy are released by on tone ofexploding TNT how may fissions occur and how much

Thus mass decay per hour can be given as

would be

consumed intheexplosion ofwarhead ?Anenergy of200 MeV Aw - —Y (Einstem s massenergyformula)

is released by fission of one atom

c

Solution /Sm =

10''x60x60J

(3xl0%/s)^

Energy released by war head =3000 tons ofTNT -(3.5x10®) (30000) = 10.5 Xl0'2j

Aw^4x 10"^kg Aw =4x 10"^ gm

Number of fusions in war head

# Illustrative Example 4.26

10.5x10'^

A reactor is developing nuclear energy at a rate of

32,000 kilowatts. How many kg of

undergo fission per

second ? How many kgof would beused up in 1000 hours ofoperation ?Assume an average energy of200MeV released per fission. Take Avogadro's number as 6 x

and

lMeV=1.6x 10-"j.

(200xl0^)(1.6xl0"^^) ^ =>

V=0.1279kg

# Illustrative Example 4.28

Calculate theenergy released bythe fission of 2 gm ofin kWh. Given thatthe energy released per fission is 200MeV.

Solution

Solution

Power developed by reactor E = 32,000 kW

^

The number ofatoms in 2gm of^^jU are

E = 3.2 X 10'watts

A

2x6.025x10^^

235

atoms

Energy released by rector per sec in eVis

p.J:^MeV/s 1.6x10"'^

Energy released per fission V=200 MeV

Pf=2x lO^OMeV/s

=>

A = 200x 1.6x io->3j

As in eachfission even200 MeVenergyisreleased, sonumber offission events occurring in the reactor per second are

=>

7^=3.2 xlO-"j

„

2x10^"^ 200

"

Energy released by 2 gm o f i s

^ 2x6.025x10^^ . ,, E= 235 x(3.2xl0-ii)j

Nuclear Physics and Radioactivity

146

^ 2x6.025x10 23 235

# Illustrative Example 4.30

(3.2x10"") 3600x10^

kWh

In a nuclear reactor, fission is produced in 1 gm of

(235.0349 amu) in assuming that j^Kr (91.8673 amu) and "^Ba

= 4.55 X lO^kWh. ^Illustrative Example 4.29

In neutron-induced binary fission of^92^ (235.044 amu) two

(140.9139amu) are producedin all reactionsand no energy is lost, write the complete reaction and calculate the total energy produced in kilowatt-hour. Given 1amu= 931.5MeV. . Solution

stable end products usually formed are42M0 (97.905 amu)and

"JXe usually formed (135.917 amu). Assuming that these isotopes have come from the original fission process, find (i) what elementaryparticles are released (ii) mass defect of the reaction (iii) the equivalent energy released.

The nuclear fission reaction in the above process is

2gU+;n-> •> +3> +3'n The sum of the masses before reaction is 235.0439+1.0087 = 236.0526amu

Solution

The sum of the masses after reaction is

(i) The reaction is represented by

140.9139+91.8973+ (1.0087) =235.8375 amu Mass defect,

From the reaction, it is obvious that the total Z value of two

Am=236.0526-235.8373 = 0.2153amu

Energy released in the fission of^^^U nucleus isgiven by

stable fission product is (42 + 54) = 96, which is 4 units more

AE = Am X 931.5 MeV

than Z value of left hand side. = 0.2153x931.5^200 MeV

This showsthat the originalunstableproductsmust haveemitted 4 p particles.Again mass number on right hand side is 2 units less than on left hand side i.e., two neutrons are also produced. Now the reaction can be represented as

-gu + in ^ fjMo + 'I^Xe + 4„?e + 2 >

Number ofatoms in 1 gm of

are

6.02x10 N=

23 -

= 2.56 xlO^'

235

Energy released in fission of 1 gm of £: = 200x2.56x 10^' MeV = 5.12x 10^3 MeV

(ii) Mass defect,

=(5.12x10^)x(1.6x10-'3) A/n = mass of L.H.S. - mass ofR.H.S.

Mass of

L.H.S. =(235.044 +1.009) = 236.053amu

Mass of

R.H.S.=(97.905 + 135.9]7)+4(0.00055)

= 8.2xl0'®joule 8:2x10

10

3.6x10^

+ 2(1.009)

kWh

= 2.28x lO^'kWh

=235.842 amu

# Illustrative Example 4.31 Thus mass defect is

Aflj =(236.053-235.842)amu = 0.211 amu

The equivalent energy released is

A deuterium reaction that occurs in an experimental fusion reactor is in two stages :

(i) Two deuterium (^D) nuclei fuse together toform a tritium nucleus, with a proton as a by-product written as D (D, p) T. (ii) A tritium nucleus fijses with another deuterium nucleus to

A£' = Am X 931.5 MeV = 0.211 X 931.5 MeV = 196.54 MeV

form a helium jHenucleus with neutron asa by-product, written as T(£), n) jHe.

J_Nudear Physics and Radloactiwty

147;

Compute (a) the energy released in each oftwo stages (b) the fission reactor is 10%. To obtain 400 megawatt power from the energy released inthecombined reaction perdeuterium, and(c) power house, how much uranium will berequired per hour? what percentage of the mass energy of the initial deuteriumis (c=3xioWs). released.

Given

2D=2.014102amu =3.016049 amu

'^He=4.002603 amu -

Solution

Powerto be obtainedfi-om power house= 400 MW

In this caseenergyobtained per hour= 400MW x 1hour

'H = 1.007825 amu

=(400 X10^ watt) X3600 s

'n = 1.00665 amu

= 144x lO'Oj

Solution

Here only 10% of output is utilized. In order to obtain

(a) The reaction involved in the process is

power house is given by

144 XlO'® joule ofuseful energy, the output energy from the

^D +fD->?T +;p +E, E =

(144xl0*®)xl00 10

Mass defect of the above reaction is

= 144x 10"J

Am = [2(2.014102)-(3.016049+1.007825)]

Let, this energy is obtained Ifom a mass-loss ofAm kg. Then = 0.00433 amu

(Am)c^ = 144x lO^'j

Energy released in the process is £•] =0.00433 x931.5MeV=4.033MeV

or

Aw=

The reaction involved in the process is

144x10"

^

(3x10^)^

= 16x 10 ^kg

^

=0.16gm.

i'T + jD^2^He+Jn +E2 Mass defect of above reaction is

Am = [C3.ai6049 +2.014102)-(4.0a2603 +1.008665)]

Since 0.92 milli gram (=0.92 x gm) mass islost in 1gm uranium, hence for a mass loss of0.16 gm theuranium required is given by , , Am

= 0.01888amu

Energy released in the process is

^2 = 0.01888 X931.5MeV=17.586 MeV

1x0.16

= 174gm.

0.92x10"^

Thus torun the power-house, 174 gm uranium isrequired per hour.

Total energy released in both processes # Illustrative Example 4.23 = 17.586+4.033=21.619 MeV

(b) Energyreleasedper deuteriumatomis

atmosphere is 2cals/cm^/min on asurface normal to the rays of sun.

21.619

^, = ^—=7.206 MeV (c) % of rest mass of

The energy received from thesun by earth and itssurrounding

released

(a) What is thetotal energy received injoules byearth andits atmosphere.

(b) What is the total energy radiated in J/m by sun to the 7.206

/= 2.014102x931 = 0.385%

universe ? Distance ofsun to earth is 1.49 x 10^ km.

(c) Atwhat ratein mega-grams perminute must hydrogen be consumed in the fusion reaction to provide the sun with the

# Illustrative Example 4.32

energy it radiates ?

In the process of nuclear fission of 1 gram uranium; the mass lostis 0.92milligram. Theefficiency ofpower house run bythe

Takemass of hydrogenatom = 1.008145amu. Take mass ofHe atom = 4.003874 amu.

Nuclear Physics and Radioactivity

i148

# Illustrative Example 4.34

Solution

(a) LetD bethe diameter ofearth.Theneffective areaofearth A nuclear reactor generatespowerat 50% efficiency byfission

of2^2^ into two equal fragments of'{^Pd with the emission of

receiving radiation normally is given as

two gamma rays of 5.2 MeV each and three neutrons. The

average binding energies per particle of

and '.J^Pd are

7.2 MeVand8.2MeVrespectively. Calculate the energyreleased in one fission event. Also estimate the amount to

^4^2

71(1.27x10'')' , 2

consumed

per hour to produce 1600megawatt power.

km

=•|(1.27)^x iQi^cm^

Solution

Energy released in one fission =Binding energy oftwo 'Ip'd

Energyreceived byearth per minute

nuclei - Binding energy of

=1^(1.27)^xI0'^jx(2x4.2) J/m

nucleus - energy oftwo emitted

gamma rays

Here binding energy of

= 10.645 xlO'«J/m

nucleus is

(A£)y=72 X235 = 1692 MeV

(b) The areaofthe surface surrounding sun at a distance equal Binding energy oftwo '{gPd nuclei to earth distance = 47rc^ 2{SE)p^=2x8.2x116=1902.4MeV

= 4jr(1.49xl0Yx lO^x iC^cm^

Energyof twoemittedgamma raysis

=4p(1.49)^x lO^^cm^

2£^ = 2x5.2=10.4MeV

The energy isreceived atthe rate of2cal/cmVmin on this surfece.

Total energy released is one event is

This amount of energy must be radiated by sun

£=1902.4-1692-10.4

=>

The energy radiated by sun in J/min is

£=200x(1.6xl0-'')

£ =2 X4.2 X[471(1.49)2 ^ |o26] J/min =>

-2.3444 X l(p J/min.

£:=200MeV

£ = 3.2xl0-"j

Hence, the number of fission per second required to produce

(c) Weknowthat the fusion reaction in sum is

1600 X10^ J ofenergypersecond (1600 MW) is given as

4H->He + Q N=

1600x10^ 3.2x10

-11

= 5x lO'^s"'

The mass defect of this reaction is

So, werequire 5 x lO'^ nuclei of Am=4x 1.008145 -4.003874

persecond. Themass of

these atoms will be

=0.028706 amu.

235 m =

»23

x(5xl0'^)

6.02x10'

The energy released in one reaction is

=>

2=0.028706 X931.5 MeV

Thus amount of Now mass of hydrogen required

m= 195.2 X10^gm/3 consumed per hour is

m= (195.2 Xicr^)x3600

4.032580 X1.6598 x 10"^'* x 2.3444 x 10^® ,-I9

gm/m

=>

w = 70.27 gm/hr

0.028706x931.5x1.6x10

Since reactor efficiency is 50%, hence consumption of

=3.6673 X10^2 gm/m

= 3.6673 X10'®megagm/m

hour is given by /w=70.27x2 = 140.5gm

per

.Nuclear Physics and Radioactivity

149

# Illustrative Example 4.35

electrical energy obtained fi-om the fission of one

nucleus

IS

In the fission of by a thermal neutron, two fission fragments of equal masses and sizes are produced and 4 neutrons are emitted. Find the force between the two fission

fragments atthemoment theyareproduced. Given

1.1 fermi.

£,=3.2xI0-"x^=8.0x10-'2j. . Therefore, the number offissions of

Solution

7.884x10'^

N=

required peryear is

Since 4 neutrons are produced, the mass number of each

240-4 fi"agment (A) = —^— =118. The atomic number ofeach

=9.855 X10^5

8.0x10"'^

Total mass number of^IJPu +neutron (thermal) =239 +1 =240. Number ofmoles of

9.855x10 n

26

= 1.636x 10^

=

6.023 xlO^^

fi"agment = 94/2 = 47.Therefore, charge ofeach fi^agment is q =47 X1.6 X10-'^ = 7.52 x lO-'^C

required inone year

will be

Therefore, mass of

required peryear is

m=1.636 X10^ X235 =3.844 XI0®g

The radius ofeachnucleus of the fi-agment is

=384.4kg

R=R,iAf'

=>

R= lAx l0-'5 X(1 18)'/3 (As 1fermi =10"^^cm) # Illustrative Example 4.37

=>

72 = 5.395 X IO-'=m

The deuterium-tritium fiision reaction (called the D-T reaction)

Distance between the centres of the two fi-agments at the

is most likely to be the basic fusion reaction in a future

moment they are produced is

thermonuclear fusion reactor is

fH +5H~>^He +Jn +2

r=2x 5.395x 10-15 =>

7- = 10.79x ir'^m

The electrostatic between them is, F =

(a) Calculate the amount of energy released in the reaction,

given ?w (JH) =2.014102 amu, m(]H) =3.016090 amu, m(Jn) = 1.008665 amu andm(2He)=4.002603 amu.

,2

(b) Find the kinetic energy needed to overcome coulomb

4k €o ^2

repulsion. Assume the radius ofboth deuterium and tritium to

1

Q-

(7.52xl0-y (10.79x10"'^)^

beapproximately 1.5 x 10"'® m. (c) Towhat temperature must the gasesbe heated to initiate the

fusion

reaction?

Take

Boltzmann

constant

A:= 1.38X

=>

F=4.37x10^N Solution

# Illustrative Example 4.36

(a) Theamount of energy released is given by A nuclear reactor using

generates 250 MW of electrical

power. Theefficiency ofthereactor (i.e. efficiency ofconversion of thermal energyinto electrical energy) is 25%. What is the amount of used inthereactor peryear ?Thethermal energy released per fission of

is 200 MeV.

Solution

Rate ofelectrical energy generation is250MW = 250 x 10^W

(or Js~'). therefore electrical energy generated in 1year is (250 X10®Js"') X(365 x24x60 x60 s) =7.884 x 10'® J

2 =[{m (JH) +m(®H)} - {m (^He) +m(^n)}] x931.2 MeV =[{2.014102+3.016090}-{4.0092603 + 1.008665}] x931.2MeV = 0.018924 X931.5 = 17.62 MeV

(b) In order to fiise, the two nuclei (D and T) must almost toucheach other. Then the separation between them is equalto twice the radius of each which is given as / = 2/-=2x i.5x io-'5

= 3.0xl0-'®m.

Thermali'energy fi-om fission ofone nucleus is200 MeV = In order to startthefusion reaction, deuterium and^tritium have 200x l.6x 10~'^ =3.2x 10""J.Sincetheefficiencyis25%,the to be heated to a very high temperature. At such high

hSO

"

Nuclear Physics and Radioac^vlt^;

temperature thetwo gases areionized, i.e. theonly electron in Wefindtheexcitationenergyfrom the atomic masses. Athermal the atomsofeachgas is removed, sothat the twonuclei havea

neutron hasa negligible kineticenergy (About 0.03 eV).

positive charge of1.6 x10"'^ C, \.e..q^^q2= 1.6 x 10"'^ C. The

potential energy when they are almost touching is

=[m(rt) +

- M(^^^U)]c^

2

=[1.0087amu+235.0439amu-236.0456amu]c^ 9X10^ X(1.6 X10"'^)^

=>

U=~

=>

^

3.0x10"'^

=0-0070 X931.5 =6.5 MeV

^

£'(239u+)= [^(„) + M(23^U) - M(23^U)]c2

(7=7.68 xlO^^J

u*to overcome Coulomb n } u The kinetic energy must bejust»enough

=[1.0087amu+238.0508amu-239.0543amu]c^ •'

repulsion between positively charged nuclei. Hence kinetic

=00052x931 5=48MeV

energy must be equal to potential energy calculated above.

Thus we have

Thus KE =7.68 X1o'"^ J

, , , • willjustovercomethe • .L Ifthisenergyissupplied,thetwonuclei repulsion and stay in touch.

has more excitation energy than

when both

are produced by thermal neutron absorption. This iswhy more easily undergoes thermal neutron fission. . ^ # Illustrative Example 4.39

(c) Let The the required temperature of the nucleito initiate

the hision reaction . Then the kinetic energy ofthe system at Calculate, the ground-state Q value ofthe induced fission temperature Tcan be given as reaction in the equation

KE=\kT

«+

Where k is the Boltzmann constant. Hence to start fusion ""

+ dermal. A thermal neutron is in thermal

reaction this kinetic energy will be sufBcient to fuse then equilibrium with its environment; it has an average kinetic together thus for this temperature is given as energy given by 3/2 kT.

Given masses are m («) = 1.0087amu; M( U) = 235.0439 amu;

j= ^37: =3x1.38x10"" 2x7.68x10'^ = 3.7xlO^K

# Illustrative Example 4.38

M(^Zr) =98.916amu;M('^^Te)= 133.9115amu. Solution

Kinetic energy ofa thermal neutron canbeneglected; even for

atemperature of10^ K, the thermal energy isonly 130eV. The Q Calculate the excitation energy of the compound nuclei produced when

^

and

value of the above reaction is given by the equation

absorb thermal neutrons. Given

0=AOTx931.5MeV

masses are

^

11^^235^^ =235 0439 amu, M(n) = 1.0087 amu.

M("^U) =238.0508 amu,

defect ofthe reaction, given as Am = [M ("^U) + m(«)] - [M (^Zr)+ M('^''Te) +3 m(«)]

=[235.0439- 98.9165-133.9115-2(1.0087)] amu

M(236u) =236.0456 amu, =0.1985 amu

M("\l) =239.0543 amu Thus energy released is Solution

_

^

2 = 0.1985x931.5= 184.90MeV

The two reactions are

n+

^ ^ 238^

Even with a thermal neutron of negligible kinetic energy a

239y*

tremendous amount ofenergy is released.

^Nuclear Physics and Radioactivity

151

# Illustrative Example 4.40

Substituting given values ofmasses into the expression for 101 gives

Show that

does notdecay be emitting aneutron orproton.

Ie I= (14.014403 -140.11180) w.

Given masses are

M

= 230.033927 amu; M

= 229.033496 amu; M

(^^a)=229.032089 amu; m(n) =1.008665 amu, m(p) =1.007825 amu.

=>

|ei=0.003223 x 931.5MeV

=>

|2l=3.00MeV

Substituting this value Solution

m=M('H), A/=M('X) The corresponding decay equations would be

(i)

intoEquation-(4.56), weget:

^55u^« + 22^^and

(ii)

^

1.007825+13.003355

=

+

.

11003355

= 3.00 = 3.23 MeV.

In first equation the energy released can begiven by ^ = [230.033927-229.033496-1.008665] x 931.5MeV =-7.7MeV.

# Illustrative Example 4.42 The nuclear reaction,

Because g < 0, neutron decay is not possible spontaneously.

Similarly insecond equation theenergy released can begiven as

e=[230.033927- 229.032089-1.007825] x 931.5 MeV =-5.6 MeV

As 0 < 0, proton decay is alsonot possible spontaneously.

is observed to occur even when very slow-moving neutrons (M„ = 1.0087 amu) strike a boron atom at rest. For aparticular reaction in which = 0, the helium (M^^ = 4.0026 amu) is observed tohave a speed of9.30 x 10^ m/s. Determine (a) the kinetic energy ofthe lithium = 7.0160 amu), and (b) the ^-value ofthe reaction. Solution

# Illustrative Example 4.41

Compute the minimumkineticenergyofprotonincidenton nuclei at restin thelaboratorythatwillproducetheendothermic

reaction '^C(p,

« + '°B^5Li +^2He

Given masses are

(a) Since the neutron and boron are both initially at rest, the total momentum before the reaction is zero and afterward is also zero. Therefore,

M{'^C) = 13.003355 amu, M('H) = 1.007825 amu

We solve this for and substitute it into the equation for kinetic ^ergy. We can use classical kinetic energy with little

m(«) = 1.008665amu,

error, rather than relativistic formulas, because

v^e 9.30 X10^ m/s is not close to the speed oflight cand

M('^N) = 13.005738 amu

will be even less since A/y >

Thus we can write:

Solution A/.:

The thresholdenergyof the incidentprotonsin the lab frameis given by

^Li = m + M

•\Q\

^th = M The magnitude ofthe Q value ofthe reaction is

...(4.56)

Q = '"final -

initial

=[MC^^N) +mJ - [MC^^C) +M('H)]

2A/,

We putin numbers, changing the mass in u tokg andrecalling that 1.60 X10"'^J= 1MeV:

(4.0026)-^ (1.66 X10~^^ )(9.3Q x10^) 2(7.0160)(1.66 xlO"^"') A:Li = 1.64x 10-'3J= 1.02 MeV

Nuclear Physics and Radioactivity:

•152

(b) We are given the data

(v)

]D +

K^ = K^=0, Q = Kn + K, •-He'

so

1

^He 2"^He^e

where

=y(4.0026)(L66x 10-2^) (9.30X 10^)2

A fiision reaction of the type given below

\T+ Ip +_AE

is most promising for the production of power. HereD and T stand for deuterium and tritium, respectively. Calculate the mass of deuterium in gram required per day for a power output of

lO^W. Assume that efficiency ofthe process to be 50%. Given •2t-,

mass of

= 2.87 X10"'^J= 1.80MeV

mass of\T=3.01605 amu.

2 = 1.02MeV +1.80 MeV=2.82 MeV.

Hence,

=2.014102amu.

mass of \p=1.007825 amu. and 1 amu =931.5 MeV.

Web Reference at www.phvsicsgalaxv.com

Age Group - High School Physics | Age 17-19 Years

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Section - MODERN PHYSICS

Topic - Nuclear Reactions Module Number - 1 to 9

(vi) The binding energies per nucleon for deuteron (^H) and helium (^He) are 1.1 MeV and 7.0 MeV respectively. Calculate the energy released when two deuterons fuse to form a helium

nucleus (^He). Practice Exercise 4.4 [23.6 MeV]

(i) Find the amount of energy produced in joules due to fission of 1gram of uranium assuming thatO.l percent of mass is transformed into energy.

Takel amu=1.66x 10~^^kg = 931.5MeV

(vii)

Find the energy released when

nucleus undergoes

fission by athermal neutron into 3gKr and 'sgBa. Given, mass of^g2U =235.043925 amu, mass ofneutron = 1.008665 amu, mass of3gKr =91.8973 amu and mass of^5^63= 140.9139 amu.

Mass ofuranium = 235 amu.

Take 1amu = 931.5MeV.

Avogadro number = 6.02 x 10^

[200.64 MeV]

[8.978 X 10'° J]

(viii) How much energy is needed to remove a neutron fi-om

the nucleus ofj^Ca ?Given

= 1.008665 amu, atomic mass of

(II) What is the power output of reactor if it.takes 30 daysto useup 2 kg offuel, and if eachfissiongives 185MeV

2iCa =40.962278 amu and atomic mass of^JCa =39.962591

of usable energy ?

amu. Take 1amu = 931.5 MeV

[58.52 MW]

[8.363 MeV]

(III) Assuming that 200 MeV ofenergy is released per fission of uranium atom, find the number of fission per second required to release one kilowatt power.

(ix)

[3.125 X 10'^ fissions per second]

amu and 7.016004 amu respectively.

(iv)

It is proposed to use the nuclear fusion reaction

Find the Q value ofthe reaction p + \\-^ '^He + '^He.

Determine whether the reaction is exothermic or endothermic.

Theatomicmassof'H, '^He and'Li are 1.007825 amu,4.002603

[Exothermic, 17.347 MeV]

JH-h jH-j-^He in a nuclear reactor of200 MW rating. If the energy fi"om the above reaction is used with a 25 % efficiency in the reactor, how many grams of deuterium fuel will be needed per day. (The

masses of^H and ^He are 2.0141 atomic mass units and 4.0026 atomic mass unit respectively and take 1amu = 931.5MeV).

4.8 Properties of Radioactive Radiations We know that an unstable nuclei achieves stability by radioactivity due to emission of a, P and y-radiations. Now in this section we will discuss how these are emitted and the

characteristics of these radiations one by one. [120.31 g]

'Nuclear Physics and Radioactivity

15^

4.8.1 Alpha Decay

and from energy conservation we can say that the released

We've already discussed that the range of attractive nuclear forces amongnucleons isveryshort and that of the electrostatic

energy Q is distributed in both, hence we have ...(4.60)

repulsiveforces betweenprotonsis unlimited. This is the reason

why nuclei with mass number above 210 aresolarge thatthe short range nuclear forces thathold thenucleon together are

Q~ T2 ^ y ^

not sufficient to counterbalance the mutual repulsion between the protons. Alpha decay occurs in such nuclei as a means of increasingtheir stabilitybyreducingtheir size.

2=

IT m

V

1+ m.

m,.

force inthenucleus. The reason isvery high binding energy of a-particles. To escape from nucleus, a particle should have sufficiently high kinetic energy and only a-particle mass is

OT. m.

sufficiently smaller than the masses of its constituent nucleons for such energy to be available.

Theequation ofthenuclear reaction for a-decaycanbewritten

a

my +/«£

In this casea question ariseswhyonlya-particles are emitted why notindividual protons areemitted todecrease therepulsive

Whenever an a-particle is emitted, the mass number of the daughter element is less then 4 and atomic number less then 2.

'a

m.

E

-

•A-A

\Q

...(4.61)

Equation-(4.61) shows that the kinetic energy of ejected a-particle in the decay process is

A-A

times the g-volunie

of the nuclear reaction in the process.

as

zZjY + 2^6 + energy

...(4.57)

It is seen that all the a-emitter radioactive element have mass

number^ > 210 sowe can say that most ofthe energy produced Inequation-(4.57) we can see thatduring a-decay some energy during decay appears as the kinetic energy of the emitted is released, this shows that the ^-value of the a-decay is positive and the total released energy is shared by both, the daughter nucleus and the emitted a-particle as shown in figure-4.11

a-particle.

4.8.2 Beta Decay

Similar to emission of a-decay, p-decay also makes the composition ofnucleus more stable. There are three fundamental beta decay processes are observed. There are

(1) Beta Minus Emission (P"): P" are the electrons,which areemitted from thenucleus when insidea neutron decays to a

(a) Figure 4.11

proton and an electron, this electron is emitted from the nucleus

The energyreleased in the above process or the g-value ofthe above nuclear reaction can be given as

with some kinetic energy and it is called P" decay, theprocess involved insidethe nucleus in p" decay is + v.(P emission)

...(4.58) Here

m'X

In reaction represented by equation-(4.62) also includes a particle with P" emission it is v (nu-bar), denotes an

mass ofparent nucleus X

m.

-> mass ofdaughter nucleus Y

m.

mass ofa-particle (^He)

antineutrino. About this we'll discuss in next section.

During a-emission linear momentum of system must be conserved, hence if the speeds attained by a-particle and the daughter nuclide Yin the aboveprocessare and then we must have m

V

y y

=m V a

a

...(4.62)

...(4.59)

(2) Beta Plus Emission (p"'"): P"'^ are the positions, which are

emitted from the nucleus when inside a nucleus a proton transforms to a neutronand position. The position emitted from the nucleus, is called P"*" decay. Theprocess involved inside the nucleus in p'^-decay is » « -H

-I- V(P"^ emission)

...(4.63)

Nuclear Physics and Radioactivity ,!

i154

In the above equation-(4.63) the reaction involves another (2) Law of Consen^tion of Linear Momentum particle v (nu)with emission, called neutrino. About thisalso During p-decaywhen the directionsof emitted p-particles (e~ we'll discuss in next section. or e^) andthat ofrecoiling nuclei areobserved, theyare almost When a proton inside a p"^ decay inside a nucleus, the parent never exactlyoppositeas required for linear momentum to be radionuclide Xtransforms to a daughter nuclide according to conserved as shown in figure-4.12. It seems that one more particle is to be emitted in other direction to conserve linear the reaction momentum but no other particle was detected during beta A Z-1

\X

/

+

+

V

...(4.64)

emission.

Here position emission leads to a daughter nucleus of lower atomic number Z by 1 and leaving the mass number A unchanged.Also remember that a proton can not decayinto a

Parent Nucleus

neutron outside a nucleus because of its smaller mass. This is

possible only inside a nucleus.

(3) Electron Capture or K-capture: This is a third kind of

P-decay itproperties are similar top"^ emission inwhich anucleus

p particle

pulls in or capturesoneof the orbitalelectronsfromoutsidethe nucleus and inside the nucleus a proton absorbs this electron and transforms into a neutron. This process is called electron

Daughter Nucleus

capture, or K-capture, sincethe electron is normally captured Figure 4.12

from the innermost or K-shell. The process ofelectron capture inside a nucleus can be given as

(3) Law ofConservation of Angular Momentum \P

-r

-> « + V

...(4.65)

As in this process also a proton transforms into a neutron, neutrino is also emitted and we can say that basically this

process resembles with p"^ emission but no particle or

are emitted from the nucleus. For a radioactive

Inside a nucleus, every nucleon-neutrons protons, electrons or

positionseach have a spin angular momentum± • If we once again look into the equations of P-decayprocesses, these are

element Xdecays to a daughter element Y, by electron capture, p decay n

the reaction can be given as X

-> z-xY+v

...(4.66)

A 1 momentum . ±2 Angular

Here also similar to p"^ decay the atomic number of daughter nuclei decreases by one and mass number remain unchanged. 4.8.3 Apparent Violation ofConservation Laws in b-decay As we've discussed that by beta decay also the nucleus becomes more stable. During beta decay it is observed that the basic conservation laws ofenergy, linear momentum and angular momentum are all apparently violated as discussed now.

P"^ decayp

...(4.67)

p + e

^± 2 \ h ^ 2 \ 2^ h ...(4.68)

n + e*

Angular momentum ± 2 2^ 2 2^ ^ 22i In both of above equations we can see that when a nucleon decays in the two on left hand side of equation before decay its

angular momentum is either +

or-

but afterdecay

on right hand side ofequation the total angular momentumwill (1) Law ofConservatiMi ofEnergy

be±

iTt

or 0.

We know when a radionuclideX-decays to a daughter nuclide

Twith P-decaythe amount of energy released or value of the nuclear reaction can be calculated by using mass defect ofthe reaction. But when the emitted P-particles are observed they do not have enough energy to account for the total energy released. If p-particle carries away only a part ofthe energy, where the remaining energy is going. This was a serious puzzle among physicists in late 1920 s. It was very difficult to accept that law of energy conservation is not valid in this case.

Thus here we can see that on the two sides ofthe equation-(4.67) and (4.68), angular momentum are not equal. For an isolated system like nucleus, we can not expect this to happetl. But again this one can not accept that the law of conservation of angular momentum is not applicable here. The above three conservation laws apparently violet in the

process of p-decay. This remain a series problem among

'Nu(^ar Physics and Radioactivity

155

physicists until 1930 when Pauli proposed that during p-decay 4.8.5 Mass Defect Calculation For b-decay one more particle is emitted, called neutrino. Existence of

neutrino wasverified experimentally in 1956. Letsdiscuss Pauli's Neutrino Hypothesisin detail.

We've discussed that there are three fundamental P-decay processes. Here mass defect involved in .the nuclear reaction

are very small as mass of P-particle (e~ or e^) is very small. Generally the masses we use for calculation ofmass defect in a

4.8.4 PauIPs Neutrino Hypothesis

nuclear reaction are themasses ofatoms where weneglect the

As discussed above Pauli suggested that during emission of a

masses ofelectrons which is included in the mass ofatom.

p-particlean additional particleisalsoemitted, namedneutrino and this hypothesis made all the above conservation laws

working finely for the process of p-decay. According to Pauli the particleneutrino has the following characteristics:

Here for the P-decay process the mass of p-particle itself is sameas that ofan electron, we can not neglectthe weightofall

the orbital electrons present inan atom thus for an atom ^^ -Z

(i) Its charge is zero

(mass of electron) thus for calculation of mass defect for a P-decay process, we first find the mass ofnuclein involved in

(iO It is an energy particle like a photon.

the reaction then wefindmassdefect. Lets discuss this one by

(iii) Its rest mass is zero like a photon.

one for each type of P-decay.

(iv) It carrieslinear momentum like a photon.

(i) Mass Defect for P" Decay:

(v) It carries angular momentum like other nucleons ±

1 h

.

2 27c

(vi) During P-decay, p^emission is accomplished byemission

For an element

when undergoes p~ emission, the decay

equation is written as

of a neutrino (v) where as P"-emission is accomplished by •r'fiY +' -I ^te + \

!a

emission ofan antineutrino (v )

(vii) It is extremely difficult to detect a neutrino because it interacts very weakly with matter. It can be concluded on basis of experiments that an average neutrino can penetrate about lO'^ km distance in the most dense material lead without

interacting with it.Even though more then 10^ neutrinos pass through our bodyeverysecond withoutany effect. It isdifficult

' Z+l'

Here the mass defect of the above reaction can be written as

Am =

ZmJ- ((Mj.- (Z+ 1)

+ mj

or Am = M^- My Thus 2-value of this reaction can be given as

AE^{M^-My)C^

but in Japan Super Kamiokande neutrino detector was made to detect neutrino.

(ii) MassDefectof P"^ Decay: Thus the new reactions for p-decay processes can be rewritten as

For an element

when undergoes P"^ decay, the decay

equation can be written as

P~decayrt

> p + e~

P^ decayp

>n +

\

..".(4.69)

z-J, + ^^e + V

:a

+w

... (4.70) Here the mass defect of the above reaction can be written as

Now using the above two equations if we again look into the

Am ={Mj^-Zm^)-

energy, momentum and angular momentum of the nucleons

involved in the decay process all conservation laws now seems to appear valid due to the presence of this additional particle in the reaction.

or

mj

Am-My-My-2m A ' y e

Thus 2-value of this reaction can be given as

AE =(Mj^-MY-2m)C^ Herewe'vediscussed that the emittedP-particles carriesenergy lessthen that evolved in the process. Theremainingenergy is (iii) K-Capture: carried by the neutrino. Thus we can say that for every emission

the sum ofenergiesof P-particIe and neutrino(or antineutrino) remains constant in the decay.Similarly the direction of emitted neutrino will account for the conservation oflinear momentum

and in the equations for decay processes, if we account for the angular momentum ofneutrino (or antineutrino) then on botli side of equality angular momentum will get conserved.

Foran element

undergoes K-capture, the decayequation

can be written as

lA

Here the mass defect of the reaction can be given as

Am = (M^- Zm^) - (My- (Z-1) m)

Nuclear Physics and Radioactivity {

•J56 We know that

or

Thus g-value ofthe above reaction is given as

AE =

My-

(i) The emission of a-particle (jHe) decreases the charge number by two and mass number by four.

C?

Thus emissionof a, a-particles reduce the charge number by

4.8.6 Gamma Decay

2a and mass number by 4a. Just like an atom, nucleus can also exist in different energy states higher then its ground state. An excited nucleus is

denoted by an asterisk after its symbol like Such excited nuclei return to their ground state byemitting photons ofenergy equalto the energydifferencebetween the final and initial states in the transitions involved. These photons have range of energies upto several MeV are traditionally known as gamma rays (y-rays).

Whenever a radionuclide imdergoes a or p-emission, the daughter nuclide may remain in excited state just after decay which after some time (its life time) transit to ground state with release ofy-ray photon.

(ii) The emissionof P-particleincreasethe chargenumber by one and leaving the mass number unchanged. Thus emission of h, P-particles increase the charge number by 6x1 = 6. Thus

Applying the law of conservation of charge number and mass number before and after the decay, we have 92 = 82 + 2o-6 and

Lets consider an example P-active Mg which decays to Al. Look at the following reactions: Mg

>A1* + P + V

... (4.71)

Al*

>A\ + Y

...(4.72)

In the above reaction we can see when P" is emitted from Mg nucleus some part of the released energy is absorbed by the daughter nucleus & gets in excited state Al* and remaining

energyis carried by p" & v .After a short durationAl* releases the excess energy in form of y-ray photon and transit to its ground state Al as shown in equation-(4.72). Some times due to moreexcitatioii energyoneor may be twogamma decays takes place for daughter nucleus to reach the ground state. Some times the energy ofy-ray photon from the nucleus which it releases at the time oftransition from higher state to groimd state, is absorbed by the atomic electrons around it. This process is similar to photo electric process in which this atomic electron is emitted with a kinetic energy equal to the nuclear excitation energy minus the binding energy of electron in the atom. This phenomenon we call internal conservation.

U^^gPb + adHe) + 6C?P)

92'

238 = 206+4a

How many alpha and beta particles are emitted when uranium

(^92^) decays to lead Solution

Let a and b, the number ofa and p particles are emitted when

decays to ^"^Pb.

...(4.74)

From equation-(4.74), a = \

From equation-(4.73), 6 = 1

# Illustrative Example 4.44

Thenucleus ^^Ne decays byp-emission into thenucleus ^^Na. Write down the P-decay equation and determine the maximum

kinetic energy of the electrons emitted. Given m (foNe) = 22.994466 amu. and m(f^a)=22.989770 amu. Ignore the mass ofantineutrino (v). Solution

The p-decay of^^Ne is represented bythe equation

foNe->55Na +e-+V+e Mass defect for the above reaction is

Am=[Oe)-M(fjNa)] # Illustrative Example 4.43

...(4.73)

= [22.994466-22.989770] amu = 0.004696 amu

Now Q-value of the above reaction is given as e = Amx931.5MeV = 0.004696x931.5 MeV =4.374 MeV

^ _ 157j

INuclear Physics and Radioactivity

The energy Q released is shared by ^^Na nucleus and the (a) Find the values ofA and Z in the above process. electron-antineutrino pair. Since ^^Na is massive, most ofthe (b) The alpha particle produced in the above process is found kinetic energy is carried by the e - v pair. The electron will

carry the maximum energy ifthe antineutrino carrieszero energy. Thus the maximum energyof the electrons emitted is 4.374 MeV.

to move in a circular track ofradius 0.11 m in a uniform magnetic field of3 Tesla. Find the energy (in MeV) released during the process and the binding energy of the parent nucleus X. Given that:

# Illustrative Example 4.45 m (Y) =228.03 amu;

A radioactive source in the form ofmetal sphere of diameter

m(Qn) = 1.009 amu

1memits beta particleataconstant rate of6.25 x 10'® particle

m(^He) =4.003 amu;

per second. If the source is electrically insulated how long will it take for its potential to rise by 1.0 volt, assuming that 80% of the emitted beta particles escape from the source ?

m(jH) =1.008 amu. Solution

Solution

Let t be the time for the potential of metal sphere to rise by one volt. Then upto this time p-particles emitted from sphere are

Given nuclear reaction is

9^X^^^|Y +a 9^X^228^Y +^He

A^=(6.25xl0'®)xf Number of P-particIesescaped in this time are

(a) According to principle of conservation of charge number, we have for above equation

iV^ =(80/100)x(6.25xlO'0)r = 5xlO>^'f

92 = Z+2 1

o

=>

Thus charge acquired by the sphere in t sec

2 = (5xl0i«f)x(l.6xi0-'^ = 8 X10'^ / coulomb

... (4.75)

Z=90

And according to principle to conservationof mass number, we have for above equation ^=228 + 4 = 232

{Qemissionof p-particle leadsto a charge e on metal sphere) =>

A =232

The capacitanceC of a metal sphere is given by (b) For circular track ofa charge particle in uniform magnetic

C = 47i€Qxr

field, we have

fUxlO^J ^ 1" 10

\

/

farad

...(4.76)

Here momentum of a-particle is given as

g=Cx7(Here- r=lvolt)

p = mv= 2e,x 5 x /•

/j =2 X(1.6 Xl(r'9)(3) (0.11) kg-m/sec

•>-12-'

=>

(8x10-9)/ =

= qvB

mv = qBr

-12

18

We know that,

.2 mv

2

10'

18

xl

Solving it for t, we get

=>

p = 1.056 X10"'^kg-m/sec

K.E. ofa-particle is given by / = 6.95 p sec E

# Illustrative Example 4.46

=

(1.056x10"'^)^ 2x4.003x1.66x10"^"^

= 0.0839 Xicr" joule A nucleusX, initiallyat rest,undergo^ alpha-decay according to the equation.

g^X-^^^|Y +a

^-ll

0.0839x10"

1.6x10"*^ = 5.244 MeV

MeV

Nuclear Physics and Radioactivity'

:158

The parent nucleus is at rest and hence its momentum is zero. Now the sum ofmomenta ofdaughter nucleus Tand a-particle must be zero i.e., the momentum ofdaughter nucleus 7 is equal and opposite to momenturn ofa-particle.

Mass defect of this reaction can be given as

Am = [M (2^; Po) - M(2°^ Pb) - M(^He)] Am = 209.98264 - 205.97440-4.00260 = 0.00564 amu

Thus K.E. ofdaughter nucleus 7is .2

7 = £v=

^ 2My

Thus

(1.056x10"'^)^

value of the nuclear reaction is

e=0.00564 X931 MeV= 5.25MeV

2x228.03x1.66x10"^'^

= 0.001473 X10-"joule=0.092 MeV In above reaction, the total energy released is in the form of kinetic energy ofparticles hence we have

=>

. =5.25 X(1.6 xl0-'3)

^

=8.4 XI0~^^joule

The decay constant Xis given byfor ^-'^o is given by

Ej =E^+E^= 5.336 meV

Total energy

,

Aw = Mass of

0.693

0.693

= 138.6days =0-005day

Mass defect of the reaction is

(Mass of 7+ Mass of a-particle)

Let Mgm be required toproduce 1.2 x ]o'joule perday. Number ofnuclei in A/gm

amu= A/^-(228.03+4.003)

N=

(6xlO^^)A/ 210

=(232.0387) amu

Now mass defect of parent nucleus

is given as

Now decay rate by N nuclei per day or the activity ofN nuclei are

=92m^ +(232-92)w„-M^

dN dt

=(92 X1.008+140 X1.009 -232.0387)-

0.005 X

•

= 1.9573 amu

per day

... (4.77)

Thus energy produced per day

=> Binding energy of parent nucleus X is

(A£);^.= 1.9573 X931.5MeV

= XN

= 0.005XX(8.4X 10-")J

...(4.78)

Thisshould beequal to 1.2 x IO'joule.

= 1823.225 MeV.

Hence 0 . 0 0 5 x (8.4 x ir'^)= 1.2 x IQ^ 210

# Illustrative Example 4.47 =>

A/= 1 gm.

... (4.79)

Polonium (^g^Po) emits ^He particles and is converted into lead Theefficiencyis 10%.Hencethe quantityrequiredwill be 10 gm. (^gjPt)). This reaction is used for producing electric power in a space mission ^'®Po has half life of 138.6 days. Assuming an Fromequation-(4.77), activityis efficiency of 10% for the thermoelectric machine, how much

^'®Po is required to produce 1.2 x 10^ J of electric energy per

4 = 0.005 X

(6x10^^)10 210

day at the end of 693 days. Also find the initial activity of the material.

=>

(Given :masses ofnuclei M(^'''Po) =209.98264 amu, M(^®^Pb)

Let.4Q be the activitybefore693 days,then from decayequation

=205.97440 amu, M(^He)=4.00260 amu)

yl = y X10^'per day

we have 693

Solution

=

(2) 138-6

The nuclear reaction for the above process can be expressed as

mPo

2gPb + ^He

Aq=

j ^32=4.57 X102" perday.

Nuclear Physics and Radioactivity

""

#Illustrative Example 4.48

"

;

First we will determine the disintegration energygofequations-

~

"

(4.80)and(4.81):

Find whether alpha decay or anyofthe beta decay areallowed

for ^ggAc. Given masses are

M(^2^Ac) =226.028356

(-» =226.017388amu.

,

M

M(2^Fe) =54.938298 amu,

=222.017415 amu, M

M(-'Ra) =226.025406amu,M{,^He)=4.002603amu.

M(gMn) =54.938050 amu and

=0.000549a™a

For ^ decay the g-value ofreaction is

Solution

e=[54.938298-54.938050

Our first step will be to write the reaction, then to find the

-2 x0.000549] ^931.5 MeV

disintegration energy g.lfg>0,the decay is allowed. Alpha decay: ^oqAc-> ^slpr + a

'

=_079MeV

xt . Negative value of^ implies thatpositron decay IS notpossible

Q= [Mfi^Ac) - M(2gFr) - M(^He)]c^

-spontaneously For electron capture the g-value of reaction is

=5.50 MeV(Alpha decay is allowed)

e=[54.938298 -54.938050] x 931.5MeV

p" decay: ^Ac =^^JXh + p" + v

=0.23 MeV

2=[Mfi^c)-MOh)]c2

D

this case.

= 1.12MeV (P~ decayis allowed)

P^ decay: ^^^c

.

Positive valueof Q implies that electron capture is possible in

^^a + p^ + v

#Illustrative Example 4.50

Q [Nl( g^Ac) M(^gpla) —2mJ

Asample of'®F is used internally as amedical diagnostic tool

=-0.38MeV (P decay is not allowed) Electron capture : ^J^Ac + e"^ ^^^Ra + v

to look for the effects of thepositron decay {T,^ = 110 min).

How long does it take for 99% ofthe'¥to decay ? ^

.

Solution

e= [M(™Ac) - M(™Ra)]c2 Radioactive decay equation is

=0.64MeV (Electroncaptureis allowed)

In above analysis it is clear that during a-decay, g-value is maximum thus chances ofa-decay are maximum.

.

}ct,Y daughter product is plutonium (Pu). The grand daughter (A) a,p,Y (O Y,a,P P) a,Y,P, product can be expressed as : ' (A)

(B)

(Q 239pu

(D)

4-38 Graphiteand heavywater are twocommon moderators

241 Pu 94ru

used in a nuclear reactor. The function of the moderator is :

4-31 The half-life of a radioactive substance depends upon : (A) Its temperature (B) The external pressure on it (Q The mass of the substance

P) The strength of the nuclear forcebetweenthe nucleonsof its atom

(A) Toslowdown the neutrons to thermal energies p) To absorb the neutrons and stop the chain reaction (Q To cool the reactor

P) To control the energyreleased in the reactor 4-39 Alphaparticlesare fired at a nucleus. Whichof the paths shown in figure-4.14 is not possible ?

4-32 The half-life period of a radioactive elementXis same as the mean-life time ofanother radioactive element Y. Initially both

Nucleus

ofthem have the same number ofatoms. Then:

(A) XandThavethesamedecayrateinitially (B) X and Ydecay at the same rate always (Q Twilldecayatafasterratethan.^ p) ZwilldecayatafasterratethanT

Figure 4.14

(A) 1

4-33 The radioactive decay ofan elementA"toelements Yand K is represented by the equation Ay

Ay

2+r ^

A+4 y

z-i ^

^

4-40 The decay constant X, of a radioactive sample: (A) Decreases with increase of external pressure and

A-A y

z-i ^

The sequence of the emitted radiations is: (A)a,p,y P) P,a,Y (Q Y,a,P (D) P.Y'Ct

4-34 The nuclear reaction 'JJCd ->

temperature

P) Increases with increase of external pressure and temperature

can occur with

the:

(A) Electron capture (Q Proton emission

P) 2 P) 4

(Q 3

p) Positron capture p) a-particle emission

4-35 Two radioactive elements Xand Thave half-life times of

50 minutes and 100 minutes, respectively. Samples X and Y initiallycontain equal numbers ofatoms.After 200 minutes, the ratio

(Q Decreases with increase of temperature and increase of pressure

p) Is independent of temperature and pressure

4-41 When high energy alpha particles (jHe) pass through nitrogen gas, an isotope of oxygen isformed withthe emission ofparticles namedx. The nuclear reaction is

-i-^e ^ 'JO + x What is the name ofx ?

(A) Electron (Q Neutron

number of undecayed atoms of X . IS number of undecayed atoms of Y

(A) 4

P) 2

(Q 1/2

P) 1/4

4-42 When boron ('jB) isbombarded byneutron, a-particle is

4-36 When a P"-particle is emitted from a nucleus,^the neutron-proton ratio: (A) Is decreased (Q Remains the same

P) Proton p) Positron

p) Is increased p) First (A) then (B)

emitted. The resulting nucleus has a mass number: (A) 11 P) 7 (Q 6 p) 15 4-43 Consider a-particles, p-particles and y-raj^, each having an energy of 0.5 MeV. In increasing order ofpenetrating powers, the radiations are :

4-37 In the given reaction : A-A Z-1

K

A~A y Z-1 ^

(A) a,p,y (Q p,Y,a

•

-

P) a,Y,p p) Y,p,a

,

;Nuc[ear"iPhysics,and, Radioactivity

167

4-44 Aradioactive nucleus emits a beta particle. The parent 4-48 Figure 4.15 shows aroughly approximated curve ofbinding and daughter nuclei are:

(A) Isotopes (C) Isomers

energy per nucleon with mass number. W, X, Yand Z are four

(B) Isobars (D) Isotones

nuclei indicated on the curve. According to this curve, the process that would release energy is ;

4-45 Fast neutrons can easily beslowed down by: (A) The useof lead shielding (B) Passingthem through water (Q Elasticcollisions withheavynuclei P) Applying a strong electric field

4-46 A nucleus ruptures intotwo-nuclear parts, which have their velocityratio equal to 2 :1. What will be the ratio of their nuclear size (nuclear radius)?

(A) 3"^-A

(B) 1:3"2 Figure 4.15

4-47 Fornucleiwith^> 100,marktheINCORRECTstatanent;

(A) The bindingenergypernucleon decreases on the average as A increases

(B) If the nucleus breaks into tworoughly equalparts,energy is released

(Q Iftwonuclei fiise to form a bigger nucleus energyisreleased P) The nucleus with 2> 83 are generallyunstable

(A) Y-^2Z (Q W->2Y

P) W^X + Z P) X^Y + Z

Nuclear Physics and Radioactivity i

i168

NumericalMCQs Single Options Correct 4-1 Two identicalsamples (samematerialand sameamount)P

andQofa radioactive substance having mean life Tareobserved

(B) 10:11 P) 81:19

(A) 19:81 (C) 15:16

to have activities Ap & Ag respectively at the time of observation. If P is older than Q, then the difference in their

4-8 Half-lives of two radioactive substances A and B are

ages is:

20 minutesand 40 minutesrespectively. Initiallysamples^ and B haveequal number ofnuclei. After 80minutes, theratioofthe remaining number of^ andP nuclei is: (A) 1:16 P) 4:1

(A) Tin

A.

f

(B) Tin .

A. \

N

(C) j In

(0 1:4

(D) T

P) 1:1

A 4-9 A100 ml solution havingactivity 50 dpsiskeptin a beaker.

4-2 Bihasa halflife of5 days. Whattime istaken by(7/8)*^ part It is now constantly diluted by adding water at a constant rate of the sample to decay ? (A) 3.4 days (Q 15 days

of 10 ml/sec and 2 ml/sec ofsolution is constantly being taken out. Find the activity of 10 ml solution which is taken out, assuming halflife to be effectivelyvery large :

(B) 10 days (D) 20 days

1/2

1/4

4-3 A certain radioactive substance has a half life of 5 years.

(A) 10

P) 10

Thus for a nucleus in a sample of the element, probability of decay in 10 years is : (A) 50% (C) 60%

44

A-

1/2

1/4

(B) 75% P) 100% >,1

B-

X2

^

(Q 50

P) 50

'-'f

4-10 A radioactive source has a half life of3 hours. A fi^eshly

C

prepared sample of the same emits radiation 16 times the

^=0

No

0

0

t

N, •]

N. "2

N, "'3

In the above radioactive decay C is stable nucleus. Then : (A) rate of decay of A will first increase and then decrease P) number of nuclei of B will first increase and then decrease

(Q if ^2 > X.,, then activity ofB will always be higher than activity ofA

p) ifXj» Xj, then number ofnucleus ofCwill always be less' than number of nucleus ofB.

4-5 The half-life of a radioactive substance is 10 days. This

permissible safevalue. Theminimum timeafterwhich itwould be possibleto work safelywith the source is : (A) 6 hours (Q 18 hours

p) 12 hours p) 24 hours

4-11 The halflife period ofa sample is 100 second. If we take 40 gram of the radioactive sample, then after 400 second how much substance will be left undecayed ? (A) 10gram p) 5 gram (Q 2.5gram p) 1.25gram

means that:

(A) The substance completely disintegrates in 20 days P) The substance completely disintegrates in 40 days (C) 1/8 part of the mass of the substance vvill be left intact at

4-12 The kinetic energy ofa 300 K thermal neutron is: (A) 300eV P) 300eV (Q 0.026eV p) 0.026MeV

the end of40 days

p) 7/8 part of the mass of the substance disintegrates in

4-13 A samplehas two isotopes ^^^A and ^B having masses

30 days

50 g and 30 g respectively. A is radioactive and B is stable. A decays to by emitting a particles. The halflife of.B

> C (stable)

At a certain time, the activity of nucleiA is 'x' and the nett rate

the energy Q released is : (A) IMeV (Q 23.8MeV

(B) 1.44sec

(2)(Q 0.69

2H + 2H

(B) 34.7min (D) 1hour

ofincrease ofnumber ofnuclei B is

Theactivity ofnuclei B

(B) 11.9MeV

at this instant is :

P) 931MeV

(A) y

(B) x-;-

(Q y-x

p) X

4-19 A radioactive substance is beingproduced at a constant rate of 200 nuclei/s. The decay constant of the substance is ls~'. After what time the number of radioactive nuclei will become 100. Initiallythere are no nuclei present: 1

(A) 1 sec

(B)

(Q In (2) sec

(D) 2 sec

In (2)

sec

4-26 The activity ofa radioactive sample decreases to one third of the original valuein 9 years. Whatwill be its activity after further 9 years ?

(A) ^(/3

(B) A^6

(Q A^9

(P) A^IS

4-27 The radioactivity ofasampleis^atatime/,and Tata time/j-Ifthemean lifeofthe specimen isx,thenumber ofatoms

4-20 What is the rest mass energy ofan electron ? (B) 0.51 MeV (A) leV (Q 931MeV (D) None of these

that have disintegrated inthe time interval (/^ - /j) is: (A)X/,-y/2 (Q iX-Y)/x

(B) X-Y p) (X-Y)x .

Nuclear Physics and Radioactivity ;

4-28 A radioactive nucleus can decay by either emitting an a

particle orby emitting a p particle. Probability ofa decay is75% whilethatofpdecayis25%.Thedecayconstantofa decay is X. .

X, and thatof p decay is a.^. — is: (A) 3

(B) Y

(Q 1

P) cannot be said

4-35 Therateofdisintegration ofaradioactive substance falls from 800 decay/min to100 decay/min in6hours. The half-life of the radioactive substance is ;

(A) 6/7hour (Q 3hrs

(B) 2hrs P) Ihr

4-36 The activity of a radioactive sample decreases to one-

tenth of the original activity in a periodof oneyearfurther. After further 9 years, its activitywill be:

(B) fA

A

4-29 Part ofthe uranium decay series is shown U238 . 92

90^"

(Q

10

P) None of these

10

„Ra226

Th^^o

90

Howmany pairs of isotopes are there in the above series' (A) 1 •" (B) 2 ^ (Q 3

• P) 0

4-37 Halflife of a radioactive elements is 12.5 hour and its

quantity is256 gm. After how much time itsquantity will remain Igm?

(A) 50hrs (Q 150hrs

p)' 100hrs P) 200hrs

4-30 A radioactive element^with halflife of2 hours decays

givinga stable element 7? Afterhowmanyhoursthe ratioofX 4-38 An unknown stable nuclide after absorbing a neutron to Twillbe 1:7?

(A) 6 hours (Q 10hours

p) 8 hours , . P) 12hours

4-31 The energy released by the fission ofone uranium atom is 200 MeV. The number of fissions per second required to produce 3.2 MW ofpower is :

(A) 10^' (Q lO's

P) 10^'' P) 10'^

emits an electron, and the new nuclide splits spontaneously into two alpha particles. The unknown nuclide is: (B) 3Li' (A) Pe' (D) cB 10 (Q 4-39 A radioactive material is made at the constant rate of

10'^ nuclei per second andthe material is gettingdecayed with decay constant 0.0123 month~k Activity ofmaterial after a very longtime (ifinitiallytherewasno radioactive material)is: 10^

dps

P) l.bxio^dps

4-32 Nuclei X decay into nuclei 7 by emitting a particles. Energiesof a particle are foundto be only 1 MeV& 1.4MeV. Disregarding the recoil of nuclei 7. The energy of y photon

(Q lO^'dps

emitted will be:

4-40 The halflife ofa certain radio isotope is lOminutes.The

(A) 0.8MeV (Q IMeV

•

0.0123

p) None of these

P) 1.4MeV

number of radioactive nuclei at a given instant oftimeis 1Ol

P) 0.4 MeV

Then the number ofradioactive nuclei left 5 minutes later would be:

4-33 An elementXdecays, first by positron emission and then

twoa-particles are emitted in successive radioactive decay. If

(A) ^

10

P) lO'^

the product nuclei has a mass number 229 and atomic number 89, the mass number and atomic number ofelement Xare:

(A) 237,93 (Q 221,84

10®

P) 237,94 . p) 237,92

4-34 The decay constant of a radioactive substance is 1 per month. The percentage of radioactive substance left undecayed after two months will be:' (A) 25% P) 50%

(Q 66%

(Q V2 X10'

' P) 87%

4-41 A uranium nucleus (atomic number 92, mass number 238) emits an alpha particle and the resultant nucleus emits a

P particle. The atomic and mass numbers respectively of the final nucleus are:

(A) 90,240 (Q 91,234

P) 90,236 P) 92,232

andiRadioactfvity

4-42 A radioactive nuclide having decay constant Xis 4-49 In which ofthe following process the number ofprotons produced at the constant rate of n per second (say, by in the nucleus increases. bombarding atarget with neutrons). The expected number A^of (A) a - decay

(B) p"- decay (D) k-capture

nuclei in existence tseconds after the number is-N^ is given by: (Q P"^-decay

(A) N=

(B) N= ^

(Q

+

4-50 Aradioactive nucleus 'X' decays toa stable nucleus 'Y'. Then the graph ofrate offormation ofA' against time V vdll

+(^0

be:

.

4-43 The equation

4

->

+ 2e^+ 26MeV represents

(A) p-decay

(B) ^decay

(Q Fusion

(D) Fission

4-44 At time t=0, Aj nuclei ofdeeay constant A,j &Aj nuclei ofdecay constant ^2 are mixed. The decayrate ofthe mixture at

(A)

(B)

(Q

(D)

timeis:

(A) AiA2e

—(A,] —A2)/

(B)

will be decreasedby a fraction :

(Q. iN{k^e ^i'+A2A.2e ^2')(D) Ajl, 4-45 The number ofa and p"emitted during theradioactive decay chain starting from (A) 3a&6p(Q 5a&4p-

4-51 Suppose thespeed oflight were halfofthepresent value, theamount ofenergy released in the atomic bomb explosion

Ra andending at (B) 4a&5pP) 6a&6p-

Pb is:

(A) i

(s) t

(Q f

(D) I •

4-52 The graph shows how the count-rateA of a radioactive

4-46 Two identical nuclei A and B of the same radioactive

source varies with time /. A and t are related as : (assume In

element undergo p decay. Aemits a P-particle and changes to

(12)= 2.5)

A'. B emits a p-particIe and then a y-ray photon immediately

In A

afterwards, and changes to R': (A) A' and B' havethe sameatomicnumberand mass number p) A' and B' have the sameatomicnumberbut differentmass

\

numbers

(C) -A'and 5' have differentatomic numbers but the same mass

\

number

P) A'andR'areisotopes.

10

•

'^N 11

20

30

Time (s)

Figure 4.16

4-47 In nuclear fission 0.1% mass is converted into energy. How much electrical energy can be generated bythefission of (A) A = 2.5 e-^^' Ikgoffuel?

.

(A) IkWh (Q 2.5 kWh

P) lO^kWh P) ZSxlO^kWh

4-48 Ifthe binding energyper nucleon in jLi and ^He nuclei is respectively 5.60 MeV and 7.06 MeV, then the energy of the reaction

jLi+ Jp(A) 19.60MeV (Q 8.46 MeV

(C) A = 2.5e-^-^'

4-53 The ratio of molecular massof tworadioactive substance

is y and the ratio oftheir decay constant is —.Then the ratio oftheir initial activityper mole will be: (A) 2

(B) f

(q|

(D) I

2 ^He is P) 12.26MeV P) 17.28MeV

P) A = ne^^' P) A = l2e-^-^'

Nuclear Physics and Radioactivityj

4-62 A radioactive element X converts into another stable ofthe a radioactivematerial of half-lifeof 30 minutes decreases element Y. Halflifeof^is 2 hrs.Initially only^is present. After time /,theratio ofatoms ofA"and fis found tobe 1:4, then / in to 5 dpsafter2 hours. Theinitial countrate was :

4-54 Thecount rateofaGeiger Muller counter for theradiation

(A) 80 second"^ (Q 20 second"'

(B) 625 second-' (D) 25 second"'

hours is :

-

(A) 2 (Q between 4 and 6

(B) 4 P) 6

4-55 A radioactivenuclide can decay simultaneously by two

different processes which have individual decay constants A,,

4-63 Two radioactive materials A, andAjhave decay constants

and Xj respectively. The effective decay constant ofthe nuclide 10 A, and Arespectively. Ifinitially they have the same number of nuclei, then theratio ofthenumber ofnuclei ofA,tothatofX^

is Xgiven by:

(A)^=V^ (C) X=^i\ +X,)

(B)

+^

will be l/e after a time:

(B)

IOA

(D) A,=A,j + X2

11

(Q

(D)

lOA

1

llA

J_ 9A

4-56 Ana-particle ofenergy 5 MeV is scattered through 180" bya fixed uranium nucleus. The distance ofclosest approach is 4-64 A radioactive materialdecays bysimultaneous emission oftwoparticles with respective halflives 1620 and BID years. ofthe order of: (A) lA (B) 10^'Ocm Thetime inyears after which one-fourth ofthematerial remains (Q lO-'2 cm

P) lO-'^cm

is:

(A) 1080 (Q 3240

P) 2430 4-57 Halflivesoftwo isotopesAand Tofa materialareknown P) 4860 tobe2 X10^ years and4 x 1years respectively. Ifa planetwas formed with equal number of these isotopes, then the current 4-65 A starinitially has 10^ deuterons.ltproduces energy via

age ofplanet, given that currently the material has 20% ofAand the processes 80% of Tby number, will be:

(A) 2 X10^ years

p) 4 x 1years

(C) 6 X10^ years

p) 8x10^ years

]R+ ^n^]U+p and

>^e + n

4-58 A nucleus raptures into two nuclear parts which have where the masses ofthe nuclei are : m (^H) = 2.014102 amu, their velocities in the ratio of2:1. What will be the ratio oftheir nuclear sizes (radii) ?

(A) 2'/^: I

P) 1

(Q3"2:1

P)1:3"2

4-59 In which ofthe followingcase the total number of decays will be maximum in a time interval of/=0 to /=4 hr. The first term

m{p) = 1.007825 amu, m(«) = 1.008665 amu and mCHe) = 4.002603 amu. Iftheaverage power radiated bythe star is.lO'^ W, the deuteron supplyofthe star is exhausted in a time ofthe order of:

(A) lO^s

P) 10®s

(Q lO'^s

P).10 16

represents the number of nuclei at time / = 0 and the second

4-66 Masses oftwo isobars ^Cuand^n are 63.929766 amu

represents the half lifeofthe radionuclide:

and93.929145 amurespectively. Itcanbeconcluded from these

(A) N^4hr (Q 3No,2hr

(A) Boththe isobars are stable

P) 2No,3hr P)4No,lhr

4-60 There are two radioactive substances A and B. Decay constant of B is two times that of A. Initially both have equal

number of nuclei. After k halflives ofv4 rate of disintegration of both are equal. The value of«is : P) -2 (A) .l (Q 4 P) All of these

data that:

P) ^Zn isradioactive, decaying to^Cu through p-decay (C) ^Cu is radioactive, decaying to ^Zn through y-decay p) ^Cu is radioactive, decaying to ^Zn through P-decay 4-67 A radioactivesubstanceA decaysinto another radioactive

substance Y. Initially only A was present A^ and A^ are the disintegration constants ofA and Y.

and

arethe number of

nuclei ofAand Yat any time /. Number ofnuclei N^. will be 4*61 An atom ofmass number 15 and atomic number 7 captures

maximum when:

analpha particle andthen emits aproton. Themass number and (A)

(A) 14aiid2 respectively (Q-17 and6 respectively

(Q

P) 16and4 respectively p) 18and8 respectively

A,

N.

atomic number ofthe resulting atom will be:

A.-A,

P)

A -A„

P) A^A,=A,A,

A -A.

^Nuclear Physics and Radioactiwty

173

4-68 Aparticle ofmass Matrest decays into two particles of

masses/Kj and/«2, having non-zero velocities. The ratio ofthe (A) — In de-Broglie wavelengths ofthe particles, X/ is: ^ (A)

(B)

(Q 1.0

P)

lA.

P) T

•4

T

4-69 A radioactive nucleus is being produced at a constant rate a per second. Its decay constant is X. If

are the number

ofnuclei attime /=0, then maximum number ofnuclei possible are:

2A,

4-75 The half life of a radioactive material is 12.7hr. What fraction of the original activematerial would becomeinactivein 63.5hr.

(B) A^o+X

(A) 1/32 (C) 31/32 ,

(D)

4-76 In the nuclear fusion reaction,

a

(C) A'.

P) nn

li^

p) 1/23 p) 23/32

4-70 AradioactiveisotopeATwith half-lifeof 1.37 XlO^years

given thattherepulsive potential energy between thetwo nuclei

decays to Ywhich is stable. A sampleof rocks from the moon

is~ 7.7 XlO"'"* J, the temperature at which the gases must be

was found to contain both the elements ATand Ywhich were in the ratio 1 :7. The age ofthe rocks is :

heated to initiate the reaction is nearly [Boltzmann's constant A= 1.38 X10-23 J/K]

(A) 1.96 X1QS years (Q 4.11 XlO^years 4-71

(B) 3.85 x 10^ years p) 9.59 x IQ^years

atoms ofa radioactive element emit

(A) lO'K (C) lO^K Particles

per second. The decay constant of the element is (in s"') A^,

(Q A^,ln(2)

(D) iV,ln(2)

p) ID^K P) lO^K

4-77 The activity ofaradioactive sample is/1, attimef, and^j at time Ifthe mean lifeofthe sample isi, then the number of nuclei decayed in time is proportionalto : {h)A.L-AJ,

p)

(Q {A,-Ap{t^-t,)

P) {A,-Apr

4-72 Sun radiates energy inalldirection. The average energy

5-78 Nuclei ofradioactive element^ areproduced atrate^2'at

received at earth is 1.4KW/m-. The averagedistance between

any time t. The elementA has decayconstant X. Let A^be the

the earth and the sun is1.5 x 10"m. Ifthis energy isreleased by conservation of mass into energy, then the mass lost per day

numberofnucleiofelement^atanytime/.Attime/=/o, ^

bythesun isapproximately (use 1day= 86400 sec)

(A) 4.4x lO^kg (Q 3.8x lO'^kg

p) 7.6x IQi^kg p) 3.8x io"^kg

minimum. Then the number ofnuclei ofelement/^ at time/= is:

•

2/f, —Xtn

4-73 Power output of reactor ifittakes 30 days touse up 2kgoffuel, and ifeach fission gives 185 MeV ofusable energy, is:

(A)5.85MW (C) .585MW

P) 58.5KW p) None of these

4-74 A radioactive material of half-life Twas produced in a nuclear reactor at different instants, the quantity produced second time was twice of that produced first time. If now their

present activities are A^ and A^ respectively, then their age difference equals:

(A)

(Q

P)

Iq

2/ft —Xtn

Xiq

P)

4-79 Suppose aradioactive substance disintegrates completely in 10days. Each dayit disintegratesat a constantrate which is twicethe rate of the previousday. The percentageofthe material

leftto be disintegrated afterpassing of 9 days is : (A) 10 (Q 25

p) 20 ' p) 50

Nuclear Physics and Radioactivity ,

174

AdvanceMCQswith One or More OptionsCorrect 4-1 When a nucleus with atomic number Z and mass number^

4-7 Two identical nuclei A and B ofthe same radioactive element

undergoes a radioactive decay process :

undergo similar p decay. Aemitsa p-particle andchanges to.4'. B emits a p-particle and then a y-ray photon immediately

(A) BothZ andA will decrease, if the process is a decay (B) Z will decrease butA will not change, if the process is

afterwards, and changes to B':

decay

(A) A' and B' may have the same atomic number and mass

(Q Z will increase butA will not change, if the process is p"

number

decay

p) A' andB' mayhavethe sameatomic number but different

p) Zand ^ willremainunchanged, ifthe process is ydecay

mass numbers

(Q A' andB' mayhavedifferent atomic numbers butthe same 4-2 When the nucleus ofan electrically neutral atom undergoes

mass number

a radioactive decayprocess, it willremainneutralafterthe decay P) 'and 5'may are isotopes ifthe process is : 4-8 A and B are isotopes, B and C are isobars. All three are (A) An a decay (B) A p" decay radioactive: (Q Ay decay P) A K-capture process (A) A, B and C must belong to the same element P) A, B and C may belong to the same element 4-3 Which ofthe following assertions are correct ?

(A) A neutron can decay to a proton only insidea nucleus P) A proton can change to a neutron only inside'a nucleus

(Q It is possible that AwillchangetoB through a radioactive

(Q An isolated neutron can change into a proton p) An isolated proton can change into a neutron

P) It is possible that B willchangeto C through a radioactive

4-4 Disintegration constant ofa radioactive material is X: logg2 (A) Its halflife equal to —— 1

decay process decay process

4-9 A nuclide A undergoes a-decay and another nuclide B undergoes p-decay: (A) All the a-particles emitted hyA willhave almostthe same speed

p) Its mean life equals to —

P) The a-particles emitted by A may have widely different

(Q At time equalto mean life, 63% of the initial radioactive

speeds

K

material is leftundecayed

p) After 3-halflives, yrd ofthe initial radioactive material is

(Q All the p-particlesemittedby5 will have almostthe same speed

p) The p-particles emittedby5 will have differentspeeds

left undecayed.

4-10 Which of the following is correct for a nuclear reaction?

4-5 A nucleus undergoes a series of decay according to the scheme

Atomic number and mass numbers ofE are 69 and 172

(A) P) (Q P)

Atomic number ofif is 72 Mass number of5 is 176 Atomic number ofD is 69 Atomic number ofC is 69

(A) A typical fission represented by 92U"^^ + 3gKr^^ +Energy p) Heavywateris usedas moderator in preference to ordinary water

(Q Cadmium rods increase the reactor power when they go in, decrease when they go outward

P) Slower neutrons are more effectivein causing fissionthan faster neutrons in case of

4-6 The half lifeof a radioactive substance is Tq. At f= 0, the number of active nuclei areN^. Select thecorrect alternative.

(A) The number ofnuclei decayed intime internal 0- Ms N^e'^

4-11 The decay constant of a radioactive substance is

0.173 (years)"'. Therefore

p) The number of nuclei decayed in time interval 0 - / is

(A) Nearly 63% of the radioactive substance will decay in

W„(l-e-^0

(1/0.173) year

(Q The probability that a radioactive nuclei does not decay in interval 0 - ds

P) The probability that a radioactive nuclei does not decay in interval I - e~^

p) Halflife ofthe radio active substance is (1/0.173) years. (Q One-fourth of the radioactive substance will be left after nearly 8 years p) All the above statements are true

iNuclear Physics and Radioactivity

175j

4-12 Which ofthe following statement(s) is (are) correct ?

4-17 Assume that the nuclear binding energy per nucleoh

(A) The rest mass of a stable nucleus is less than the.sum of (B/A) versus mass number (A) isas shown in the figure-4.18. the rest masses of its separated nucleons. Use this plot to choose the correct choice(s) given below. (B) The rest mass ofa stable nucleus isgreater than thesum of the rest masses of its separated nucleons. BIA

(Q In nuclear fission, energy isreleased by fusing two nuclei 8

ofmedium mass (approximately 100 u)

P) Innuclear fission, energy isreleased by fi-agmentation ofa 6-

very heavy nucleus.

4i

4-13 Aradioactive samplehas initial concentrationWq ofnuclei (A) The number ofundecayed nuclei present in the sample

2

decays exponentially with time

(B) The activity (R) ofthe sample at any instant is directly proportional to the number ofundecayed nuclei present inthe

100

sample at that time

Figure 4.18

(Q The number ofdecayed nuclei grows linearly with time . P) The number ofdecayed nuclei grows exponentially with time

,

.

200

A

(A) Fusion oftwo nuclei with mass numbers lying in the range of 1

+ a. The atomic masses needed are as ^^P =31.974 amu and that of

=31.972 amu.

follows:

238pu

Ans. [1.86 MeV]

234y

238.CM955amu

234.04095 amu

4.002603 amu

4-52 The selling rate ofaradioactive isotope is decided by its

Neglect any recoil ofthe residual nucleus.

activity. What will bethesecond-hand rate ofa one month old

Ans. [5.58 MeV]

800 rupees?

4-46 Calculate the g-volume in the following decays:

Ans. [1 87 rupees]

(a)'^O -> '^F + e+V

4-53 In an agricultural experiment, asolution containing 1mole

^(^2"

days) source if it was originally purchased for

(b)^^Al,^ ^^Mg + e^ +v

ofa radioactive material (t^^ - 14.3 days) was injected into the

The atomic masses needed are as follows :

roots of a plant. The plant. The plant was allowed 70hours to settle down and then activity was measured in its fruit. If the

'^0

i^F

activity measured was 1 pCi, what percent of activity is

19.003576 amu

18.998403 amu

transmitted from the root to the fruit in steady state ?

^^Al 24.990432 amu

25jyjg 24.985839 amu

Ans. [1.26 X lo-'i %]

^

Ans. [(a) 4.816 MeV, (b) 3.254 MeV]

4-54 Calculate the energy released by 1g ofnatural uranium assuming 200 MeV is released in each fission event and that

the fissionable isotope 4-47 Findthe maximum energythatabeta particle can have in

has an abundance of 0.7% by

weight in natural uranium.

the following decay

,

176lu

F.

Ans. [5.7 X lO^ j]

Atomic mass of "^Lu is 175.942694 amu and that of'^®Hf is 4-55 A radioactive nucleus X decays to a nucleus Twith a decay constant = 0.1 sec~^ Yfurther decays to a stable 175.941420 amu.

nucleus 2 with a decay constant Xy= 1/30 sec"'. Initially, there are only nuclei and their number is Vg = 10^®. Setup the rate equations for the populations ofX, Fand Z. The population of

Ans. [1.182 MeV]

4-48 A radioactive sample has 6.0 x 10'^ active nuclei at a

the Ynucleus as a function of time is given by Ny (/) = {®2cp (- Xyt) - exp (- X^f)}. Find the time at

certain instant. How many of these nuclei will still be in" the

which Ny ismaximum and determine the population ofA'and Z

same active state after two half-lives ?

at that instant.

Ans. [1.5 X io'8]

[(')

4-49 The number of atoms in an ancient rock equals the number of ^°^Pb atoms. The half-life of decay of is

ciN^

(iii)A'^= 1.92 X N. = 232 X 10'^]

dN

~jr" ^ =5.76 x IQ'

dN, dl

= X N, (ii) 16.48 s,

4.5 X10^ y. Estimate the age ofthe rock assuming that all the 4-56 Calculate the 0-value of the fusion reaction ^'^^Pb atoms are formed from the decay of ' ''He+ ''He = ^Be.

Ans. [4.5 X lO^y old.]

Is such a fusion energeticallyfavourable ?Atomic massof ^Be is 8.0053 amu and that of'He is 4.0026 amu.

4-50 Find the energy liberated in the reaction

Ans. [- 93.1 keV, no]

223Ra ^ 209pt, + 14(3

223j^a 223.018amu

20

+ |3"+ v.

Where represents a mercurynucleus in an excited state at energy 1.088 MeV above the groundstate. Whatcan be the maximum kinetic energy of the electron emitted ? The atomic mass of '^^Au is 197.968233 amu and that of

is

197.966760 amu.

to find the wavelength of the Ka X-ray emitted following the electron capture. Ans. [(a) Neutrino, (b) 20 pm]

4-94 A radioactive isotope is being produced at a constant rate dNIdt= /? in an experiment. The isotopehas a half-life Show that after a time t »

the number of active nuclei will

Ans. [0.2806 MeV]

become constant. Find the value of this constant.

4-89 A nucleusat restundergoesa decayemittingan a-particle

A

ofde-Brogliewavelength,a, =5.76x 10"'^m. Ifthemassofthe daughter nucleus is 223.610 amu and that of the a-particle is 4.002 amu, determine the total kinetic energy in the final state. Hence, obtain the mass ofthe parent nucleus in amu

(lamu = 93I.470MeV/c2) Ans. [6.25 MeV. 227.62 amu]

,

r ^6/2 ,

t 0.693 ^

4-95 Consider the situation ofthe previous problem. Suppose

the "production of the radioactive isotopestarts at 7= 0. Find the number of active nuclei at time t.

Ans. [yd -e-^')]

.^Nuclear^Physics and Radioactiwty

4-96 A body of mass is placed on a smooth horizontal surface. Themass ofthe body is decreasing exponentially with disintegration constant A,. Assuming that the mass is ejected backward with a relative velocity ii. Initiallythe body was at rest. Find the velocity ofit after time t.

; \ 4-102 A number

. 185;

of atoms of a radio active element are

placedinsidea closed volume. The radioactive decay constant

for the nucleus ofthis element isA,,. The daughter nucleus that form as a result of the decay process are assuthed to be radioactive toowitha radioactive decay constant Determine the time variation ofthe number ofsuch nucleus. Consider two

Ans. [«>,/]

limiting cases A.j»

4-97 Radioactive isotopes are produced in a nuclear physics

and A.j« X^.

Ans. [

]

experiment at a constant rate dN/dt = R. An inductor of

inductance 100 mH, a resistor ofresistance 100Hand a battery are connected to form a series circuit. The circuit is switched on

4-103 A sample of 100millicurie of krypton gas consists of a

at the instantthe production of radioactive isotope. It is found

mixture ofthe active isotope ^^Kr &the stable isotope ^K.r. If

that i/N remains constant in time where i is the current in the circuit at time t and N is the number of active nuclei at time t.

the volume ofthe mixture is 10 cm^at STP & half-life of^Kr is

Find the half-life of the isotope.

10 years. Calculate the %byweight of®^Kr present inthe mixture. Ans. [0.632 %]

Ans. [6.93 x 10^ s]

4-104 Astationary ^g^b nucleus emits an a - particle with

4-98 A charged capacitor of capacitance C is discharged through a resistance R. A radioactive sample decays-with an average life t. Find the value of R for which the ratio of the

kinetic energy = 5.77 MeV. Find the recoil velocity of a daughternucleus. What fraction ofthe total energyliberatedin this decay is accoimted for by the recoil energy of daughter

electrostaticfield energystored in the capacitorto the activity

nucleus ?

ofthe radioactive sample remains constant in time.

Ans. [3.4 X iqS mis, 0.020]

Ans. [2t/C]

4-105 Energy evolved from the fusion reaction is to

4-99 A smallbottlecontains powdered beryllium Be& gaseous, be 21H = 2He+ Q usedfor the production ofpower. Assuming radon which is used as a source of a-particles. Neutrons are produced when a-particles of the radon react with beryllium. The yield of this reaction is (1/4000) i.e. only one a-particle out

of4000induces thereaction. Findtheamount ofradon (^^^Rn) originally introduced into the source, if it produces 1.2 x

neutrons persecond after five days. [T,/2 of Rn = 3.8days]

the efficiency of the process to be 30 %. Find the mass of

deuterium that will be consumed in a second for an output of 50 MW.

[Massof ^He =4.002603amu; fH=2.014I02amu] Ans. [2.9 X 10"^ kg]

Ans. [2.1 X 10"^ g]

4-106 Findthe amount ofheatgenerated by1.00mgof a ^"'Po

4-100 Theenergyofalphaparticlesemittedby^'®Pois5.3MeV.

preparation during the mean lifetime.period of these nuclei, if the emitteda- particles are known to possess the kineticenergy 5.3 MeV& practicallyall daughternuclei are formed directlyin

The halflifeofthis alpha emitter is 138 days.

(a) What mass of^'®Po isneeded topower athermoelectric cell of 1 W output if the efficiency of energy conservation is 8 percent ? (b) What would be the power output after 1 year ? Ans. [88.4 mg, 0.16 watt]

the ground state. Ans. [s 1.6 MJ]

4-107 To investigate the P-decay of ^^Mg radionuclide a counter was activated atmoment r=0. Itregistered Nj P-particles by a moment = 2.0 s and a moment = 3/,, the number of registered P-particles was 2.66 time greater. Find the mean life

4-101

decays with a halflife of4.51x10^yrs, thedecay time of the given nuclei. [TakeIn (0.88)= - 0.125]

series eventually ending at

which is stable. A rock sample

analysis shoes that the ratio of the numbers of atoms of ^®®Pb

to^•'^U is0.0058. Assuming that allthe^'^Pb has been produced by the decay of and that all other halflives in the chain are negligible. Calculate the age ofthe rock sample. Ans. [38 X 10'' yrs]

Ans. [t = 16 s]

4-108 Find the decayconstant and the mean life time of ^^Co radionuclide if its activity is known to decrease 4% per hour. The decay product is non-radioactive. Ans. [X - 1.1 X 10"^ s"'.]

Nuclear Physics and Radioactivity ;

:-i86

4-109 Taking into account the motion of the nucleus of the nucleus of a hydrogen atom, find the expressions for the electron's binding energy in the ground state. How much (in percent) do the binding energy obtained without taking into

4-113 Find the greatest possible angle through which a deuteron is scattered as result ofelastic collision with an initially

stationary proton ? Take w, & Wj as masses of a proton & a deuterium.

account the motion ofthe nucleus differ from the more accurate

corresponding valueofthisquantity. Given

ut

Ans. [S , = sin '—^ = 30°]J LPmax =0.00055, where

m and Mare the masses of an electron and a proton. Ans. [

. , 0.055%. Here h =

and n =

]

4-114 The element Curium

has a mean life of

10'^ seconds. Itsprimary decay modes arespontaneous fission and a-decay modes are spontaneous fission and a-decay, the former with a probability of 8% and the latter with a probability

4-110 An a - particle with kinetic energy

= 7.0 MeV is

of92%. Each fission releases 200 MeV ofenergy. The masses

scattered elastically byaninitially stationary ^Li nucleus. Find

involved in a-decay are as follows : (1 u = 931 MeV/c^). Calculate thepower output from asample of10^° Cm atoms.

the kinetic energy ofthe recoil nucleus if the angle ofdivergence of the two particles is 0 = 60°. Take masses of a particle and lithium as m and Mrespectively. Ans. [r =

[1 +(M- m)^ / 4ffjA/cos^ 0]

Ans. [3.32 x 10"^ Js"']

4-115 Nuclei of radioactive element.^ are being produced at a constant rate a. The element has a decay constant X. At time

= 6.0 MeV]

/ = 0, thereAq nucleiofthe element. 4-111 A neutron collides elastically with an initially stationary deuteron. Find the fraction of the kinetic energy lost by the neutron. Take masses of neutron and deuteron as m and M

respectively. (i) in a head-on-collision.(ii) in scattering at right angles. Ans. [(a) T) = 4 Mm {m + M)^ = 0.89, (b)

(a) Calculate the number N ofnuclei ofA at time t.

(b) If a = 2NqX, calculate the number of nuclei ofA after one half-life of^ and also the limiting value ofiVas / ^ oo.

Ans. [(a) Af =-^ [a - (« - XN,)

(b) 2%]]

2m

{m+M)

4-116 The mean path length ofa-particles in air under standard

conditions is defined bythe formula i? = 0.98 x 10"^' 4-112 Find the energy of the reaction (a , p) if the kinetic energy ofthe incoming a-particle is =4.0 MeV & the proton outgoing at an angle 0 = 60° to the motion direction of the a-particle has a kinetic energy = 2.09 MeV. Ans. [Q = {1 + Tip T;- (1 - -nj T^-2

cos 9 = - 1.2 MeV,

cm,

where Vq (cm/s)isthe initialvelocity ofan a-particle.Usingthis formula, find for an a-particle with initial kinetic energy7.0 MeV. (a) Its mean path length. (b) The average number of ion pairs formed by the given a-particle over the whole path R as well as over its first half. Assuming the ion pair formation energy to be equal to 34 eV.

wheren

Ans. [(a) 6.1 cm (b) 2.1 x 10^ and 0.77 x IQ^]

Geometrical Optics FEW WORDS FOR STUDENTS

Image

The topic of ray optics or geometrical optics is concerned with the analysis ofpropagation of light in a medium by considering it as the propagation of a 'Ray\ A ray defines

Mirror

the path,.along with lightpropagates and it can be assumed as an infinitesimally thin pencil oflight moving in a specific

direction at a given point in space. In this topic ofray optics we do not bother about the nature oflight and we ignore the wave and photon character oflight. Here wefocus on how light behaves on a large scale.

CHAPTER CONTENTS

5.1

Understanding a Light Ray andLight Beams

5.12

Total Internal Reflection

5.2

Reflection ofLight

5.13

Refraction by a Prism

5.3

Understanding Object and Image in Geometrical

5.14

Refraction by a Thin Lenses

Optics

5.15

Analysis ofImage Formation by Thin Lenses

5.4

Reflection and Imageformation by a Plane Mirror

5.16

5.5

Field of Viewfor Imageformed by a Plane Mirror

5.6

Characteristics ofImageformed by a Plane Mirror

5.7

Understanding Shadow Formation

5.8

Spherical Mirrors

5.9

Analysis ofImageformation bySpherical Mirrors

5.10

Refraction ofLight

5.11

Refraction ofLight by Spherical Surfaces

Optical Power ofa Thin Lens and a Spherical Mirror

5.17

Lenses and Mirrors submerged in a Transparent Medium

5.18

5.19

Displacement MethodExperiment to measure focal length ofa Convex Lens ' Dispersion ofLight

5.20

Optical Aberrations in Lenses and Mirrors

5.21

Optical Instruments

COVER APPLICATION

Figure-(a)

Figure-(b)

Figure-(a) shows the experimental setup of refraction of light through a trihedral prism and understanding of minimum deviation of light and figure-(b) shows the ray diagram of light refraction through the prism.

Geometrical Optics,

188'

In this chapter of light, we will involve only geometric considerations for light propagation and all laws are formulated using the concepts of geometry thats why, ray optics is also called ^Geometrical Optics\ The basic concept we use is that a light ray travels along a well-defined path. Ray optics in reality is an approximation but it is very important when character and effects of light are studied at macroscopic level. Topic of ray optics is very important in understanding image formation by mirrors, lenses and it helps in studying the working ofdifferent types of optical instruments. Ray Optics is a very important branch of physics but many aspects of light cannot be explained on the basis of ray optics likeinterference, diffraction and polarizationoflight which require microscopic level understanding of wave theory oflight and such concepts and theories we will study in next chapter of ''WaveOptics'.

5.1.1 DifferentTypesofLightRays

When light rays incident on an optical device like a spherical mirror, a spherical surface, a lens or an optical instrument then in reference to the optical device light rays are categorized in two ways paraxial rays and marginal rays. Lets discuss these in detail.

Paraxial Rays: These are the light ray^ which incident on an optical device very close to its principal axis and make very small angle with it or which are nearly parallel to principal axis of the device. In case ofsome lenses and mirrors paraxial rays are shown in figure-5.3. Most ofthe analysis ofimage formation in geometrical optics, we will study in this chapter are restricted to paraxial rays only.

5.1 Understanding a Light Ray and Light Beams In previous grades you have studied that light travels in a straight line that's what we call '^LawofRectilinear Propagation ofLight'. This is the concept based on which we say that an object placed in the path oflight coming from a source produces a sharp shadow on a flat surfece or a screen. This law ofrectilinear propagation is considered valid only at macroscopic level or large scale. Light does not travel in a straight line when the size ofobject or obstacle is ofthe order of wavelength oflight. Such concepts we will study in microscopic analysis oflight under wave optics and not considered here.

1

h^,

0|, 02 are very small

As already explained that a light ray can be imagined as a very thin pencil oflight travelling in a specific direction. It is represented by drawing a straight line with an arrow in the direction oflight propagation as shown in figure-5.1. A bundle of several light rays is called a light beam which is shown in

figure-5.2. For analysis of ray optics it is very important to understand different types of light rays and light beams in reference to an optical device (generally a mirror or a lens) or any optical instrument on which the light rays incident as the concept of image formation by any optical device involves a light beam through which light rays incident on the optical device.

hy h^, 63, 04are verysmall Figure 5.3 Figure 5.1

Figure 5.2

Always remember that light rays as such do not exist in reality and by any method light rays cannot be isolated experimentally from a light beam,these exist only in theoreticalunderstanding.

Marginal Rays: These are light rays which makes large angles to the principal axis of the optical device falling on it. Figure-5.4 shows some marginal rays in different cases of optical devices. Images formed by marginal rays are blurred and distorted so

•Geometrical Optics 189

these rays are not generally used in image formation while propagation and all light rays appear to converge at apoint in analyzing the cases under geometrical optics. space which is called the point of convergence of the light beam. Agenera! convergent beam oflight is shown in tigure-5.6.

Figure 5.6

Generally such abeam is incident on an object (or optical device) which is placed between the light beam and its point of convergence as shown in figure-5.7(a). Ifthe object (oroptical

device) is placed beyond the point of convergence then the incident beam will be diverging not converging as shown in figure-5.7(b)

hy hy

9,, 0^ and 0^ are large Figure 5.4

Converging incident rays falling on object (a)

5.1.2 Different Typesof Light Beams

From a source oflightalways several rays areemitted in form of

a beam. A single light ray cannot be considered as only ray emitted from a lightsource. Depending upon the behaviour of

light propagation, light be.jm is categorized in three ways Convergent Beam, Divergent Beam andParallel Beam oflight. Lets discuss these in detail.

Diverging incident rays falling on object

Divergent Beam of Light: This isa light beam in which the

(bj

diameter of light beam increases in the direction of light propagation andall light rays appear tobecoming from a point opposite to the direction of propagation of light. A general

Figure 5.7

divergent beam oflight isshown infigure-5.5(a). From a luminous pointsourceof light diverging beam of light is emitted in its surrounding in all direction as shown in figure-5.5(b)

Parallel Beam ofLight: This isa light beam in which all rays constituting the beam move parallel to each other and the

diameter ofthe beam remains same throughout the propagation oflight. Aparallel beam oflight isshown in figure-5.8. i\

/•

I, Figure 5.8

5.2 Reflection of Light (a)

(b)

Figure 5.5

ConvergentBeamof Light: Thisisa lightbeam inwhich the diameter of light beam decreases in the direction of light

Whenever a lightrayincident on a boundary oftwomedia then

apartoflight isbounced back into thesame medium and a part goes into the other medium as shown in figure-5.9. This phenomenon ofbouncing oflightenergy intothe medium from

Geometric^

;i90

(

which light incident on the boundary is called 'Reflection of gets reflected in alldirections from the point sothe light spot on the surface can be seen from all directions. Due to diffused

Light'.

reflection only when we puta parallel light beam ofa torch ona wall, thespot oflight formed can be seen from any location in theroom as lightrays in the beam arereflected in alldirections randomly in theroom from thesurface where theincident beam

N

Medium-1 Medium-2

falls.

Figure 5.9

Transmission of light energyinto the other medium is called 'Refraction ofLight'.Inthissection wewill study reflection of light indetail then later we will cover refraction oflight. Reflection oflightfrom a surface or a medium boundary isclassified intwo ways. These are 'Regular Reflection' or 'Specular

Surface with

irregularities Diffused reflection

Figure 5.11

Reflection' and 'DiffusedReflection'. 5.2.1 Regular or Specular Reflection

Whena lightbeam incident on a perfect planesurface then the reflection is called Regular or Specular Reflection. In such a case the reflected light beam has high intensity only in one

Rough surface

direction which is the direction of propagation of the reflected beam as shown in figurerS. 10. Figure 5.12

5.2.3 How we see an object in our surrounding

Perfectly

plane surface

When from any luminoussourceoflight, light raysincidenton an object, due to the roughness of the surface of object these raysare reflected from the object surface in diffused mannerin all directions so wherever an observer is situated, she will be

able to see the face of object in front of her eye as shown in Regular reflection

figure-5.13

Figure 5.10

Angle ofDeviation:'Infigure-5.10 thetotal angle bywhich the light ray is-'-rotated which is shown as 5 is called angle of deviation, the angle between the direction of propagation of initial ray and final ray.

5.2.2 Irregular or Diffused Reflection

When a light beam incident on a surface which is rough or havingirregularities thenthe lightrays in the incident beam of light will be reflected in irregular behaviour as shown in figure-5.11. Eachlightrayofthe beam isreflected from"the local point on the surface on which it incidents and different light rays are reflected in random directions depending upon the irregularities onthe surface. Thefigure-5.11 isa highly enlarged viewofa rough surface, if we lookonto it normallyit lookslike figure-5.12 whereon a point ifa narrowlight beam incident,it

Figure 5.13

For understanding of observer's view it is very important to understand how our eyes perceive the shape, size and colour

^Geometrical Optics 191 •

ofan object. Everyobject in our surrounding which is not point Second Law ofReflection : The angle ofreflection is equal to

sized (these are called extended objects) can be considered asa the angle ofincidence soaccording tothis law we have combination of several points on its surface and from each

point on the surface ofobject, light rays falling on it will be irregularly (diffused) reflected in all directions so from any side ofobject onwhich anylightis falling, it is visible toobserver's

eyes inwhich irregularly(difl^ised) reflected light rays from the surface are incident as shown infigure-5.13.

...(5.1)

I =r

5.2,5 VectorAnalysis of Laws ofReflection

Laws ofreflection can be analyzed with the help ofvector algebra also by considering unit vectors in the direction of incident

Thelight rays which areirregularly reflected from thesurface of rays, reflected rays and normal to the boundary. Keeping above

object carry information ofcolor and its brightness and from all points ofsurface and boundaries ofthe object these rays fall

two laws of reflection we relate these unit vectors with the

anglesof incidence and angleof reflection.

into observer's eye simultaneously due to which observer is able tosee thewhole object, itsspecific size, shape andcolour. In the figure-5.16, reflection ofa light ray incident on aplane surface is shown. If we consider i, r and n as unit vectors

along thedirection ofincident ray, reflected rayand normal to thesurface asshown then first we can write components of i Laws of Reflection governs the way light is reflectedfrom the and r intermsoftheunit vectors along thenormal and along surface of a boundary of two different media/ There are two the surface. Here we are considering i as a unit vector along 5.2.4 Laws of Reflection

reflection laws which explains how light is reflected from a

the surface so these are given as

surface.

When a light ray isincident on a surface then the angle itmakes with thenormal tothesurface iscalled 'angleofincidence' and after reflection, theangle which thereflected raymakes with the normal is called 'angle ofreflection'. Irithe figure-5.14 these angles are represented byangle i and angle r respectively.

i = (sin 0)i - (cos 0)h

...(5.2)

r = (sin 0) t + (cos 0) n

...(5.3)

['NrHnNhl=l

medium-l

(lsin0)

(lcos0)

medium-I!

Figure 5.14

First LawofReflection: Inthereflection process theincident ray, thereflected rayandthenormal at thepointof incidence lie insame plane. This plane iscalled 'PlaneofIncidence' asshown

(1 cos 0)

(1 sin 0) Figure 5.16

Now from equations-(5.2) and (5.3), subtracting these we get

in figure-5.15.

r=i+ (2cos 0)«

... (5.4)

From the dot product of i and h we have Plane of Incidence

i. « = - cos 0

...(5.5)

From above equations we get r = i -2{i .h)h

•

...(5.6)

Above equation-(5.6) is called the equation of lawsofreflection which account for both the laws as vector form includes the

plane ofincidence as well as equal angle ofincidence and angle Figure 5.15

ofreflection as shown in figure-5.16.

^192-'

Geometrical Optics

y ' _ _

and finally after reflection through the second plane mirror placed along xz plane theunit vector oftherayisgiven as

# Illustrative Example 5.1

Arayoflight is incident on the(y-z) plane mirror along a unit

vector e, = +^7+-^^. Findtheunitvectoralongthe reflected ray.

1 ^ F'

e, =.

41

1 \ r 7ry+-?r^

43' 43'

# Illustrative Example 5.3

Solution

A ray of light is incident on a plane mirror along a vector

The normal unit vector to the (y - z) plane mirror is along

i-\-j-k. Thenormal onincidence point isalong 1+ j. Find a

X direction so we have

unit vector along the reflected ray. h = i Solution

xii)

Ascomponent of incident rayalongthe normal getsreversed while the component along the surface remains unchanged. Thus the component of incident ray vector A= i+j-k Plane Mirror (yz plane) Figure 5.17

parallel tonormal, i.e., i+j gets reversed while perpendicular

Then the unit vectoralong reflected ray is given as ^2 and the

to it,.i.e. alongthe surface, -k remains unchanged. Thus,the reflected ray can be written as

relation between Cj and e, is given as

-^2 ~ ^1

R=-l-]~k

•n)h

&=( ' ; _ 2(-4)i 'S S' S' 1 : e-, =

1

The unit vector along the reflected ray is given as

1

,

7r'

R

'~R~

-i-j-k

43

Or we can directly say that the component of the incident ray

along the normal gets reverses i.e. along unit vector i which

r =

0+J + k)

directly gives the final result.

.5.3 Understanding Object and Image in Geometrical

# Illustrative Example 5.2

Optics ^

Two plane mirrors arecombined toeach other as such oneis in (y-z) plane and other is in (x- z) plane.A ray of light along In analysis ofvarious situations andcases ofgeometrical optics, it is very important to have a clear understanding of the terms is incident on the first mirror. Find

'Object^ and ''Image'' in the given situation otherwise lot of

the unit vector in the direction of emergence ray after

confusion arise and even simple problems seem very complex. First of all students must keep one thing in mind that these terms Object' and Image' in geometrical optics are considered

vector

S

43

+

43 '

successive reflections through these mirrors.

with respect to a specific optical device which is generally a Solution

Just like the case of previous problem here we can see that

mirror or a lens or it can be any type of optical instrument or human eye also. Lets first define an object and an image.

after reflection from the mirror alongyz plane the component of

the incident ray along x direction gets reverses so the unit vector of the ray after reflection from first mirror is given as \

I

.-

5.3.1 Object in Geometrical Optics

For any given optical device, an object is considered as the intersection point ofall the incident rays falling on the optical

"Geon^trical Optics 193

device. For existence ofobject for adevice it is not necessary that these incident rays really intersect. Objects can be point sized or extended, as shown in figure-5.12 and 5.13 ifobject is extended then every point on surface ofobject isconsidered as apoint object and combination ofall these point object is taken as the extended object.

Virtual Object

There are three wa)^ in which object is classified in geometrical optics depending on the type oflight rays incident on the optical device. Lets understand these cases ofvarious incident rays

Optical Device Figure 5.19

on a device for a point object.

Case-Ill: ParaUel Incident Rays on anOptical Device Case-I: Diverging Incident Rays on anOptical Device

When light rays incident on an optical device are parallel then

When light rays incident on an optical device are in diverging in this case we consider object is located at infinity as parallel manner then their intersection point will be in the direction

rays can only be considered intersecting at infinity. So when

opposite to the direction of propagation of light as shown in

object is located at infinity, we do"not talk about its nature whether realor virtual. Figure-5.20 below shows the case when

figure-5.18. This point is always regarded as a"Real Object' for the optical device irrespective ofwhether light rays are really parallel incident rays incident on an optical device. intersecting on this point (actually coming from this point) or not. The type of object (in this case a real object) is defined

only by the type of light rays in the vicinity ofthe optical device and not from the actual source from where light rays are coming.

Object is at infinity Optical Device

Figure 5.20

Real Object

5.3.2 Image in Geometrical Optics Optical Device

For anygiven optical device, an image is considered as the Figure 5.18

Note: Most ofthe irregularly reflected light rays from objects

intersection point ofall the reflected or refracted rays from the

device depending whether the device is areflecting or refracting

one. Based on the type ofreflected or refracted rays coming in our surrounding fall inour eye indiverging manner asshown from the device, image can also be classified in three ways in in figure-5.12 so for our eye (or for an observer's eye) what we geometrical optics in reference to the device producing the see inour surrounding can be taken asa "Real Object'. image. Case-II: Converging Incident Rays on anOptical Device

Case-I: Diverging Reflected or Refracted Rays from the

When light ra>^ incident on an optical device are in converging

Optical Device

manner then their intersection point will lie in the direction of When light rays coming from the optical device are in diverging propagation of light behind the optical device as shown in manner then their intersection point will lie behindthe device figure-5.19. These lightrays cannot intersect reallyin such a opposite to the direction of propagation of light as shown in case and in this case this point O is regarded as a "Virtual figure-5.21. This point isalways regarded as a ' VirtualImage'

Object' for the optical device. In this case also the type of for this optical device. Always remember that for diverging light object (virtual object in this case) isdefined only by the type of rays it is not possible to reallymeet or intersectin the direction light rays in the vicinityof the optical deviceand not fromthe of light propagation so such reflected or refracted rays are actual source from where light rays are coming.

considered to produce a virtual image.

Geometricai Optics!

;i94

Diverging Reflected or

Refracted rays

Ifwe look atthe figure-5.24, here for optical deviee-1,1 isa 'Real Image" as reflected or refracted rays from this device are converging but before actually intersecting at I light rays incident on another optical device-2 so for this second device-1 will actlike a' Virtual Object", as shown infigure-5.24.

•e?:-.--

I

^^30°J

=>30° (clockwise)

Figure 5.62

[12]

Geometrical Optics j

l20;e

(v)

A point source of light B is placed at a distance L in

front ofthe centre ofa mirror ofwidth d hung vertically on a wall. A man walks in front ofthe mirror along a line parallel to the mirror at a distance 2L from it as shown in figure-5.63. The greatest distance over which he can see the image of the light

(ix)

A plane mirror placed along xz plane is moving with

velocity -3f +5y-4/fc. A point object in front of the plane mirror ismoving with velocity -2z- 4y+4^. Find velocity of image.

[-2/ +14j+4i ]

source in the mirror is:

(x) Two plane mirrors are inclined to each other as shown in figure-5.66. Arayafler the three successive reflection falls

j

on the mirror Mj and finally retraces its path. Calculate the angle between the two plane mirror.

11

Figure 5.63 [3d]

(vi) In a 3D coordinate systema plane mirror is placed parallel to XYplane and above the mirror a point object is moving at

velocity Vq =5i -3J-Ilk m/s.Ifmirrorisalsomovingparallel toitselfat velocity = 2z"+y-3yt, find thevelocity ofimage

Figure 5.66 [15°]

produced in mirror.

[5/-3j +5^ m/s]

5.8 Spherical Mirrors

(vii) Two planemirrorsA/, and are inclined at angle0 as shown in figure-5.64. Aray oflight 1, which is parallelto Mj

A spherical mirroris a part ofhollow glassspherepolished on

strikes

and after two reflections, the ray 2 becomes parallel

to Mj. Find the angle9.

one ofits surface. Ifinner surface ofsphere is polished then its outer surface becomes reflecting and it is called a convex mirror as shown in figure-5.67. Reflecting face //'

\\

Policed face

(a)

Figure 5.64

Figure 5.67

(viii) An object moves with 5 m/s towards right while the

Ifthe outer surface ofthe hollow spherical part is polished then its inner surface becomes reflecting and it is called a concave mirror as shown in figure-5.68.

W]

mirror shown moves with 1 m/s towards the left as shown in

Reflecting face-

figure-5.65. Find the velocity ofimage. object »

>

5 m/s

1 m/s

j: mirror

(b) Polished face

Figure 5.65 [7 m/s]

(a)

Figure 5.68

jGeometrical_Optics

207

5.8.1 Standard terms related to Spherical Mirrors For concave or convex mirrors there are some specific terms

and their definition associated which are useful in analyzing image formation by these mirrors so first we will discuss all these terms one by one. Focus of concave mirror

(i) Pole or Optic Center of Mirror : It is the center of the

(a)

physical sphericalmirror as denotedbypointP in figure-5.69. (ii) Center of Curvature of Mirror : It is the center of the

spherical shellof whichmirroris a part. It isdenoted bypoint C in figure-5.69 and 5.70

(lii) Principal Axis of Mirror: It is the line joining the optic center of mirror and its center of curvature as shown in

P

'

*

\

figure-5.69. (iv) Radius of Curvature of Mirror ; It is the radius of the spherical shell ofwhich the mirror is a part which is the distance OC and it is also denoted by'/?' as shown in figure-5.69.

(v) Focal point of Mirror : It is the mid point ofline joining the optic center and center of curvature and denoted by 'F". This is a point where it is considered that all paraxial light rays falling on the mirror which are parallel to principal axis will converge in case of concave mirror as shown in figure-5.70(a) and appear to diverge from this point in case ofconvex mirror as shown in figure-5.70(b). It is also called 'Principal Focus' of the spherical mirror. (vi) Aperture of the Mirror : It is the diameter of the actual mirror cross section through which light rays incident will fall on the mirror surface. This is shown as 'tf in figure-5.67(b) and 5.68(b) and asAB in figure-5.69(a) and 5.69(b)

Focus of convex mirror

(b)

Figure 5.70

5.8.2 Focal Length of a Spherical Mirror

It is the distanceofPrincipalfocus of mirror to the OpticsCenter of the Mirror. Generally concave mirrors are also referred as

'Converging Mirrors'' because these converge all parallel rays incident on it. This convergenceoccursat focalpoint if incident rays are parallel to the principal axis of mirror as shown in figure-5.70(a).In this manner all convex mirrors are also referred as 'Diverging Mirrors' because these diverge all parallel rays incident on it. If incident rays fell on a convex mirror parallel to the principal axis then all these rays diverge in such a manner that these appear to be diverging from the focal point of the mirror as shown in figure-5.70(b). The situation slightly differswhen incidentrays are not parallel to principal axis. In this case also concave mirror converges these rays and convex mirror diverges these rays but the point ofconvergence and divergence like in a plane normal to principal axis passing through the focus of mirror as shown in figure-5.71(a)and 5.71(b).

Principal axis

(a)

3"^ %

1^^ 1

c

% g

d

B

(b) Figure 5.69

Focal plane

Geometrical Optics !

208.

willnot be a sharp point, it will get blurred.At the end ofchapter we will discuss on the defect in image formation ifrays are not paraxial under the topicof sphericalaberration.

Focal planci

5.8.4 Standard Reflected Light Rays for Image Formation by Spherical Mirrors

Forunderstanding ofimageformation bya spherical mirror, we need to consider some standard light rays incident on the mirror and their corresponding reflected rays. If any two of these rays incidenton a spherical mirror and their correspondingreflected rays are used then image can be obtained by finding the

(b)

Figure 5.71

5.8.3 Image Formation by a Spherical Mirror using Paraxial Rays

intersection point of the these two reflected rays as we have already discussed that all the reflected rays intersect at the samepoint(image)if all incidentraysare paraxial. Letsdiscuss four standard rays used in image formation by ray diagram for spherical mirrors.

It is observed that for a specific point object placed in front of a spherical mirror all the light rays incident on the mirror are reflected such that these all reflected rays converge or appear

to diverge only fromone point which is consideredas 'Image^of the'Object" produced bythe mirror.Figure-5.72(a) and (b)shows

images /, and formed bythemirrors A/, and Here wecan see that image /, isproduced isa 'RealImage" because reflected ray are converging and image produced is a' VirtualImage" because reflected rays are diverging.

Ray-1: Incident Ray parallel to PrincipalAxis

Figure-5.73(a) and (b) shows a rays incident on concaveand convex mirrors which is parallel to principal axis. Such a ray after reflectionpasses through the principal focus of the mirror as shown. In case ofconcave mirror it actually passes through the focus and in convex mirror it gets reflected such that it appear to pass through its focal point In virtual space behind the mirror.

Object

Real imaae

Real focus

(a)

Object

Virtual

image

'V

Virtual focus

Figure 5.72

Note : Above cases which we discussed for image formation by a spherical mirror is valid only when we consider incidence ofonly paraxial rays from object to mirror. Ifincident rays are not paraxial then all the reflected rays will not converge or appear to diverge from one single point. In this chapter we will mainly consider the image formation by paraxial rays only. If rays considered are not paraxial then for a point object image formed

(b) Figure 5.73

Ray-2: Incident Ray passing through Center ofCurvature

Figure-5.74(a) and (b) shows a rays incident on concave and convex mirrors which passing through center of curvature. Such a ray falls on mirror normal to its reflecting surface and gets reflected as it is and re-traces the path of incident ray.

iGeornetrical Optics _

^

20£j

Ray-4: IncidentRayIncidentonPoleofthe Mirror

Figure-5.76(a) and (b) shows a rayincident on concave and convex mirrors onitspole. Atthis point themirror normal isthe principal axis only so the light ray isreflected asifitisreflected

,

c\

F

from aplane mirror atthe same angle from principal axis at which it incidents asshown infigure.

Real focus

Real focus

(a)

Virtual focus

(a) Figure 5.74 Virtual focus

Ray-3 : Incident Raypassing through Focus

(b)

Figure-5.75(a) and (b) shows a rays incident on concave and convex mirrors which passing through focus ofthemirror. Such

Figure 5.76

arayafter reflection gets parallel tothe principal axis ofmirror

5.8.5 Relation in focal length and Radius of Curvature of a

as shown. In case ofconcave mirror lightactually passes through ^SphericalMirror focus and then after-reflection becomes parallel to the principal axis butincase ofconvex mirror itincident onthemirror insuch Figure-5.77 shows alight ray which is parallel to principal axis a waythat it appearto pass through virtual focus ofthe mirror

ofthe mirror shown and itincident atapointyf ofthe mirror atan

and after reflection itbecomes parallel to the principal axis as

angle 0to the normal (dotted lineAC) atthis point which passes through the center ofcurvature ofthe mirror. According to laws ofreflection it will get reflected at the same angle 8and pass through apoint Bon principal axis as shown in figure. Now in A

shown in figure.

ABC we can use AC^R^lBCcos 0 R

BC =

2cos0

Real focus

\

P

N

P

Figure 5.77 s,

1

1

F

—

1

—

C

Thus focal length ofmirror can begiven here as distance PBso we have

Virtual focus

PB = R~BC

(b)

Figure 5.75

PB^R-

R

2cos0

...(5.9)

:

"

_•

Iftheincident light ray isparaxial then 0 will besmall so we can usecos 0~ 1soPB=f=R/2. Equation-(5.9) isgives thedistance PB where parallel marginal rays after reflection will intersect on

'Geome^ical Opticsj

Case-n; Objectislocated beyond CenterofCurvature

Figure-5.79 shows this situation in which object is asmall candle andwe find theimage oftipofthecandle byconsidering ray-1

principal axis. Distance PB can be considered as focal length and ray-2 as explained inarticle-5.8.4. With this ray diagram we onlywhen incident rays areparaxial.

canconclude thattheimage produced for thislocation ofobject

5.8.6 Image formationby ConcaveMirrors

(placed beyond Con principal axis) isreal, located between F and C, inverted (on other side ofprincipal axis) and smaller in size than that ofobject.

Whenever an object is specified for a spherical mirror then

using any two of the four incident rays and corresponding reflected rays mentioned inarticle-5.8.4 we can find theimage

by using ray diagram. There are some positions near to principal axis indifferent regions infront ofthemirror where ifanobject iskept, using raydiagram wecan getsome information about theimage produced. This information isvery helpftil inrough analysis of image formation. For both concave and convex mirrors we aregoing to discuss different cases for position of

Figure 5.79

object in front of the mirror and its corresponding image Case-m: Objectislocated at Center ofCurvature produced. First we will take up the cases for concave mirror in Figure-5.80 shows this situation inwhich object (candle) located which five possibilities are there for placement ofa real object at C and wefindthe imageof the candle byconsidering ray-1 in front ofthe concave mirror.

andray-4 asexplained inarticle-5.8.4. With this raydiagram we

Case-I: Objectislocatedatinfinity

can see and concludethat the imageproducedfor this location

ofobject (placed at C on principal axis) is real, located at C, We have already discussed the case when object is located at inverted (on other side ofprincipal axis) and ofsame size asthat infinity, incident rays felling on the mirror will be parallel and for of object.

parallel incident rays there are two possibilities. One is when all

¥

incident rays are parallel toprincipal axis when image produced 0

isrealand located at focus ofthemirror asshown infigure-5.78(a)

Object

and other possibility iswhen these are atsome angle toprincipal axiswhen image produced is realandlocated in focal planeas shown in figure-5.78(b). In both of these cases we have considered the image produced is highly diminished in size, realand inverted (produced onthe other sideofprincipal axis

C

Image

Figure 5.80

as that of object).

Case-IV: Objectis located at between Focus and Center of Curvature

Figure-5.81 shows this situation in which object (candle) is locatedbetweenF and C and we find the image of the candle by

considering ray-1 andray-3 asexplained in article-5.8.4. With this ray diagram we can see and conclude that the image (a)

produced for this location ofobject (placed between F and C onprincipal axis) isreal, located beyond C, inverted (on other side ofprincipal axis) andenlarged compared tothat ofobject. M

(b) Figure 5.78

Figure 5.81

[Geometrica{ Optics

^^

211:

Case-V: Object is locatedat Focus

Case-I: Object islocated at infinity Figure-5.82 shows this situation in which object (candle) is When light rays from distant object fall on a convex mirror, located atfocal point ofthe mirror and we find the image ofthe these rays diverge after reflection insuch awaythatfor incident candlebyconsidering ray-l and ray-4 asexplained inaiticle-5.8.4. rays parallel toprincipal axis image isobtained atfocal point of With this raydiagram we can see that thereflected rays are mirror as shown in figure-5.84(a) and fer incidentrays non parallel parallelandwillproduce image at infinity. Sowecanconclude with this diagram thatimageproduced will beatinfinity, inverted (on theother sideofprincipal axis) andhighly enlarged.

toprincipal axis image is obtained in focal planeas shown in figure-5.84(b). Theimage produced will bevirtual, diminished

and erected (produced on the same side ofprincipal axis where object is located).

Figure-5.82

Case-VI: Object is located between Focus and Pole

Figure-5.83 shows this situation in which object (candle) is located between Oand F and we find the image ofthecandle by considering ray-1,ray-2andray-4 as explained in articIe-5.8.4.

With this raydiagram we can see that allthethree reflected rays fi^om mirror are diverging inaway that these appear tobe coming from thepoint/behind themirror from a virtual image. So for thislocation ofobject (placed at F onprincipal axis) image is

(a)

produced behindthe mirror, virtual,erected (on the sameside ofprincipal axis) and enlarged. Focal plane (b)

Figure 5.84

Case-II: Object is placed anywhere in front ofmirror

Figure-5.85 shows a situation in which the object (candle) is placed infront ofthe convex mirror and tofind image using ray Figure 5.83

diagram we consider ray-l, ray-2 and ray-4 as mentioned in Above five cases are general cases which explains the relative article-5.8.4. We canseein thisfigure thattheimage is formed position, nature andorientation ofimage produced bya concave byback extension ofthereflected rays corresponding tothese mirror for a real object placed close to itsprincipal axisusing incident rays from theobject andtheimage is produced behind paraxial rays. While solving different questions above cases themirror between F'and O, virtual, diminished and erected (on gives an idea about the estimation of image produced upto thesame sideofprincipal axis as that ofobject). some extent before going for the exact process of image formation.

5.8.7 Image formation by ConvexMirrors

Unlike to the different cases of imageformation bya concave

Image f

mirror for different positions of object in front of it, in case of

convex mirror there areonlytwo possibilities ofimage formation for a real object placed in front ofit. Figure 5.85

Geometricai Optics ^

:212

5.8.8 How an observer sees image of an extended object in a spherical mirror

Figure-5.86(a) shows an object (candle) placed in front of a concave mirror placed between its pole and focus. This we have already discussed in case-V of article-5.8.6. From the tip and bottom points of the objectsAB the light rays are incident on the mirror gets reflected and falls into the eye of observer. Here observer simultaneously sees the image oftop and bottom points of objectas A' and B' and simultaneouslyfor all points in between which constitutes the full image of candle yl'5'. For objectsplaced close to principal axis we consider all rays incident

on mirror are paraxial so we can find the image of the tip ofthe object in any case and drop the normal from this point to the principalaxistogetthe fullimagein all suchcases. Figure-5.86(b) shows the similar case for an observer seeing the image produced by a convex mirror.

Field of view to

see the image (b)

Figure 5.87

5.8.9 How image produced by a spherical mirror can be obtained on a screen

(a)

A screen is generally an opaque and rough white surface. Any point of which if light rays fall are diffused reflected in all direction from the point of screen. If a light beam falls on the screen than it makes a light spot on screen which can be seen from any direction in front of it as everypoint in the spot light rays are diffused reflected by screen in all directions as shown in figure-5.88. In this figure the light spot formed due to the

light beam can be seen by all the observers Oj,

and O3

shown who are looking at the spot from different directions.

(b) Figure 5.86

Ifobject is point sized and placed on the principal axis then its image will also be produced on the principal axis as shown in figure-5.87(a) and it can also be seen by the observer as shown. Figure-5.87(b) shows the field of view by the shaded region bounded by the reflected rays from the edges ofthe aperture of the mirror from which the image can be seen as reflected rays exist in this region only so to view this image observer eye must lie in this region.

03 Figure 5.88

Figure-5.89 shows a case when image formed by a concave mirror for an object (point light source) is obtained on a screen.

IGeometrical Optics 213

We can see in figure-5.89(a) when screen is placed exactly at the location ofimage produced then all thelight rays reflected

from the mirror will be meeting at point/on screen and abright and sharp image can be obtained on the screen. Ifwe displace the screen slightly toward the mirror, sharp image will not be

obtained on screen and ablurred spot is produced due to light

Blurred image

rays falling onthescreen before meeting atpoint/as shown in

figure-5.89(b). Ifwe displace the screen slightly behind the image then also we can see that a blurred spot will be formed on

(b)

screen rather than sharp image asshown infigure-5.89(b).

Blurred image

(c)

Sharp image on screen

Figure 5.90

(a)

With the above analysis it is clear that sharp image can be obtained on a screen only if screen is placed exactly at the location ofimage. Anther fact about images obtained onscreen is that onlyreal images can be obtainedon screen becausereal

images are produced byconverging rays. Diverging rays falling Image gets blurred (b)

ona screen can only produce brightness onscreen asnosharp boundaries can bemade bydiverging rays onany surface. 5.8.10 Sign Convention

For agiven spherical mirror, object and its corresponding image may lieanywhere either in front ofthe mirror ifimage isreal or behind the mirror ifimage is virtual. For mathematical analysis ofimage formation by a spherical mirror, we associate a sign convention with thegiven situation ofmirror and inanalysis of image formation all the distances ofobject, image, focal length

Image gets blurred

orradius ofcurvature ofmirror, wemeasure, willbeconsidered (c)

Figure 5.89

Figure-5.90 shows the formation ofimage ofan extended object (candle) on ascreen. Here in figure-5.90(a) the screen is placed exactly at the location ofimage in which thesharp image of candle is obtained on screen which can be seen from any direction ifwelook atscreen. But ifscreen isslightly displaced along principal axis inanydirection, theimage will get blurred as shownin figure-5.90(b) and5.90(c).

with proper '+' or

signs according tothe sign convention

used.

There are several sign conventions invented inpast tobe used in geometrical optics for different cases of reflection and

refraction. We are discussing here two sign conventions both of which are very popular for analyzing image formation in differentcases of reflection and refraction. In this bookwe will be using the first convention however ifstudents wish then for

each case studied they can use second convention and.verify that the results obtained are same. It is most important to understand and follow a sign convention properly.

FirstConvention: Incident RayReference Sign Convention

This sign convention is also called 'New Cartesian Sign (a)

Convention'. Ifwelook atthesituation shown infrgure-5.91(a). It shows a concave mirror and in front of it a pointobject O is

Geometrical Opticsj

placed of which thepointrealimage 7isproduced. Similarly in fig;ure-5.91 (b)shows a convex mirror forming pointsized virtual image /ofthe pointobject O placed in frontof it.In general for any case ofgeometrical opticsit is a general trend to represent the distanceofobjectfrompole of mirror bya letter 'w' and the distance ofimage fromthe pole of mirror bya letter 'v' and focal length of the mirror can be denoted by the letter'/'. Afi

Important pointsof measurements aboutthis sign convention are-

•

All distances are measured from the pole of spherical

mirrors.

• The origin of the Cartesian coordinatesystemis considered at the centre of the optical device (pole of spherical mirror in this case).

/

•

—

c

0

»

\

F

r -«—^

• Distances measured toward right ofthe optical device are takenpositiveand distances measured towardleft ofthe optical device are taken negative as in general we consider positive x-axis toward right of origin and negative x-axis toward left of origin.

(a)

• For the size of object above principal axis its height is considered positive and below the principal axis it is considered negative.

In figure-5.91 (a) all distancesu, vand/are taken negativeand in figure-5.91(b) distance u is taken negative and distances/and v are taken positive.

Figure S.9I

Note : In cases of spherical mirrors both of the above sign conventions gives same signs for the situations when object is

Important points of measurements aboutthis sign convention

placed on the left of mirror facing its reflecting surface and different signs when object is placed on the right of mirror facing its reflecting surface.

are-

•

All distances are measured from the pole of spherical

.

5.9 Analysis ofImageformation by Spherical Mirrors

mirrors.

• Distances measured in the direction (or along) ofincident ray are taken positive and those measured opposite to the direction ofincidentray are taken negative.

• For the size of object above principal axis its height is considered positive and below the principal axis it is considered negative.

In image formation by spherical mirrors for a specific object, we mainly analyze four specific characteristics of image which includes ''Exact Location ofImage', 'Nature ofImage' (real or virtual), 'Orientation of Image' (erected or inverted) and 'Magnification ofImage' (enlarged or diminished with exact size ofimage). Lets discuss all these characteristics one by one. 5.9.1 Mirror Formula for Location ofImage

In figure-5.91(a) all distances m, v and/are taken negative because these are measured from pole ofmirror in direction opposite to the incident ray and in figure-5.91(b) distance u is taken negative and distances/and v are taken positive. Second Convention

Cartesian Coordinate System Sign

Mirror Formula is a mathematical relation between object distance, image distance and focal length ofa spherical mirror

from its pole. This formula is used in analj^is of image formation and used in finding the'Exact Location o/7wa^e'produced by a the mirror for a given object.

Convention

For the same situation shown in figure-5.91(a) and 5.91(b) in this sign convention we consider a simple two dimensional coordinate system associated with the given spherical mirror for all measurements of distances.

Figure-5.92 shows a concave mirror in front of which a point objectis placed at a distance u for which point image/is obtained at the location shown in figure. In the figure only one light ray from object is taken which incident at point and after reflection

IGeometrical Optics

215

it passes through I. The line joining point X and center of curvature of mirror C is the normal to the mirror at point X at which angle ofincidence and angle of reflection is same.

Note: It isadvisable to alwa>^ drawraydiagramwhilesolving a questionof reflection by a spherical mirror to get the proper understanding of the given situation and image formation by reflected light rays. Also students must note that above relations ofMirror formula is only valid for paraxial ra>^.Also students must note that above relations ofmirror formula is only valid for paraxial ra}^.

5.9.2 Analyzing Nature of linage Produced by a Spherical Mirror Figure 5.92

Forparaxial rays inthegiven figure wecanconsider angles 0], ©2 and 63 are verysmall so we can use h

h

h

a, = — ; a, = "T and a, = — ^

u

R

V

Now from ACXO we have ttj = aj + 0 and in ACXI we have a3= 02+ 0 and from theserelations wehave

By mirror formula when we analyze the location of image then with the location we get a clear idea whether reflected rays are converging or diverging. If image is produced in front of the mirror then reflectedrays willbe converging and image produced is real and ifimage is produced behind the mirror then reflected rays will be diverging and image produced is virtual. Figure-5.93 shows the situation of real and virtual image formation by converging and diverging raj^.

2a2 = a[ + a3

Now substituting the Values of

a2 and

Spherical mirror (Concave or Convex)

in terms of h we

get

i R~ u

V

Real image

As R=f12we can write •" ^

1-i

1

f

V

u

...(5.10) Spherical mirror (Concave or Convex)

Above equation-(5.10) is called 'MirrorFormula' which is valid for both concave and convex mirrors and in this relation values

ofu, Vand/are substituted with proper signs according to any ofthe sign conventions stated in article-5.8.8. In this book we will use Cartesian Coordinate S)^tem Sign Convention whereas in many books New Cartesian Sign Convention is used but it hardly makes any difference from student point ofview. For all the questions we are going to take up in this book, students can also try solving these questions by the first sign convention also for getting a clear picture ofanalysis. Equation-(5.10) can be rearranged in three forms and used as per requirement for calculation ofw, v of/as given below. uv

/=

...(5.11)

-SpJct

Diverging reflected rays

r--''

Virtual image

Polished face

Reflecting face Figure 5.93

5.9.3 Magnification Formula for Size and Orientation ofImage Figure-5.94 shows the image produced for an extended object (candle) by a concave mirror. In this case we are using ray-1, ray-2 andray-4 as mentioned in article-5.8.4 for image formation.

u + v

Now in this figure ifwe analyze in AAPBand AAPB'for paraxial

v =

ijf u-f

...(5.12)

u

V/ V-/

...(5.13)

-

Polished face

Reflecting face

rays we can get the relation in height of object hQ{AB) and height ofimagehfA BO perpendicular to principalaxis. -hi

...(5.14)

1216

GeometriGa! Optics

real object or virtual image ofvirtual object) then image will be produced on other side of principal axis (w is negative) and if nature ofobject and its corresponding image produced by any optical device is opposite (virtual image of real object or real image of virtual object) then image will be produced on the same side ofprincipal axis (w is negative).

5.9.5 Longitudinal Magniffcation of Image Figure 5.94

Lateral magnification formula forspherical mirrorgivestheimage height above the principal axis ofmirror and in this section we will discuss on how to calculate the image width along the

Magnification ofimage is given as Height of Image m=

Height of Ojbect

...(5.15)

from equation-(5.14) we have h; m

...(5.16)

=

Equation-(5.16) is called''Lateral Magnification^ ofthe image which relates the objectand image height with their locations. Ifmagnification ismore than1then image formed will beenlarged and ifit is less than 1 then image formed will be diminished. For

m = 1 image height is sameas that of object height. Negative sign in equation-(5.16) shows the orientation of image with respect to object. Ifin a case m is positive then it indicates that

principal axis of mirror. The relation in objectand image width along the principal axis of mirror is called 'Longitudinal Magnification^ as given below. Longitudinal magnification ofimage is given as m, =

Width of Image along Principal Axis Width of Object along Principal Axis

Figure-5.95 shows image formation of an object located at a distance x from the concave mirror of focal length / which produces an image of this object at a distance y which is real inverted and enlarged because object was placed between F and C. Here we can see that object edge^ was close to C so

imageis formed on the samesideofprincipalaxiswhereobject corresponding image edge^'is also closer to C. Ifwe consider is located andifmisnegative thenit indicates thatimage formed objectis of verysmall width dx and imageproduced is having a is on the other side of principal axis where object is located. width dy then from mirror formula we have Thus for an erected objectif m is positiveimage produced by themirror will beerected andifmisnegativethenimage produced 1 =1 1 f

will be inverted.

X

y

Differentiating this expression we get

Lateral magnification formula canalsoberearranged in different ways usingequations-(5.14), (5.15) and(5.16) for itsuses asper requirement ofthegiven parameters ina specific question. Below arethe given forms oflateralmagnification formula.

0=—\r dx—\dy X

y.

From this relation we can get the'Longitudinal Magnification' as

u

u-f

f

...(5.17)

Note: As already discussed about the mirror formula that it is

m.

dx

..-.(5.18)

applicable onlywhen paraxial rays areforming theimage sothe magnification formula asexplained above will also bevalid only when paraxial rays are in consideration.

5.9.4 Relationin Nature and Orientation ofImage Forall the opticaldevices in geometrical opticswecan findthe orientation ofimage produced using themethod explained here. This method is onlyusing the basic geometry in finding the

B' .4'C Object

Figure 5.95

image orientationon principal axis. For small width object ifimage is produced by a spherical mirror

Always remember if nature of an object and its corresponding (concave or convex) then image width can be calculated by use image produced byanyoptical device aresame (real image of ofthe equation-(5.18). But ifobject size is large then this relation

[Ge^efrical ^Opfics 217

cannot be used and in that case we need to calculate the image ofboth edges ofthe object along principal axis and take the difference as explained in figure-5.96. Here we first need to

consider the distance ofedge5 ofthe object asx, and find the corresponding image ofthis edge at 5'located at adistance>'j

using mirror formula then we consider the distance ofedge^ of the object asx^ ~ +%and find the corresponding image^^ 'at location 3^2 using mirror formula. Now the width ofimaged 'fi' can be obtained as

(b)

Figure 5.97

In figure-5.97(a) image edge I. can be obtained by lateral magnification as I. {where m=- v/«) inboth sides ofthe square edges same I. will be there so final image produced is a

magnified squarereal image and similar to this in figure-5.97(b) for convex mirror imagewill be adiminished square virtual image. Figure 5.96

The area ofimage produced is given as A, =If =nP-ll =m^A

In the same manner we can find the height ofthe image at the (-^0" ^0 ofobject surface) edges and by separately using the lateral magnification at both theedges. Case-II: Object is placed in the plane ofprincipal axis 5.9.6 Superficial Magnification byaSpherical Mirror

Figure-5.98(a) shows a small square object ABCD of edge ^0 (^0

which is placed at a point between F and C of a

As we have studied the lateral and longitudinal magnification, concave mirror ata distance u from the mirror. The image is we can easily find out the magnification in area for the image produced AB'CD'ata distance vfrom the mirror asshown and produced by a spherical mirror. There aretwo cases in which

in this case the image produced will be in shape ofarectangle

area ofimage is calculated which are given here one by one. In because for the edges BC and AD ofthe object which are both ofthese cases we are considering object size very small perpendicular to the principal axis, we use lateral magnification compared tothe distance of object fi-om themirror. and for the edges AB and CD which are along the principal axis, Case-I: Object isplaced normal toprincipal axis

Figure-5.97(a) shows asquare object

weuse longitudinal magnification.

which is placed at So in this case the image height and width are given as

a point between F and C of a concave mirror at a distance u

from the mirror. The image isproduced as^'B'C'D'at adistance v

h. = mlr.

from the mirror as shown and the image will also be ofsquare shape as both edges ofthe object are perpendicular to principal axis ofmirror so in both we use lateral magnification and size of both will be same. Similar situation for aconvex mirror isshown

w

where lateral magnification isgiven asm=~vlu

in figure-5.97(b) -

^D'

C

T'lF "mi.

Cr' B'

A'

CA

Image

B

Object

B''

Image

(a) (a)

F

Geometrical Optics | i18.

„

+ve Virtual object

\ V-> +ve Virtual image \

V

C'WWD'

\

/

/

>C^2/)

B' A' /

/ /

Real image , v -¥ +ve Virtual object u -> -ve

(/./)

/ /

/

*-u

for Convex Mirror

(b)

Figure 5.98

u -) +ve v~¥-ve

As Iq and w. are small we consider almost same height for image

Virtual object Real image

edges A'D' and B'Cand similar case for a convex mirror, is shown in figure-5.98(b)

(b)

Figure 5.99

5.9.7 VariationCurvesofImageDistancevsObjectDistance

We have discussed themirror formula which relates theimage

#Illustrative Example 5.11

distance from pole ofmirror for a given object distance. The

Apoint object is placed atadistance 30 cm in front ofaconcave

mirror formula is given as

mirror offocal length 20cm. Find thenature and location of i_l

i

f

V

u

image obtained. Solution

v =

In this case for mirror formula we use / M=-30cm

As focal length of a given mirror is constant and it can be

positive or negative depending upon the sign convention and type ofmirror. The above function can be plotted as shown in figure-5.99(a) or(b) for a concave and convex mirror facing toward left along negative x direction which we generally

/=-20cm

Usingmirror formula v =

consider.

• ~ve

• +ve

uf u-f

+30X+20 -30-(-20)

30x20

Virtual image Real object

v=

V

=-60cm

Theimage formation is shown in raydiagram andwecan see thatimage produced isreal and located ata distance of60 cm

/ /

/ ✓ /

from the poleofmirrorin front ofit.

✓ ✓ ✓

' ✓

✓

- 20cm

✓

fv^ -ve

/ /

/ ✓

/

✓

U\-3 ) 2/)

/ /

+ve

for Concave Mirror

\ \

u —> -VA V—> —ve \

30cm

Real image) 60cm

Real object

(a)

Figure 5.100

[Geometrical Optics 219

# Illustrative Example 5.12

Solution

Athin rod oflength c//3 isplaced along theprincipal axis ofa For image produced by concave mirror, bycoordinate sign concave mirror offocal length = t/suchthatitsimage, which is convention for mirror formula, we use real andelongated, justtouches therod. Find thelength ofthe u = + 35cm,/=-i-25cm

image ?

Now by mirror formula we have Solution

uf

35x25

u-f

10

= + 87.5 cm

v =

Given that image istouching theobject that means this point Thus imageproducedbyconcavemirror/is locatedat a distance 87.5cm from Pwhich isat87.5 —35 =52.5cm from the object O.

must be at the center of curvature of the mirror as shown in

figure-5.101. The image of^ is formed at the same position (since is at Center of curvature). For image of 5 we will use

-87.5 cm •

mirror formula.

35 cm-

Centre of curvature

Separation between O & /is

Figure 5.101

= 87.5-35 = 52.5 cm

Applying mirrorformula forimageoiB, according tocartesian coordinate signconvention, weuseu = +Sdl3 and/= + d

Position of

from object is at O distance 52.5

= —^ =26.25 cm

i.l=l V u

•52.5 cm'

Figure 5.103

Distanceof

f

from poleof mirror is P = 35+26.25 = 6I.25 cm.

v'^ 5d

d

# Illustrative Example 5.14

5d

An observer whose least distance of distinct vision is 'd,

v =

views his own face in a convex mirror ofradius ofcurvature V.

Hence length ofimage obtained is

Prove that the magnification produced can not exceed

A'B'=—-2d =^ 2

2

d +\ld^ +r^

# Illustrative Example 5.13 Solution

In figure shown find the distance from pole P of the concave

For clear vision, the distance between object and-image 01 mirror shown in figure-5.102, at which when a plane mirror is mustbe morethan d. Ifin the figure shown distance OP bex placed, image produced byboth mirror for the object O will then by mirror formula, we have

coincide.

#

/=25 cm

M, -35 cm-

Figure 5.102

I" Figure 5.104

^

Geometrical OpticsJ

......

Here bycoordinate signconvention, for maximum magnification

plane mirror M2 fi'om the object is30cm. We know that in a

we use 01 = d, so we use

plane mirror theimage isformed behind themirror at thesame distance as the objectin front of it. It is also giventhat there is

v = +{d-x);u = -x;f= +- (forjc>0)

^

1

1

2

d~x

X

r

x-d+x

2

{d-x){x)

r

no parallax between the images formed by the two mirrors, thus the imageproduced by the convex mirror is formed at a distance of 10 cm behind the convex mirror which will coincide

with the imageproduced by plane mirror at a distance 30cm behind at position Q as shown in figure below.

- 2jt^ + 2dx = 2rx —dr

Forming a quadratic in x we have 2 values

d-r±yjr^ +d^ x

=

h«-10cm

discarding the negative value we have

d-r-i-\lr^+d^ • x

=

Figure 5.105

Forreflectionat convexmirror using coordinate sign convention for mirror formula we have

M= 50cm, v=- 10cm

Magnification produced by the mirror is

m

=

—

d-x d +r~ylr^ +d^ d-r +ylr^ +d^

Using mirror formula, we have

1 =1 +1 f

Rationalising the above equation, we get

m =

30cm

20cm

J_ _ J

(d +r)-{^lr^+d^f {d-r +^r^ +d^ ){d +r +ylr^ +d^) 2rd

u

V

I_ _

/ " 50 10 "7 50 50

So radius of curvature ofthe mirror is

2d'^+2d{4r^+d'^) Thus the maximum magnification produced by the mirror can be given as

d +yjr'^ +d^

4

\

50x2

R = 2f=- — c m the radius of curvature ofconvex mirror is 25 cm.

^ Illustrative Example 5.16

Using a certainconcave mirror, the magnification is found to # Illustrative Example 5.15

An object is placedin fi-ont of a convex mirror at a distance of 50 cm a planemirroris introduced in fi-ont ofthe convex mirror

be 4 times as great when the object was 25 cm fi:om the mirror as it was with the object at 40 cm from the mirror, the image, being real in each case. Find the focal length of the mirror.

covering the lower halfofit. Ifthe distance between the object

Solution

and the plane mirror is 30 cm and it is found that there is no parallax between the imagesformedby the twomirrors for the same object. What is the radius of curvature of the convex

We use magnification

mirror?

V m= -

u

j

Solution

Let O be the objectplaced in front of a convexmirror A/, at a distance of 50 cm as shown in figure-5.105. The distance ofthe

/ =

7

w-/

In first case when u = 25 cm, magnification is m

•

/ 25-/

...(5.19)

jGeometrical Optics

22/]

In second case when m= 40 cm, magnification is

-.=36

Now using mirror formula we have

J

...(5.20)

1_ J_

15~70

V,

As itisgiven that w, = 4ot2 we use / 25-/

/ 40-/

= 4

c , • we get Solving

v, = 210 cm

Thus final image / is formed in fi"ont of concave mirror at.

Solving we get/= 20cm.

distance of(210/l 1) cm and it is real.

# Illustrative Example 5.17

Magnification Wj for first reflection is

Aconcave and aconvex mirror each 30 cm in radius are placed opposite toeach other 60 cm apart on the same axis. An object 5cm inheight isplaced midway between them. Find the position and size of the image formed by two successive reflections,

12. 30

Magnification

for second reflection is

consider first reflection at convex and then at the concave mirror.

Solution

V2

(210/11)

~ «2 ~

70

3

"IT

Final magnification = w, x w, = —x— = — ' 2 3 11 11

The ray diagram of image formation is shown in the below figure-5.106.

Thus size ofthe final image=5x^ ^ cm. 5.9.8 Effect ofMoving Object and Spherical Mirror on Image When object ormirror isinmotion thedistance between object and mirror changes which affects the position and size ofimage. To find the image velocity and for analysis ofimage's motion we can diflferentiate the mirror formula and find the rate at which

r = 30cm

r - 30cm

Figure 5.106

distances between object orimage and mirror ischanging. Ifwe consider a; and y as object and image distance fi-om pole of mirroroffocal length/then bymirrorformula wehave

For first reflection at convex mirror, using coordinate sign convention we have for mirror formula

Mj=-30cm, / = +15cm

1

f.

y

X

Differentiating the above relation with respect totime, we get

Now using mirror formula we have 1

1-1

0=-J- .

^ dt

• 1

1

dt

•+ —

15 or

-30

Vj =+ 10cm

dx .

Where — is therelative velocity ofobject with respect tothe

Thus image produced will be virtual. This is formed at/ behind the convex mirror at adistance of10 cm. The image /, acts as a mirror and — isthevelocity ofimage with respect tomirror. real object forconcave mirrorbecause its diverging light rays fall on the concave mirror. X

Now for second reflection atconcave mirror, using coordinate sign convention for mirror formula, we have 10 = +70cm, / = + 15 cm

v, = -

y 2

...(5.21)

Geometrical Optics |

Where m is the lateral magnification produced bythe mirror.

Ifpolice jeep is at a distance behind the thiefscar then we

Theexpression ofimage speed as given in equation-(5.21) is valid onlyfor the velocity component ofthe image and object along the principal axis. If the object and mirror is in motion along the direction normal to principal axis we can directly

can use m = - c/ so we have

differentiate theheightofobject andimage above principal axis

^

which are related as

Thus distance of image from mirror is

10

1

10-M)

10

t/=90m

h: = mh v = -mu =

Differentiating this with respectto time we get dh;

= 9m

Now rate at which magnification is changing is given as

dh„

— = m —T"

dt

dv

dt

j

dm v,-^=wv ON

Here we can use

dh

=v.^and

u

du V—

dt

dt

dt

d\

=VQ^which are the velocity

components ofimage andobject respectively in direction normal to the principalaxis. ^

dm

(-90)(-lxl0-^)-9(l)

~dt'

90^

dm

-81

~dt

10x8100

n-i

= +l X 10"^ s"'

§ Illustrative Example 5.18

it Illustrative Example 5.19

A thief is driving away on a road in a car with velocity of 20 m/s.Apolicejeep is chasing him,whichis sightedbythief A Convex mirror ofradius ofcurvature 7? is fixed on a stand at in his rear view mirror, which is a convexmirror of focal length rest with total mass m which is facing a block ofequal mass m 1Om. He observesthat the image of the jeep is moving towards as shownin figiire-5.107 and iskepton a fiictionless horizontal him with a velocity of 1cm/s. Ifthe magnification ofthemirror surfece. Theseparation between theblock and mirroris IR and block is moving at a speed vtoward the mirror. Consider elastic for the jeep at that time is 1/10. Find: collision between block and stand, find the speed of image

(a) The actual speed of the jeep.

(b) The rate at which magnification is changing.

after time 3i?/v.

Assumethat policejeep is on axis of the mirror. Solution

(a) The velocity of imagewithrespect to mirror is related to velocityof objectwith respect to mirror is given as

-IR-

Figure 5.107 Solution

=>

- 1 X 10"^= After elastic collision block will come to rest and mirror with

stand starts moving at the same speed v and the collision occur at time IRIv and after a fiirther time RJvwhen total time elapsed

Velocity of object with respect to ground is given as Vn,n=Vn,^-¥V.ml G

is 3i?/vtheseparation between blockandmirrorwill be R and at this instant the position of image will be given by mirror formula with

Vo/a = ^+20=(+21 m/s)/ (b) The magnification produced by the mirror is m =

/ f-u

1 10

u=-RJ=+RJ2 Now using mirror formula we get

uf _ ~Rx{RI2) _ R v =

u-f ~ ^R-RI2 ~ 3

fGjeometrical Optics

2231

At this instant magnification is

^

V _ 7g/3 ~ u

Same effect can be there on image ifa spherical mirror iscut and displaced. In figure-5.109(a) aconcave mirror isforming image/ ofan object O. If it iscut atcenter andthetwo parts and aredisplaced indirection normal toprincipal axisbya distance Xas shown in figure-5.109(b) theneach part ofthemirror will behave like a separate mirror and produces its own separate image ofthe given object. Due tothis cutting and displacement ofmirror parts, object distance from each ofprincipal axis of these parts has become x/2 and by these two mirror parts two

1

-^~3

Speed of imagewithrespect to mirroris given as

/i\2

il v=i 3

9

V

images are produced at a distance y above and below the two principal axisin thesame plane ofprevious image as shown in figure-5.109(b) wherey isgiven as

10

Speed of image with respect to ground is v + —= — v

5.9.9 Effect ofshifting Principal Axisof a Mirror

y=

.X

Figure-5.84 shows a point object and its corresponding point image produced by a concave mirror. If in this situation the

mirror is displaced upward bya small distance Ax in upward direction then with mirror its principal axis will also shift as

cut here

shownin figure-5.108 and in final state object willbe located at a distance Ax below the principal axis of mirror. Due to this image will be displaced up and finally obtained above the

principal axis at a distance Ax + Ay from its initial position. Here (a)

Ay can be directly given as Ay = w Ax

X

(b) Figure 5.109

(a)

Note: If in figure-5.109(a), a small section at its center of width

Xiscutand removed as shown infigure-5.110 then image will remain at the same position where it was because now the two

parts ofmirror will behave asa single mirror with same principal axis.

(b)

Figure 5.108

Figure 5.110

Geometrical Optics t

\22A

5.9.10 Image formation of distant Objects by Spherical Mirrors

incident rays offigure-5.112(a) andmeet atpointO,andproduce theimage I2 at this location as shown.

For distantobject, weknowthat a sphericalmirror produces its image in focal plane. The size of image can be calculated by analyzing its angular width with respectto the pole of mirror. Figure-5.111(a) shows the formation of image of Sun by a concave mirror. S is the solar disc and here 6 is the angle

subtended by solar disc at any point on earth surface which will be same at the pole of mirror as shown. Here we consider twolight raysfromthe edgesofthe solardiscwhichincidentat pole of mirror at angle 0 and these get reflected at the same angle and producethe real image of Sun on focalplane. Here we can see that the image produced is inverted and its diameter can be given as =/0 as 0 is a very small angle and image is obtained at a distance/from the pole. Figure-5.111(b) shows the formation ofimage ofthe Sun by a convex mirror. The only difference here is that virtual erected image is produced at focal plane as shown. A

(b) Figure 5.112

•

S/

Here we can state ifan object is placed at the location ofimage produced by a spherical mirror then for this position of object, image is produced at the location of initial object.This concept is called 'Reversibility ofLight' for image formation by mirrors. Real image X 1-^

# Illustrative Example 5.20 •/—H

A concave mirror of focal length 20 cm is cut into two parts from the middle and these two parts are moved perpendicularly bya distance 1 cm fromthe previousprincipal axis/fS. Find the distance between the images formed by the two parts ?

(a)

Virtual

10cm

image

Figure 5.113

(b)

Figure 5.111 Solution

5.9.11 Concept ofReversibility of Light

Figure-5.112(a) shows an object in front ofa concave mirror of which image is produced at point In this case we have

discussed that allparaxial lightra>^ from O, gets reflected from themirrorand afterreflected raysintersect at point/, andproduce image oftheobject. Ifweplaceanother object O2 attheposition /, as shown in figure-5.112(b) then according to laws of reflection all light rays from this object will incident on the mirror following the same path offigure-5.112(a) in opposite direction and gets reflected from mirror and follow the path of

Consider the figure-5.114 in which we take

is the principal

axisforthemirrorM, &P2 is theprinciple axisfor themirror M2 and/,, I2are the imagesproduced bymirrorsM, and A/j. Using coordinate convention for mirror formula here we use w= -10cm and/ = -20cm Now by using mirror formula, we have

=>

1

J_ _ J_

-10

V ~ -20 V = + 20 cm

'Geometrical Optics 225

Thus image will be produced by both mirrors at a distance Web Reference at www.phvsicsgalaxv.com

20cm behind the mirrors as vis'+ ve'. Now the magnification produced ism=- 20/(-l0) =+ 2thus image will be produced on the same side ofprincipal axis ataheight 2(1 cm) =2cm from

Age Group - High School Physics ] Age 17-19 Years

the principal axis.

Topic - Geometrical Optics I-Reflection ofLight

Section - OPTICS

Module Number - 24 to 40

Figure-5.114 below shows the two images /, and /j produced by the two mirrors. Practice Exercise 5.2 10cm

P,-

/,_

Q

A--

20cm

2cm

P

(i) Find the distance from a convex mirror, offocal length 60cm where an object ofheight 12 cm should be placed sothat

its image is produced at 35 cm from mirror. Also find the height ofimage. [- 84 cm, 5 cm]

Figure 5.114

The heights of the two images are given as

hj -2 cm below the principal axis P,

and hj^ =2cm above the principal axis Thus the separation between the images is given as

(ii) A 2cm high object is placed on theprincipal axis of a concave mirror ata distance I2cm from the pole ofmirror. Find

the location ofthe image and focal length ofmirror ifthe image height is 5cm and it is inverted. [30 cm, 8.6 cm]

7,72=2cm (Hi)

# Illustrative Example 5.21

An object is located at a distance 30cm onprincipal axisfrom the pole ofaconcave mirror offocal length 20cm. Suddenly the mirror is displacedby a distance 1.5cm in the direction normal

to its principal axis. Calculate the displacement of image

A pointobject on theprincipalaxisofa concave mirror

offocal length 20cm, ismoving ata speed of5cm/s atan angle 45° to the principal axis asshown in figure-5.115. Initially object islocated ata distance of25cm from thepole ofmirror. Find the velocity components ofimage along and normal to principal axis at this instant.

produced by the mirror due to this.

/= 20cm 5cm/s

Solution

^5° By using coordinate convention if we consider the object to be located to the left of the mirror, we have

-25cm-

w= -30cm and /=-20cm Figure 5.115

By mirror formula we use v/ i/ =

v-f

Magnification is w =

(iv)

-30X-20 -10

=-60 cm

-60

A man uses a concave mirror for shaving and sees his

2 timesenlarged imagein mirror when his faceisat a distance 40cmfrom mirror. Findfocal length ofmirror. [80 cm]

-30

If mirror is displaced up by a distance 1.5cm then object

(v)

distance from principal axis of the mirror becomes 1.5cmand

concave mirror ata distance of12 cmfrom thepole. Ifthe image is inverted, realand5cm high, find thefocal length ofmirror. If

the new position of imagewillbe at a height-2 (1.5)=- 3cm

from the principal axis which was earlier produced on principal axis.

Thus displacement of image is 3cm.

A 2cm high object is placed on the principal axis of a

the object starts movingat a speed of 1.2cm/s toward the mirror find the speed ofimage and its direction of motion. [7.5 cm/s]

Geometrical Optics

'226

(vi) Figure-5.116 shows two spherical mirrors Mj and on same optical axisat a separation of50 cm. Apointobject O is placed midway between mirrors onoptical axis. Find location & nature of its image after two successive reflections first at A/, then at Mj. •/f^ = 20 cm

Denser medium

.

= 30 cm

Rarer

medium

Angle of deviation 5 = 02 - 0| (b) Figure 5.117 f*

50cm

,

H

5.10.1 Absolute Refractive Index of a Medium Figure 5.116

When a light ray enters a mediumthen how its speed changes is measuredby ^Refractive Index" of the medium. It is defined as the ratio ofspeedof light in the medium to the speedoflight in free space andthis is alsocalled absolute refractive indexof

[75 cm]

the medium.

5.10 Refraction of Light

Absolute Refractive Index ofa Medium

Whenever a lightrayis incident on theboundary oftwodifferent media, a part of it is transmitted into the second medium and whenever a light enters in a transparent medium, it suffers a sudden change in velocityof light which we call refi^action of light. In the process ofrefraction the direction of propagation oflight ray changesifit incidentson the surfaceat someangle of incidence to the normal to boundary as shown in

Speed of Light in free Space Speed of Light in the Medium ...(5.22)

Wherec is the speedoflight in free space and v is the speedof light inthegiven medium. Thus themedium inwhich light travels

figure-5.117(a). This case corresponds to the situationwhen a faster has lower refractive index and is called 'Rarer Medium'" light ray travellingin a rarer mediumincidenton the boundary compared to another mediumin whichlight travelsslower and of a denser medium after entering into the denser medium light has higher refractive indexwhichis called'Denser Medium'. ray bends toward normal, which you might have covered in yourprevious classes also. Figure-5.117(b) showsthe situation 5.10.2 Relative Refractive Index of a Medium when a light ray travellingin a densermedium incidenton the Manytimerefractive index ofa medium is specified asrelative boundary of a rarer medium and after entering into the rarer refractive index with respect to another medium. This can be mediumlight ray bends awayfromthe normal as shown. defined by the relation similar to equation-(5.22). Relative refractiveindex ofa medium-1 with respect to another medium-2 is given as Rarer

Relative R. 1ofmedium-1 with respect to medium-2

medium

Speed of Light in Medium - 2

Speed of Light in Medium-1 Denser

medium

...(5.23)

Here 2M.1 isthe way how we denote refractive index ofmedium-1 Angle of deviation 5=01-0,

with respect to medium-2. So if we define refractive index of medium-2with respectto medium-1,we write 1

(a)

11^2'

21^1

Geometrical Optics 227

Here for the two media if their absoluterefractive indices are

taken as and

we can write )x, = — and ji,= — V2

to medium-2 the product ofrefractive index ofthe medium'and

thesine ofthe angle made by thelight raywith normal in that medium remain constant. In this case we have

p,sin0, = P2 sin02 =Constant

5.10.3 Laws of Refraction

When a lightraypropagates through aninterface oftwo different media then its behaviour is governed by certain laws called lawsofreffaction which are also known as 'Snell'sLaws'. There are two laws ofrefraction which are described here.

...(5.24)

Even for several media separated byparallel interfaces asshown in figure-5.119, the same relation holds valid.

(1) sinOp = pjsin0[ = p^sinGj^ p3sin03 = p^sin0^^ ... = constant

First LawofRefraction : In therefraction ofa light rayat a medium interface the incident ray, refracted rayandnormal to theinterface lieinsameplane called 'Plane ofIncidence' onthe

medium interface. Figure-5.118(a) shows the incident ray, refracted rayand normal and inthis figure the plane ofpaper is considered as plane of incidence. Figure-5.118(b) shows a perspective view oftherefraction ofthesame lightrayinwhich plane of incidence is shown by the grey shaded plane which

Medium-!

Ml

Medium-2

M2 x!

03^ Medium-3

makesthe understanding better. :.Medium-4

^4

Or..

'co

Normal

%

Figure-5.119

Medium-1 (ji,)

Asshown in above figure wecanseethatforparallel interfaces

Medmm-2 (ji,)

when light comes out inairagain it comes out atthesame angle 00 andemerges outparallel to theincident ray. 5.10.4 Vector form ofSnell's Law of Refraction

The second law ofrefraction relates the refractive index and the

(a)

angle light raymakes with thenormal inthatmedium. The angle can be expressed by using unit vectors in the direction of

incident ray, refracted rayand normal to" the boundary. If in figure-5.120 we consider unit vectors /, r and n along the direction of incident ray, refracted ray and normal as shown Plane of Incidence

then the equation-(5.25) can be written as

p, (/ X«) = p^(rx it)

(b)

Figure-S.118

Second Law of Refraction : For the situation shown in

figure-5.118(a) fortherefiaction oflightraygoingfrom medium-1

Figure 5.120

...(5.25)

Geometrical ^OptlcsJ

228 .

Above equation-(5.25) isused for analyzing law ofrefraction in vector analysis whenever in different questions light raysare specified in vector form orvectors aregiven along thedirection of incident and reflected rays.

5.10.5 Image Formationdue to Refraction at a Plane Surface

When all lightrays from an object falling onthe planeinterface of two different media as shown in figure-5.121 these all rays are refracted in such a way that after refracting these rays will

appear tobecoming from a point / which istheimage ofobject as seen bv the observer shown in figure.

Dividingequation-(5.28) byequation-(5.27) we get ...(5.29)

Equation-(5.29) iscalled asrefraction formula for image formation by refraction oflight at plane surfaces. Whenever an object is placed at a distance u from a plane interface oftwo transparent media then theimage ofobject asseen from theother media will appear at a distance v as shown in figure-5.121 which can be obtained by using this formula given in equation-(5.29) and thisimage distance isalsocalled Apparent Depth' ofthe object. 5.10.6 An Object placed in a DenserMediumisseen fromAir

Medium-2

Medium-1 1^1

Observer

Objccl

Figure-5.123 shows an object placed inwater(denser medium) having refractive index p anditis seen byanobserver from air. Ifobject isplaced ata depth hbelow theair-water interface then the image of the object is seen.by the observer at a depth h' (apparent depth) givenby refractionformula as l^l

^2

Imase

Observer

Figure 5.121

If we consider a specific paraxial ray from object incident as shown in figure-5.122 which incident on the interface at an angle 0 and gets refracted at angle (jithen by Snail's law we can

water (u)

write

...(5.26)

p, sin 0 = Pj sincj) Medium-1

Medium-2

Object Figure 5.123

Here w= /?, v=

p, = p and Pj = 1sowehave h__h

It" 1 Figure 5.122

=>

h'=-

...(5.30)

P

Forparaxial rayson the interfacewe can considerangles0 and (f) are very small so we can use sin 0 ~ 0 and sin ({) ~ (j) so from

If the distance on the interface AP is taken as r then for small

Equation-(5.30) is the relationused for 'Apparent Depth ofan object placedin a densermedium seenfromair. In figure-5.123 we can see that object appear to be closer to the interface compared to the actual depth of the object which generally happens when we see an object placed inside water level at some depth. The shift of object due to refraction can also be

angles 0 and (J) we have

calculated and given as

above equation-(5.26) we have •

Pie=P2(|)

r = iiQ = v(j)

...(5.27)

... (5.28)

Shift

h AS = h - - = h P

1-1

...(5.31)

jGeon-ietricai Optics' 229

Above equation-(5.30) and equation-(5.31) are valid only for # Illustrative Example 5.22 paraxial rays. In case ofplane surface we consider it only for

observation ofobject along the normal or with the light rays at aconvergingbeam oflightrays incident on aglass-air interface near normal incidence. ^^own in figure-5.125. Find where these rays will meet after S.10.7 AnObject placedinAir and seenfroma DenserMedium

refraction.

Figiire-5.124 shows an object placed in air having refractive and itisseen by an observer from water (denser medium) with

glass (P = 3/2)

refractive index p. Ifobject isplaced ata depth h above theairwater interfece then the image of the object is seen by the observer at a depth h' (apparent depth) given by refraction

20 cm

Virtual object

30cm

formula as u

V

Fi

F2

Figure 5.125 Solution

* Image /f'v

/' ' I I

For plane surface we can use refraction formula as

«\ I \

t\ \ Object T

ilj V,

h'-

' / /

' /'/ ij'i ir't

'/ '

"

\ N \ \

I 5

1 iair(l) water (ji)

F2

2

i/ = + 20

/ /

B.

h.

u

V

1

3/2

20 "

Observer

V

v = + 30cm

# Illustrative Example 5.23

Figure 5.124

Aconcave mirror isplaced inside water with itsshining surface

HereM = /i, v= h\ pj= 1andp, - ft sowe have

upwards andprincipal axisvertical as shown in figure-5.126. Rays are incident parallelto the principal axis of the concave mirror. Find theposition ofthefinal image.

A! 1

=>

ft

h'=\ih

...(5.32)

Equation-(5.32) isthe relation used for'Apparent Depth' ofan object placed,in airasseen from adenser medium. Infigure-5.95 we cansee thatobject appear tobedisplaced farther away from the interface compared to the actual depth of the object this generally happens when an underwater diver sees an object

Air Water 4/3 •'Ocm

placed outside in air level at someheightabove thewaterlevel,

R = 40 cm

the object appear tobe further away from water level compared

Figure 5.126

to its actual height. The shift of object due to refraction can also be calculated and given as Solution

Shift

A5=pA-/i = /i(p-l)

...(5.33)

The incident rays will pass undeviated through the water In thiscase also above equation-(5.32) andequation-(5.33) are surface and strike the mirror parallel to its principal axis. also valid only for paraxial rays as these are obtained bythe Therefore for the mirror, object isatoo its image/J (in figure-5.127) refraction formula which we derived using lightrays with small angle of incidence or near normal incidence.

will be formed at focus which is 20 cm from the mirror. Now for

the interface betweenwater and air, d= 10cm

Geometrical Optics

!230

# Illustrative Example 5.25

t

Figure-5.129 shows a concave mirror offocal length Fwithits principal axis vertical. Inmirror atransparent liquid ofrefractive index p isfilled upto height d. Find where on axis ofmirror apin

Water 10 cm 4/3

should beplaced sothat itsimage will be formed on itself.

30 cm

^ = 40 cm

Figure 5.127 d

d'=-r^

10 = 7.5 cm

4/3 - 1

# Illustrative Example 5.24

Figure 5.129

Abird in air is diving vertically over atank with speed 6cm/s. Solution The base ofthe tank is silvered. The fish in the tank is rising

For image of prism to be produced on itself we use

upward alongthe sameline with speed 4 cm/s. (Take: ^^=4/3). Find:

h\\.+ d=R

(a) The speedof the image of the fish as seendirectlybythe

R = 2F

bird.

2F-d

(b) The speed of the image of the bird relative to the fish

h It

looking upwards. # Illustrative Example 5.26

Solution

(a) Velocity offish in air = 4x—=3t

The^-Tplane is the boundary between two transparent media. Medium-1 with z > 0hasa refiractive index V2 andmedium-2

Velocity offish w.r.t. bird = 3+6=91

with z < 0 has a refractive index

A ray oflight in medium-1

given by the vector ^ =6VJ + 8>/3 j -10^isincident on the planeof separation. Findthe unit vector in the direction ofthe refracted ray in medium-2. Solution

See figure-5.130

Figure 5.128

I

Figure 5.130

(b) Velocity ofbird in water =6x^ =8i Given that

w.r.t. fish = 8 + 4=121

AB = 6^/3/ + sVsj'-lO^

IGeometrical Optics From figure

231

consider object is placed at a distance h from the front face of

AB = AM + MB

Thuswehave

= eVJf + SN/Ijand ^ =-10/t

theslab, first image /j due tofirst refraction will beproduced at a distance \ih from the surface as shown which is given by

The angle between the two vectors can be obtained byusing

equation-(5.32). If glass slab thickness is t then for the second

the following formula

refraction image will act likean object located at a distance (/ + p/j) andfinal image/2 isproduced at a distance (t + \xh)/^ at a distance from this face of glass slab which is given by equation-(5.33). Sofor the observer looking at the object from the other sideof glassslab the shift of object observed will be

AB-MB cos 1 =

_AB\\MB\

Hence

^= I

11-A | cos;

given as

10 cos / =

V36x3 + 64x3 + 100

t + \xh

Shift

2

F

cos i = —andsin i = V3 / 2

1-1

...(5.34)

By using Snell's law, we have

...(5.35)

Equation-(5.35)is the expressionused for calculation of shift of objectdue to refraction by a parallel sided glass slab.

|a, sin 1= ^2 sin r

=5«

V2 sin/ = >/3 sinr

=>

^/2x(^^/2) = V3 sinr

glass (u)

• 1 and^ cosr=—7= 1 => r = 45® smr=-7=

=»

^/2

consider vector BN.

BN =fiA'^cosr(- k) + BNsinre where e is unit vector along CD

.

bVJf-f-sVJ?

10V3

6t

8-,

"io'"'io-^

unit vector along reflected rays is —

5D

1

.

1

—i H 10

1

I0V2 _

1

" 10^^ V2

j 10

(a)

1

(-10/fc)+7^7-j= (6/+8y)

IOV2

[6/ +8y-10;t] „

.

.

-^^[3/ +4J~5k]

Final Image

(b)

S.10.8 Shift of image due to Refraction ofLight by a Glass Slab

Figure 5.131

When a light ray from an object O is incident on a glass slab at some angle / it gets refracted inside the slab at angle r and from its other face it emerges out in air at the same angle i as shown in figure-5.131(a). In the whole process light ray suffers two refractions at the two parallel surfaces of the glass slab. If we

Figure-5.131(b) shows the ray diagram of formation ofimage which is seen by an observer from the other side ofglass slab. The expression ofshift ofobject as given in equation-(5.35) is also valid only for near normal viewing as the result is obtained for paraxial ra>^ only.

Geometrical Optics t

232

5.10.9 Shift due to Refraction of Light by a Hollow thin walled Glass Box placed inside a Denser Medium

Figure-5.132 shows a hollowboxmadeup ofthin wallsofglass is placed in water and a light ray from an objectO falls on the first interface ofthe box in which the light enters into air from water as very thin walls of glass can be neglected because these thin walls of glass will not produce any shift or lateral displacement in path oflight ray.

5.10.10 Lateral Displacementof Light Ray by a Glass Slab

Figure-5.133 shows therefraction ofa lightraythrough aparallel sided glass slab which incidents on the slab at an angle of incidence z and enters into the slab at angle of refraction r

finally emerges out in direction parallel to the initial ray. If refractive index of the glass used is p then at the point A in figure we can apply Snell's law as ...(5.38)

sm z = p sm r

Water (h)

A

0

/,

11

^

r air(l)

Figure 5.133 h

Hi /+-

In the above figure in right angled triangle A/4CD we can use

M

....

As we can see in the above figure, a light ray from object O located at a distance h from the box gets refracted from first face

ofbox andproduces an image/, at a distance

hoUow ho.x

/+—

Lateral Displacement oflight ray

A^= CD ACsin(i-r)

which willact

as an object for second refraction at second face of box and finally the image is produced at a distance m{t+ /i/p) which is seen by the observer looking at object located on the other side as shown. We can see that the final image of object appears to be shifted away from the box and this shift can be given as h

CD

smO-r)=^

Figure 5.132

-{h+t)

in bxAEC we havecos r =

t

AC t

...(5.40)

AC = cosr

From equation-(5.39) and equation-(5.40) we get lateral displacement oflight ray is given as r.sin(z-z')

A5.hollow

Note : If the light ray from an object placed in a medium with

refractive index p, passes through a slab ofthickness t made up ofanother medium ofrefractive indexpj andfrom othersideof slab in the same medium of object an observer is situated and sees toward object then its final image appears to be located with a '5/z/y?' which is given by the general formula used for

cosr

Ifincidence angle z is very small (for near normal incidence), then we can use

sin(z'-r) ~ i-r cos r = 1

A = /(z-r) For very small i, from equation-(5.38) we can use

shift due to a parallel sided slab

z= pr

1_£l

...(5.37)

In equation-(5.37) ifA5 is positive then shift is taken toward the slab and ifit comes negative then it is taken away from the slab. With a careful analysis of this formula you can get equation(5.35) and equation-(5.36) by properly substituting the values

of p, and P2 in it.

...(5.41)

A =

...(5.36)

box

...(5.39)

(as for small 0, sin 0 = 0)

z

Thus in equation-(5.41) for small angles we can write sin (z- r) = (z-r) and cos r = 1, we get A = ti

i-i

Geometrical Optics 233

5.10.11 LateralDisplacementofaLightRay due to Refraction 5.10.12 Concept ofReflection bya Thick Mirror by Multiple Glass Slabs

Aplane mirror isathin glass slab polished atone ofitssurfaces

When several parallel sided glass slabs are placed adjoining to due to which the other side becomes reflecting. Ifthe glass each other then due to refraction while passing through these used in making mirror is thick then it also shifts the image before

slabs thelight rayisdisplaced laterally which can be calculated

reflection from therear polished surface aswell asafter reflection

bysummingup the individual displacement ofthe light which and before coming out in air. Light passes through the glass of occurs when the ray passes through the slabs independently- mirror twice as shown in figure-5.135{a). Figure-5.135(b) shows as shown in figure-5.134. athick plane mirror ofglass thickness /placed in front ofapoint object O. A light rayfrom Ofirst gets refracted from the front

glass face

then gets reflected from mirror A/(polished face)

and after reflection again gets refracted from the front face S^ before coming out in air and finally produces the image L. Here /] and/j are the intermediate images produced by the light ray.

(a)

Figure 5.134

Asshown infigure-5.134 when a light rayisincident from airon

the surface ofthe first slab and ifonly this slab ispresent inair,

(h + t/ix)

the light will come out in air along the dotted lineL, as shown in

figure and the displacement ofthe light would have been Aj and ifadjoining next slab isplaced at a negligible separation with first slab then it enters into it and comes out along the dotted lineLj and the lateral displacement ofthe light ray would have been A, +A^. In the same manner ifthird ormore slabs are (b) Figure 5.135 placed all the lateral displacements produced by each slab independently for the angle ofincidence i are added together. If r,, r^, are the angles which light rayismaking with thenormal Final position ofimage produced can be directly calculated by in each ofthese slabs then the total lateral displacement oflight considering the shift inrear surfece as observed by the observer. Due tothickness ofglass the polished face appear to be at an ray is given by

apparent depth r/jiasshown in figure due towhich thedistance between object and this newposition M' of the mirror will be

Ay.— A| + A2 + Aj +...

(h +t/\x) and now the image will be produced exactly atthe same distance behind the mirror. So the separation between object

_ f.sin(/-rj) r.sin(f-r2)-_ r.sin(f-r3) =>

^^+ ^ ^+^

cosr,

C0S/'2

(5.42)

COSr3

Fornearnormal incidence oflight this equation-(5.42) will reduce to the equation given below

Aj, /,/• 1— +v 1—+V

1—-

.

F3.

+

.(5.43)

and image can be given as

Lor2\h^-

...(5.44)

Students are advised to verify the separation given in equation-(5.44) using calculations in stepped manner byfinding the locations ofimages /j and then step by step.

J2^4

Geometrical Optics

.

# Illustrative Example 5.27

Considerthe situation shownin figiire-5.136. Aplanemirror is

fixed at a height h above the bottom of a beaker containing water ofrefractive index m uptoa height hj.Find theposition

Beaker

ofthe image ofthe bottom formed bythe mirror. Glass

Coin

Figure 5.137

^

h

o

+ t2 f,1

I

f

3^ +10

= 10 1—

l

^2^

4j

2)

[

1--

3j

Hencethe coin appearsto raised by 5.8 cm. Now the apparent depth of the coin is 20 - 5.8 = 14.2cm. Figure 5.136 Solution

(b) As is evident from figure CD and EF are the refracting surfaces. At surface CD, the ray passes from glass to water and at surface EF, the ray passes from water to air.

Due to water in beaker the bottom of beaker appears to be

shifted upward and its apparent depth is given as h^lp. Thus the distanceof refracted image of bottomfromthe plane

mirror is/? -/ij + hj\i. Plane mirror produces the image atthe same distance behind it so final image of bottom after reflection

from the plane mirror is produced at a distance h-h^+ h^l\x

3/2

Case-{i)

4/3

Critical angle at this surface is given by

C, = sin-' (8/9) orC, =62M5' Case-(ii) Similarly,

above the mirror.

# Illustrative Example 5.28

C2 = sin-'(3/4)orC2 =48°36'.

A rectangular glassblock of thickness 10 cm and refractive As wesee that criticalangleis smallerin case(ii),total internal index1.5placed over a smallcoin. Abeakerisfilled withwater reflection occurs at the upper surface DF earlier as the eye is of refractive index 4/3 to a height of 10 cm and is placed over the glass block.

moved away from the normal.

(a) Findtheapparent position ofthe object when it isviewed

# Illustrative Example 5.29

at near normal incidence.

(b) iftheeye isslowly moved away from thenormal ata certain position, the object is found to disappear, duetototal internal

Aglass slab ofthickness 3 cmandrefractive index 1.5 isplaced

reflection. At what surface does this happen and why?

a point object be placed if it is to imageon to itself?

Solution

Solution

Let ABCD be a glass block, (p = 1.5) placed over a coin as shown in figure-5.137. Let a beaker containing water upto a

Theglassslabandthe concave mirrorare shown infigure-5.138.

height 10 cm placed over glass slab.

(a) As the ray from coin passes from glass slab to water it movesawayfrom the normal as it enters a rarer medium from a denser medium. Similarly at surface EF, the ray again moves awayfrom the normal. Whenviewed fromN, the coin appears

to beatly Apparent shift in multiple slabs is given by

in front of a concave mirror of focal length 20 cm. Where should

Let the distance of the object from the mirror be x. The slab causes a shift in position of object which is given as s- t

i-i

= 1 cm

The direction of shift is towards the concave mirror so the

apparentdistance ofthe object from the mirror is (a: -1) cm.

Geometrical Optics

235;

To produce the image onitselfthereflected rays must retrace Thus equivalent focal length ofthe combination mirror is given the path of incident rays so the object should appear at the as centre ofcurvature ofthe mirror.

/•

= 2

Slab

60 ~ 60

feq 30

Icm

0

1P

Je O

P

As image formed by the mirror liquid system coincides with

the source, the location ofobject isat 2f^ •R-

60

^

.

w= 2/ =

Figure 5.138 60

Thus we use

x-\ =2/= 40cm

According to the given condition, — + 30 is the distance of [I

theimage formed bythemirror itself, thus using mirror formula X = 41 cm from the mirror

we have

11-1

# Illustrative Example 5.30

V

A concave mirrorhasthe form of a hemisphere witha radiusof R = 60 cm. A thin layer of an unknown transparent liquid is pouredjnto the mirror. Themirror-liquid system forms onereal image and anotherrealimageis formed bymirroralone ofthe source in a certain position. Image produced bycombination coincides with thesource andthatproduced bymirror alone is

located at a distance of / = 30cm from the source away from mirror. Find therefractive index p ofthe liquid.

1 =>

u

f

Ixu

1

-60

30

—I

V

''

60 v =

60

p-2

Image distance from the mirror is

60

2-p

, thus form the already

obtained condition we use 60

60

— + 30 = -

Solution

=>

\i 2-p |i^+ 2p-4=0

For concave mirror with unknown liquid, equivalent focal

^--l±V5

length of the combinedmirror is given as

J

2^1

•feq

fi

fhi

Where

P-i

p=- I + -v/s (as pcannot be negative) # Illustrative Example 5.31

Inprevious illustrations ifimage formed bythemirror coincides with the source and that produced by the combination is produced at a distance 30cm from thesource away from mirror then findthe refractive indexof the liquid.

60

A

Solution

Mirror produces its image on source when the source is located

at the centerofcurvature thussourceposition mustbe at 60cm from the pole ofmirror.

Now we use mirror formula for calculation of image distance Figure 5.139

for mirror liquid combination

1 +1 = ±

and focal length ofmirror is

^

"

fe

60

/«^Y=30cm

1 J_ V

-60

-iL 30

Geometrical Optics

236

60 v =

5

60

J(\0'+h^) _J_

1-2^1

1.5

According to the given condition we use 100+ /i^

- ^ +30 =60

25 +h^ ~ 9

2^-l

:=>

16

h - 8.45 cm

|i=1.5

# Illustrative Example 5.33

# Illustrative Example 5.32

A person looking through a telescope Tjust sees the points

Asmallobject isplaced 20 cmin front ofa block ofglass 10cm

on the rim at the bottom of a cylindrical vesselwhen the vessel

thick and its ferther side silvered. The image is formed 23.2 cm behind the silvered face. Find the refractive index ofglass.

is empty. When the vessel is completely filled with a liquid (jo, = 1.5), he observes a mark at the centre B, of the bottom without moving thetelescope or the vessel. What is the height

Solution

of the vessel if the diameter of its cross section is 10cm.

In figure-5.141 ABCDht theblock ofglass whose side

is

silvered as shown.

Solution

A

B'

B

In the figure-5.140 shown if we consider the position of telescope is at 7" when the point>1 is visible on the rim. Now the same arrangement and position of telescope due to refraction of the ray at point D on water surface.

C

D

when the vessel is filled with a liquid, point B is observed at

C

I*—-V—♦p—-H

(10-.r)

K

20cm

4*— 10cm —*k

23.2cm

-H

Figure 5.141

Here in figure we use sin I =

BC

BC

BD

^[(SC)'+(CZ))']

Suppose it appears that the image isformed due tothe reflection at image ofsilvered face 5'C'and ifthe apparent depth ofglass block is X cm then we use

5

20+.v = 23.2 + (10-x)

sin I =

.•i:=5.6cm.

Again

ZNDT^ZADC^Zr

sin r =

AC

AC

AD

^l(ACf+(CD)-]

As we know the refractive index is the ratio of real depth and apparent depth, we have real depth P-'-

10 sm r =

Vao'+A')

apparentdepth

10 = — = 1.51.

6.6

# Illustrative Example 5.34

Now using Snell's law, we have

py sin / =

sinr

A light rayfrom air is incident ona glassplate of thickness t and refractive index p at an angle of incidence equal to the

angle of total internal rellection of glass. Compute the displacement ofthe rayduetothis plate in terms ofthickness and refractive index ofglass p. Solution

Figure-5.142 shows the ray diagram of the light ray passing through the slab. Here the angle of incidenceis given as '

3cm

5cm

Figure 5.140

sin i = — (As i is equal to critical angle) 4

1Geometrical Optics

237

A J

IrC"v

1 t

I i

r 1

\

«

Figure 5.143

v/

Solution Figure 5.142

BySnell's law,

The image formation of the dust particle is shown in figure-5.144

sin;

sinr

= 1^

1

sin/ sm r

^\

In figure wecan seethat displacement of the light ray is given

f

\v'/

as

CE-x = BCsin (i-r) In figure we also have Figure 5.144

BF cos r =

Here we first consider the image formed byconcave mirror.

BC

For which weuse/? = - 40 cmand w= - 5 cmfor using in the

t

BC =

mirror formula, whichgives

cos r

I x

sin (C-r)

=

II

cosr

V

_2

t

[sin Ccos r-cos C sin r]

-40

cosr

= / sin C

, I

sin/- cosC ; X sin C

cos/-

R

I_ -5

40

40 v=—cm = 6.67cm

=>

D

As Vis positive, the image F, is formed below the mirror (virtual). The reflected rays are refracted at water surface. The

V(n'+i) 1-

depth ofpoint Pj from thewater surface is 6.67 + 5.00= 11.67. Due to presence ofwater, theimage P, will beshifted up and the final position of image is given byitsapparent depth given as

h

Image position = ~ ~

# Illustrative Example 5.35

11.67

=8-77 cm

Thus final image is formed 8.77cm below the water surface.

A concave mirror ofradius40cm lies on a horizontal table and

water in filled in it upto a height of 5.0cm as shown in figure-5.143. Asmali dust particle floats on the water surface at a point P vertically above the point of contact of the mirror with the table. Locate the image of the dust particle as seen from a point directly above it. The retractive index of water is 1.33.

Web Reference at w\v\v.phv$icsgalaxv.coiTi

Age Group - High School Physics | Age 17-19 Years Section-OPTICS

Topic- Geomciric.il Optics II - Refraction of Light Module Number - 1 to 8 and 25 to 32

^ Geometrical

|238 Practice Exercise 5.3

opposite to him. He sees thatthereflected andrefracted rays

(i)

comefrom the samepoint which is the centreof the canal. If the 17ft mark and the surveyor's eye are both 6ft above the waterlevel, estimate the widthofthe canal,assuming that the

A converging light beam is incident ontwo glass slabs

of different materials placed in contact with each others (as shown in figure-5.145.) Where willthe raysfinally converge?

refractive index ofthe water is 4/3. Zero mark is at the bottom of the canal. [16 feet]

H= 3/2

H= 2

(vi) A ray oflightis incident on a parallelslabof thickness i and refractive index p. If the angle of incidence is 0 then for small 0 show that the lateral displacement of light ray will be Air

r0(p-l) 6cm

A =

4cm

— 14cm

Figure 5.145

(vii) How muchwatershould befilled in a container ofheight 21 cmsothat it willappearhalffilled whenviewed alongnormal

[8 cm to the right of second slab]

to water surface. Take refractive index ofwater

(ii)

Find the apparent depth of an object O placed at the

bottom ofa beaker as shown in which two layers oftransparent liquids are filled:

= 4/3.

[12 cm] '

(viii) A glassplatehas a thickness t andrefractive indexp. A light ray is allowed to incident on the plate from air. Findat whatangleofincidence willthe raysrefracted andreflected by the plate be perpendicular to each other ? For this angle of incidence, calculate the lateral displacement of the ray. Kn -1)

h, - 25 cm

H, = 1.5

A, = 15 cm

1^2= 2.5

Figure 5.146 [22.67 cm]

(ix) A man standingon the edgeof a swimming pool looks at a stone lying on the bottom. The depth of the swimming pool is equal toh.Atwhat distance from thesurface ofwater is the image of the stone formed if the line of vision makes an angle 0 with the normal to the surface ? H Acos 0

(H^-sin^ 0)^'

1

(iii) A point object is placed33 cm from a convex mirror of curvature radius 40 cm. A glass plate of thickness 6 cm and index2.0 is placed between the object andthe mirror, closer to

(x)

the mirror. Find the distance offinal image from the object ?

embedded into the river stands vertically with Im of it above

[42 cm]

the water surface. If the angle of inclination of sun above the horizon is 30°, calculate the length of the post on the bottom

(iv) A light ray falling at 60° angle with the surfece ofa glass slab of thickness Im and is refracted at angle 75° with the surface. Calculate the time taken by the light to cross the slab.

In a river 2m deep, a water level measuring post

surface ofthe river, (p for water = 4/3) [3.44 m]

[|xlO"® S]

(xi) A concave mirror of radius of curvature one metre is placed at the bottom of a tank of water. The mirror forms an

(v)

80cm and 40 cm ofthe water in the tank.

A surveyor on one bank of a canal observes the image

image ofthe sun when it is directly overhead. Calculate the distance of the images from the mirror for different depths,

of the 4 inch mark and 17 ft mark on a vertical staff, which is

partially immersed inthewaterand heldagainstthebankdirectly

[47.5 cm, 57.5 cm]

;Geometrical: Optics

.

2391

(xii) A small object is kept at the centre of bottom of a

undeviated attheinterface. Always for image formation we can cylindrical beaker of diameter 6 cm and height 4cm filled consider one rayalong the optical axis and one other rayin completely with water (ti=4/3). Consider thelight ray from an surrounding toproduce image. Infigure-5.147(b) we can see all

object leaving the beaker through a comer. Ifthis ray and the paraxial rays from object after refraction meet at the point I rayalong the axis ofbeaker isused tolocate the image, find the which is the image produced due torefraction oflight at the apparent depth in this case.

spherical surfece.

[2.25 cm]

Inabove situation explained infigure-5.147, image produced is real image as refracted rays are converging. Depending upon the refractive indices ofthe two media refracted rays may be

5A1 Refraction of Light by Spherical Surfaces

diverging as shown infigure-5.148(a) and(b)inwhich thefinal image produced is virtual.

Figure-5.147(a) shows a spherical interfece separating two different media 1and 2with refractive indices p, and Here C is the center of curvature of the spherical surface, R is the radius ofcurvature ofthesurface, P isthecentral point onthe

Medium-l

Medium-2

(Denser)

(Rarer)

Mi

M2

surface called optic center of the surface and line XX' is

considered asthe Optical Axis' for thegiven setup. Ingeneral the optical axis for the refraction surface can also be called as

principal axis, theterm which we generally usefor spherical mirrors and lenses.

Medium-l

Medium-2

(Rarer)

(Denser)

Ml

(a)

M2 Medium-l

Medmm-2

„ (Denser)

Object

(Rarer)

(a)

Medium-l

Mediumt2 (Denser)

(Rarer)

(b)

Figure 5.148

Note: Infigure-5.148 when object isplaced indenser medium and light rays are refracted toararer medium ftie imageproduced can never be real, it will always be virtual because refracted

rays insuch a case will always bediverging butwhen object is placed inrarer medium asshown in figure-5.147 and light rays are refracted to denser medium imageproduced can bereal or (b)

virtual depending upon the refracted rays are converging or

Figure 5.147

diverging. Infigure-5.147 refracted rays shown areconverging. Students are advised to thinkand draw ray diagrams for the

If we consider a light ray from an object placed in medium-1 situation for virtual image formation in this case on their own incident on the spherical interface at an angle of incidence i carefully. gets refracted atan angle rand therefracted rayinother medium meets theoptical axisatpointI. We canalsoconsider a ray of Similarsituation canalsobeanalysed bystudents for a concave light from the object along the optical axis which passes surfacefeeingthe objectas shown in figure-5.149.

Geometrical Optics

240

UsingSnell's law at point Awe have

p, sin / =

Medium-2

Medium-I

(Rarer)

' ;(Denser)

Forparaxial rayswecan usesmall angles sowe have

p|/=p/

I

Object

...(5.45)

I

C

In the geometry of figure shown wecan relatedifferent angles from properties of triangles as ...(5.46) (a)

and

/~0 + a

...(5.47)

Combining above three equations we get Mcdiuni-1

Mcdium-2

(Rarer)

(Denser)

.p,(0+ a)-p2(ct-^)

p,0+P2=(l^2~F|)aObject '

C

...(5.48)

From tlgure-5.148 for small angles we can use the length of normal/JyV/on optical axis as AM= h^i/Q = J?a = v(^ (b)

... (5.49)

If we use co-ordinate sign convention for this situation as

Figure 5.149

explained in article-5.8.8, we can consider P as origin of ft

co-oridinate systemall distances toward left as negativeand all Note: In tlgure-5.149(b) when object is placed inrarer medium distances toward right as positive so here u will be taken andlight rays arerefracted toa denser medium image produced negative, v and R are taken positive. Sowith sign convention can never be real, it will always be virtual because refracted the angles 0, a and ^ are given byequation-(5.49) as rays in such a case will always be diverging whereas in ^ h h , , h tlgure-5.149(a) when object is placed in denser medium and 0=—;«=•;: ana -u R V lightrays arerefracted to rarermedium image produced canbe realor virtual depending upon the refracted raysare converging Substituting these values in equation-(5.48) we get or diverging. In this tlgure-5.149(a) refractedrays shown are F2 "1^1 Bconverging. Studentsare advisedto think and drawray diagrams ...(5.50) V n R for virtual image in this case on their own caretully. Above equation-(5.50) is called ^Refraction Formula for 5.11.1 Analysisof Image formation by Spherical Surfaces SphericalSurfaces'andit is used for finding theexact location ofimageproduced dueto refraction oflight through a spherical Figure-5.150 shows an object placed on the leftof a spherical surface. surface from which a light ray is consideredwhich incidentson the surfaceat pointAand after refraction meetat point/ on the # Illustrative Example 5.36 optical axis as shown.

Mcdium-l

Mediuin-2

(Rarer)

(Denser)

Figure-5.151 shows a glass sphere of radius 10 cm. Along its diameterfrom one side a parallel beam of paraxial rays incident on it. What should be the refractive index ofglass so

that after refraction all rays will converge at opposite end B. Obicci

R = i Ocm

Fiaure 5.150

Ficure 5.151

IG^eometrlc^t Optics

-241

Solution ' 1--

I

For refraction formula, we use

V

2x40

10

J__

u =00

^

v=2R = + 2Q cm

1_

V~ 90 20 ~ 80 v=-80cm

^=+10cm

Thus final image is seen by observerat a distance 80cm from the poleof curved surface and it is a real image. # Illustrative Example 5.38

Using refraction formula, we have 1^2

A glass rod has spherical ends as shown in figure-5.153. The refractive index of glass is p. The object O is at a distance 2R from the surface of larger radius of curvature. The distance

Hi _ H2 -Hi R

=>

_H

H-1

between apexes ofends is 3i?. Findthedistance ofimage formed

20

10

ofthe point objectfrom right hand vertex. What is the condition to be satisfied ifthe final image obtained is to be real?

.

ji = 2}i—2 |r=2

Glass

# Illustrative Example 5.37

r

h—2^—^

3^

Figure 5.153

Figure-5.152 showsaglass hemisphere Mofp= —and radius 10 cm. A point object 0 is placed at a distance 20 cm behind the flat face which is viewed by an observer from the curved side.

Solution

Find location of final image after two refractions as seen by

For first refraction at surface A, we use u = -2R'm refraction

observer.

formula as

l^_Hi_ ^ H2-H1 v'

u

R,

iL+J__(H-l) v' h = 20cm

2R

...(5.51)

R

Solving for v', we get 2pjg

i? = ID cm

V

-

Figure 5.152

...(5.52)

(2h-3)

For refraction at surface B, we use u = -\'3R-

Solution

2p^

(2p-3)J

m

refraction formula as

After first refraction at flat surface image is produced at a distance given as 3

H/2 = —^20 =30cm 2

1 -

For second refraction at spherical surface, for refraction formula

V

+

H2

H]

V

u

H2'-Hi

1-H {RI2)

H

2\iR

...(5.53)

2p-3

we use

w= + 40 cm; i?= + 10cm; ^X•^ =

3

Pj ^

Substituting values in refraction formula, we get _i H2 ~ Hi V

u

R

H= -

3R-

2\xR

2p-3 Solving equation-(5.53) for v, we get

(9-4p)j? a0n-9)(ti-2)

Geometrical Optics

1242

The imagewill be real if abovevalue of v is positivefor which finallyrefracted rayswillbe converging, so condition forreal

and

1--

cos r =

2 n2

image is

Substituting these values in equation-(5.54), we get

(9-4p)i? (10p-9)(^i-2)

>0

On simplifying wegetthat it happens when refractive indexof

Sin— =

2

—

R'

1-

'-F

glass is between 2 and 9/4. Illustrative Example 5,39

p/2

sin— =

p/?

A ray oflight passesthrough a transparent sphereof refractive index p and radius R. Ifb is the distancebetweenthe incident rayanda paralleldiameter ofthe sphere, show thatthe angleof deviation 9 is given by the expression

^

2

ixR' b

=i>

sin —= (/j/p7?^) [(p-

sm— =

sm— =

p/?^ [(p2^2_^2)l/2_(^2_£,2y/2]^

~ {R^ - b^y^] # Illustrative Example 5.40

Solution

The ray diagram of the light ray passing through the sphere is

A parallel beam of light travelling in water (refractive index = 4/3) is refracted by a spherical air bubble of radius 2 mm,

shown in figure-5.154

situated in water. Assuming the light rays to be parallel

(a) Find the positionof the imagedue to refractionat the first surface and the position ofthe final image.

(b) Draw a ray diagram showing the positions of both the images. Solution

Figure 5.154

(a) For refraction at first surface, we use ,

From above figure, the angle of deviation is given as

Mi = (4/3); w= oo andi?j = 2mm .

0 = (/ _ r) + (/ - r) = 2/ - 2r = 2(/ - r)

So the position of image due to refraction at first surface is given by refraction formula as Taking sine on both sides of above equation, we get

V

1 Vj

sin— = sin (i-r) 2

sin— = sin I cos r - cos i sm r

...(5.54)

Solvingwe get

w

(4/3) 00

1^2-i^i R

1-(4/3) 2

v, =-6mm.

2

Hence the image is formed at a distance of 6mm to the left of

Also from the figure we have

first surface. The ray diagram is shown in figure-5.155.

b

sini - ^

cos / =

'-F

By Snell's law we have sini smr =

p/?

Figure 5.155

fGebrfieWMr Optics

;243J

(b) For refraction at second surface we use

magnification gives the orientation ofimage whether image formed is on the same side ofoptical axis as that ofobject or

—(6 + 4)=- 10 mm,

1^1 ^ ^'^2 ~ (4/3)

^

erected (positive) or itison the other side ofobject or inverted

and R2 =-2 mm I (4/3)-l

(negative)

V (-10) " (-2)

,• . . .

• ;

5.11.3 Longitudinal Magnification oflmage

Solving we get V= - 5 mm

Figure-5.157 shows an object ofwidth dc which is placed on

Sothe final imageis produced at a distance of5 mm to the left optical axis of the refracting surface Sal a. distance x from the

optic center P. After refraction its image / is produced at a

of the second surface.

distance^ from the optic center which is ofwidth dy as shown 5.11.2 LateralMagnification ofImage byRefraction

in figure.

Figure-5.156 shows an object ofheight habove the optical axis

ofwhich image produced is ofheight h'. As we have already studied that for areal object ifimage produced isreal, it will be

inverted or on the other side ofprincipal axis. In the figure we have produced image bytwo paraxial light rays, one which is passirig throughcenterofcurvature ofthe interface which will get refracted undeviated and other which is incident on the

optic center ofthe surface at an angle ofincidence iwhich gets refractedat an angle r. Figure 5.l'57

In above figure the distances ofobject and image from the optic center ofthe surface are related by the refraction formula given as

. • 'iiz. _ Hi' = V

.

w

••

.

R

,

Here we use M=—X, =+7? and v=+y which gives if^ ^

^ M'2 "1^1 X

Figure 5.J56

The angles iand r are very small angles as light rays are paraxial

R

-^:-(fy-~dx^O y

h

~h'

~u

V

...(5.57)

Differentiating the above equation we get

so these are related by Snell's law as ...(5.55)

-

X

. ...(5.58)-

From above equation-(5.58), we get longitudinal magnification as

Bygeometryin this figure wecan use/= — andA-= —so we get

m, =

dx

II

V

Here we iised image height h' and object distance ii negative according to the sign convention.

_ ^|V

As already discussed that above relation given by equation-(5.59) is only valid for paraxial rays. Here negative sign shows lateraiinversion ofthe image.

5.11.4 Effect ofmotion ofObject or Refracting Surface on

Thus lateral magnification ofimage canbegiven as _

...(5.59)

Image ...(5.56)

Similar tothecase we've discussed incase ofspherical mirrors, here also for small velocities ofthe object orrefracting surfaces, As already discussed that above relation given by the velocity magnification along and perpendicular tothe optical equation-(5.56) isonlyvalid for paraxial rays. Thesign oflateral axis can be given by the expressions oflateral and longitudinal h

P2"

GeojTieiri^ljJipt^ magnifications. Figure-5.158 shows an object moving with some velocity at a direction shown in such a way that its velocity component along theoptical axis with respect totherefiracting

_ J__J

!_

V ~ 10 20 ~ 20

v=+20cm surface is v^.aiong^^^ direction normal to the optical axis is then the image velocity components can be directly Longitudinal magnification for refraction is given as V

o-normal

given as

3l2x{20y ^/•normal

^ ' ^o-iwrmal

^i-along

' ^o-ahng

m

=

m

=

ix{i5y 1.5x400

8

225 Vn-normal

Jmage

V:-along ?!

Image size

—xl = -r mm.

3

3

^ Vg-along Object V:-normal

Lateral magnification m

# Illustrative Example 5.42

A long cylindrical tube containing water is closed by an equiconvex lens offocal length 10 cmin air. Apoint source is placed along the axis of the tube outside it at a distance of

HiV

Longitudinal magnification

, P2"

21 cm from the lens. Locate the final image of the source. Refractive index of the material of the lens =1.5 and that of water = 1.33

Figure 5.158

Glass

Air # Illustrative Example 5.41

^

-Water-

Source Hi

Figure-5.159 shows a smallobject Mof length 1mmwhich lies along a diametrical lineof a glass sphereof radius 10 cm and

h

"21 cm-

Figure 5.160

3

p = —which is viewed byan observer asshown. Find the size

Solution

of object as seen by the observer.

We consider first refraction at air-glass interface, using refraction formula we get ^ Vi

u

R

Substituting m=- 21 cm; w,= 1;/K2 ~ ^ i? = 10 cm

/? = + i?,weget

!:£ J_ _ M

...(5.60)

IX" R

Figure 5.159

The image formed bythe first surface will act as the object for Solution

the second surface and the radius of curvature of the second surface will be taken as - i? for second refi-action as the lens is

For refi-action at glass-air interface, we use

u = + 15cm; R= + 10cm; p, = 3/2 and Pj" ^ Substituting values in refi-action formula, we get V

^

M

R

1 3/2 ^1-3/2 V 15 ~ 10

equiconvex. For glass-water interface, weuserefi-action formula with Wj= 1.5 and^2= 1.33 weget 1.33

1.5

1.33-1.5

0.17

-R

R

...(5.61)

Adding equations-(5.60) and (5.61), we get 1.33

1

V "^21

0.67

...(5.62)

i'Geometrical 6'ptics 245

The focal length ofthe lens in air is 10 cm. So by using lens sin

maker's formula we have 1

^1 = (1.5-1) 1)

10 or

^(1-cos^r) 1/2

I

1-^

7!2

'y 1 .7!'^7!

73x0.1 I

M 2R

i?=2.5cm

4R

^ = —-—7(4-3) =0.086 metre Total height oflight ray above the plane mirror is

...(5.63)

Substituting the value of^ in equation-(5.62) and solving it, we get

H= 7? + /i = 0.1 + 0.086 = 0.186 metre From figure we have X = 7? cot /

V = 70 cm

Thusfinal image is formed 70cminside thetube andit isreal.

also we use

(7 = 7! +a:= 7?(1 + coti)

=>

(7=7!(l+cot2r) =7!ri +^^l V s\n2r)

# Illustrative Example 5.43

2cos^ /•-!

(7=7! 1+2 sin r cosr

A cylindrical glassrod of radius0.1 m and refractive index

/

lies on horizontal plane mirror. Ahorizontal ray oflight moving perpendicular to the axis of the rod is incident on it. At what

uV(H?74)

leaves the rod at aheight of0.1 mabove the plane mirror? At what distance a second similar rod, parallel to the first, be placed on the mirror, such that the emergent ray from the second rod is in linewith the incident rayonthe firstrod ?

(2n^/4)-]

(7= 7! 1 +

height from the mirror should the ray be incident so that it d= R 1 +

1/2

75/2

= 0.1

l+-j= 75

.

Distance of the second rod from the first rod is

Solution

2(7 =

.

75

= 0.3155 m.

The situation and raydiagram isshown infigure-5.161. #Illustrative Example 5.44

\

\

I I'' / H

\

-v ^

Aquarter cylinder ofradius Rand refractive index 1.5 is placed

\ fl

on a table. Apoint object P is kept at a distance mR from it.

To

Find the value ofwfor which aray from Pwill emerge parallel to the table as shown in figure-5.162.

I*-/?

'

i«—^

Id

4

Figure 5.161

From figure we have i = 2r and we have

h

\*-—mR

AO sin i

Figure 5.162 Solution

= R sin i

= 7?sin 2r=2R sin rcosr By Snail's law we have

...(5.64)

To solve this problem, we use the principle ofreversibility of light.

sini

p. -

H

sin2r

—

= 2 cos r

To analyze the situation we consider that the light isincident

sinr

from therightparallel tothehorizontal surface oftable asshown

From equation-(5.64) we have

and we find where it converges on the table after emerging from the plane surface. This must be point P ifpath oflight ray

h-2R sin r cosr

is reversed as specified in the question.

Geornetric^;jOpticsj

246

diameter ofthesphere from the sideon which it lies. How far fromthe surfacewill it appear.Refractiveindex ofglass is 1.5. [2.5 cm behind the surface] •

C

' K

B

\A

(iii) Arayoflight refracted through a sphere, whose material has a refractive index p in such a waythat it passes through

♦K- R -H

mR

Figure 5.163

the extremities oftwo radii which make an angle 0 with each other. Prove that if a is the deviation of the ray caused by its passage through sphere

Usingrefraction formula for curved surface, \ye have V

u

R

cos(0 - a) = p cos 0/2

According to the given problem we use

u=oo;

1,5; Pj = 1and^ = +i?

(iv) A ray is incident on a glass sphere as shown in figure-5.165. The opposite surface of the sphere is partially

Substitutingthe values in refraction formula, we get 1.5

1

1.5-1

V

CO

R

:=>

silvered.If the net deviationofthe ray transmittedat the partially silvered surface is 1/3^ ofthe net deviation suffered by the ray,

,

V=3R ^

When the lightrayincident ontheplane surfece, it gets refracted and produces an imageat a distance 2R/ii= 2R/1.5 = 4R/3.

reflected at the partiallysilvered surface (afteremerging outof the sphere), findthe refractive indexof the sphere. Partially silvered

From the given condition ofwe can use 4R mR= —

4m =

Figure 5.165

Web Reference at www.phvsicsgalaxv.com

[S ]

Age Group - High School Physics ] Age 17-19 Years Section - OPTICS

(v)

Topic- Geometrical Optics II- Refraction of Light •

near normal incidence. Show that the image in terms of the

A parallel incidentbeam fallson a solidglass sphereat

Module Number -16 to 24 and 33

index of refraction p and the sphereof radiusR is givenby 2(p-l)

Practice Exercise 5.4

(vi) (1) A spherical surface S separates two media I and 2 as shown in figure-5.164. Find where an object 0 is placed in medium-I so that the light rays from object after refraction becomes parallel to optic axis of this system. •s

h = 4/3

/

.

^

~

H2 = 3/2

A hollow sphere of glass of refractive index p, has a

small mark on it interior surface which is observed from point

outsidethe sphere on the side opposite the centre. The inner cavity is concentric withthe external surface andthe thickness of glass is everywhere equalto the radius ofthe inner surface. Prove that the mark will appear nearer than it really is by a

distance(p - 1)i?/(3p- 1), whereR is the radius of the inner surface.

u - 240 cm Mark

Figure 5.164 [ 240 cm]

(li)

A glass sphere of radius 5cm has a small bubble at a

distance 2cm from its centre. The bubble is viewed along a

iR

Figure 5.166

^eometriparpptics' "' •247

(vii) Atransparent sphere ofradius Rhas acavity ofradius R

2 as shown in figure-5.167. Find the refractive index ofthe Fish

sphere ifaparallel beam oflight falling on left surface focuses at point P.

Figure 5.169

[3.84 mm/s]

(xii) Figure-5.170 shows an irregular block of material of Figure 5.167

refractive index ^/2. Arayoflight strikes the face AB as shown in figure. After refraction itis incident on aspherical surface CD ofradius ofcurvature 0.4m and enter amedium ofrefractive

index 1.514 to meet PQ at E. Find the distance CDupto two places ofdecimal.

(viil) Alight ray incident at apoint on the surface ofaglass sphere ofp = Vs atan angle ofincidence 60°. Itisreflected and refracted at the farther surface of sphere. Find the angle between reflected and refracted rayat this surface. [90°] u= i.514

(ix)

E

Q

Aglass sphere (p = 1.5) with a radius of15.0 cmhasa

tiny air bubble 5 cm above its centre. The sphere is viewed looking down along the extended radius containing the bubble. What isthe apparent depth ofthebubble below thesurfece of the sphere ?

Figure 5.170

[6.06 m]

[8.57 cm from top]

(x) Figure-5.168 shows aglass hemisphere placed on awhite horizontal sheet. Avertical paraxial light beam ofdiameter d incident on the curved surface ofhemisphere asshown. Find 5.12 Total Internal Reflection thediameter ofthelight spot formed onsheet after refraction.

Figure-5.171(a) shows alight ray refracting through an interface oftwo media. The incident light ray is travelling in denser medium and after refraction itbends a\yay from normal as we know and this figure also shows the reflected ray which is always present

in all cases ofincidence and it is reflected at the same angle of white sheet

Figure 5.168

incidence. If we increase the angle of incidence as shown in figure-5.I7I(b), theangle ofrefraction also increases. As the angle ofrefraction is more than the angle of incidence it will

approach to 90° at some specific value ofias shown. This angle ofincidence atwhich the refracted ray grazes along the interfece ofthe two media is called 'Critical Angle' for this pair oftwo media. Figure-5.171 (c) shows alight ray incident on the medium

(xi)

Figure shows afish bowl ofradius 10 cm in which along boundaryatan angle slightlygreater than critical angle at which

a diametrical line a fish Fis moving atspeed 2mm/sec. Find the no refracted rayexist and only reflected ray exist. This iscalled

speed offish as observed by an observer from outside along

Total Internal Reflection' and it occurs when a light ray

same line when fish is at a distance 5 cm from the centre of

travelling in adenser medium incident on a boundary ofararer medium atan angle greater than the critical angle.

bowl torightofit as shown in figure-5.169.

Geor^trical O^icsJ^ 1248

5.12.1 Refraction ofLightRaysfrom a Source in a Denser Medium to Air

Figure-5.172(a) shows apoint isotropic source oflight placed inside water at a depth h below the air-water boundary. The Rarer Medium (|i,)

light rays which incident close to normal as shown will get

• Denser" Medium (1^2)

refracted toairand diverge. Ifwe see aspecific light ray which isincident ontheair-water interface at critical angle will graze

along the interfece and all the rays from source which incident atangle greater than this angle will get totally reflected into water. In this way we can say that light rays from source will emerge out in air only from a circular region as shown by

(a)

perspectiveview in figure-5.172(b). Here we can see that all the light rays from the source on the surface ofaconical region of halfangle 0^ will graze radiallyalong the air-water interface and all light rays from source felling on interface within this conical region will escape to air and all the light rays falling on interface Grazing Refraction

outside this conical region will get totally internally reflected and will not come out in air.

•'

>

r = Atan 6,

water

(b)

Source

Rarer Medium (u.)

(a)

/•= Atan 9r.

Denser Medium (u,)

(c) , Figure 5.171 •"

Source

The critical angle can be given bySnell's law attheboundary as P2 sin 0c

(b)

Figure 5.172 sin U.= ^2

Infigure-5.172(a) inDSOA wecanwrite

0 =sin-'l — >2

For a light ray passing through a denser medium having

r = htan 0,

...(5.65)

refractive index pincident ontheboundary ofair(^3;^= 1) inthe surrounding ofthe medium then critical angle for this medium Equation-(5.65) gives the radius ofthe circle ontheinterface just above the point source oflight from which light rays escape can be given as to air. Byusing the concept ofsolid angle ofthis cone wecan also find the fraction of total light power of the source which

gets refracted to airin this case. The solid angle ofthe cone

IGeqmetrrcai'Optics

shown infigure-5.172(b) with halfangle 6^ is given as n = 2;i(l-cos0^)

=>

n-2jt(l-^l-sin^0^ ) r

=>

'

When a light ray incident on the boundary oftwo media from the side of denser medium in a grazing manner at incidence angle 90° as shown in figure-5.174 then the light ray passes as it is incident on the boundary as the incidence angle is more than critical angle so it gets intemally reflected.

fl = 27r 1- l-'—r (As sincos0= —)

- ' I V It' J

Rarer Medium

At the source all light rays are emitted isotropically in total solid angle An steradian in which total power of light source is distributed uniformly. If we consider the source power is P watt then the light power which is escaping out from water to air is the power within the shaded conical region as shown in figure-5.172(b) which is given as

Denser Medium

Figure 5.174

5.12.3 Refraction by a Transparent Medium of varying

•^to air

^^

Refractive Index

,-„-±

The fraction of light power of source which escapesfrom water

When a light ray incident on the surface of a medium having continuously varying refractive index with distance then the path of light ray is not a straight line. Figure-5.175 shows a

is given as

medium in which we consider that its refractive index increases

^ ^toair 4^ ^

/=

"-7

5.12.2 Cases of Grazing Incidence of Light on a Media Interface

We know that path of a light ray in cases of reflection and refraction is retractable and we have studied that a light ray

travelling in denser medium when incident on a boundary of rarer medium at critical angle will graze along the boundary as shown.in figure-5.173(a). So by the concept of reversibility of light we can state that a light ray travelling in a rarer medium when incident on the boundary ofa denser medium at angle of

with x-coordinate of the position. To understand the path of light shown in this figure we consider an elemental slab of medium at a positionx and having thicknessdx which is shown in figure by a shaded region. When a light ray incident on this slab at an angle 0 then it will slightly bent toward normal as refractive index ofthe medium slightly increases on the right side ofslab. In the same sense as continuously refractive index of the medium is increasing to the right, the light will continuouslybend toward normal to each elemental slab which is resulting in a curved path as shown in the figure.

Glass slab having

incidence approximately 90°whichis calledgrazing incidence of light as-shown in figure-5.173(b), the light will enter in the denser medium at critical angle0^.

where/is an increasing function

Rarer Medium

Grazing Refracted Ray

: . Denser Medium

Figure 5.175

In above figure, insidethe mediumfor all parallel normals,we

can relate theangles cfjj and (jjj by Snell's law at points AandB where refractive indices are pj and pj as given below.

(a)

Grazing Incidence

|j.] sin(90°-(j),)= ^2sin(90°-(l)2)

Rarer Medium

Denser Medium

=>

jUj cos4>, = }i2 cos 42 =>

# Illustrative Example 5.48

1 sin i > — V-

From figure, we have

A rod made of glass (p = 1.5) and of square cross-section is bent into the shape shown in figure-5.182. A parallel beam of light falls perpendicularly on the plane flat surfece^f. Referring to the diagram, d is the width ofa side and R the radius ofinner semi-circle. Find the maximum value of ratio {d/R) so that all

sm I ~

PO

R

OQ

(R + d)

R

1

(R + d)

1.5

iGeometrrcalJDptics

:253:

l.5R>{R + d) 0.5R>d

or

Substituting the values of a and weget d/R

f=sin-' [ff sin {45°-sin-Vw,/ff)}]

178° clockwise

(ii) For the ray to pass through the diagonal face undeviated, theraymust strike normally on theface, i.e..

#Jllustrative Example 5.55

A right angle prism (45°- 90°- 45°)ofrefractive index n hasa

sin I

plate ofrefractive index w,(«, < w) cemented to its diagonal face, The assembly is in air. Aray is incident on AB (figure-5.201)

sin 45°

= 1.352

or

sin / = sin45° ^ (1.352)

or

sin I

1.352

A

= 0.9562

/ = sin-' (0.9562) = 72.9°

-

/

# Illustrative Example 5.56 n

B

In figure-5.203, light refracts from material 1into athin layer of

Figure 5.201

material2, crossesthat layer, and then is incidentat the critical

(i) Calculate the angle ofincidence ziAB for which the ray

angle on the interface between materials 2 and 3.

strikes thediagonal face at thecritical angle. (ii) Assuming«= 1.352,calculatetheangleofincidenceat/45

for which the refracted ray passes through the.diagonal face undeviated. «,= 1.60

Solution Figure 5.203

(i) In the given situation we consider the critical angle for face ACbe0^ Then bySnell'slawwehave sinQ^= {n^/n)

(a) What is the angle 0 ? ...(5.93)

Let the required angle ofincidence be / and angle ofrefraction at faceAB is r asshown in figure-5.202. From figure, or

...(5.94)

p, sin/, = |i, sin/2 = ^^sin 0^, Forinterfaceof medium 1and 2, we use

Forrefraction at/'(using Snell's law), wehave sin / = « sin r

Solution

(a) By Snell's law we have for the two interfaces

r+0^=45°

r = (45°-0^)

(b) If0isdecreased, isthere refraction oflight into material 3?

...(5.95)

(1.6) sine = (1.80)

1.30 1.80

,(\3) (b) If 0 is decreased, then angle of incidence on the second

interfece will decrease from the critical angle. So light refraction will occur into medium-3. Figure 5.202

Geometrical bpticP

[260_^ , # Illustrative Example 5.57

Soco-ordinates of image formed afterreflection from concave

Findthe co-ordinates ofimage ofthe pointobject 'O' formed

mirrorare |fr^175cm,—cm [.

after reflection from concave mirror as shown in figure-5.204

assuming prism to be thin and small in size and ofprism angle 2°. Refractive indexofthe prismmaterialis 3/2. /= 30 cm

'X

(0,0)'

Monochromatic light is incident on a plane interface AB between two media of refractive indices «, and «2 («2 ^ "i)

anangle ofincidence 0as shown infigure-5.206. The angle 0 is infinitesimallygreater than thecritical angle for thetwo media so that total internal reflectiontakes place.Now if a transparent slab DEFG of uniform thickness and refractive index is introducedon the interface(as shown in figure), show that for

•20cm

K5cm-f

Illustrative Example 5.58

Figure 5.204

any value ofWj all light will ultimately be reflected back again intomedium II. Consider twocases (i)«3 < «, and (ii)n2>ny

Solution

Medium III («,) D\

When light rays passes through prism, allincident rays will be

"

1 1

deviated by the deviation angle given as

Medium III ("3)

i

Si

^2

1

IF

G\

J Medium II (wj)

5=(^-lM=(f-l]2° =r=^rad

Si

Asprism isthin,object and image will beinthesame plane as

Figure 5.206

shown in figure-5.205 below. Solution

(i) Inthis case asW3«,wehave

« «2- if"we consider 0^. be the critical angle for medium and ^2 thstiwe have sin 0-=

Figure 5.205

Thus in this casethe shift of image due to prism in y-direction is given as

Shift^=a5=5[y^Q^

It is given that 0 is infinitesimally greater than 0^^. Here we

consider refraction from «2

where

< ^2 ^itd if0^^' be the

critical angle for this interface,then we have sin 0-'= —

c

W2

With the given conditions wehave 0^^' < O^- and so 0 > 0(-.'. Thus the raywillbe totallyreflected backintothe medium n^-

Now thisimage will actas an object for concave mirror. Sofor

(ii) In this case when ^2 ^ "r fti this case we have

mirror formula we use

M= - 25 cm and/= - 30 cm uf

= 150 cm

v =

sin Qr'~ — «2

Butasperthegiven conditions we have 0^'> 0^^. and 0< 0^'so theray incident at 0, willberefracted intomedium ^3. If 0' be the angle of refraction,then by Snell's law, we use

Also,

m=- "

= + 6

sin0'

u

Thus distance of image from principal axis is given as n TZ

^

W3

n

——x6 = — cm 36

sin 0

6

sin0'=sin0x

...(5.96)

{Geometrical Opt^

261

As 0 isslightly greater than G^-we can use , Web Referenceat www.physicsgalaxv.com sin0 =

; Age Group- High School Physics | Age 17-19,Years «2

Section-OPTICS

Topic -Geometrical Optics 11 -Refraction of Light

Thus fromequation-(5.96), wehave

; Module Number - 9 to 15 and 34 to 44

=>

n,

rio

n,

«2

"3

«3

sin0'=—X ^=-!-

Thus the light ray refracted into

will be incident at the interfece

DE at an angle 0' which is slightly greater than the

corresponding critical angle. Thus the ray will be totally

reflected back into

and finally transmitted to n^.

Practice Exercise 5.5

(i)

A glass prism ofangle 72° and indexof reflection 1.66is

immersed in a liquid ofrefractive index 1.33. Find theangle of minimum deviation for aparallel beam in light passing through the prism.

# Illustrative Example 5.59

[22° 22 T

Aprism with prismangle 60° and refractive index -JTfl isgiven.

(11)

In figure-5.208 shownfind the angle of incidence ofthe

Findtheminimum possible angle ofincidence, sothatthelight light ray on faceABof the prism forwhichlight willreach face ray is refracted from the second surface. Also find 8

. max

AC at incidence angle 60°. /I = 90°

Solution

Asweknow thatminimum andmaximum incidence angles are corresponding tothe maximum deviation of light raythrough the prism. Figure-5.207showsthe ray diagram for the case of minimum incidence angle.

Figure 5.208 [90'

(111)

A ray of light is incident on a glass slab at grazing,

incidence. The refractive index of the material of the slab is

given by p,= -s/Cl +y) .Ifthe thickness ofthe slab is d, determine the equation of the trajectoryof the ray inside the slab and the coordinates of the point where the ray exits from the slab. Take Figure 5.207'

the origin to be at the point of entry of the ray.

According to Snell's law, we have b= vl

l^sin/^j„= Jj sin(A-Q A point source of light is placed directly below the

sm i^.^= ^jj (sinA cos C-cosA sin Q sin lmin=^/J

sin60Jl-y-cosbOJy

surface ofa lake at a distance h from the surface. Find the area

on water from which the light will come out from water. TZ/l

(M -1)

L. =30'

(v) A ray of light incident normally on one face of the faces ofa right angledisosceles prism is foundto be totallyreflected.

6^ = i^^ + 90°-A = 30°+ 90°-60° = 60°.

What is the minimiun value ofthe refractive index ofthe maierial

mm

Geometrical OpticsJ

262

of the prism? When the prism is immersed in water,trace the path of the emergentraysfor the same incident ray, indicating values ofemergence angle,

= 4/3)

[1.414, 48°35']

(vi) Figure-5.209 shows a triangular prism ofrefracting angle 90°. A ray oflight incident at face ABat an angle 0 refracts at point 2 with an angle of refraction 90°. (a) What is therefractive indexof the prism in termsof 0 ? (b)What is the maximumvalue that the refractive index can have ? What happens to the light at O ifthe incident angle at Q is (c) increased slightly and (d) decreased slightly?

(ix) A light ray composed of two monochromatic components passes through a trihedral prism with refracting angle A = 60°. Find the angle AS between the components of the ray after it passes through the prism if their respective indices ofrefraction are equal to 1.515 and 1.520. The prism is oriented to provide the least deflection angle. [0.5° approx]

(x) A parallel beam of light falls normally on the first faceof a prism ofsmall angle.Atthe secondface, it is partlytransmitted and partly reflected.The reflected beam strikes at the first face again, and emergesfromit in a directionmaking ao angle 6°30' with the reversed direction ofthe incident beam. The refracted

beam is found to have undergone a deviation of 1°15' from the original direction. Find the refractive index ofthe glass and the angleoftheprism.

Figure 5.209

[(a) n/i +sin"0 ; (b) V2 ; (c) No grazing emergence; (d) Total Internal ReOection]

(vil)

On one face of an equilateral prism a light ray strikes 3

normally. lfitsp= —, find the angle between incident ray and the ray that leaves the prism. [60°]

(viii) A long rectangular slab of transparent medium of thickness d is placed on a table with length parallel to x-axis and width parallel to the^-axis. A ray oflight is traveling along j'-axis at origin. The refractive index p ofthe medium varies as

p= Vr+^^ .The refractive index ofthe air is 1.

(xi) A ray oflight is incident at an angle of60° on one face of a prism which has an angle of30°. The ray emerging out ofthe prism makes an angle of30° with the incident ray. Show that the emergent ray is perpendicular to the face through which it emerges and calculate the refractive index of the material of prism. [73 ]

(xii) The refracting angle of a glass prism is 30°. A ray is incident onto one of the faces perpendicular to it. Find the angle 5 between the incident ray and the ray that leaves the prism. The refractive index ofglass isn= 1.5. [18.6°]

(xiii) A ray oflight travelling in air is incident at grazing angle (incident angle = 90°) on a large rectangular slab ofa transparent medium ofthickness t = 1.Om(see figure-5.211). The point of incident is the origin A(0,0).

(a) Determine thex-coordinate ofthe pointy4, where the ray intersects the upper surface of the slab-air boundary. (b) Write down the refractive index ofthe medium at.^.

Air

1= I.Om

Ax,.v,)

/(x,y)

Glass slab Air

A{0. 0)

Figure 5.211 Figure 5.210

[2d In (2), VJ 1

The medium has a variable index ofrefraction fi(y) given by

l)'/2

:^Ge6metricai Optics where A: = 1.0 m"^^.The refractive index of air is 1.0.

Convex and Concave lenses shown in figure-5.212 are basically (a) Obtain a relation between theslope ofthetrajectory ofthe lens families and these are further categorized in three ways. rayat point 5(x,y) in themedium andtheincident angle atthat Convex lenses are of three typesas given below point.

(b) Obtain an equation for the trajectoryy{x) ofthe rayin the medium.

(c) Determine the coordinates (x,, y,) ofthe point P, where the rayintersects the upper surface ofthe slab-air boundary. (d) Indicate the path of the ray subsequently.

1. Biconvex Lenses : Theseare the lenses bounded bytwo intersecting sphericalsurfaces having their center ofcurvature on different sidesofthe lensas shown in figure-5.213.

2. Plane Convex Lenses : These arethe lenses bounded by one spherical surface intersecting with one plane surface as shown in figure-5.213.

[(a) e = cot-i

•Ah)x4k = 47""; (c)4, 1; (d)

"li

3.

Concavo Convex Lenses: These are the lenses bounded

by two intersecting spherical surfaces having their center of curvatureon same side of the lens as shownin figure-5.213. 5,

5.14 Thin Lenses

Lensesare the transparent medium bounded bytwo spherical surfaces. There are two types of lenses in general based on thickness ofthe lens at its center, thin and thick. Both lenses

differ in context of image formation and analysis of image characteristics. One byonewewill discuss upon both butmore

Biconvex lens

important is to understand ^Thin Lenses^ which are most commonly used lenses in practice.

Figure-5.212 shows the wayhowlenses are madebyusing two spherical surfaces. Figure-5.212(a) shows a case when the two

surfaces bounding a region in between intersect at the edges and figufe-5.212(b) shows the case when the two surfaces

bounding a region in between do not intersect at the edges. First one is called ^Convex Lens^ and second one is called 'Concave Lens'.

Piano convex lens

Concavo convex lens

S-,

Figure 5.213

In the manner described above for convex lenses, concave lenses are of three types as given below

Convex lens

1. Biconcave Lenses: These arethe lenses bounded bytwo non-intersecting spherical surfaces having their center of

(a)

curvatureon differentsidesofthe lens as shown in figure-5.214.

2. Plane Convex Lenses ; These are the lenses bounded by one spherical surfece not intersecting with one plane surface as shown in figure-5.214. 3. Concave lens

(b) Figure 5.212

Concavo Convex Lenses: These are the lenses bounded

bytwo non-intersectingsphericalsurfaceshaving their center of curvatureon same sideof the lens as shownin figure-5.214.

Geometrical Optics-j

Due to the above behaviour all convex lenses placed in rarer

surrounding medium are called ConvergingLenses\^^ all concavelensesplaced in rarer surroundingmediumare called 'Diverging Lenses'.As shown in figure-5.215 wecan seethat the converging lenses have a 'Real Focus' on the side oflens where refracting raysexistanddiverging lenses havea' Virtual Focus' on the side ofthe lens where incident ra>^ exist.

Biconcave lens

The behaviour of both type oflens frmilies get interchanged when the surrounding is denser than lens material. This is explainedin figure-5.216(a) and (b)

Piano concave lens

Convexo Concave lens

Figure 5.214

5.14.1 Converging and Diver^ng Behaviour of Lenses F'

In general lenses having made up of material denser then their surrounding like most common is the uses ofglass lenses in air. In such cases when a parallel beam of light incident on the lenses, all convex lenses converges the beam after refraction to a point and all concave lenses diverges the beam after refraction which appear to come from a point behind the lens as shown in figure-5.215(a) and (b).

•

Focal point (focus) of convex lenses in denser medium (a)

/V 4'

\/

Focal point (focus) oflens

oflens

' Focal point (focus) oflens

(a)

Focal point (focus) of concave lenses in denser medium

Figure 5.216

Note: Behaviour oflenses - converging or diverging is defined only for parallel incident rays on it, there should not be any misconception developed that a converging lens always Focal point (focus) oflens

Focal point (focus) oflens

converges all type of light rays. Its defined for parallel rays which a converging lens converges and a diverging lens diverges when the surrounding medium on the two sides of lens is same.

5.14.2 Priiuary and Secondary Focus of a Lens Focal point (focus).of lens (b) Figure 5.215

When light rays are refracted from a lens then there are two focal points defined for a lens, primary and secondary for both concave and convex lenses. Figure-5.217(a) shows primary

fGeometrical Optics

-

'26^

focus of a convex lens. This is the point on one side of lens

where ifan object is placed then after refraction all light rays

5.14.3 Standard Reflected Light Rays forImage Formation by Thin Lenses

will become parallel to the principal axis of the lens. Figure-5.217(b) shows the primary focus of a concave lens. There are three standard incident paraxial rays and their

This isthe point toward which ifa converging beam oflight is corresponding refracted rays after double refraction through a incident on lens then after refraction these light rays become thin lens, using which we can roughly analyze thelocation of parallel to principal axis of the lens. image on the principal axis ofthe lens. Any two ofthe three standard rays we can use for finding the relative position of

image produced and understanding ofthese rays also helps in

Real

object

analyzing the ray diagram for image formation by a lens in different situations. We will discuss these rays one by one.

Ray-1: Incident onthelens parallel totheprincipal axis Figure-5.219(a) and(b) shows aparaxial light rayincident on theconvex and concave lens which isparallel toprincipal axis which after refraction from the two surfaces ofthe lens passes

(a)

through the primary secondary focus of the lens. In case of Virtual object

(b)

convex lens itactually passes through thefocal point whereas in case of concave lens it appears to pass through the focal point as shown.

.

Figure 5.217

Figure-5.218(a) shows parallellight raysincident ona convex lens which afterrefraction throughthe lens meetat the focus on the other side of lens, this is called secondary focus ofthe convex lens. Similarly when parallel raysincident ona concave

lens as shown in figure-5.218(b), after refraction light rays (

(

diverge in such a waythatthese appear tocome from the focal II. point located behind thelens, thisiscalled secondary focus of (

I,

It

II

s

the concave lens.

1

(b) Figure 5.219

Ray-2: Incidentonthelens directly at theopticcenterofthe Real Image /

H

(a)

lens

Figure-5.220(a) and(b) shows aparaxial light rayincident on theconvex andconcave lens at the optic center ofthe lens. As lensis thinatthecenter it behaves like a glass slabas shown in insetview so the light ray passesundeviatedwith small lateral

^

"

displacement which can be ignored and considered the ray passing straight if incident on the optic center of the lens.

virtual Image

(b)

Figure 5.218

Note : For thin lenses the focal length of the lens for both

primary and secondary focus is same. This we can also prove afterunderstanding the calculation offocal lengthofthe lensin coming sections. (a)

Geometrical Optics J

•266

Case-I: Obj ect Islocated at infinity

Asalready discussed while analyzing reflection cases that there are two possibilities of an object at infinity. One is when all incident rays areparallel toprincipal axiswhen image produced is real and located at focus of the mirror as shown in

figure-5.222(a) andother possibility iswhen incident rays areat someangleto principal axiswhen image produced is real and located in focal plane as shown in figure-5.222(b). In both of these cases we have considered the image produced is highly diminished in size, real and inverted (produced on the other side of principal axis as that of object).

(b)

Figure 5.220

Ray-3 : Incident on the lens along the direction of primary focus

Figure-5.221(a) shows a paraxial raypassing through theprimary

2F

focus ofthe convex lens arid then incident on the lens, this ray after refraction becomes parallel to the principalaxis ofthe lens.

Similarlyfora concave lens,a light ray which isincidenton the lens along the linepassing through its primaryfocus as shown in figure-5.221 (b)becomes parallel after refraction.

Focal

plane

(b)

Figure 5.222

Case-n: Object is located beyond 2F point F,

Figure-5.223 shows this situation in which object is a small candle and we find the image oftip of the candle by considering

(b)

Figure 5.221

ray-1 and ray-2 as discussed in article-5.12.3. With this ray diagram we can conclude that the image produced for this location ofobject(placed beyond 2Fpoint on principal axis)is

Now using these casesof paraxial incident and refracted rays

real, located between F and 2F, inverted (on other side of

we can discuss various cases of image formation by convex and concave mirrors for different positions ofobject.

principalaxis)and smallerin size than that ofobject.

5.14.4 Image Formation by Convex Lenses

Whenever an object location is given for a convex lens then using any two of the three incident rays and corresponding refracted rays discussed in article-5.12.3 vye can find the image location using ray diagram. There are some positions near to principal axis in different regions where if an object is placed, using ray diagram we can get some informationabout the image formation. This information is very helpful in rough analysis while solving different types of questions based on image formation. For both convex and concave lenses we are going to discuss different cases for position of object in front of the lenses and its corresponding image produced. First we will take up the cases for convex lenses in which there are five possible cases we will discuss for a real object in front of the convex lens.

Figure 5.223

Case-m: Object is located at 2F point on Principal Axis

Figure-5.224 shows this situation in which object (candle) is located at 2F point and we find the image of the candle by considering ray-1 andray-2 as discussed in article-5.12.3. With this ray diagram we can see and conclude that the image produced for this location of object (placed at 2F point on principal axis) is real, located at 2F point on other side of lens, inverted (on other side ofprincipal axis) and ofsame size as that of object.

^Geometrical Optics

267

produced is located behind the lens, virtual, erected (on the same side ofprincipal axis) andenlarged. IF 2F

F

\1 Figure 5.224 f >

Case-IV : Object is located between F and 2F points on

11 X

. , li 2F

f

11 X

2F

•.s

F

KT

Principal Axis

Figure-5.225 shows this situation in which object (candle) is located between Fand 2Fpoints and we find the image ofthe candle by considering ray-1 and ray-2 as discussed in article-5.12.3. With this ray diagram wecan seeand conclude

Figure 5.227

Above sixcases aregeneral cases which explains therelative position, nature and orientation ofimage produced byaconvex

that the image produced for this location of object (placed between Fand2Fpoints on principal axis) isreal, located beyond 2Fpoint, inverted (on other side ofprincipal axis) and enlarged

lens for a real object placed close to its principal axis using

compared to that of object.

give an idea about the estimation ofimage produced upto some

paraxial rays. While solving different questions above cases extent before going for theexact process of image formation.

IF

2F

F

Figure 5.225

Case-V: Object is located at Focus

5.14.5 Image formation by Concave Lenses

Unlike to the different cases of image formation bya convex lens for different positions of object in front of it, in caseof a concave lens there areonlytwo possibilities ofimage formation for a real object placed in front of it.

Case-I: Objectlslocatedatinfinity

Figure-5.226 shows this situation in which object (candle) is located at focal point ofthelens and we find the image ofthe When light rays from distant object fall on a concave lens,

these rays diverge afterrefraction in such awaythat for incident candle by considering ray-1 and ray-2 as discussed in rays parallel toprincipal axis a image is obtained at focal point article-5.12.3. Withthis ray diagramwe can seethat the refracted ofmirror as shown in figure-5.228(a) andfor incident rays non raysare paralleland willproduce imageat infinity. Sowecan

conclude with this diagram that image produced will be atinfinity, inverted (on the other side ofprincipal axis) arid highly enlarged.

parallel to principal axis image is obtained in focal plane as shown infigure-5.228(b). The image produced will bevirtual,

diminished and erected (produced on the same side ofprincipal axis where object is located).

Figure 5.226

Case-VI; Objectislocated between Focus and OpticCenter

Figure-5.227 shows this situation in which object (candle) is located between O and Fand we findthe imageof the candleby considering ray-1 and ray-2 as discussed in article-5.12.3. With

this raydiagram wecan seethat theserefracted raysfrom lens arediverging in a waythat theseappearto becomingfrom the point/behind the lens from a virtual image. So forthis location of object (placed between O and F on principal axis) image

(b).

Figure 5.228

Geometrical Optics 1,

|268

Again with same sign convention we can use « = v =/ Figure-5.229 shows a situation in which theobject (candle) is |i, = )x,li2=l and ^ =^2 and we get placed infront oftheconcave lens and tofind image using ray = ...(5.98) diagram we consider ray-1, ray-2 and ray-3 as mentioned in / V, F,

Case-II: Object is placed anywhere in front ofthe lens

article-5.12.3. Wecan seein this figure that the imageis formed

byback extension ofthereflected rays corresponding tothese incident rays from theobject and theimage isproduced behind the mirror betweenFand (9, virtual, diminished and erected(on the same side of principal axis as that of object).

Note : In above equations-(5.97) and (5.98) we have not

considered the signs of u, and R^ as the lens can be of different types among thesixtypes asexplained in article-5.12. Adding the two equations, we get 1

1

R

R•2j

...(5.99)

Above equation-(5.99) is called 'Lens Makers Formula'' used to calculate the focal lengthofa thin lens.With this expression we can see that focal lens of thin lens for both primary and

secondary focus remain same on thetwo sides ofthelens. The

Figure 5.229

aboveformula is usedwhen a thin lensis kept in air and ifa thin lensofrefractive indexp. iskqjt in a surrounding medium having

5.14.6 Focal length of a thin lens

refractive index

then the above formula can be modified as

Figure-5.230 shows a thin convex lens onwhich parallel light givenbelow in equation-(5.99). ra>^ incident from leftandaftertworefractions at itsspherical 1 ( II

surfaces ofradius i?j and

light rays converge to its focal

1^ R2 ,

/

point at adistance called the^Focal Length' ofthelens and it is denoted by. Wewill findoutthis focal length/by usingthe analysis of refraction of light at the twosurface.

\

-(s-ioo)

5.14.7 Focal length ofdifferent types ofstandard thin lenses

Byusing Lens Makers Formula wecanfindthefocal length of any thin lens but for quick reference we are giving here the

52(^2)

directrelationsfor magnitudeof focallength of somestandard shaped lenses.

' .

1. Equiconvex or Equiconcave Lenses : These lenses have same radiiofcurvature ofthetwo surfeces soweusei?j = ^2 F andforthisequation-(5.99) willreduce tothe form given below

Figure 5.230

because the two center ofcurvatures ofthe surfaces are on the

For first surfece ifwe use the refraction formula as given below

Fli?2

Hi _ B. _ \

u

opposite sides ofthelenses andthese will have different signs. /=

R

Here wesubstitute w= inf,pj= 1,^2= M' H-1

F=i?jandweget ...(5.97)

...(5.101)

Nowusingi?j = i?2 = R, weget R

f-

2(n-l)

...(5.102)

Figure-5.230 shows thatfirst image/, isobtained afterrefraction Equation-(5.102) gives only the magnitude of the lens. When from first surface ata distance v,,which canbetreated asobject we use this focal length in image formation by a lens then it is for the light raj« inside the lens falling on the secondsurface. substituted with appropriate sign. After refraction from second surface the final image is produced

at Fas all refractedrays converge at focal point oflens. Soifwe again use the refraction formula for refraction at the second

2. Piano-Convex or Plano-Concave Lenses : These lenses

surface

forthis equation-(5.99) willreduceto the formgivenbelow. 1^2

Hl U

R

have oneplane surface sowecantakeR, = inf and R^—R and

[Geometrical Optics

269

3. Concavo-Convex or Convexo-Concave Lenses: Such lenses have different radii ofcurvature and their centers ofcurvature

lie on the same side oflens so the signs ofthe two radii i?, and

As we have

= (11-1)

J

1_

Ri

Rt

willbesameandthe equation-(5.99) willreduce tothe form given below

1

...(5.104)

/=

Above equations-(5.101) to (5.104) students can keep on tips for quickreference forcalculation offocal lengthmagnitude of anythin lens while solvinga problemofimage formation.

5.15 Analysis of Image Formation by Thin Lenses

1

__1_ "/

...(5.108)

Equation-(5.108) is called Lens Formula' used to find the

location of image produced bya thin lensfor a specific object forparaxial ra}^incident on the lens.

Aswehavealready discussed that for thinlenses magnitude of primary and secondary focal lengths of lenses are equal but their signs are different as these exist on two different sides of

the lens.Forthin lenses when a light rayincident on it then in For finding the exact location of imageproduced bya thin lens above lensformula wealways useits secondary focal length. If made upofamediumofrefi'active index p,wecanstartbyusing weuseincidentrayreference sign convention then always the refraction formula twice for refraction at the two surfeces ofthe focal length of a converging lens is positive and that of a lens. Figure-5.231 showsa biconvex lens and a point objectis diverging lens is negative whereas in cartesian coordinate placed at a distance u from the optic center of the lens on its system sign convention it depends upon the direction from principal axis. Ifweanalyze the imageproduced duetorefraction which light ray incident on the lens.As alreadydiscussed that

atits first surface 5'j which isatlocation /j asshown infigure,

we will use the refraction formula for this. 1^2

B.

...(5.105)

R

U

Here we use m= «, v= v,^

= 1, ^2 = ^ and R= R^ (for first

surface) and we are not using signs as in general case lenses can be of different types and object can also be at different locations. This gives p

1

Vj

u

in this chapter we are using cartesian coordinate sign convention in solving different questions ofgeometrical optics. If students wish then theycan alsosolve each problem using the other convention also.

5.15.1 Lateral Magnification in Image Formation by a Thin Lens

Figure-5.232 shows a small object (candle) placed on the principal axisof the lensat a distance u from the optic centerof

p- 1 .

...(5.106)

the lens and it producesan image/ which is real and inverted.If

theheights ofthe object and image above theprincipal axisare h and ^'respectively then from the figure we can see that the angular sizes of both the objectand imageare same on the two

SiiRi)

sides of lens and it is equal to 9. 0

I

••

h

V Figure 5.231 Figure 5.232

Now for the light ray inside the lens which falls on the second

surface 52will again suffer refraction for which again weuse the

refraction formula as given in equation-(5.105) with «=Vp v=-v, Pj = p, |i2= 1 and R - R^. This gives

l_ii V

Vj

triangles formed by objectheight and image height on principal axis we have h

l-fi

...(5.107)

R-.

u

/j' =

Adding equations-(5.106) and (5.107), we get ^

1

/

Where

m

V

— \h = mh

Height of Image (A') Height of Ojbect (h)

v n

- Geometrical Optics

'270

Herem = (v/m) is called 'Lateral Magnification^ by a thin lens used to find the size of image and its orientation. In this relation of lateral magnification u and v are substituted with proper signs according to sign convention whichcan resultin positive and negative values of m. Positive value of m denotes that image is erected or on the same side of principal axis and negative value of m denotes that image is inverted or on the opposite side of the principal axis of the lens.

For small width object if image is produced by a thin lens (converging or diverging) then image width can be calculated by using the equation-(5.109). But if objectsize is large then this relation cannot be used and in that case we need to calculate

the image of both edges ofthe object along principal axis and takethe difference as explainedin figure-5.80 in ariticle-5.9.5 for spherical mirrors. — 5.153 Variation Curves ofimage Distance vs Object Distance

5.15.2 Longitudinal Magnification by aThinLens

for a Thin Lens

Lateral magnification formula for thin lenses gives the image height above the principal axis of mirror and in this section we will discuss about the image width along the principal axis ofa thin lens. The relation in object and image width along the principalaxis ofmirror is called'LongitudinalMagnification'' as given below.

We have discussed the lens formula which relates the image distance fi-om pole of mirror for a given object distance. The lens formula is given as

l_i

1

f~v~u ...(5.110)

v =

Longitudinal magnification ofimage is given as

+ 1

/ As focal length ofa given lens is constant and it can be positive

Width ofimage along Principal Axis m,

2f

Width of Object along Principal Axis

or negative depending upon the sign convention and type of lens used. Tfie above function given in equation-(5.110) can be plotted as shown in figure-5.234(a) and (b) for a converging and diverging lens with negative x direction on left and positive x-direction on right which we generally consider.-

f

. V

1 Figure 5.233

Figure-5.233 shows image formation of an object located at a distance x from the convex lens offocal lengthfwhich produces an image ofthis object at a distance^'which is real inverted and enlarged because object was placed between F and 2Fpoints. Here we can see that object edge A was close to C so corresponding image edge A' is also closer to C. Ifwe consider objectis of verysmall width dx and image produced is having a width dy then from lens formula we have

Real Image of / Real object /

Real Image of virtual object

II

+

u = -j

u

0

u

Virtual image of real object

i_i =1 V

!

(a)

f

V

Here by coordinate sign convention we use ti= -x,f= +f and v = +y

Diveiging Lens Image of Real

-L -1

J_

f

-X

y

/

Differentiating this expression we get . y--f

0^ ——:rdx- —ydy X

u-+f

0

y

Virtual image of real object

y

Virtual object

/

From this relation we can get the 'Longitudinal Modification'' as

m, =

dx

2

..2 = -m'

...(5.109)

(b)

Figure 5.234

Virtual image of virtual object

Geometrical Optics

271

5.15,4 Effectofmotion ofObject and Lenson Image

When object or lens is in motion the distance between object andlens changes which affects theposition and size ofimage. Tofind the imagevelocity and for analysis of image's motion we can differentiate the lens formula and find the rate at which

Solution

Let theradius offirst surface be/?, and refractive index oflens be p and parallel rays be incident on the lens. Applying refraction formula at first surface

distances between image and lensis changing. Ifweconsider a: andy as object and image distance from poleofmirroroffocal length/then by lens formula we have f

y

Differentiating the above relation withrespect to time, weget dx

\

dt

1

p-1

Vi

00

/?.

...(5.112)

Refraction formula at second surface

X \

p

H

2-p

V

V|

-20

...(5.113)

Addingequation-(5.112) and (5.113),weget

dy dt

y

2

ili-l + ^ V]

dx .

Where — is the relative velocity of object with respect to the

CO

R-1 ^2-p

V V]

/?,

-20

dy

lens and — is the velocity of image with respectto lens. 11

^

dx

yR,

v„ = -

li-1

2-p

20

20

dy

' Vq- ^-v-Cusing —- v^and —=v.) .2 *^0

-20

m-Vr

1 . .1 2 1 1 1 -2 =7(,na,r)+--=~--=-

...(5.111)

v=40cm

Where m is the lateral magnification produced by the mirror. The expression of image speed as given in equation-(5.111) is 10 20 valid only for the velocitycomponent of the image and object along the principal axis of the lens. Ifthe objectand mirror is in Thus all parallelraysincidenton the lenswill focus at a point motion along the direction normal to principal axis we can 40cm from the lens in the medium with refractive index 2. directly differentiate the height of object and image above # Illustrative Example 5.61

principal axis which are related as

h. = mhQ

A biconvex lens has focal length 50 cm and the radius of

Differentiating this with respectto time weget dhj

curvature of one surface is double that of other. Find the radii ofcurvature ifrefractive index of lens.material is 2.

dh. ~m

dt V.

m

dt •mv

Solution

OiV

dh:

Here we can use —7^= v.., and dt

dt

-

which are the

velocity components of image and object respectively in direction normal to theprincipal axis.

/?l/?2 f=

(p-l)(/?,+/?2)

And we are given with/?-= 2R.

# Illustrative Example 5.60 Focal length ofa thin lens in air, is 10 cm.

For biconvex lens focal length of lens is given as

/? = 20 cm

/=

Now medium on one side ofthe lens is

replaced by a medium of refractive index p = 2. The radius ofcurvature of surface of lens, in contact with the medium, is 20 cm. Find the point on /,ir= principal axis where parallel rays Figure5.235

incident on lens from air parallel to axis will converge.

50 =

2/?.

2/?.

(p-l)(3/?,)

3(p-l)

2^1 3(2-1)

R^ =75 cm /?2-150cm

Geometrical Optics

# Illustrative Example 5.62

Solution

A convexlensoffocal length20 cm is placedat a distance5 cm

Due toreflection bya mirror, image ofobject isformed onitself when reflected ra>^ fells normally onthe mirrorandretrace the

from aglass plate =2J

^ ^

path ofincident rays. For this the image produced by thelens

placed at a distance 30 cmfrom lensonthe othersideofglass plate. Locate the final image produced bythis optical setup.

must be formed at the center of curvature of the mirror as shown in ray diagram-5.237. A = 20 cm

Solution

A = 10 cm

Figure-5.236 shows the optical setup described in question

/? = 20 cm

and the ray diagram for image formation. Scm

h

3cm

»+•

"

60cm

Figure 5.237 Image

For lens formula, we use 30cm 60cm

1cm-H

/=-20cm v=-15cm

Figure 5.236

For lens formula to be used in refraction by lens, we use w = - 30cm

__L i _zi 15 X ~ 20

/= + 20cm

^

i _ J__J_

i-1 =1

X ^ 15

V u f J

60

'm

x = 60cm

1_

v"^30 ~ 20 20x30 v=——=60cm

Shift of image due to refraction by the glass slab is given as S=t

4-3

20

i-i

# Illustrative Example 5.64 A convex lens is held 45 cm above the bottom of an empty

tank. The image of a point on the bottom of a tank is formed 36 cm abovethe lens. Now a liquid is poured into the tank to a

depth of40cm. It is found thatthedistance oftheimage ofthe same pointon the bottom ofthe tankis 48 cm above the lens. Find the refractive index of the liquid. Solution

S= 3| 1-— = 1 cm Thus position offinal image = 60+ 1= 61 cm.

When the tank in empty and point object O is placed at the bottom ofthe tank, then for lens formula we use M= - 45 cm and v = + 36 cm

# Illustrative Example 5.63

A diverging lens of focal length 20 cm is placed coaxially 5 cm toward left ofa converging mirror offocal length 10 cm. Where would an object be placed toward left of the lens so that a real image is formed on object itself.

Using lens formula, we have

l_i _ J_ V « "/ Mv

/= « —V

-45x36

1620

-45-36

81

= 20 cm

jGeome^ical Optics

,

;

273

When the liquid ispoured in thetankto a depth 40 cm, then image is produced at 48cm above the lens so again for lens formula we use v'=+48cm

If u' bethedistance oftheobject O' for image to be produced

V^.

at 48cm from lens then we use I/,

J___l_

1

J

~V /

48

1_

Plane mirror

«. = 15cm

20

Figure 5.238

Solving we get

The image /j acts as the object for theplane mirror andafter reflection oflight ra}^ from the plane mirror final imageproduced

H'=-34.29cm

is

The distance between liquid surface and lens is 5 cm

In a planemirror, the image formed is at same distance

at which objectis kept fromit and sizeremain same. Sothe final image is produced at a distance 30 - 10 = 20cm as shown in

Distance of O' from the surface is 34.29 - 5 = 29.29 cm which is the apparent depth of object below water surface at which when it is placed, image is producedat the specifiedlocation.

the reflectedraysbytwicethe angle at which mirror isrotated.

The refractive index of liquid is given as

# Illustrative Example 5.66

Real depth

40

Apparent depth

29.29

figure-5.238 atanangle 90® totheprincipal axis asmirror rotates

A point object is placed at a distance of 0.3m from a convex lens of focal length 20cm which is cut into two halves each of

= 1.37

whichis displaced by 0.5 mm as shown in figure-5.239. Find the position of the image. If more than one image is formed,

# Illustrative Example 5.65

find their number and the distance between them.

An object of height 4 cm is kept to the left of and on the axis of a converging lens of focal length 10 cm at a distance of 15 cm from the lens.Aplanemirroris placedinclined at 45® tothe lens

L0.5 mm

axis, 10cmtotheright ofthelens.Findtheposition andsizeof the image formed by the lens and mirror combination. Trace the path of the rays forming the image. Figure 5.239 Solution

Solution

LetAB be the objectplaced at a distance of 15 crh fromthe lens We have studied thateverypartofa lens behaves like complete as shown in figure-5.238. We shall calculate the position of lens and produces a separate image soin this case two images image formedbythis lens in absenceofplane mirror. areobtained. Theimage formation is shown in figure-5.240.

For lens formula weuse «j=- 15 cm, and /=+ 10 cm 1

j0.5mm

113-21

^

vi ~ 10 15 ~ 30 ^ 30

or

v,=+30cm

So the image would be formed at 30 cm from the lens to the Figure 5.240

right ofit

The light rays from object passesthrough optical centres Magnification by lens is m, = — = — = 2 *' ' «i 15 ^

and OjUndeflected.The images of(9 are produced at/j and/2 due to upper and lower part of lenses. To find the location of

Sizeoftheimage = 2 x4 = 8cm

The image produced bylens is /, which is shown by figure-5.238.

images we use refraction formula as

in

1-1=1 V

u

f

Geometrical Optics

i274

For refraction formula we use w= - 30cm and/= + 20cm

I ^

I J_

V~

1_ J ~

l_

Am:

If d is the distance between the two images 7, and Ij then separation between these two images can be calculated from

Substituting the given values, we have

M+ V

Am

O1O2 30 + 60

Solving we get

(neglecting higher terms)

(v-fY

triangle AOO, O2 and A7/j I2, inwhich by similarity, we use

0.05 + 0.05

(V-/).

fAv

v=60cm

'

Av 1—

(V-/)'

30 20 ~ 60

d

-1

/'Av Am =

(25)^x18

(25)'xl8

» 0.5 mm.

(500-25)2

# Illustrative Example 5.68

30

(7=3 mm

Determine the position of the image produced by an optical system consisting of a concave mirror with a focal length of 10 cm and a convergent lens with a focal length of 20 cm. The

# Illustrative Example 5.67

distance'from the mirror to the lens is 30 cm and from the lens

A thin converging lens of focal length/= 25.0cm forms the image of an object,on the screen, at a distance 5cm from the lens. The screen is then drawn closer by a distance 18cm. By what distance should the object be shifted so that its image on

to the object is 40 cm." Find the image produced after one refraction from lens and one reflection from mirror. Solution

the screen is sharp again?

Object is placed at distance 2/from the lens so its image is also formed at distance 2/on other side. For mirror this image acts as an object which is located at a distance of 10cm and we applymirror formula in this case for which we use m=+10cm

Solution

Since the image is formed on the screen, it is real 1

Now

1

and/=-10cm

1

—+ ~ = "7 V

M

, /

1 =>

l_i

or

/ u =

U

V-/

V

A V-/

]_ J ...(5.114)

/(v-Av) (v-Av)-/

1

V

J

r

1_

v"^+10"-10

=>

In the second case, let the image is formed at v - Av. Let the corresponding position of object be m- Am. Now. M+ Am =

^

1

- +

v=-5cm

Raydiagram forimageformation is shown infigure-5.241 below. -40 cm-

...(5.115) -40 cm-

The shift of the object m+ Am - m= Am

Subtracting equation-(5.114)from equation-(5.115),we get Am =

Am =

/(v-Av)

fv

(v-Av)-/

v-f

H-5cm-

/(v-Av)(v-/)->((v-Av)/) {(v-Av)-/}(v-/)

-fAv Am =

(v-Av-/)(v-/)

/^Av

Am =

(V-/)'

!-•

Av

(V-/).

Figure 5.241

iGeometrical Optics

# Illustrative Example 5.69

which gives

U'=2{00)

= 2(^-I)a/• A small angled prism(refractive index p and angle a) and a convex lens are arranged as shown in figure. Apoint object O Thus image position is3/ontheright side ofthe lens along the is placed as shown.

axis, and 2(p- l)a/transverseto axis.

(a) Calculate the angle of deviation of the rays hitting the # Illustrative Example 5.70

prism at nearly normal incidence

(b) If the distance between object, prism and the lens are

shown in the figure, locate the position of the image both along and transverse to the axis.

Astrong source oflight when used with aconvex lens produces a number of images of the source owing to feeble internal reflections and retraction called flare spots as shown in

figure-5.243. These extra images areF'j,

Solution

position of

. IfF^ is the

flare spot, then show that

The optical setup described inquestion isshown infigure-5.242

(« + l)p-l

below.

L

^4

F

/

/

^

7,f.

»

Figure 5.243

Figure 5.242

Solution

Figure change fromprintouts...

(a) The deviation produced bythe prism

Light converges at

after two refractions and one reflection

from the lens. So we use

5=(|i-l)a

(b) The prism forms image ofthe objectat O'. 00' = bf={si-\)af

J__i. h

"

fm

Where focal length ofequivalent independent lens isgiven do

The image O'becomes objectfor lens. 3/ Now using lens formula, we have u= —^ so we use

fe

V u~ f ^=2(p-iy

J_ I_ _ J_ V V "/

—~—+ F,

1 _ J_ V"3/

1

ir

2 2(p-l)/

2p-l

Fx (M-1)/ For F^, there are three refractions and two reflections

v=3/ Also

f

V

i- _ -L A

OO' ~ u

F2 ~fx ^7^

IL IL 2

= 2 — +

1

1

/ Rn ^ f^R

Geometrical Optics^

5276

_1

By lens formula,

4

1 1=1 —, we u have _3_ '

V u

2 (li-l)/

3(^-l) + 2

3^1-1

(li-1)/

(li-1)/

1

1

V

-0.5

f

v=-0.6m

1

Web Reference at www.phvsicsgalaxv.com

(li-1)/

Age Group - High.School Physics | Age 17-19 Years

By using lens formula

Section - OPTICS

_1_' V w "/' 1

1

Topic - Geometrical Optics III -Thin Lenses

h = + 15 cm,/= + 30cm

where

J_ we have

which gives

1

Module Number -1 to 14

1_

V +15 ~ +30

Practice Exercise 5.6

v=+ 10 cm

(1) A point sourceof light is kept at a distance of 15 cm from a converginglens, on its opticalaxis. The focal length of

The plot of rays is shown n figure.

the lens is 10 cm and its diameter is 3 cm. A screen is placed on

# Illustrative Example 5.71

the other side of the lens, perpendicular to the axis of the lens,

A small fish, 0.4 m belowthe surfeceofa lake, isviewedthrough a simpleconverging lensof focal length 3m. The lens is kept at 0.2 m above the water surface such that the fish lies on the

at a distance 20 cm fi-om it. Find the area of the illuminated part of the screen ?

[(Tt/4) cm^l

optical axis of the lens. Find the imageof the fish seenby the observer. The refi-active index ofwater is 4/3.

(ii) A 5.0 diopterlensforms a virtual imagewhich is 4 times the objectplaced perpendicularlyon the principal axis of lens,

Solution

find the distance of object fi^om lens.

The apparent position ofthe object O fi"om the surface ofwater is

(Hi) A point object is placed at a distance of 25 cm fi-om a convex lens of focal length 20 cm. If a glass slab of thickness t and refiactive index 1.5 is inserted between the lens and the

object, the image is formedat infinity. Find the thickness t ? [15 cm]

'

0.2m

(iv) A convex lens offocal length 20 cm and a concavelens offocal length 10cmare placed10cmapartwiththeirprinciple axis coinciding. A parallel beam of light of diameter 5 mm is incident on convex lens symmetrically. Prove that emerging beam will also be parallel & find its diameter. [2.5 mm]

Figure 5.244

The distance

PI = 0.3 + 0.2 = 0.5 m

For convex lens, we use

M=-0.5m,/=+3m

(v)

A convexlens of focal length 20 cm is placed 10 cm in

fiont of a convex mirror of radius of curvature 15 cm. Where

should a point object be placed in front of the lens so that it produces image on itself? [100 cm]

iGeometrical Optics

__277J

(vi) A converging lensof focaMength 20 cm is separated 8 (xii) An object is kept at a distance of 16 cm from a thin lens cmfrom a diverging lens offocal length 30cm. Aparallel beam andtheimage formed isreal. If theobject is kept at a distance of light falls on converging lens and after passing through of 6cmfromthe samelens,the image formed is virtual. If the diverging lens focussed atpoint P. Find thelocation ofpoint P. sizeofthe images formed in above twocases are same,"findthe Repeat the calculation forthe casewhen theparallelbeam first focal length ofthe lens? falls on diverging lens. [ 11 cm]

[42.2 cm from convex lens]

(vii) Two symmetric double convex lensesA and B have same focal length but the radii of curvature differ so that R, = 0.9R„. A

D

Ifrefractive index ofA is 1.63 find the refractive index of.^. [1.7]

(vii!) An object is placed at 20 cm left ofthe convex lens of

(xiii) An object is placed midway between the lens and the mirror as shownin figure-5.246. Themirror'sradius ofcurvature

is 20.0 cm and the lens has a focal length of - 16.7 cm. Considering only the rays that leaves the object and travels first toward themirror, locate the final image formed bythis system. Is this imagereal or virtual? Is it upright or inverted? What is the overall magnification ?

focal length 10 cm. If a concave mirror of focal length 5 cm is placed at 30 cm to the right of the lens, find the magnification and the natureofthe final image. Drawthe ray diagramand locatethe position of the final image.

Y /

[At the object and of same size]

(ix) In the figure-5.245 it is shown, the focal length/of the two thin convex lenses is the same. They are separated by a horizontal distance3/and their optical axes are displacedbya vertical separation 'if {d^is we can ignore such coefficients and most commonly we use the two term form ofCauchys Equationgiven as p=^+ —

(xi)

"t

...(5.147)

with pj = 1.70 andother diverging with pj = 1-51. Both lenses

With the above equation-(5.147) wecanseethatlightwith lower

have same curvatureradius of their surfacesis equal to 10cm. The lenses were put close together and submerged into water. What is the focal length of this system in water.

travels slower. So if we compare violetand red light in same mediumthen we can seethat violetlight travelslowerthan red

wavelength has higher refractive index in a medium and hence

light in the medium. [33.3 cm]

Thisdifference in speed causes lightra)«ofdifferent wavelength (xii) A point object is located at a distance of 100 cm from a screen. A lens of focal length 23 cm mounted on a movable frictionless stand is kept between the source and the screen. The stand is attached to a spring ofnatural length 50 cm and spring constant 800 N/m as shown in figure-5.277. Mass ofthe

to refract differently when these travel from one medium to another and that is the cause of the phenomenon called 'Dispersion ofLight\

stand with lens is2 kg. Howmuch impulseP shouldbe imparted

Figure-5.278 showsa lightray of whitelight whichincidenton a transparent medium. As we know white light consist of all

to the stand so that a real image of the object is formed on the

colours sowecanconsider this lightrayas overlapping of light

screen after a fixed time gap. Also find this time gap also.

rays of all colours making the white light and for each colour

Geometrical Optics

1292

therefractive index ofthemedium will bedifferent and rayof colours but when this refracted beam emerges out oftheslab in each colour inwhite light will bend atdifferent angles and split airthen all the light rays inside the refracted beam will come out whenthe lightrayentersintothe medium as shown in figure.

at same angle ofemergence which isequal toangle ofincidence

so all'the rays become parallel after coming oiit and form a parallel white light beam with edges of violet and red colour only. Wecan alsoseethat dueto dispersion the emergent beam oflight is slightlythickerthan the incidentlight beam. glass

Figure 5.278

The situation shown in figure-5.278 is an ideal situation which doesn't happen in general practice because of the dimensions of the light ray. Actually a light ray is a theoretical concept which does not have any dimensional thickness but in practice we consider a very thin light beam as a light ray. Look at the figure-5.279 which is a magnified viewofthefigure-5.278 in which a light beam (enlarged view of light ray) incident on the medium. In this beam we can consider several light rays incident on the medium boundary at the same angle ofincidence. Due to dispersion each light ray will split into its colours and you can see that except the light rays at the boundaries of the beam all the colours in between will merge into each other and produce white light again. Only the edges of light beam will have violet colour at the left edge in the medium and red colour at the right edge in the medium. This is the reason why at the time of refraction, colours are not seen in the refracted light,beam.

Figure 5.280

5.19.2 Dispersion of White Light through a Glass Prism When a beam of white light passes through a glass prism then due to two refractions of beam of white light the colours in

beamgets separatedfromeachother and at somedistance from theprismafterrefraction ifweplacea screen, coloured spectrum is obtained as shown in figure-5.281(b).

Fkedfaeatn \ .

(a) y R f RV Ry

White Light Figure 5.279

5:19.1 Dispersion of White Light through a Glass Slab

Figure-5.280 shows a light beam incident on a parallel sided glass slab at some angle of incidence. Due to the concept explained in previous section we can see that ihe edges of the refracted beam (slightly divergent) are having violet and red

(b) Figure 5.281

[Geometjical Optics 293

Figure-5.281(a) shows how the coloured beams get separated beam. If the range ofwavelength is large then we take the after second refraction. We can see in this figure that at first difference ofrefractive indices corresponding to the wavelength

refraction at surface^5 ofthe prism, white light gets refracted limitsin the lightbeam. into the prism and the middle region between the violet and red,

edges will be white because ofmixing ofall the coloured light rays inthis region asexplained infigure-5.280 but atthe second refraction when all these rays incident on the face AC of the prism then each colour at theincident region 'w«' ofbeam on

face will emerge out atthe same angle like all violet rays in this region 'mn' will emerge out atsame angle as aparallel beam and all red rays ofthis region emerge out as separate parallel

beam because both colours will have different emergent angles. This situation is unlike to the case discussed in figure-5.280 when all coloured light rays emerge from the glass slab atsame angle and produce an emerging white light beam due to

Figure 5.282

overlapping of ail colours in same direction.

The concept ofdispersive power is used in taking corrective measures in image formation by refracting devices due to get separated after some distance and produce aspectrum on a dispersion. We'll discuss it later in the topic of 'Chromatic Figure-5.281(b) shows practical situation inwhich allthe colours

screen. •

Aberratioh\

5.19.3 Dispersive Power of a Prism Material

5.19.4 Dispersion Analysis for aSmall Angled Prism

Dispersive power ofa substance is its ability to disperse light Figure-5.283 shows a white light incident on a small angled passing through it.Dispersive Power ofa material is defined as

prism with prism angle^^. In general for awhite light we consider

relative deviation oflight beam from its mean path asafianction

its wavelength limits are bounded by violet and red rays and.

ofrefractive index ofthe material. For agiven material itis given mean wavelength is considered asyellow light. Iftherefractive

indices ofthe prism material for these colours are given as pp

as

...(5.148)

03 =

In equation-(5.148) angle 0is the mean deviation oflight for all the wavelengths present initand af0 isthe change indeviation

p^and Pyrespectively and 5p 5^^ and 5yare the corresponding deviation angles when these colours pass through the prism then therelation among these are given below. ?>y=A{^y-\)

due to variation ofrefractive index d\x which is corresponding

...(5.151)

to the wavelength change c(K in the beam. The dispersive power

for aprism material can be calculated by passing alight through

...(5.150)

by =A{^y-\)

...(5.152)

a smallangled prismas shown in figure-5.282. If weconsider a

light consist ofvery small wavelength range liol +dland itis passed through the small angled prism ofprism angle A as

shown, the mean deviation oflight for the wavelength Iis given as

e=^(n^i) r

/0 =Ad\x=

d\x Ll-I

Now from equation-(5.148) thedispersive power ofmaterial is given as

dQ 03 =

Figure 5.283

d\}i p-1

..(5.149)

Above equation-(5.149) gives the dispersive power ofa prism material which ishaving the wavelength range dkinthe light

In above figure mean deviation of the incident light ray is considered corresponding to themean wavelength present in

the light so here mean deviation ofincident light is taken as 8y

^^Geom'^ricaLOptit^l 294

-

The 'Angular Dispersion' for the light passing through the For theabove case Which we call thesetup ofprism combination prism is defined as the angle bywhich light will totally disperse. for 'Deviation without Dispersion' or 'Achromatic Prism Combination'. Inabove case thetotal dispersion iszero hence

Hereangulardispersion is given as

the angular dispersion by the two prisms must be equal. IfX)

D—5j/^ 5j^ —A(p-f/-* 1)— =>

and D'are the angular dispersion by the first and second prism

D=

...(5.153)

thenthe condition of deviation without dispersion is given as D=D'

For the prism shown in figure average dispersive power isgiven

...(5.155)

as

D co=

5y

\Xy

Ap

P/j

Py 1

...(5.154)

Pavg 1

In above case thetotalmean deviation ofthewhite lightcan be given as

Note: In somecasesif meanrefractive indexcorresponding to

yellow light isnot given then students can use py-

5™» = 5,-5/

^ 5.19.6 Direct Vision Prism Combination

5.19.5 Achromatic Prism Combination

When two prisms of materials having different dispersive powers are placed in relative inverted positions with their prism angles are chosen in such a way that dispersion produced by one prism gets compensated by the other one then a ray of white light passing through this combination gets refracted

When two prisms ofdifferent materials are placed in contact with relative inverted positions and their refractive indices and

prism angles are chosen in such awaythat emerging light beam will not suffer any mean deviation but having dispersion in samedirection then such a prismcombination is called 'Direct Vision Prism Combination'.

and deviated but in emergent beam no colors are seen or no

overall dispersion of light beam take place then such a Figure-5.285 shows combination oftwo prisms with inverted combination of prisms is called 'Achromatic Prism relative positions. The prism angles ofthese prisms are^ and ' Combination'. and having refractive indices for red and violet lights |ij,and

}i^'respectively. Ifthese values ofprism angles and refractive Figure-5.284 shows combination oftwo prisms placed incontact with opposite positions oftheir prism angles. The prism angles ofthese prisms are^ and^'.These prisms materials have their

indices are adjusted such that the mean deviation for yellow colour by first prism is exactly equal to that bysecond prism then the total deviation of the white light incident on this

refractive indices p^ and Pj^for first prism and p^'and p^'for combination willbezeroandthedispersed lightbeam will emerge the second prism corresponding to red and violet light out in the same direction of incidenceas shown in the figure. respectively. Ifthe values ofprism angles and refractive indices are adjusted such that the angular dispersion produced byfirst prism is equal to that produced by the second prism then the ray ofwhite light incident on the combination as shown in Yellow White Light figure will emerge out as while light only and will be free from colours. This happens because thenetdispersion produced by the first prism in the light is exactly nullified by the second

prism because itis placed inverted in respect ofthe first prism. Figure 5.285

'mean

White light

In above case ofdirect vision prism combination ifS^and fiy'afe the mean deviation oflightbythetwoprisms thenthe condition for this top happen is given as

A{v^y-\)= A'{\x'-\) Figure 5.284

...(5.157)

[Geomeffical-'Optics^'';''f;; ,

1295:

In this case ifD and £)'are the angular dispersions produced by the two prisms.then the net angular dispersion ofthe emerging light beam from theprism combination isgiven as

...{5.158)

5.20 Optical Aberrations in Lenses and Mirrors

Optical aberrations are the defects in image formation byoptical devices dueto deviation in performance ofthe device because

of non-paraxial rays or polychromatic light used in optical systems. Aberrations occur because all light rays from one point object do not converge (or diverge) at a single point after undergoing reflection orrefraction through the optical device inuse.- Inoptical systerhs the image produced gets blurred due toaberrations which causes unclear image fcamation bydifferent

(b) Figure 5.286

5.20.2 Methods to Reduce SphericalAberrations There are differentwaysby which correctivemeasurescan be

taken to reduce the spherical aberrations in image formation.

Some specific methods are given here for basic understanding, detailed analysis is not covered here.

corrective measures must be taken to avoid or minimize

(i) Use ofStops : Stops are the opaque planes having asmall aperture used to cut-offmarginal rays toincident on the optical

aberrations.

device as shown infigure-5.287(a) and(b).

optical instruments. While designing optical instruments

Optical aberrations are classified in two broad categories in general. These are ^SphericalAberrations' and 'Chromatic Aberrations'. Spherical aberrations are due to the curvature of

optical devices which causes marginal rays not to meet at the point of image formation by paraxial rays; and chromatic aberrations are due tothe presence ofseveral wavelengths of light in theincident beam which causes different wavelength rays to refract differentlyfrom the optical device. Chromatic

aberrations areconsidered only in refracting optical devices as reflecting devices are free from chromatic aberrations.' 5.20.1 SphericalAberrations

(a)

Due to large aperture ofmirrors and lenses when marginal rays • incident on the outerpart ofthe mirrorsor lenses, thesedonot converge at the point where paraxial rays are meeting and

forming theimage. Figure-5.286(a) and (b) shows theparallel light rays incident on a convex lens and a concave mirror. All

paraxial rays inthese cases will converge atfocal point butthe rays which are incident on outer edge' of the mirror or lens converge before focal pointand causes image to get blurred.

Due tothis reason a mirror ora lens having large aperture foils to produce sharp image ofan object or point image ofa point object.

•

• .• :

(b).

Figure 5.287

(ii) Using Parabolic Mirrors : For focussing parallel rays, spherical mirrors can be repla!ced by parabolic mirrors as

parabolic'mirrors focuses all parallel light rays foiling on itparallel (a)

toprincipal axis ata single focal point asshown'in figifre-5.288'.

Georrietrical-Optics' 1296

i; i

i i

Figure 5.290

In above figure, the focal lengths/^and/^ are given as Figure 5.288

1

'1

__1_^

A

^2 ,

1

"sing this combination

for image formation, spherical aberrations are minimized.

I '

...(5.159)

fy

(iii) Using Convex Lenses at some separation : When two convex lenses areseparated bya distance equal tothedifference

in their focal lengths {d= \f^ -/2I)

' 1

= (lifl-l)

/r

...(5.160)

Fromabove equations-(5.159) and(5.160) weget

Using Crossed Lenses : Crossed lenses are specially designed lenses with their radii ofcurvature ofthe two surfaces chosen for a specific case of image formation in such a wayto

ir" A We can multiply the RHS numerator anddenominator bythe

minimizesphericalaberrations.

term (py-1) where ^yis the refractive index for yellow (mean) colour in white light.

5.20.3 Chromatic Aberration in a Lens

As we have alreadydiscussed that chromatic aberrations are

due to presence of different colours (wavelengths) in the incident light beam which are refracted bythe optical device. Mirrors are free fromchromaticaberrationsas all coloursfollow

same laws ofreflection. Figure-5.289 shows that a parallel beam

of white light falling on a concave mirror always produces a bright white spot atits focus whereas for a lens its focal length depends upon the refractive index oflens which isdifferent for different colours.

1

J__,

/

/r

_L

-L

fv

/r

JJ

(Py-1)

Rj

(11

Ir "fv _ i\^y

White light.

(Mr-l)

(IVZM x(n -1) -L-1

1

fvfR

L

I Ad ^2 »J

^ 1

(liy-1)

fy ...(5.161)

Ir fy ^fy Asfor mean colour wecan use/^yj; = fy

Equation-(5.161) iscalled 'Longitudinal Chromatic Aberration' by a lens for white light. In general when a light consist of wavelengths ofsmall range from A. toX, + then corresponding

Bright white spot

variation in refractive index ofthe material oflens will be from p.

Figure 5.289

tod\i. Inthiscase thefocal length ofthelens for thelight can be

Below figure-5.290 shows thesituation when a parallel beam of white light incident on the convex lens, due to dispersion it splits into colours and produces several coloured images because focal length of all colours of this lens given by lens

maker's formula will be different. As )iy>

bylensmaker's

formula we get/^

(5000)^

= 1.232

sin /2 = 1.232x0.8=0.9856

i2 = 80°16' Deviation of beam Q is given as

5^ =/2-e =80°16'-53°=27°16' (c) As the two beams have light wave of different wavelengths, these are non coherent light waves. When two or more non coherent light waves superpose each other on a point, the average resulting intensity is the sum ofindividual intensities of the component waves. Thus here the resulting intensity of the light at focus is 47+7 = 57.

(iv) A crown glass prism of angle 5° is to be combined with, a flint prism in such a way that the mean ray passesundeviated. Find (a) the angle ofthe flint glass prism needed and (b) the angular dispersion produced by the combination when white light goes through it. Refractive indices for red, yellow and violet light are 1.514,1.517 and 1.523 respectively for crown glass and 1.613,1.620 and 1.632 for flint glass. [(a) 4.169°, (b) 0.0348°]

(v) Fora crown and flint glassfor CandF'lines p^= 1.515 and p^= 1.523 andP(,=1.644, p^= 1.664 respectively. Calculate the angle of flint glass prism which may be combined with crown glass prism having refi^acting angle 20p such that the combination is achromatic for C and Frays.

Web Referenceat www.phvsicsgalaxv.com

(vi) The prism ofa spectrometer has a refracting angle 60° and is made ofglass whose refractive indices for red and violet are respectively 1.514 and 1.530. A while source is used and the instrument is set to give minimum deviation for red. Determine (a) angle ofincidence, (b) the angle ofthe emergence for violet light and (c) the angular width ofspectrum.

Age Group - High School Physics | Age 17-19Years

[(a) 49°12', (b) 50°38', (c) TZb"]

Section - OPTICS

Topic - Dispersion of Light Module Number -1 to 12

(vii) An equiconvex lens of crown glass and an equiconvex lens of flint glass make an achromatic system. The radius of curvature of convex lens is 0.54 m. If the focal length of the combination for the mean colour is 1.54 m and the refractive

Practice Exercise 5,8

indices forthe crown glassare

= 1.53 and \Xy= 1.55, find the

dispersive power of the flint glass.

(i)

Two thin prisms are combined to form an achromatic

combination. For first prism ^ =4°, = 1.35, py=1.40, = 1.42. Forsecond prism p'^= 1.7, p'y.= 1.8 and p'^= 1.9find theprism angle ofsecond prism and the net mean derivation. [1.4°, 0.48°]

(ii) The index of refraction of heavy flint glass is 1.68 at 434 nm and 1.65 at 671 nm. Calculate the difference in the angle ofdeviation ofblue (434 nm) and red (671 nm) light incident at 65° on one side ofa heavy-flint glass prism with apex angle 60°. [2.8°]

(Hi) The dispersive power of crown and flint glasses are 0.03 and 0.05 respectively. The refractive indices for yellow light for these glasses are 1.517 and 1.621 respectively. It is desired to form an achromatic combination of prisms ofcrown and flint glasses which can produce a deviation of 1° in the yellow ray. Find the refracting angles ofthe two prisms needed. [4.8°, 2.4°]

[0.055]

(viil) How would you use two planoconvex lenses of focal lengths 0.06m and 0.04 m to design an eye-piece fi-ee fi-om chromatic aberration. What will be its focal length and magnifying power for normal vision ? Will it be a positive or negative eye-piece ? [0.048 m, 5.2, Negative]

(ix) An achromatic lens-double is formed by placing in contact a convex lens offocal length 20 cm and a concave lens offocal length 30 cm. The dispersive power ofthe material of the convex lens is 0.18.

(a) Determine the dispersive power of the material of the concave lens.

^

(b) Calculate the focal length ofthe lens-doublet. [(a) 0.27, (b) 60 cm]

[^bmetrlcal Optics' 5.21 Optical Instruments

-3or

Posterior chamber

Conjunctiva Sclcra

Optical Instruments are the devices vvhich utilize the

phenomenon ofreflection and refraction for image formation of various objects for their study indetail. All optical instruments aregenerally categorized intwo groups. One group ofdevices are those which produce real images of an object which is projected on ascreen oraphotographic plate and such images can be viewed simultaneously by many observers. Many different types ofProjectors and Spectroscopes belong tothis categorywhich we will not discuss in details as their construction

and working is not in scope ofthis book. The other group of devices are those which forms virtual image ofan object and only one observer can see the image. The virtual image formed by theinstrument istransformed by observer's eye into a real

Anterior chamber

Chomid

Retina

Tear?lm

N^treous

Cornea

Ciliary body, Optic nerve Canal of Schiemni Meibomian

glend Epithelium

Stroma

Endothclium

image on its retina. Some such optical instruments are also called as ' Optical Aids'" and such instruments are also used to

correct defects in human eye. Most commonly used optical instruments in this group are spectacles, microscopes and

Bowman's

Dcscemet's

membrane

membrane

telescopes which we study in detail. 5.21.1 TheHumanEye

In all types of optical instruments the observer's eye is an essential partasthe optical instruments we are going to study in coming sections, the analysis is based on observation of final image byobserver's eye so a good knowledge of human

eye is very important however in previous grades you might have studied it; Still for a quick reference we arediscussing the samehere again. Figure-5.293(a) showsa sectionalviewofthe (b) human eye which isnearly spherical in shape having an average Figure 5.293 diameter of2.5cm for an adult and figure-5.293{b) shows the front view ofa normal human eye. Therough outer protective When light rays enters into the eye through the cornea and and stiffskin ofeye ball is called 'Solera". Aithefront portion gets refracted from the eye lens and produces the image on a of eye, the sclera extends into a thin transparent membrane thin layer ofsensory cells inside the sclera. This thin lining is

which is called 'Cornea" in front ofthe eye lens. Justat theback called 'Retina" which acts asa screen for image formation. The

ofcornea isthe 'Iris", which isa textured pattern muscular ring.

surface ofretina ishemispherical inshape and it contains light

Iris can be of different colours which causes different color of

receptor cells called 'Rods" and 'Cones". These cells sense the

eye indifferent people. Atthe center ofiris there isan aperture imageproducedon retina and transmit it to human brain via the through which light incident on theeye lens. This aperture is 'Optic Nerve" shown in figure-5.293(a). called 'Pupil" anditsdiameter isvariable andit gets broader or shrink toadapt tochanging light intensity falling on eye. Behind the iris there is a crystalline 'Eye Lens" which is biconvex and made up of fibrous jelly. The eye lens is soft at the edges and

For an object to be seen sharply, the image must be formed

exactly at retina. For different positions of object, eye adjusts

to different object distances bychangingthe focal length of its lensj the distance between lens and retina does not change. hard atthecenter. This eye lens isheld bycircular ringshaped The focal length of eye is adjusted by varying the radius of 'CiliaryMuscles" at theouter edge in front oftheeye ball. The curvature of the lens by the ciliary muscles. This process by' region between theeye lens andcornea contains a liquid called which ciliary muscles changes the curvature of eye lens to 'Aqueous Humor" and behindthe lens eyeis filled witha water obtain sharp image of different .object is called based jelly type liquid called ' VitreousHumor".

'Accommodation".

- ^Geornetricaf'Optics-]

Human eye has limited capacity ofaccommodation which isin therange between ^Near Point' and 'Far Point'oftheeye. For a normal eye itsfar point is infinity. When eye focus.an object located at infinity then in this state ciliary muscles are fully relaxed. When eye focuses on an object placed at a short distance then the state at which ciliary muscles are fully

to eye itappears large and ifitisdisplaced away its size appear small dueto decrease in angular sizeof the object which it is

subtending on eye. Figure-5.295(a) and (b) shows an object of size 'h' which islocated atdistances x, andxjffom an observer's

eye. The angular size ofthe object as seen by eye in the two situations is given as

•

h 0, = —

contracted and eye lens produces sharp image on retina, this distance is called near pointoftheeye. Thisminimum distance at which an eye can see objects distinctly and clearly without getting tired is called the 'Least Distance ofDistinct Vision'

and

1 Xi

.

^ h 0, = —

X2'

In-

which is about 25cm for a normal eye. The range of

accommodation (near point tofar point) ofahuman eye gradually diminishes with ageandwith lifespan nearpoint andfarpoint

(a)

changes which results in defects of human eye called 'Presbyopia' and 'Myopia' which you have already covered in your previous grades. (b) 5.21.2 Camera

Figure 5.29^

Aphotographic camera isanoptical device which issimilar to Here angle 02 is less than 0, in the two positions ofthe same human eye in which image ofan object isproduced bya convex object and as we know the object which is located far away lens (camera lens) on photographic film instead of retina. fi-om us appear smaller thats why here angle 02 is less than Figure-5.294 shows the structural diagram ofa camera inwhich angle 0j. through the aperture light passes into the camera for short duration and produces a diminished real image on the photo sensitive film. The duration for which aperture opens and

5.21.4 Simple Microscope

captures the image decides the illumination ofthe image on N'Simple Microscope' or aG/as5'is aconverging film. lens of small focal length. When a magnifying glass, is held close to the object and.observer's eye is placed on the other side then in this situationthe size of virtual imageproducedis Shutter

larger then object and its distance isalso ferther firom the object as shown in figure-5.296(b). In this figure the angular size of image is same as that ofobject which is 0 as shown.

i (a)

Aperture

r

h'

Figure S.294

1 ^

5.21.3 Angular Size of Objects and Images

.f ^

%]•

h-w-A:

•v = D-

4

Whenthe observer seesan object directlyor byusing an optical instrument then the size of image appears to observer's eye is

analyzed byangular sizeofimage which also depends upon the distance where the image is located. Foran object located close

(b)

4

Georri'etricd-Opticsj

•

magnified inverted image i4'5' at.a distance vas shown. This Thus magnifying power ofthe compound microscope in normal image A'B' is obtained atadistance from the eyepiece which conditions is given as is less than focal length of the eyepiece so that it produces a

virtual magnified image CD atadistance equal tonear point of the observer's eye which is kept close to the eyepiece on the

=>

other side for viewing the image.

^n=-7.

14- —

...(5.176)

fs.

If in a compound microscope final image is produced at far

Eyepiece (X).

point ofthe eye then the magnification ofeyepiece is given by equation-(5.173)as

Objective (/J

D

fE

Thus magnifying power ofthe compound microscope for final imageat far.point ofeyeis given as

...(5.177) Figure'5.297

Figure-5.298 shows the situation when theimage produced by the objective A'B' isatthe focal point ofeyepiece due towhich thefinal image seen bytheobserver isatitsfarpoint (infinity).

The distance between two lenses, objective and eyepiece of

the'compound microscope is called Tube Length. We can'

Eyepiece (X)

X

5.21.8 'RibeLengthofa Compound Microscope

calculate tube lengthofa compound microscope in twd.cases

4\

when itproduces final image atthenear point and atfar point of

Objective (X,)

the observer's eye.

CaseT: When Image is produced at NearPoint

Thissituation isshown infigure-5.174 in which thetube length can be-given as

ufo

Figure 5.298

, %

...(5.178)

li-fo

Inequation-(5.178) all values are substituted inmagnitude as

5.21.7 Magnifying PowerofCompound Microscope

signs are already considered with symbols.

Overall angular magnification ofthecompound microscope is

the product oftwo factors. First isthe lateral magnification of CaseTI: 'When Image is produced at Far Point the objective, which gives the linear size ofits image ^'5'and this situation isshown infigure-5.175 in which thetube length second factor is theangular magnification oftheeyepiece which is given byequation-(5.173) andequation-(5.174) in the two cases ofimageformation at near point and far point ofthe eye.

w/o u-fo

The lateral magnification of the objective is given as -0 = - - = "

fo u-fo

...(5.175)

Fora compound microscope innormal conditions when image

is produced at near-point then magnification of eyepiece is given by equation-(5.173) D

can be given as

...(5.179)

In equation-(5.179) allvalues aresubstituted in magnitude as signs are already considered with symbols. 5.21.9 Refracting^^tronomicalTelescppe

A telescope-is-used;to view heavenly bodies. As heavenly bodies are very far away fi-om earth,: theysubtend very small angular size .on the observer's, eye looking at these so these

^Geprnetrical Optics 305

bodies appear very small in size. A telescope produces a For the normal adjustment oftelescope for final image obtained magnified image ofsuch bodies which is having large angular at infinityas shown in figure-5.299 the angular size ofimage can size so that eye can see more details ofthese bodies in enlarged be taken as p whichis given as' view ofimage. Atelescope which uses a lens as an objective is CdWoA

h

Refracting Telescope'.

...(5.181)

A simple astronomical telescope consists of two lenses- From equations-(5.180) and (5.181), the magnifying power of objective and eyepiece. The objective is a convex lens oflong telescope for normal adjustment is given as

focal length and itforms the real image ofdistant object infocal plane ofthe objective. Figure-5.299 shows the optical system ofa simple refi-acting telescope. The image AB formed'by

M

^

= -

a

objective acts as an object for eyepiece and based on distance ofeyepiece from thefocal plane ofobjective itproduces avirtual

and enlarged image ofAB either atnear point ofobserver's eye orat far point(infinity).

...(5.182)

' /£. In equation-(5.182) a negative sign is inserted as the image produced by objective is inverted. For the situation when

telescope produces final image atnear point ofobserver's eye asshown infigure-5.300 the image size p' isgiven as ...(5.183) , «£

From equation-(5.180) and (5.183), themagnifying power of telescope for image at near point ofobserver's eye isgiven as Figure 5.299

^

Above figure-5.299 shows the normal adjustment of lenses in

telescope inwhich the final image isproduced at infinity and observer's eye is in relaxed state which is the most common

adjustment for a telescope used in general and figure-5.300

Mp

is obtained at anear point Dfrom eyepiece offocal lengthy^ so bylens formula

is given as %

...(5.185)

Ur.—

at near point ofthe eye.

H*—fa—\_

Eyepiece - r •

...(5.184)

Infigure-5.-300, the distance w^is such that image byeyepiece

shows the situation when final image isproduced by eyepiece ^Objective (/q)

a

Substituting the value of

from equation-(5.i85) in

equation-(5.184) we get fo D/e

fo f1 ,

Figure 5.300

5.21.10 Magnifying Power ofa Refracting Telescope

...(5.186)

In equation-(5.186) a negative sign is inserted as the image

In all cases of telescopes always object is considered located produced by objective is inverted. far away fi-oih the device so we onlydiscuss in terms of the angularsizeof object and image. As shown in figure-5.299 or 5.21.11 T\ibe Lengthofa RefractingTelescope figure-5.300 the angular size of object can be taken as a and first image AB is produced in the focal plane of objective, if The distance between two. lenses, objective and eyepiece of the telescope is calledits' Tube Length'. We can calculate tube AB = h then we can use length ofa refi"acting telescope in two cases when it produces final image at the near pointand at far point of the observer's _h_ fo

...(5.T80)

eye.

Geometrical Optics 306

Case-I: When Image is produced at Near Point

5.21.13 Terrestrial Telescope

Thissituation isshown infigure-5.300 inwhich thetube length

Thisisalso an 'Astronomical Telescope'' which produces erected

can be given as

images. This is similar to a normal refracting telescope but it uses anerecting lens between objective and eyepiece asshown j

=/• + —

in figure-5.302. The disadvantage ofthis telescope isits large ...(5.187)

In equation-(5.187) all values are substituted in magnitude as

tube length duetowhich it is rarely used. Objective

signs are already considered with symbols. Case-II; When Image is produced at Far Point Erecting Lens (/,)

Thissituation isshown infigure-5.299 in which thetube length can be given as

Figure 5.302

5.21.12 Reflecting Telescope

In a reflecting telescope the objective isreplaced with a large concave mirror. This is an 'Astronomical Telescope' which is used for astronomical observations. To study the details of distant astronomicalbodieslarge apertureofobjective has many

advantages. Large aperture causes brighter image with fine details intheimage produced andmounting oflarge sized mirror ismuch easier compared tomounting ofa large lens. Figure-5.301 shows theworking ofa reflecting telescope innormal adjustment

for obtaining the final image atinfinity. We can see that thefinal image in a reflecting telescope is formed in theregion where incoming rays are traveling in tube of telescope by placing a

Thetube length ofa Terrestrial Telescope is L=fQ + 4/,

Where/, is the focal length ofthe erecting lens. 5.21.14 Galilean Telescope

This isone among theoldest known telescopes made by Galileo Galilei in 1609.This uses a concave lens as eyepiece as shown

in figure-5.303 which shows the normal adjustment of the telescope. The advantage ofthis telescope isshorter tube length and it gives erected image without using theerecting lens.

small mirror at an angle 45® to the axis of tube and placing eyepiece at the side of the tube.

H

fo-

Figure 5.303

# Illustrative Example 5.87

Figure 5.301

A short-sighted man, the accommodation of whose eye is between 12cm and 60 cm wears spectacles through which he can see remote objects distinctly. Determine the minimum distance at which the man can read a book through his

Refracting telescope uses a thin lensas objective and lenses of spectacles. largeaperture cannot beproperly manufactured. If made then due tochromaticand spherical aberrations distorted imagesare Solution formed. Duetouseofparabolic mirrors asobjective inreflecting

telescope these arealmost free from chromatic and spherical aberrations.

As perthe given situation, a man can manage to see objects clearly ifplaced between 12cm and60cm (accommodation of

{Geometrical Optics 307

eye). If v is the distance between eye lens andretina then the 54

focal length of eye lens when an object is placed at 60cm distance is given by lens formula, used as

13.5

1 -60

M=

...(5.189)

L

25

X (54/131) TT 2.5 =-327.5

Ifheuses spectacles with focal length ofitslens f then hecan # Illustrative Example 5.89 see far objects clearly, that means for far objects the combination of eye lens and spectacles lens produces the Atelescope has an objective offocal length 50 cm and eyepiece

image at retina at distance v from eye lens then for the of focal length 5 cm. The distance of distinctvision is 25 cm.

combination ofthe lenses, we use lens formula as

i-1 =-L 1

The telescope is focussed for distinct visionon a scale 200 cm awayfromthe objective. Calculate ...(5.190)

^ ^ ///

From equation-(5.189) and (5.190), weget

(i) The separation between the objective and eyepiece, (ii) Themagnification produced.

/=-60 cm

Solution

For the near point ofthe eye at 12cm ifthe focal length ofthe The situation isshown infigure-5.304 with raydiagram,

eye lens is^^' then bylensformula we use 1 -12

...(5.191)

/;

(i) Ifthe separationbetween the two lenses bex then for lens formula forrefraction at objective lenswe use

WQ =-200cm and /Q =+ 50cm

Ifthe minimum distance at which the person can read a book

clearly isplaced at a distance/) from eye then with spectacles

From lens formula, we have

lenses, we use lens formula as

]_

1

1

J—— = J-

1_

^0

"o «o

To

...(5.192)

V ~D~ f' -60

J

1_

—+—

Fromequation-(5.191) and(5.192), weget

/o

50

«0

4-1

/) = 15cm.

Illustrative Example 5.88

3

200 •

^0

200

200

200

Vo =+- 3

cm

The focal lengths ofthe objective and eyepiece ofamicroscope Thus a real image is formed at a distance of 200/3 from the are4 mmand25mmrespectively, andthelength ofthetube is objective. This image acts as object for the eyepiece. For 16 cm. If the final image is formed at infinity and the least

refraction through eyepiece, we use

distance of distinct vision is 25 cm, then calculate the magnifying power of the microscope.

^0" =_

X-

3

v^ = -25 cmandj^= +5 cm

Solution

J

1-

Magnificationof microscope is given as

I

200 "l ~ 5

25

X--

M=^

£

"o

Here

v^, = 16-2.5= 13.5 cm

and

j^=+4mm =+0.4cm

Objective

*

200cni

Using the lens formula, we have

^0

^0

/o

Figure 5.304 X

Geometrical Optics' 308

Let Vo'be the new distance ofthe image formed bythe objective thenbyusing lensformula again, wehave

_i _L--1

200 ^ ~ 5 25 ~ 25 1

X--

_J_ J_ _ _i 1 - -L

3"^5~15

Vo'

6x-400=25

VQ'=+15cm

6x=425 . .

Thus the image isshifted fi-om the objective through adistance

425

15 cm-6cm = 9cm. Sothe eyepiece shouldbe movedaway

x= ~-r~ =70.80.cm 6

from the objective by 9cm to refocus the image atsame position. 200

(ii) Magnification of Objective

3x200

V.

3

25x6

Magnification of Eyepiece =

= 6

#Illustrative Example 5.91 In a compound microscope the objective and eyepiece have focal lengths of0.95 cm and 5 cm respectively, and are kept at a distance of 20cm.The final imageis formed at a distance of

Total magnification^

25cmfrom eyepiece. Calculate theposition oftheobject and

^ ^ 2-

the total magnification. ' Solution

#Illustrative Example 5.90

Acompound microscope is used to enlarge an object kept ata

From the lens formula for eyepiece, we use

distance of 0.03 m from its objective which consists ofseveral

Ve =-25cm andX"•'"5cm • e .

convex lenses in contact and has focal length 0.02m. Ifa lens

1 __i___i

offocal length 0.1 misremoved from theobjective, find out the distance by which theeyepiece ofthemicroscope must moved

^ " V, X " 25 5 " 25 =>

to refocus the image.

L-i = -A

Wp =-(25/6)cm

For the objective, we use Solution

X=0.95cm and VQ=20-(25/6) =95/6cm

Sol. Let Vq be the distance ofthe image formed by the objective Using lens formula, we have alone so by lens formula for objective, we use

Vo

"0

Vq

Wf)

/o

Vq

/o

Mq

/o

_1

1__^_1_

Vq

Wo

\

I

Vq

3 2

/o Wo

6

5

1

6-100

0.96

95

_94

Vp=+6cm Wq

Theimage isformed at 6 cmbehind theobjective. If /q' bethe new focal length of the objective when a lensof focal length

95 95

«D='5Jcm

0.1 m(lOcm) is removed from it,then forcombination, we can use

/o'

_1_

1

_1_

/o

10

10

/;=+(5/2)cm

Total Magnification

M=^ Wo

1.^ L

(95/6) M=

-(95/94)

1.^ 5

=-94

{Geometrical Optics

309:

# Illustrative Example 5.92

If/be thefocal length ofthe correcting lensfor seeing distant objects then by lens formula for combination, we have

The focal lengths of the objective and the eye-piece of an astronomical telescope are 0.25 m and0.02m, respectively. The ...(5.194) V CO /^ / telescope is adjusted to view an object at a distance of 1.5m from theobjective, thefinal image being 0.25m from theeye of Subtracting equation-(5.193) from equation-(5.194), weget the observer. Calculate the tube length of the telescope and /=-2m

the magnification produced by it.

Thus power ofthe lens required is Solution

1

1

P = — = — =-Q.5 dioptre For objective, by lens formula, we have

J Vn

I_ -1.5

l__

.Web Reference at www.Dhvsicsgalaxy.com

+0.25

I Age Group- High School Physics | Age 17-19 Years

;r=="333 =>

Section - OFnCS

Topic - Optical Instruments Module Number - 1 to 20

Vg=+0.3m

For eyepiece, by lens formula we have Practice Exercise 5.9

1

1

1

-0.25

-u,

+0.02

=>

1

I

1

Mg

0.25

0.02

(!)

Find (i) the size ofthe picture on the screen and (ii) the ratio of illumination of the slide and of the picture on the screen.

Mg=+0.01852m

The tube length of the telescope is given as

[2401]

1 = 0.3 + 0.01852 = 0.31852 m.

(ii) The focal length of the objective of a microscope is = 3mmandof the eyepiece^^ = 5 cm.An object is placed at a distanceof 3.1 mm fromthe objective. Find the magnification ofthemicroscope fora normaleye, ifthefinalimageisproduced at a distance 25 cm from the eye (or eyepiece). Also final the separation between the lenses of microscope.

Magnificationby objective is V,

A projector lens has a focal length 10 cm. It throws an

image ofa 2cm x 2cm slide on a screen 5 metre from the lens.

0.3

'"»=^=i?=" Magnification by eyepiece is 0.25

[180, 13.46 cm]

Total Magnification

= m, x Wj = 27.39

a Illustrative Example 5.93

A short-sighted person cannot see objects situated beyond 2m from him distinctly. What should be the power of the lens which he should use for seeing distant objects clearly? Solution

(iii) The optical powers of the objective and eyepiece of a microscope are equal to 100 D and 20 D respectively. The microscope magnification is equalto 50when imageisproduced at near point of eye. What will be magnification of the microscope be when the distance between the objective and eyepiece is increased by 2 cm ? [62]

(iv)

If^ be the accommodate focal length of the eye-lenses for 2m, then by lens formula, we have 1 -2

fe

...(5.193)

Where v is the distance between eye lens and retina.

The focal lengths of the objective and the eye-piece of

an astronomical telescopeare 0.25m and 0.025 m, respectively. The telescope is focussed on an object 5m from the objective, the final image being formed 0.25 m from the eyeofthe observer. Calculate the tube length of the telescope and its magnifying power.

[0.2859m, 11.6]

Geometrical Optics 1

(v) An astronomical telescope consisting of two convex lensesoffocal length 50 cm and 5 cm is focussed on the moon. What is the distance between the two lenses in this position ?

If the telescope is then turned towards an object 10 m away, how much would the eye-piece have to be moved to focus on the object without altering the accommodation of the eye? Calculate the magnification (angular) produced by the

(vii) An astronomical telescope in normal adjustment has a tubelength of93cmand magnification (angular) of30. If the eye-piece is to be drawn out by 3 cm to focus a near object, withthe final imageat infinity, findhowfar awayis the object and the magnification (angular) is this case. [27.9m, 31]

telescope in the two adjustments. [10]

Advance Illustrations Videos at www.phvsicsgalaxv.com

(vO The eyepiece and objective of a microscope, of focal lengths 0.3 m and 0.4 m respectively, are separatedbya distance of0.2 m. The eyepiece and the objective are to be interchanged such that the angular magnification of the instrumentremains same. What is the new separation between the lenses ? [0.2575 m]

AgeGroup - Advance Illustrations Section - OPTICS

Topic - Geometrical Optics Illustrations - 46 In-depth IllustrationsVideos

Geometrical Optics

311

Discussion Question Q5-1 What is the function of the circular stop at the focal

Q5-16 How focal length of a spherical mirror changes when

plane of the objective of a telescope? Give reasons.

placed in different media.

Q5-2 Abeam ofwhite light passing through ahollow prism Q5-17 An air bubble inside water broadlybehaves as aconcave gives nospectrum. Isthis true orfalse? Give reasons. Q5-3 The magnifying power of a telescope in normal adjustment is greater than that when it is focussed for least

js

qj. f^igg .

Q5-18 Why does an aeroplane flying at a great altitude not cast a shadow on the earth ?

distance of distinct vision. Is this true or false? Give reasons.

Q5-4 Can youthink ofa specific opticalsetupwith a trihedral prism (no other device) in which a light ray passes undeviated through the prism. Think and draw the ray diagram. Q5-5 Can a real imagebe photographedbya camera ? Q5-6 Can you obtain image produced by a convex lens on a

Q5-19 The magnifying power of a telescope in normal adjustment is greater than that when it is focussed for least distance of distinct vision. Is this true or false?

Q5-20 When sunraj^ passthrough a smallholein thefoliage at the top of a high tree, they produce an elliptical spot of light on the ground. Explain why. When will the spot be a circle?

screen without using any other device.

Q5-21 Ifa mirrorreverses right and left, whydoesn't it reverse Q5-7 It is difficult to thread a needle with one eye closed.

up and down?

Why?

Q5-22 Is it possible to photographa virtual image ? Q5-8 If a single lens is used to forman image, it is better to use a lensof largediameter,in which the outerparts near the rim are blocked off. Explain.

Q5-23 Is it possible for a given lens to act as a converging lens in one medium, and as a diverging lens in another?

Q5-9 What is the bestposition of the eyeforviewing an object

Q5-24 A cameralens is markedJJ\.8. What is the meaningof

through a microscope?

this mark?

Q5-10 Can the optical length between two points ever be less than the geometrical path between these points?

Q5-25 Some motor cars have additional yellow headlights.

Q5-11 Can two lenses of the same material produce achromatismwhen placed in contact ? Explain.

Q5-26 Why are lenses often coated with thin films of

Q5-12 Whyis the objective of a telescope of large focal length and large aperture?

Q5-13 A iiver inside the sea observes a ship on the water

Why?

transparent material ?

Q5-27 If there are scratcheson the lens of a camera, they do not appear on a photograph taken with the camera. Explain. Do the scratches affect the photograph at all?

surface. Does he finds the ship taller or smaller then its actual height above the water surface. Give reason and draw ray diagram to support your logic.

Q5-28 The sun seems to rise before it actually rises and it seems to set long after it actually sets. Explain why.

Q5-14 A plane projectoris projectinga sharp still image on a

Q5-29 Doesthe focal length of a lens depend on the medium

screen, the image consists of objects of several colours. Can

in which the lens is immersed? Is it possible for a lens to act as

you comment on sharpness of all these coloured objects in image.

a converging lens in one medium and a diverging lens in another

Q5-15 Why does the moon, purely white during the day,have

Q5-30 Explain why the use of goggles enables an underwater swimmer to see clearly under the surface ofa lake.

a yellowish hue after sunset?

medium?

Geometrical Optics i

;312

ConceptualMCQs Single Option Correct 5-1 A beaker containing liquid is placed on the table underneatha microscope which can be movedalong a vertical scale. The microscopeis focussed, through the liquid onto a mark on the table when the reading on the scale is a. It is next focussed on the upper surface of liquid and the reading is b.

5-4 Two planemirrors oflengthI. are separated bydistance/, and a man

is standing at distance L from the connecting

lineofmirrors asshown infigure-5.306. AmanAf, iswalking in a straightlineat distance 2Lparallel tomirrors at speed u,then man at Owillbeabletoseeimage ofM, for total time:

More liquid is added and the observations are repeated. The corresponding readings are c and d. The refractive index of

IM,

I

liquid is: d-c-b+a

d~b

(A)

(B)

d-c~b + a

'd^b d-c-b+a

b-d

(D)

(Q d-c-b+a

5-2 An insect ofnegligible mass is sitting on a block of mass M, tied with a spring of force constant k. The block performs simple harmonic motion with amplitude^ in front of a plane mirror placedas shown in figure-5.305. The maximum speedof insect relative to its image will be:

-21-

Figure 5.306 3L

4L

(B)-

(A) '///////A

9L

6L

P)

(Q —

u

5-5 A point source has been placed as shown in the figure-5.307. What is the length on the screen that will receive reflected light from the mirror ?

Figure 5.305

H

(C)

^

n

'7777777777777: -H-

5-3 Inside a solid glass sphere ofradius R, a point source of light is embedded at a distance x(x < K) from centre of the

4

-IH-

Figure 5.307

sphere. The solid sphere is surrounded by air of refractive index 1.0. The maximum angle ofincidence for rays incident on

(A) 2H

the spherical glass-air interlace directly from the point source

(Q H

(B) 3H (D) None of these

is:

(A) cos-' —

(B) sin -1 A

(Q cos-\/-

P) sir-' dj

R

5-6 A thin lens offocal lengthfand its aperture has a diameter d. It forms an image of intensity I. Now the central part of the aperture upto diameter {dH) is blocked by an opaque paper. The focal length and image intensity would change to

(A)/2,//2

'CB)///4

(Q 3flMI2

P)/3//4

jGeometrical Optics

313

5-7 A ray of light fallson a plane mirror. When the mirror is turned, about an axiswhich is at right angletothe planeofthe mirrorthrough 20°, theangle between the incident rayandnew reflected ray is 45°. The angle between the incident ray and original reflected ray was; (A) 65° (Q 25°or 65°

(B) 25° (D) 45°

5-10 Aninfinitely long rectangular'strip isplaced on principal axis of a concavemirror as shown in figure-5.310. One end of the strip coincides with centre of curvature as shown. The

heightofrectangular stripis verysmallin comparison to focal length ofthemirror. Then theshape ofimage ofstripformed by concave mirror is similar to a:

5-8 In the figure shown-5.308, lightis incidenton the interface

between media 1(refractive index p,)and 2 (refractive index Pj) at angle slightlygreaterthan the critical angle, and is totally reflected.The light is then also totallyreflected at the interface

between media 1and 3(refractive index P3), after which ittravels in a direction opposite to its initial direction. The media must have a refractive indices such that:

Figure 5.310

(A) Rectangle (Q Triangle

(B) Trapezium (D) Square

media 1

5-11 A concave spherical surface of radius of curvature 10cm separates two mediums X and Yof refractive indices 4/3 and 3/2 respectively. Centre ofcurvature ofthe surface lies in the medium X. An object is placed in medium X: (A) Image is alwa>^real (B) Image is real ifthe object distance is greater than 90 cm (Q Image is always virtual (D) Image is virtual only if the object distance is less than

media 3

90 cm

Figure 5.308

5-12 In the figure.shown-5.311 a slab of refractive index —is

(A) p,pf

5-9 Figure-5.309 shows a spherical cavityin a solid glassblock. The cavity is filled with a liquid and from outside an observer sees the distance AB which is the diameter of the cavity and it

moved at speed 1m/s towards a stationary observer. A point 'P' is observed by the observer with the help of paraxial rays through the slab. Both 'O' and observerlie in air. The velocity with which the image will move is:

appear as infinitely large to the observer. Ifrefractive index of

liquid is p. and that of glass is p,, then — is : ^2

Glass

Figure 5.311

(A) 2 m/s towards left

(B) —m/s towards left

(Q 3 m/s towards left

(D) Zero

5-13 Light passes from air into flint glass with index of refraction n. The angle of incidence must the light have for the component of its velocity perpendicular to the interface to Figure 5.309

(A) 2 (Q 4

(B) 1/2 (D) None of these

remain same in both mediums is:

(A) sin"'« (Q COS"' n

(B) sin"'(l/«) (D) tan"' n

Geomefricaj^ pities j

1^1:

5-14 Inthefigure-5.312M, and are two fixed mirrors shown. Ifthe object' O" located between thetwo mirrors moves towards theplane mirror, then the image I which is formed after two successive reflections first from Mj & then from

5-17 Apoint object ismoving along principal axis ofaconcave

respectivelywill move:

the number oftimes at which the distance between object and its image is 40 cm are.

mirrorwithuniform velocity towards pole. Initially theobject is at infinite distance from pole on right side of the mirror as

shown inthe figure-5.315. Before theobject collides with mirror,

object 0 z

6M,

Figure 5.312 Figure 5.315

(A) Always towards right (B) Always towards left (Q Depends on position of O P) Cannot be determined

p) Two times P) Data insufficient

(A) Onetime (Q Three times

5-15 A person AB of height 170cm is standing in fi-ont of a plane mirror. His eyes are at height 164 cm. Atwhat distance

5-18 In the figure shown-5.316, the maximum number of

firom P should a hole be made in the mirror so that he cannot

reflections light rays will undergo are: M,

see the top ofhis head :

B

Figure 5.316

P

Figure 5.313

(B) 161an

(A) 167cm (Q 163cm

(B)3 P)1

(A) 2 (Q 4

P) None of these 5-19 A convex lens is cut into two parts in different ways that

5-16 In the figure shown-5.314, blocks P and 0 arein contact but do not stick to each other. The lower face ofP behaves as

are arrangedin four manners, as shown. Which arrangement will givemaximum opticalpower?

a plane mirror. Thesprings are in theirnatural lengths. The system is released from rest.

(A) I

Ua

(Q

Figure 5.314

Then the distance between Q and its image when Q is at the lowest point first time will be: 2mg

(A)

K

^mg (Q

K

4OTg

(B) P)0

K

5-20 A parallelbeam of light passes parallelto the principal axis and falls on one face ofa thin convex lens offocal length/ and after two internal reflections fi"om the second face forms a

real image. The distance of image fi-om lens if the refi-active index ofmaterial oflens is 1.5 :

(A)//7 {€) If

.

P)//2 • p) None of these

jGeometrical Op^cs

5-21 An object and aplane mirror are shown in figure-5.317.

5-24 The position ofareal point object and its point image are

Mirror is moved with velocity Vas shown. The velocity of asshown in lhefigure-5.320./15 istheprincipal axis. Thiscan be achieved by using :

image is:

Or

• Object (fixed)

T a

A

[ 1

^

a

i

Mirror

Figure 5.320

Figure 5.317

(A) 2Ksin0

(B) IV

(Q 2Fcose

P) None of these

5-22 A prism ofangle ^4 and refractive index 2 is surrounded

bymedium ofrefractive index >/3 .Aray is incident on sidePg at an angle of incidence / (0 < / < 90®) as shown in the figure-5.318. The refracted ray is then incident on side PR of prism.The minimum angle^4 ofprism forwhichrayincidenton

sidePQ does not emergeout of prism from sidePR (for any value of/) is:

(A) Convexmirror (Q Planemirror only

(B) Concave mirror

p) Convexmirror only

5-25 The image of the moon is produced bya convex lensof focal length/ Thearea ofimage is directly proportional to: (A)/ (B)f (Q 1// P) yf ' 5-26 A converging lens of focal length 20 cm and diameter 5 cmis cutalongthe VmeAB. The part ofthe lensshown shaded inthe diagram isnowused to form an image ofapoint/*placed 30 cmawayfrom it on the lineA7, whichis perpendicular tothe plane ofthe lens. The image ofP will be formed:

Normal

T 2 cm

2^

a Incident ray 5 cm

X

•30 cm

Figure 5.318

(A) 30® (C) 60®

(B) 45® (P) \2QP

5-23 Two plane mirrors are joined together as shown in figure-5.319. Two point objects O, and are placed symmetrically such that AO^ = AO2. The image of the two objects is common if:

Figure 5.321

(A) 0.5 cm above AT (C) on AT

P) 1 cm belowAT P) 1.5 cm belowAT'

5-27 A man stands on a glass slab of height h and inside an elevator accelerated upwards with 'a'. If is refractive index of glass then the bottom ofthe slab appears to have shifted with respect to the man by a distance ; (A) less than P)' greater than (Q equal to h/\i P) can't be said

5-28 When anobject isata distance m, and

from theoptical

centre ofa lens, a real and virtual image are formed respectively,

with the same magnification. The focal length oflens is: (A) (u. + M,) Figure 5.319

(A) 0 = 60= (Q 0 = 30=

(B) 0 = 90= (P) 0=45=

(Q

«|+«2

P) «,+ y «i —U-

P)

Geometrical

5316

5-29 A liquidof refractive index 1.33 is placed between two 5-32 Aplane mirror having amass mistied tothe free end ofa identical plano-convex lenses, with refractive index 1.50. Two massless spring of spring constant k. The other end of the

possible arrangement P and Qareshown infigure-5.322. The spring is attached to a wall. Thespring with the mirror held vertically tothefloor onwhich itcanslide smoothly. When the system is: spring isatitsnatural length, themirror isfound tobemoving at a speed of v cm/s. The separation between the images of a man standing before themirror, when themirror isin itsextreme positions:

Wall

Figure 5.324 '

Figure 5.322

(A) Divergentin P, convergent in Q (B) Convergent in P, divergent in 2 (Q Convergent in both QD) Divergent in both

\m

V

(A) Vjy \m

(Q 2vJj

m

m

(D)4vj-

5-30 Twoparticles^&Pofmass Wj and^2 respectively start moving from O with speeds v, and V2. A moves towards the 5-33 Anobserver cansee, through a pin-hole, thetop endofa planemirrorand B moves parallelto mirrorhorizontally. The' thin rod of height h, placed as shown in the figure-5.325. The mirror is iny-z plane.The absolute-speed ofimageof centre of mass of the system (image of ^4 + image ofB) is:

beakerheight is 2h and its radiusis h. Whenthe beakeris filled with a liquid up to a height 2h, he can see the lowerend of the rod. Then the refractive index ofthe liquid is:

Figure 5.325

Figure 5.323

(A) 5/2

(B)

(Q -JJ/T.

(D) 3/2

WiVi

(A) Zero

(B)

m2

5-34 A ray oflight isincidenton the leftvertical face ofa glass cube ofrefractive index

(Q

2^2

(D) None of these

m,

5-31 A mango tree is at the bank of a river and one of the

as shown in figure-5.326. The plane

ofincident is the plane of the page, and the cubeis surrounded

by liquid of refractive index p,. What is the largest angle of incidence 0j for which total internal reflection occurs atthetop surface?

branch oftree extends over the river. A tortoise lives in'river. A

mango falls just above the tortoise. The acceleration of the mango falling from tree appearing to the tortoise is (Refractive index ofwater is 4/3 and the tortoise is stationary): (A) g

(B) f

4g (Q —

(D) None of these Figure 5.326

. Geometrical Optics

(A) sin 1 =

317

P) 45=

(Q 60°

p) 60°-sin \lj

(B) sin 1 = /

(Q sinl

(A) 30°

(D) sin 1 =,

\2.

-^1 +1 >^2

5-35 Twoplane mirrors are inclined to each other at angle 0. A ray oflight is reflected first at one mirror and then at the other. Find the total deviation ofthe ray: (A) 360°-2e (B) 36O'' + 20 (Q 180°-28 P) 18O'' + 20 5-36 Two plane mirrors are inclined to each other such that a ray oflight incident on the first mirror and parallel to the second is reflected fi^om the second mirror parallel to the first mirror. Determine the angle between the two mirrors: (A) 60° (B) 30° (Q 90° (P) 180°

5-39 The mirror of length L moves horizontally as shown in thefigure-5.329 with a velocityv.The mirror is illuminatedbya point source of light 'P' placed on the ground. The rate at which the length ofthe light spot on the ground increases is : r*—L—H

bvWWWN

V,

Wall

Figure 5.329

p) Zero P)3v

(A) V (Q 2v

5-40 Monochromatic light rays parallel tox-axis strike a convex 5-37 A slab of high quality flat glass, with parallel faces, is placed in the path ofa parallel light beam before it is focussed to a spot by a lens. The glass is rotated slightly back and forth from the dotted centre about an axis coming out of the page, as shown in the diagram. According to ray optics the effect on the focussed spot is:

lenses ofrefractive index 0.5. Ifthe lens oscillates such that

AB tilts upto a small angle 0 (in radian) on either side ofy-axis, then find the distance between extreme positions ofoscillating image:

Figure 5.330

Rotating glass Figure 5.327

(A) There is no movement ofthe spot (B) The spot moves towards then away from the lens (C) The spot moves up and down parallel to the lens P) The spot moves along a line making an angle a (neither zero nor 90°) with axis oflens

5-38 Aparallel glass slab ofrefractive index >/3 isplaced in contact with an equilateral prism ofrefractive index -Jl. Aray is incident on left surface of slab as shown. The slab and prism combination is surrounded by ain The magnitude ofminimum possible deviation ofthis ray by slab-prism combination is:

(A) /secO (Q /(secO-1)

P) / sec^O P) The image will not move

5-41 The focal length ofa concave mirror is/and the distance from the object to the principal focus is x. Then the ratio ofthe size ofthe image to the size ofthe object is: (A)

if + x) f

/ P) -

X

(D) ^ 5-42 A light ray gets reflectedfrom a pair ofmutually-Lmirrors, not necessarily along axes. The intersection point ofmirrors is at origin. The incident light ray is alongy = x+2. Ifthe light ray strikes both mirrors in succession, then it may get reflected finally along the line: (A)y-2x-2 p)y = -x + 2 (Q y = -x-2 p)y = x-4

Figure 5.328

Geometrical Optics

1-318

5-43 A ray is incidenton the first prism at an angle ofincidence 53® as shown in the figure-5.331. The angle between side CA and B'A' for the net deviation by both the prisms to be double of the deviation produced by the first prism, will be: A

i=53°

Figure 5.331

, 2

, 2

(A) sin-' 2 +53'

(B) sin-' 2+37'

(Q cos-'2+53®

(D) 2sin"' -j

♦

+

*

*

+

IGeometrical Optics

NumericalMCQs Single Options Correct 5-1 An equilateral prism deviates a raythrough 45® for two angles of incidencediffering by20®. The refractive index of the prism is:

(A) 1.567 (Q U

(B) 1.467 (p) 1.65

5-2 Two plane mirrors I, and

5-5 Two plane mirrors AB andAC are inclined at an angle 0= 20°. Arayoflight starting from point Pis incident at point 0 on the mirror AB, thenatR onmirror ACand again onSon AB. Finally the ray STgoes parallel to mirror^C. The angle which theraymakes with the normal atpoint Qonmirror AB is:

areparallel toeach other and

3m part. A person standings m fromthe right mirror looks into this mirror and sees a series of images. The distance between the first and second image is 4 m. Then the value ofx is:

y2^/777Z777777Z^7777^^^ A

PR

Figure 5.334

(A) 20® (Q 40®

vv

P) 30° .

\*—x-

P)60®

5-6 A person's eye level is 1.5m. He stands in front of a 0.3m Figure 5.332

(A) 2m (Q Im

longplanemirrorwhich is0.8m above theground. Thelength ofthe image he sees ofhimselfis :

(B) 1.5m (D) 2.5m

5-3 An elevator at rest which is at 10''^ floor of a building is

(A) 1.5m (Q 0.8m

p) 1.0m P) 0.6m

having a plane mirror fixed to its floor. A particle is projected

5-7 A plane mirror of length 8 cm is present near a wall in situation as shown in figure-5.335. Then the length of spot

with a speed V2 m/s and at45® with the horizontal asshown

formed on the wall is:

in the figure-5.333. At the very instant ofprojection, the cable of the elevator breaks and the elevator starts falling freely. What will bethe separation between the particleand is image

,

wall

1 5 = source oflight d

i

0.5s after the instant of projection ?

V77777777/. H—8cm—H

Figure 5.335

u = 'J2 m/s

(A) 8 cm (Q 16cm

P) 4 cm P) None of these

5-8 Anobject 'O' is keptinair in front ofa thinplano-convex lens ofradius ofcurvature 10 cm. It's refractive index is 3/2 and Mirror

Figure 5.333

the medium towards right of plane surfeceis water ofrefractive

index 4/3. What should bethe distance 'x' ofthe object sothat (A) 0.5m (Q 2m

(B) Im p) 1.5m

the rays become parallel finally. M

JT

»•

5-4 Aplane mirror is moving with velocity 4f+4/ +8^. A point object in front of the mirror moves with a velocity

1

'ii+Aj + Sk . Here k is along the normal to the plane mirror

«,-3/2A|

and facing towards the object. The velocity of the image is :

rt-fz

^—

Figure 5.336

(A) -3^'-4j +5A

(B) 3f+4y+llit

(Q ~4i+5j + Uk

p) 7i+9j+3k

(A) 5 cm (Q 20cm

P) 10 cm (D) None of these

Geornetrical Optics |

5-9 A diverging lens of focal length 10cm isplaced10cm in front of a plane mirror as shown in the

5-13 The curvature radii of a concavo-convex glass lens are, 20 cm and 60 cm. The convex surface of the lens is silvered. With the lens horizontal, the concave surface is filled with

figure-5.337. Lightfrom a veryfar away source falls on the lens. The

water; The focal lengthof the effective mirror is (p of glass= 1.5, pofwater = 4/3):

I

final image is at a distance:

(A) 20cmbehindthe mirror

(A) 90/13 cm (Q 20/3 cm

(B) 80/13 cm p) 45/8 cm

Figure 5.337

7.5 cm in front ofthe mirror

5-14 Aplano-convex lens, when silvered at itsplane surface isequivalent toa concave,mirror offocal length 28cm. When

(C) 7.5 cm behind the mirror (D) 2.5 cm in front of the mirror

its curved surface is silvered and the plane surface not silvered, 5-10 A fish in near the centre of a spherical fish bowl filled

it is equivalent to a concave mirror offocal length 10cm,then

with water of refractive index 4/3. A child stands in air at a

the refractive index ofthe material of the lens is:

distance IR {R is the radius of curvature of the sphere) from the centre ofthe bowl. At what distance from the centre would

(A) 9/14 (Q 17/9

the child's nose appear to the fish situated at the centre : (B) 2/? (A) AR (Q 3^ (D)4fi

5-15 A prismhas a refractive index

5-11 An objectis placedat a distanceof 15cm from a convex lens of focal length 10 cm. On the other side of the lens, a convex mirror is placedat its focus such that the imageformed bythe combination coincides with the object itself. The focal length of the convex mirror is : •15cm

10cm

H jp'

(B) 14/9 P) None

andrefracting angle

90°. Find the minimum deviation produced by the prism. (A) 40° P) 45° (C) 30° P)49°

5-16 A certainprism is found to produce a minimum deviation of 38°. It produces a deviation of 44° when the angle of incidence is either 42° or 62°. What will be the angle ofincidence when it undergoes minimum deviation ? P) 49° (A) 45°

(Q 40°

P) 55°

5-17 A light ray is incidenton a prism of angleA = 60° and

refractive index p= -Jl •The angle ofincidence atwhich the Figure 5.338

(A) 20 cm (C) 15cm

(B) 10cm (D) 30 cm

5-12 Two plano-convexlenses each of focal length 10 cm & refractive index3/2 are placedas shown-5.339. Water (p = 4/3) is filled in the space between the two lenses. The whole arrangement is in air. The optical power of the system in diopters is:

emergent ray grazes the surface is given by: (A) sin

(Q

sm

-1

V3-1-

-I

2

P) sin

-1

P) sin -1

\S 2_ 41

5-18 A transparent cylinderhas its right halfpolished so as to act as a mirror.Aparaxial light ray is incidentfrom left,that is parallel to principalaxis, exitsparallel to the incidentray as shown. The refractive index n ofthe material ofthe cylinder is :

Figure 5.339

(A) 6.67 (Q 333

(B) -6.67 (D) 20

Figure 5.340

(A) 1.2 (Q 1.8

P) 1.5 P)2.0

iGeometricai' Optics

321

5-19 Acomposite slab consisting ofdifferent media isplaced

30°.The refractiveindex ofthe liquid is :

in front ofa concave mirror ofradius ofcurvature 150cm. The

whole arrangement isplaced in water. Anobject O isplaced at

Point source

a distance 20cm from the slab. The refractive indices ofdifferent

mediaare givenin the diagramshownin figure-5.341. Find the position of the final image formed by the systemf

M=4/3

H=4/3 \

M=1.5

f-1-5;

ti'qu id-

0

, .20 cm.

Figure 5.343 45 cm

24 cm

54 cm 10 cm

Figure 5.341

2_ S

(A) ^

1

(A) To the left of Object (B) On the Object

(Q -iz

(C) To the right ofObject

5-23 In the figure-5.344 ABC is the cross section of a right angled prism and BCDE is the cross section of a glass slab. The value of 0 so that light incident normally on the face AB does not cross the faceBC is (given sin"' (3/5) = 37°):

(Q Data insufficient to calculate the image position

5-20 A luminous point object is moving along the principal axis ofa concave mirror offocal length 12 cm towards it. When its distance from the mirror is 20 cm its velocity is 4 cm/s. The velocity ofthe image in cm/s at that instant is : (A) 6, towards the mirror (B) 6, away from the mirror (Q 9, away from the mirror (D) 9, towards the mirror

B

E

IN •o

fn

II

II

^ 5^.

5-21 Twoblockseach of mass m lie on a smooth table. They are attached to two other masses as shown in the figure-5.342. The pulleys and strings are light. An object O is kept at rest on the table. The sides AB & CD of the two blocks are made

C

D

Figure 5.344

(A) 0 p^.

IGeometrical Optics

329

5-29 An object O is keptinfrontofa converging lensoffocal

(0 The average speed of the image as the object moves with

length 30 cm behind which there is a plane mirror placed at a distance 15 cm from the lens. Then which of the following

uniform speed from distance

3F

F

to — is greater than the

statements is/are correct.

average speed of the image as the object moves with same 30 cm

F

F

2

4

speed from distance — to —

(iQ The minimum distance between a real object and its real image in case of a converging lens is 4F where F is its focal

•15 cm

15 cm-

Figure 5.369

(A) The final image is formed at 60 cm from the lens towards right ofit O) The final image is at 60 cm from lens towards left ofit (C) The final image is real (D) The final image is virtual

5-30 Choose the correctalternative corresponding to the object distance 'w', image distance 'v' and the focal length 'P of a converging lens from the following.

length. (A) both are correct (B) both are incorrect (Q (i) is correct, (ii) is incorrect (D) (i) is incorrect, (ii) is correct

5-31 An object and a screen are fixed at a distance d apart. When a lens offocal length/is moved between the object and the screen, sharp images ofthe object are formed on the screen for two positions of the lens. The magnifications produced at

these two positions areM, andM^. {A)d>2f

Af

(Q

(jy)\M,\-\Mji=\

Geometrical Optics :

?330

Unsolved NumericalProblemsfor Preparation ofNSEP, INPhO & IPhO For detailedpreparation ofINPhO andIPhO students can refer advance study material on ww>v. physicsgalaxy.com 5-1 The left end ofa long glass rod ofindex 1.6350 is grounded and polished to a convex spherical surface ofradius 2.50 cm. A small objectis located in the air and on the axis 9.0 cm from the vertex. Find the lateral magnification.

ofthree thin lenseswith a common optical axis. The focal lengths

ofthelenses areequal to/j = 10cm (converging) and^ = 20cm (divering)and^ = 9cm(converging) respectively. Thedistance between the first and the second lens is 15 cm and between the

Ans. [- 0.0777] '

5-2 Focal length ofa convex lens in air is 10 cm. Find its focal

lengthin water. Given that

5-9 A parallel beamoflight is incidenton a system consisting

=3/2 and\x^ = 4/3.

Ans. [40 cm]

5-3 Find the distance ofan object from a convex lens ifimage is two times magnified. Focal length ofthe lens is 10 cm. Ans. [5 cm, 15 cm from lens] '

second and the third is 5 cm. Find the position of the point at which the beam converges when it leaves the system of lenses. Ans. [Infinity]

5-10 A ray of light is incident on the left vertical face ofglass cube ofrefractive index as shown in figure-5.370. The plane of incidenceis the plane of the page, and the cube is surrounded

by liquid of refractive index pj. What is the largest angle of incidence 0] for which total internal reflection occurs at the top surface ?

5-4 A pole 4m high driven into the bottom of a lake is Im above the water. Determine the length of the shadow of the pole on the bottom ofthe lake if the sun rays make an angle of 45° with the water surface. The refractive index ofwater is 4/3. Ans. [2.88 m]

5-5 An object is placed 12cm to the left ofa diverging lens of focal length 6.0 cm. A converging lens with a focal length of 12.0 cm is placed at a distance dto the right of the diverging lens. Find the distance d such that the final image is produced at infinity.

Figure 5.370

Ans. [ sin"

{i

111

Ans. [8 cm]

5-6 A solid glass sphere with radius R and an index ofrefraction 1.5 is silvered over one hemisphere. A small object is located on the axis of the sphere at a distance 2R to the left of the vertex ofthe unsilvered hemisphere. Find the position offinal image after all refractions and reflections have taken place.

5-11 One face ofa prism with prism angle 30° is coated with silver to make it reflecting. A ray incident on another face at an angle of 45° is refracted and reflected from the silver coated face and retraces its path. What is the refractive index of the prism ? Ans. [ V2 ]

Ans. [On the pole of mirror]

5-7 A glass sphere with 10 cm radius has a 5 cm radius spherical hole at its centre. A narrow beam of parallel light is directed into the sphere. Find the location of final image produced ? The index ofrefraction ofthe glass is 1.50.

5-12 In an isosceles prism ofprism angle 45°, it is found that when the angle of incidence is same as the prism angle, the emergent ray grazes the emergent surface. Find the refractive index ofthe material ofthe prism. For what angle ofincidence the angle ofdeviation will be minimum ?

Ans. [5cm to the left of the surface of sphere]

Ans. [yjl, 41.5P]

5-8 A source of light is located at double focal length from a convergent lens. The focal length of the lens is/= 30 cm. At

5-13 An astronomical telescope with objective offocal length 100 cm and eyepiece of focal length 10 cm is used by a

what distance from the lens should a flat mirror be placed so

that the rays reflected from the mirror are parallel after passing through the lens for the second time ?

shortsighted man whose far point is 33 cm from his eye, to form an image ofan infinitelydistantobject at his far point. Find the separation ofthe lenses, and magnification obtained.

Ans. [45cm]

Ans. [107.5 cm, 13.3]

iGeometricar optics

331

5-14 Figure-5.371 shows a right angled prismy45C having

5-18 An image Fis formed ofa pointobjectJFby a lenswhose

refractiveindex ="loweredintowater ^j.Findangle

opticaxis is^5 as shown in figure-5.372. Draw a ray diagram to locate the lens and its focus. Ifthe image Fofthe objectA'is formed by a concave mirror (having the same optic axis AB) instead of lens, draw another ay diagram to locate the mirror and focus. Write down the steps of construction of the ray

a so that the incident ray normal to face AB will be reflected at face BC completely.

diagrams.

•y

Figure 5.372

Ans. [-

Figure 5.371

angles.

5-19 A thin plano-convex lens.of focal length/is split into two halves : One ofthe halves is shifted along the optical axis (figure-5.373)The sqjaration between objectand image planes is 1.8m. The magnification ofthe image formed by one ofthe halflens is 2. Find the focal length ofthe lens and separation between the two halves. Draw the ray diagram for image

Ans. [60° and 40° ]

formation.

Ans. [ a >sm | -

5-15 An equilateral prism deivates a ray through 40° for two incidence angle which differ by 20°. Find the two incidence

5-16 A ray of light strikes a glass slab ofthickness t. (i) Prove that it emerges on the opposite face, parallel to the initial ray. 1.8m—

(ii) Prove that the value of deflection of beam which passed through the plate is:

Figure 5.373

Ans. [40cm, 60cm] 1-sin 2,-i1

1-

t sin i

«^/-sin^'i

(iii) Prove that for a small angle of incidence z,, the internal shift Xis given by

5-20 The focal lengths of the objective and the eyepieceof a compound microscqje are 1 cm and 5 cm respectively.An object placed at a distance of 1'. 1 cm from the objective has its final image formed at (i) infinity (ii) least distance ofdistinct vision. Find the magnifying power and the distance between the lenses. Least distance of distinct vision is 25 cm.

1

x = ti,

1-

Ans. [(i) 16 cm, - 50; (ii) 15.17 cm, - 60] g J

5-17 A plano-convex lens has a thickness of 4 cm. When

5-21 Find the minimum size of mirror required to see the fiill image of a wall behind a man standing at the centre ofroom, where His the height ofwall.

placed on a horizontal table with the curved surface in contact

Ans. [H/3 ]

with it, the apparent depth ofthe bottom-most point ofthe lens is found to be 3 cm. Ifthe lens is inverted such that the plane face is in contact with the table, the apparent depth of the centre ofthe plane face of the lens is found to be 25/8 cm. Find the focal length ofthe lens.

placed making 10°withthe mirrorMy Findthepositions ofthe first two images formed by each mirror. Find the total number ofimages.

Ans. [75cm]

Ans. [10° and 50° from A/," and 20° and 40° from M^, 11]

where

is the refractive index ofglass with respect to air.

5-22 Two mirrors are inclined by an angle 30°. An object is

Geometrical Optfcs

1332

5-23 AB is a man ofheight 2m and Mis a mirror of length 0.5m and mass 0.1 kg. Initially top ofmirror Mand A are at the same level and the Mstarts falling freely always remaining vertical. If the level ofthe eyes ofthe man is 1.5 cm below his head .^4,

5-26 See the following figure-5.377. Which of the object(s) shown in figure will not form its image in the mirror.

•O4

find the time after which the man sees the reflection ofhis feet.

I

Figure 5.377

M

Ans. [O3]

5-27 Twoplane mirrors are inclined at an angle of 75® to each other. Find the total number ofimages formed when an object is placed as shown in fIgure-5.378.

Figure 5.374 Ans. [0.318s]

5-24 Figure-5.375 shows a point object.^ and a plane mirror MN. Find the position of image of object/I, in mirror MV, by drawing a ray diagram. Indicate the region in which observer's

eye must Be present in order to view the image.

✓

V/////////////////A

Figure 5.378

M

f N

Figure 5.375

Ans. [4]

5-28 Two plane mirrors are inclined at an angle of70° to each other. Find the total number of images formed when object is placed as shown in figure-5.379. Total images = 5

Ans. [ ef

'////////////////////.

5-25 An object is placed at.^(2,0) and MT^is a plane mirror, as shown in figure-5.376. Find the region on T-axis in which reflected rays are present.

iv

P(4.3)

Figure 5.379 Ans. [5]

5-29 There is a point object placed in front of a plane mirror. If the mirror is displaced 10cm away from the object, find the distance by which its image will get displaced. Ans. [20cml

M

^(4.2)

A

(2, 0)

Figure 5.376 Ans. [Reflected rays exist on F-axis between (0,6) and (0,9)]

5-30 A crown glass prism of refracting angle 8° is combined with a flint glass prism to obtain deviation without dispersion. If the refractive indices for red and violet rays for the crown glass are 1.514 and 1.524 and for the flint glass are 1.645 and 1.665 respectively, find the angle of flint glass prism and net deviation. Ans. [1.53"]

iGeometrical Optics

333J

5-31 An opaque cylindrical tank with an open top has a 5-35 Figure shows a torch producing a straight light beam

diameter ofS.OO mand iscompletely filled with water. When

felling on a plane mirror at an angle of 60° as shown in figure-5.383. The reflected beam makes aspot Ponthe vertical sunlight ceases to illuminate any part of the bottom of the screen as shown. If at r = 0, mirror starts rotating clockwise

the setting sun reaches an angle of37° above the horizontal,

tank. How deep is the tank ?

about the hinged with an angular velocity = 1° per second.

Ans. [4 m]

Find the speedof the spot on screen aftertime t = 15s.

5-32 In the situation shown in figure-5.380, find the velocity vector ofimage in the co-ordinate system shown infigure.

Screen '//////////////////z

5 m/s

fixe

incident

ray

10 m/s

Figure 5.380

Figure 5.383

Ans. [ -5(1 +-VJ)/ +5y m/s]

271

Ans. [ — m/s]

5-33 Find the velocity of the image of a moving object in situation shown in figure-5.38I in which object and mirror velocities in horizontal and vertical directions are shown. 5 m/s

5-36 Alight ray/is incident onaplane mirror M The mirror is rotated in the direction as shown in the figure-5.384 by an arrow atthe frequency (Wtt) rev/sec. The light reflected by the mirroris received on the wall IFwhich is at a distance 10m from

i

the axis ofrotation. When the angle ofincidence becomes 37°, find the speedof the light spot on the wall.

O 10 m/s

30 m/s 6 m/s

Figure 5.381

Ans. [70.178 m/s]

5-34 Two plane mirrors are inclined toeach other atan angle

lOm-

W

30°toeach other. Arayoflightis incident at anangle of40°to Figure 5.384

the mirror (M^). Find the total angle ofdeviation ofthe ray after the third successive reflection due to mirrors.

Ans. [lOOOm/s]

5-37 A sphericallight bulbwith a diameterof 3.0 cm radiates

light equally in all directions, with apower of4.57c W. (a) Find the light intensity atthe surface ofthe bulb, (b) Find the light intensity 7.50 mfrom the centre ofthe bulb, (c) At7.50 m, a convex lens is set up with its axis pointing toward the bulb. The lens has a circular face with a diameter of 15.0 cm and a 40>

Figure 5.382

Ans. [160° Clockwise ]

focal length of30.0 cm. Find thediameter ofthe image ofthe bulb formed ona screen kept at thelocation oftheimage, (d) Findthe light intensity at the image. Ans. [(a) 5000 W/m^; (b) 0.02 W/m^; (c) 0.125 W/m^; (d) 288 W/m^j

G^metrical OpticsJ -

5-38 A coin liesonthebottom ofa lake2mdeep at a horizontal

5-42 A small object ofheight 0.5 cm isplaced in front of a

convex surface ofglass (p = 1.5) ofradius ofcurvature 10cm. beam oflight situated 1mabove the surface ofthe liquid of Findthe height ofthe image formed inglass.

distance xfrom thespotlightSwhich isa source ofthin parallel

refractive index p=

as shown in figure. The liquid height is

2m. Findx sothat a narrowbeamoflight fromS whenincident

on the liquid surfece atincidence angle 45° fells directly on the coin.

Ans. [1cm]

5-43 Infigure-5.387 shown AB isaplane mirror oflength 40cm

placed ataheight 40 cm from ground. There is alight source S atapointon the ground. Find the minimum and maximum height ofaman (eye height) required tosee the image ofthe source if he is standing at a point P onground shown in figure. T E O

I.

com

K20cm-i

-40cm-

Figure 5.387

Figure 5.385

Ans. [320 cm]

Ans. [ I 1+

I Im]

5-44 Aplane mirror ofcircular shape with radius r =20cm is fixed to the ceiling. A bulbis to be placed on the axis of the

5-39 What should be the value of angle 0 such that light

entering normally through thesurface

ofa prism («=3/2)

does not cross the second refracting surface AB.

mirror. A circular area of radius

= 1 m on the floor is to be

illuminated afterreflection oflightfrom the mirror. Theheight ofthe room is 3m. What should be the maximum distance from the centre ofthe mirror where bulb is to be placed so that the

requiredarea is illuminated? Ans. [75 cm]

5-45 A room contains air in which the speed of sound is 340 m/s. The walls ofthe room are made of concrete, in which

thespeed ofsound is 1700 m/s. (a) Find the critical angle for total internal reflection of sound at the concrete-air boundary,

(b) In which medium should thesound be travel to undergo total internal reflection ? Figure 5.386

Ans. [(a) sin"' ]y ]; (b) air]

Ans. [ 6/3z +x=10 is incident atapoint

Ans. [5 m].

-Pon the face/4Rofthe prism.

5-51 Aray in incident normally on aright angle prism whose refractive index is ^ and prism angle a =30°. After crossing the prism, ray passes through a glass sphere: It strikes the

glass sphere at ^ distance from principal axis, as shown in

(a) ,Find thevalue ofp for which thefay grazes theface /IC

(b) Findthedirectionofthe.mallyrefractedrayifp=

3

(c) Find the equation ofray coming out ofthe prism ifbottom BC is silvered ?

the figure-5.389. The sphere is half polished. Find the total angle of deviation of the incident ray after all reflections and refractions from thisoptical setup.

(0, 0, 0)

Figure 5.391

Figure 5.389

Ans. [{a) ^ ; (b) Along -z axis; ic)yl3z +x=\0 ]

Ans. [ISO")

5-55 A convexlens of focal length 20 cm and a concavelens

5-52 Ashort-sighted man, the accommodation ofwhose eye offocal length 10 cm are placed 10 cm apart on the same optic isbetween 12 cm and 60 cm wears spectacles through which axis. Abeam oflight travelling parallel to the optic axis and he can see remoteobjects distinctly. Determine the minimum havinga beamdiameter5.0 mm, is incidenton the convex lens. distance at which the man can read a book through his Showthat the emergent.beam is parallel to the incident one. Findthe beam diameter oftheemergent beam. spectacles. Ans. [15 cm]

Ans. [2.5 mm]

Opticsl i336

5-56 A thin convex lens of refractive index ^ = 1.5 is placed 5-59 A glass rod has ends as shown in figure-5.394. The

between apoint source oflight Sand a screen A, as shown in refractive index ofglass is p. The object Oisata distance 2R the figure. Light rays from the source Sare brought tofocus on from the surface of larger radius of curvature. The distance the screen A, forming apoint image P.The distance SPisequal between apexes ofends is 3R. Find the distance ofimage formed ofthe point object from right hand vertex. What isthe condition to 50 cm. Water

p=—Iis now poured into avessel interposed on p for formation ofa realimage.

between the object and thelens, and it isobserved thatwhen Glass

the water level is 8 cm the screen has to be moved up by a

distance of6 cmin order togeta sharp image. Findthe focal

"

length ofthe lens. Screen

7:

•2R-

Figure 5.394

5-60 When the object is placed 4 cmfrom the objective of a

microscope, the final image formed coincides with theobject. The final image isatthe least distance ofdistinct vision (24 cm). Ifthemagnifying power ofthe microscope is15, calculate the

Figure 5.392

focal lengths of the objective and eye-piece.

Ans. [12 cm]

Ans. [/•(, = 3.125 cm and/^ = 7.5 cm]

5-57 Athin equiconvex glass lens (p^ = 1.5) is being placed on the top ofa vessel ofheight /i = 20 cm as shown in the 5-61 A hemispherical portion ofthesurface ofa solid glass figure-5.393. Aluminous point source is being placed atthe sphere (p= 1.5) ofradius r issilvered to make the inner side bottom ofthevessel ontheprincipal axis ofthelens. When the reflecting. An object isplaced on the axis ofthe hemisphere at air is on both the sides of the lens, the image of luminous sourceis formed at a distance of 20 cm from the lens outside the vessel. When the air inside the vesselis being replaced by

a distance,3r from thecentre ofthe sphere. Thelight from the

object is refracted at the unsilvered part, then reflected from the slivered part and again refracted at the unsilvered part.

a liquid ofrefractive index Pj, the image ofthe same source is Locate the final image formed. being formed at a distance,30 cm from the lens outside the

"

vessel. Find the Pj.

'

fi, = 1

T

2r

2()cm

Luminous

Figure 5.395

source

Figure 5.393

Ans. [At the pole of slivered face]

Ans. [1.11]

5-62 The focal lengths of the objective and the eye-piece of an astronomical telescope are 0.25 m and 0.02m, respectively. 5-58 Light passes symmetrically through a 60® prism of The telescope is adjusted to view an object at a distance of refractive index 1.54.After emergence out from the prism the 1.5 mfrom theobjective, thefinal image being 0.25 m from the light ray is incident on a plane mirror fixed tothebase ofthe eye ofthe observer. Calculate the length ofthe telescope and

prism extending beyond it. Find the total deviation ofthe light the magnificationproducedby it.

rayafterreflection fromthe mirror.

Ans. [31.85 x lO'^ m, 16.2] Ans. [0®]

•GTOmetrical Optics 337

5-63 Athin converging lens offocal length/= 1.5 misplaced

5-69 In a simple astronomical telescope the focal length of

along_v-axis such that its optical centre coincides with the the object glass is0,75 m and thatoftheeye-piece is 0.05 m.

origin. Asmall light source Sisplaced at (- 2.(5 m, 0.1 m) as

Calculate the magnifying power when the final image ofadistant

shown infigure-5.396. Where should aplane mirror inclined at object is seen (a) a long way off, (b) at a distance of 0.25 m. an angle 0 with tan 0 = 0.3, be placed such that^-coordinates Find the distance between the two lenses in each case.

offinal image is0.3 m. Also find a: co-ordinate offinal image.

Ans. [(a) 0.0417 m (b) 0.7917 m]

5-70 An astronomical telescope consisting of two convex lensesof focal length 50 cm and 5 cm is focussed on the moon.

What is the distance between the two lenses in this position? If the telescope is then turned towards an object 10 maway, how much would the eye-piece have to be moved to focus on Figure 5.396

the object without altering the accommodation of the eye? Calculate the angular magnification produced by the telescope in the two adjustments.

Ans. [5m, 4m]

Ans. [10]

5-64 A ball is kept at a height yQ above the surface of a 5-71 How would you use two plano-convex lenses of focal index ji. At ^= 0, the ball is dropped to fall normally on the lengths 6cm and 4cm todesign aneye-piece free from chromatic transparent sphere of radius R, made ofmaterial ofrefractive

sphere. Find the speed of the image formed as a function of

time for t


D

2

k where n = 0,1,2 2

# Illustrative Example 6.6

" •

The coherent point sources

and 5*2 vibratingin samephase

emit light ofwavelength k. The separation between the sources

is 2k. Consider a linepassingh through $2 andperpendicular to the line 5,5*2. What is the smallest distance from where a minimum of intensity occurs due to interference ofwaves from the two sources? H

D


d=A.9x 10^m=0.49mm

(6500xl0-»)(2)^0 0366mn, 15x10"^

# Illustrative Example 6.9

Change in fringe width

Twoslits in Young's interference experimenthave width in the ratio 1 : 25. Deduce the ratio of intensity at the maxima and

= p1- p = 0.0866 - 0.065 = 0.0216 mm

minima in interference pattern.

# Illustrative Example 6.11

Solution

In Young's double slitexperiment theslitsare0.5mmapartand

The intensities dueto separateslits are in proportionto the slit width. Theamplitude is proportional to the squareof intensity.

interference is observed on a screen placed at a distance of 100 cm fromthe slits. It is foundthat the 9'*^ brightfringeis at a distance of 8.835 mm from the second dark fringe from the

centre of the fringe pattern. Find the wavelength of light used. 502

Solution

^2 = 50,

{0^+02?

The distance ofn''^bright fringe from the central fringe is given

(a,+5iJ,)^

by

Now

XD

36a{ 16a?

36 _ 9 16 ~ 4

^nux-U = 9:4.

where p = {XDUd)is the fringe width.

For 9^^ bright fringe,

.

X5=9P

...(6.31)

^Wave Optics 353

The distance of dark fringe from the central fringe is given by X '=n

{2n-\)U) Ad A = t7sin n —

2

2 d

«--lP

For2"'^ dark fringe, Figure 6.19

...(6.32)

In above situation the phase difference between waves from

Fromequations-(6.31) and(6.32), weget

slits

and Sj at screen center C is given as

15

,

271

(p= ~

X

X(j = Dsin0

...(6.36)

Above equation-(6.36) gives theposition where path difference iszero or the position ofcentral bright fringe after changing the illuminated earlier than 5, sowaves from will lead inphase overthe wavesfrom 5", becauseof the extra path d sin0 which direction ofincident light beam ondouble slit plane. Ifthe parallel light beam incident onthedouble slitplane normally then both the wavefront illuminating slit will travel till itreaches slit 3",. the slits will getilluminated simultaneously and the point of Ifweanalyze thepath difference inthetwo waves interfering at zero pathdifference (central maxima) would be located atpoint screen center C then the two waves travel equal path after slit Candiftheincident beam isrotated downward upto the situation the inclined incidence of beam we can see that slit

will be

plane upto this point but due to initial path difference before

slitplane, the total pathdifference in waves at pointC will be d sin0.

shown in figure-6.19 thenwithabove analysis wecan statethat during rotation ofthe beam the interference pattern will shift upward bya distance given byequation-(6.35).

Wave Ophcsf .

6.4.2 Effect of Submerging YDSE Setup in a Transparent

Thetimetaken byray-1 intravelling through theglass slabis

Medium

pw

w

In a transparent medium ofrefractive index m, when a light of wavelength / enters then we know its wavelength and speed reduces which are given as

c/pj

c

Path length covered byray-2 in space while ray-1 was travelling in slab is

c V = ~ jl-

and

"k X= ~ |X

pw

/ = CX —

.,.(6.38)

= pw

c

Ifa YDSE setup is submerged ina transparent liquid asshown in figure-6.20 then in the submerged part due to decreased Thus path difference between thetwo rays after theglass slab wavelength fringe width of the interference pattern also is given as decreases which is given as

d

A= ptv- w= w(p -1) ...(6.37)

ixd

... (6.39)

If these two light waves (rays) are brought to focus of a converging lens as showi then the two waves will interfere with a phase difference given as

Fringe Pattern in Screen

Shrunken fringe pattern in water

Narrow slits

27C

...(6.40)

Water

Ifeachofthe lightwave in the two thin light beams (rays) have

Water tank

intensity 1q then at the focal point ofthe lens the resulting intensity of light is given as

Incident light beam

Figure 6.20

,, ' • f7tw(p-l) /, =4/,cos2|^ =4/,cos^

...(6.41)

6.4.4 EffectofPlacinga Thin Transparent Film in front ofone of the slits in YDSE Setup

Here p, sowecansaythat on submerging a given YDSE setup ina transparent medium, theinterference pattern shrinks. 6.43 Path difference between twoparallel waves due to a denser medium in path ofonebeam

Figure-6.22 shows a YDSE setup and in front ofslit5, a thin transparent film ofthickness t and refractive index p isplaced. Duetothisfilmthe lightcoming outfrom slit will getdelayed

bysome path asitgets slower inthe film medium. Atthe screen center where thephysical pathdifference iszerointhetwolight

Figure-6.21 shows two coherent light rays (thin beams) from a waves coming from slitiSj and S^, due tothin film an optical single source of light travelling in same direction parallel to path difference isintroduced inthepath oflight waves which is each other. Ifin pathoffirst beam a glass slabofrefractive index

p is placed which is ofwidth wthen the light ray-1 will slow

givenbyequation-(6.39).

down after it enters in slab at point Aand its speedwill reduce to c/m.Whentheray-1 whichentersin slabcomes outat point B then in this duration the ray-2 which was travelling in space would have travelled a longer path as it was travelling at speed c. C A= ?(fi- 1)

Screen

Figure 6.21

Figure 6.22

[Wave Optics

355

Optical path difference in the twolightwaves from the slitsat

this case obviously it is not necessary that at screen center

screen center C is given as

there isa bright fringe because the path difference at point Cis now given as A^=/(p - 1). Ifwe find the ^z-value'' at screen

A = /(p-l)

/ ...(6.42)

center then it is given by equation-(i) as

Atscreen center theextra path given by above equation-(a) is

introduced in the light coming from slit .Sj as it got delayed

%-l)

while travelling inside thefilm. If we tryto locate a point on screen where optical path difference iszero then we can clearly

X

state that it will be locatedabovepoint C on screen where the

...(6.47)

Saytheexpression obtained inequation-(6.47) gives a numerical value 6.23 aftersubstituting the numerical values ofallconstants

extra optical path introduced in first wave is compensated by init.Thatmeans atpoint Conscreen thepath difference inthe the extra physicalpath travelled by the second wavewhich is two waves is 6.23X or it shows that the original center bright given by equation-(6.42). If P is the point of net zero path fringe is located about 6 fringes above thepoint Cwhere path difference located at a distance Xq from thescreen center C in difference will be zero.

figure-6.13 then at point

we can use

Thus at any point on screen if you calculate the 'z-value' then

dXr,

it directly gives you anidea about thepath difference ofbright ...(6.43)

Above equation-(6.43) gives the shiftof interference pattern on screen due to insertion of thin film in front ofone of the slits

or dark fringes above or below this point which helps us in calculating the total number of fringes in any region of interference pattern.

HIllustrative Example 6.12

becausethe fringe of zero path differencewas locatedat screen

center inYDSE before the film was inserted infront of5,.At a A double slit arrangement produces interference fringes for general pointPon screen located in upper half of screen at a sodium light {X = 5890 A) that are 0.40° apart. What is the angularfringe separation ifthe entirearrangement is immersed

distancex from C, the path differenceis given as

in water ? dx

...(6.44) Solution

And at a general point P' located in lower half of screen at a

distance x from C, the path difference is given as

In double slit arrangement, the angularseparation is given by

dx

XD

...(6.45)

6.4.5 Concept of z-value in Interference Pattern of YDSE

At any point in interferencepattern on YDSE screen, we can define a numeric parameter 'z' for relating pathdifference inthe two light waves and wavelength oflight as

2d

Let angular separation in air and water be respectively. Now

2d

andp^=

X^,D

'W

"W

Ap=Zpk

,(6.48)

P.

Ao

...(6.46)

We know that

Above numerical parameter Zp is a constant which gives the multiplier of lightwavelength at anypointonscreen that gives the path difference at a the specific point on screen. This

or

Velocity of light in air Velocity of light in water

Va

f^A

'W

fXw

'z-va/we' helpsin quicklydetermining thenumberofbrightand dark fringes in any length of interference pattern of YDSE.

where/is the frequency oflight.

For example ifwe look at the YDSE Setupshown in figure-6.14

Now

inwhich a thin film isplaced in front ofslit.Sj due towhich the overall fringe pattern is shifted upward by some distance. In

and

From equation-(6.48) and (6.49), we have

(6.49)

Wave Optics [

f356

hL

A=i/=3.4x ICHmx — K

P. A=

or

3.4x10^ ,

—^=54SAX

6.2x10"

Total no ofmaxima on x aixs

by D are= 548

Illustrative Example 6.13 # Illustrative Example 6.15

A YDSE setup is immersed in water (p = 1.33). If has slit separation 1 mm and distance between slits and screen is 1.33m.

Incident lightonslits have wavelength 6300A. Findthefringe width on screen.

In YDSE setup slits are illuminated by a light of wavelength

4000A and a lightofunknown wavelength. It is observed that fourth dark fringe ofknown wavelength coincide with second bright fringe of unknown wavelength. Find the unknown wavelength.

Solution

W/L oflight in water

Solution

X

6300 ,

X = —= A ® p 1.33

For

dark fringe from centre {N-l)XD

In YDSEfringe width is given as

^NDX„D

6300x10"'® X1.53

d

1.33x10"^

A =

N=4,X=moA

for

7x4xlO~^xD

= 6.3xia^m

^4D

=0.63 mm

For

2d

bright fringe from centre NXD

^IllustrativeExample 6.14

Thereare two coherent sources 5", and

2d

^NDwhich produce waves

in same phase placed on Y axis at points (0, 2d) and (0, d)diS shown in figure-6.23. A detector Displaced at origin which moves along X direction, find the number of maxima recorded by detector excluding points x = 0 and a:= co. Given that wavelength of light produced is 6200 A and separation

^

N=2,X = 1

for

2XD

d Given that

AD

7x4xlO"^D

between sources is 0.34 mm.

2^ =>

2XD

~ d A.= 7x 10-7m= 7oooA

# Illustrative Example 6.16

In a YDSE setup, light of wavelength 5000 A is used and slit separation is 3 x 10"' m. When a transparent sheet ofthickness

1.5 X10"' m is placed overone oftheslits which has p= 1.17,

A = 548X 0

at CO

A = 548.4?.

A = 0

Figure 6.23

Solution

Path different between wavesfrom 5", and S^ at origin

find the shift of fringe pattern. Take separation between slits and screen is 1 m.

Solution

At Cpath different A = r(p-1)

!Wave Optics 357

sides ofC. At these points violet colour will be missing from

•/'A=0

white light due to destructive interference and the colourseen

lt_4 CA = X(n-l)

on screen will be reddish white (not red as it iswhite - violet) because other then violet all other colours will be present at this point asonly violet will be completely absent.

Bluish

Figure 6.24

At point P we have

Reddish A= —

dx

^

Reddish

^ 1x1.5x10"^ xQ.17 a

3x10"^

Bluish

Double slit plane

a: = 0.055 m = 8.5 cm Screen

Illustrative Example 6.17

Figure 6.26

In YDSE setup find the distance between two slits that result

Similarly ifwemove fiirther away from point Conscreen next

inthe third minimum for 4200 Aviolet light atanangle 30®.

point will be the point ofdestructive interference ofred light which ishaving the maximum wavelength inwhite light where

Take d«D. Solution

At point P on screen path different between waves from 5", and

is A = Srl2

III min

j-^2 ••m D-

red colour will beabsent and path difference atthis point will be /2. As the wavelength ofred colour isapproximately double that ofviolet colour soat this point violet color light will also have higher intensity and the colour offringe here will appear ^Bluish' (Not blue asother colours are also present here). Beyond this point onscreen allcolours will mix and nospecific colour

will be seen on screen and screen looks 'Mixed White' or 'Off White'mcoXour.

6.4.7 Effect ofChanging SUt Width inYDSE Setup

'Ot/sin 0

In a YDSE setup if each slitisproducing light intensityon screen then this intensityis directlyproportional to the width Figure 6.25

A-sin 30®

d X

5X

- ~:r (forII min) 2,

£/=5X=5x4200x lO-'Om = 2.1x10-^ m

6.4.6 Use of White Light in YDSE

When white light is used in YDSE setup, all the component waves of white light will have zero path difference at screen

ofthe slits indouble slitplane. Figure-6.27 show that a plane parallel beam oflight ofintensity/, illuminates thedouble slit plane. Ifeach slit isofwidth wand length / then the light power which pass through these slitswill be given as ...(6.50) ^siit h • Thus thepower oflightfrom each slitwill beproportional tothe

slit area given by equation-(6.50) and aswe have already studied insection ofspherical and cylindrical waves that intensity due to any light sourceis givenbythe ratio ofsourcepowerand the surface area of wavefront at any point. We know that a slit produces halfcylindrical wavefronts after double slitplane and

center Candallcolors inwhite light will inta-fere constructively and a bright white fringe is produced. But.as we move away

the wavefront W^ soata distancex from a slit thelight intensity

from the center then closest to screen c^ter first the destructive

can be given as

interference of violet color will take place as violet color has minimum wavelength in white light at a point shown in

figure-6.26 where path difference will become XyH on both

hlit

^Slit

LJw

Tixl

•Kxl

/iw

= ~-

...(6.51)

Wave Optics

i358

Thusthe lightintensityon screen at a distance D from double slit plane due to each slit is given as /jW,

...(6.52)

4lit~

Thus the light intensity by each slit on screen is directly proportional tothe slitwidth andinversely proportional to the separation between double slit plane and the screen.

N

^

Intensity

/ i AU ; Central Bright Fringe D.

h \ A'v'V'V

N

\

\

\ Screen

Intensity /,

f' / ' VV

D, B,

i i

0| Central Bright Fringe D,

1 1 1 •"

-

'

Figure 6.28

6.4.8 Fresnel's Biprism as a Limiting case of YDSE

/'/ // / ' // y

/

Fresnel demonstrated interference pattern similar to YDSE by

using a biprism. The biprism consists of two prisms of very Screen

smallrefractinganglesjoined baseto base.It ismade bygrinding

a thin glassplatewithopposite inclination onthetwohalves of the plate on one side as shown in frgure-6.29. Figure 6.27 Thin glass plate

Ifboth slits 5, and S2 are ofequal width then intensity byboth slits will be same /q on screen and the resulting intensity of bright and dark fringes will be

Shaded part is removed

and

!

From equation-(6.54) we can see that dark fringes will be perfectlyblack and separatingthe bright fringes and in such a case proper interference pattern can be seen in which we can clearly distinguish bright and dark fringes as shown in

Biprism Perspective view

(a)

figure-6.27.

Ifthe width ofthe slits are unequal as shown in figure-6.28 then

the intensities of the twoslits on screenwillbe different, say/^,

and /q2 respective then the light intensity ofbright and dark fringes on screen will be (b)

hrighl ^

-^01 -^02 ^4^01^02 •••(6-55)

^Dark ~(4^ ^ 4^ ~-^01 -^02 ~^4^01^2 •••(6-56) From above equation-(6.56) we can see that the intensity of dark fringe is not zero due to which in the interference patter we cannot clearly distinguish bright and dark fringes as shown in figure-6.28.

Figure 6.29

We've already studied that a light ray from a source S when incident on a small angled prism at near normal incident as shown in figure-6.29(b), it gets deviated by angle ofdeviation 5, and from the other side it appears to be coming from the image ofthe light source 5". The angle ofdeviation 8 is given as 5=.I(p-l)

...(6.57)

wave Optics

359'[

the slit and screen is taken as D = + Zj so the fringe width on the screen can be given in the same wave we use for YDSE setup. XD

Fringe width in Fresnel Biprism setup is p=-jSubstituting the values ofD and d in above expression we get X(a + b) P = 2aA(^-l)

...(6.59)

Figure 6.30

6.4.9 Lloyd's Mirror as a limiting case of YDSE

Now if we look at figure-6.30 in which two light rays from a source S incident on a biprism then these bends after refraction through biprism by same angle of deviation toward base of prisms such that the refracted beams appear to be coming from the two images of the source S' and S". Figure-6.31 shows the interference setup of'Fresnel'sBiprism ExperimenC in which there is only one slit kept at a distance 'a' fromthe biprism and at a distance 'Zj' fromthe biprism screen is kept. Due to refraction, biprism splits the incident light in two beams and from the other side the light beams appear to be coming from the two images S' and S" of the slit and these beams interfere and producesthe interferencepattern similar to

In 1834Lloydmade a setup to demonstrate interferenceusing a single mirror and a slit of light at almost grazing incidence of light.Figure-6.32 shows theLloyd's Mirrorsetupofinterference

inwhich light from theslit5j falls onthemirror andthe reflected light interfere with the direct light on the screen. The reflected light appears to be coming from the image of the slit as shown.For theregion ofscreen in whichthetwo lightsinterfere, interference patternis obtained whichis similarto YDSEsetup.

Region of Interference

YDSE on screen.

X

Screen

Figure 6.32

In the above setup the separation between the two slits is taken

diSd=2h and the separation between slit plane and screen is given as D. If we consider a general point P on the screen in the region of interference then the path difference at point P between the lights from the two sources is given as Figure 6.31

dx A — D

X

...(6.60)

1— 2

In above figure we can see that each light fi-om slit S falling on biprism gets deviated by the deviation angle given by equation-(6.57). For the region betweenslit plane and biprims

In above equation-(6.60), the first term is the physical path difference in the path travelledbythe two light waves reaching at point P and the second term is the extra path introduced in

for the ASOS'-wq can use for small value of5, we have

the reflected light due to its reflection from a denser medium at mirror.

— =crtan5i ad 2

In this setup at point P, a bright fringe will be obtained if the

path difference is an integral multiple of the light wavelength so

~ =aA(H'l)

we use

d = 2aA (p-l)

...(6.58)

Equation-(6.58) gives the separation between the two coherent sources S' and S" from which the interference pattern is being produced and the separation between the plane of images of

dx

X

A=— + -r =NX D

2

x.. = {2N-\)

...(6.61)

XD 2d

...(6.62)

Wave OpticsJ

J360

Similarly we can say that at point P there will be dark fringe if the path difference in the two waves at P is an odd multiple of half of the light wavelength so we use dx

X

X

shown in figure-6.34. If the angle ofconvergence between the light beams is 0 then the light beams after interferenceproduces a fringe pattern as shown in figure where the fringe width is given as

...(6.63)

A= —+ -=(27V+ 1)D 1 ^ ^2

...(6.65)

NXD

...(6.64)

^ND

From equations-(6.62) and (6.64) we can analyze that the interference pattern obtained in this setup is inverse to that of the YDSE setup. At the location of bright fnnges in YDSE, here we are getting dark fringes and vice verse. But we can see that the distance between two successive bright or dark fringes which is the fringe width remain same. Fringe Width in interference pattern obtained in Lloyd's Mirror setup is given as

XD

XD

d ~ 2h 6.4.10 Billet Split Lens as a limiting case of YDSE

This is an experimental setup shown in figure-6.33. Athin lens of small aperture is cut in two halves at the center and these halves are separated by some distance from original principal axis oflens. These halflenses produce two images of the slit as shown in figure. The light from these two images will produce interference pattern on screen in the region where the two images overlap.

Figure 6.34

The derivation of above equation-(6.65) is left for students as an exercise.

# Illustrative Example 6.18 Consider the situation shown in figure-6.35. The two slits

and ^2 placed symmetrically around the central line are illuminated by a monochromatic light of wavelength X. The separation between the slits is d. The light transmitted by the slits falls on a screen placed at a distance D from the slits. The slit

is at the central line and the slit

is at a distance z

from Sy Another screen £"2 is placeda further distance D away from £,. Findtheratioofthemaximum tominimum intensity observed on £2 if ^ is equal to : XD

(a) ^

XD

(b) V

XD

(c)

4d

Screen

Figure 6.33

This setup is similar to YDSE setup if we look at the plane of two images of the slit producing interference pattern on the screen. By using lens formula and magnification we can find

out the location ofthese images and their separation by which using the formula offringe width in YDSE setup we can find the fringe width in this setup.

Figure 6.35 Solution

6.4.11 Interference of Two Converging Coherent Parallel Beams of Light

When two monochromatic and coherent parallel light beams incident on a surface (or screen) in a converging manner as

Light from sources and get interfered and thereafter and becomes new sources. At the path difference between

the lightscoming from and£2 iszero. Therefore theyinterfere constructively and so iij = (a + fl) = 2a

^Wave Optics 361

(a) At

the path difference Ax = cisinQ = dtan 0

ax —

D

D Screen

Figure 6.36

The path difference

Corresponding phase difference = n radian

^x = S^-S^P

«4= 0

("i+atf _(2a +af

The ratio

^min

2a~0

= = 1

= X. cos 0

The maximumpath difference can be

XD dz

(b)

D

cos 0

A

= X

= X,; when cos 0 = 1 or 0 = 0°

•*max

and minimumpath difference

Corresponding phase difference = 2k radian. A_ =0; when cos 0 = Oor 0 = 90®

a^ = a + a = 2a

Now

and

Thus in between these two positions there is only one miima

(2a + 2ay

L

-for which

{2a-2a)^ Ax = —. Thus •d dz

(0)

L4J — = X cos 0

D

Corresponding phase difference ^ — radian

or

cos 0 = — 2

0 =60®

4^ = 0^ + 0^ + 2aa cos — = 2d^

a Illustrative Example 6,20

- yfla

or

/„i„

(2a +^af _ (3.414)^

In a modified Young's double slitexperiment, a monochromatic

(2a-^f

uniform and parallel beam oflight ofwavelength 6000A and

(0,586)^

=34

# Illustrative Example 6.19

intensity ^ W/m^ is incidentnormallyon two circular apertune Aand Bofradii 0.001 mand 0.002 mrespectively. Aperfect transparent film of thickness2000 A and refractiveindex 1.5for

thewavelength of 6000 A is placed infront ofaperture Aas

The two coherent sources of monochromatic light of shown in figure-6.37. wavelength Xare located at a separation X. The two sources

are placed on a horizontal line and screen is placed perpendicular to the linejoining the sources. Findposition of the farthest minima from the centre of the sources. Solution

Suppose at P the farthest minima will occur. Let it subtends an angle 0 at the centre of the sources.

Figure 6.37

Wave Optics-

[362 _

Calculate theintensity oflightreceived atthefocal point ofthe lens in watt.The lens.is symmetrically placedwith respect to

Screen

1

X

the aperture. Assume that 10% ofthe power received by each aperture goes in the original direction and is brought to the

d

-D-

focal point.

Figure 6.38 Solution

Solution

The intensities oflight from the sources

andS2 are given as

(a) Thepath difference at O is given as

X71(0.01)2= 10-5 IT

Ac=

7t;

-2D

Forthe darkfringe at 0, this pathdifference should be 10

k = — |xjt(0.02)2=4xia-5lf

X 3X

Ax=-,y,...

V. ^

The intensities ofsources after emerging from the lenses are

j^inimum value ofd, we use

/ A =0.10 X10-5 lf=l(r^ If

2ylD^+d^-2D --

/„ = 0.10x4x 10-5lf=4x 10"^^ D

The path difference produced dueto film

or

Ax =(^-l)r=(1.5-l)x2000xlO->o=10-^m ]2 ^

271

271

(b=

^

X

=

, --7

X

1+D'

D

or

1/2

,n xlO

-^ = 4

6000x10"'®

(})= — radian

Asrefraction through lensdonotintorduce anypathdifference soabove willbethe netphasedifference in thetwolightbeams

D

or

1+ -

4

2D'

D+

or

d^

X

D =-

2D

4

superposing atfocal point ofthelense. Sonetintensityreceived at F is given as

d =

or

=10*^ +4X 10^+ 2Vl0"^x4xl0^ cosj

(b) Attheabove calculated value old, first bright fringe will be obtained at a position where the path difference between the two waves will be X.

= 7x10-6 If

# Illustrative Example 6.21

5,1

Consider the arrangement shown in figure-6.38. The distance D is large comparedto the separation ^/between the slits.

(a) Find theminimum value ofd sothatthere is a darkfringe atO.

(b) Suppose ^/has this value. Findthe distance x at whichthe next bright fringe is formed.

(c) Find the fringe width.

5: 4

D

O-*-

]


0=±2.25>

/p= 7qCOs2 0

...(6.111)

Fromthe above equation-(6.111) wecanstate'"When apolarized light ispassed throughan analyzer then the intensity of light

Ifweconsider an oscillation amplitude Ashown bya bold arrow in above figure whichis at an angle0 tothe transmission axisof the Polaroid then its component A cos0 will pass and its component Asin0 willbe blocked. The resulting amplitude of

Wave Optics'

•38,8

thepolarized lightafter transmission through thepolaroid will be sum ofall A cos0 for all the oscillations ofthe field vector in

unpolarized light withvalue of0 varying from 0to2n.Ifintensity

^^foRotaton,

ofunpolarized light is Iq then theintensity ofthe transmitted

Substance

\/

light can be given as

1

Polarizer

Ij.= k ^ ^COS0 e=o

Figure 6.107

As at all angles 0 in unpolarized light, the amplitude of oscillation can be considered same with equal probability so we can write the above expression as

Ij~ (Iq cos' /, =/o(cos

There are two ways in which optically active substances are divided based on the direction of rotation of plane of

polarization. When the emerging light is observed from the average from 0=0 lo 2n

average from 6=0 to 2n

As average valueofsquareof cosine of an anglevarying from

0to 2jc is given as ^cos^ 0^ =—. ...(6.112)

It /()

With the result obtained in above equation-(6.112) we can state

"The intensity of unpolarized light after passing through a polarizer reduces to halfinpolarized light". 6.8.9 Optical Activity of Substances

directionoppositeto the directionof propagationof light then the substances which rotates the plane of polarization in clockwise direction (toward right) are called 'Dextrorotatory Substances' and those which rotate the plane ofpolarization of

light in anticlockwise direction (toward left) are called 'Laevorotatory Substances'.

The optical activity of a substance is analyticallymeasured in terms of 'SpecificRotation' ofa substance whichis defined for a given wavelength of light as the rotation produced by one decimeter long column of the solution containing unit concentration (1 gm/cm^) ofthe substance. Ifa polarized light passes througha solution ofoptically active substance having concentration C(in gm/cm^) and oflength / (indecimeter) and the angle bywhich planeofpolarizationof lightgets rotatedby 0 then specificrotation of the substanceis given as Qt—

Some specific transparent substances have property to rotate

the planeof polarization of a polarized light when the light is passedthrough it, suchsubstances are called''OpticallyActive'' substances and this ability of substances to rotate the plane of

polarization of light is called 'Optical Activity'. Quartz and Cinnabar are common examples of opticallyactive crystalsand substances which are opticallyactive in both solid and liquid state are Sugarand Camphor. Figure-6.107 shows an optically active substance on which a polarized light is incident with the field vectoroscillations in a plane which is in the plane of paper which is the plane of polarization. We can see in figure, as the light passes through the substance, the plane of polarization rotates and the light which comes out of the material remain

IC

UIllustrative Example 6.31

The axes of a polarizer and an analyzer are oriented at 30° to each other.

(a) If unpolarized light of intensityf is incident on them, what is the intensity ofthe transmitted light?

(b) Polarized lightof intensity/q is incident onthispolarizeranalyzer system. If the amplitude ofthe light makes an angle of 30° with the axis of the polarizer, what is the intensity of the transmitted light ?

linearly polarized but its plane ofpolarization gets rotated. Solution

0Rotatoiy Substance !3«~rfC-i

Polarizer

'Stance

(a) Half the light passes through the polarizer, so the intensity of the polarized light incident on the analyser is

1=

/(j, andthe intensity/'ofthe lightpassingh through the

analyzer is/'=/cos^0= —/qCOS^30° = 0.375/q.

[Wave_Optics

389

(b) After passing through the polarizer the intensity is # Illustrative Example 6.33 1=/pCos^ ®~^ cos^ 30° =0.75 7^. This light is now polarized at 30° to the axis of the analyzer, so the intensity of the light The axes ofa polarizer and an analyzer are oriented at right passingh through the analyzer is

/'= /cos^ 0=0.75 /q cos2 30° =0.563/o.

angles toeach other. Athird Polaroid sheet isplaced between them with its axis at45° tothe axes ofthe polarizer and analyzer. (a) If unpolarized light of intensity 7^ is incident on this

UIllustrative Example 6.32

system, what is the intensity ofthe transmitted light? (b) What is the intensity of the transmitted light when the

A beam ofplane polarised light falls normally ona polariser

middle Polaroid sheet is removed ?

(cross-sectional area3 x lO^m^) which rotates aboutthe axis

ofthe ray with an angular velocity of31.4 rad/s. Find the energy

Solution

of lightpassing through the polariser per revolution and the intensity oftheemergent beam ifflux ofenergy oftheincident (a) The light that passes through the polarizer has an intensity ray is 10"^ W.

of

and is polarized at 45° to the middle sheet. Thus the

light that passes through the middle sheet has the intensity

Solution

7= -7qCos^ 45° =0.257q and ispolarized 45° tothe axis ofthe If 7q is the intensity of plane polarised light incident on the polariser, then intensityofemerging light is given by

analyzer. Thus the intensity ofthe light passing through the

analyzer is 7'=7cos^ 45° =0.1257^.

7=/qCos^0 The average value ofI over one revolution can be calculated

(b) Ifthe middle Polaroid sheetis removed, wehavea crossed polarizer-analyzer and no lightgetsthrough.

as:

j 2n Web Reference at www.phvsicsgalaxv.com

I = — r 7^/0

- 2nl

;Age Group - High School Phj^sies ] Age 17-19 Years Section - OPTICS

Topic - Polarization of Light

1

Im = — f 7o cos^ 9- J "

I av

Module Number - 1 to 13

Practice Exercise-6.4

2

Intensity is given by

(i)

ona second Polaroid whose planeofvibration makes an angle

Power

^0

(a) Ordinary light incident on one Polaroid sheet falls

of 30° with that ofthe first Polaroid. Ifthe Polaroids are assumed

area

tobe ideal, whatis the fraction ofthe original lighttransmitted 10"

L = 3x10

through both Polaroids?

=y10 W/m2

(b)

If the second Polaroid is rotted until the transmitted

intensity is 10percentofthe incident intensity, whatis the new and

,= ^ =1W/ml

angle ? [(a) 3/8; (b) 63.4°]

The energyoflight passingthroughthe polariserper revolution 271

E = IxA^T=I av

5

x^x — CO

271

= _,(3,10-)x — = 10^J

(ii) Three nicolsprisms are placed such that, first and third are mutually perpendicular. Unpolarised light is incident on first nicoTs prism, the intensity of light emerges from third nicol's prism is 1/16 the intensity of incident light. Find the angle between first and second nicol's prisms. [22.5°]

Wave •{

;390

Polarized light of intensity 7q is incident on a pair of transmitted intensity is maximum. Thenobtain thetotal intensity ofthe transmitted beam in terms of/q. incidentamplitude and the axes of the first and secondsheet, respectively. Show that theintensity ofthe transmitted lightis

(iii)

Polaroid sheets. Let 0j and 02 be the angles between the

1~1q cos^ 0j cos^ (0j - Sj). (iv) A mixture ofplanepolarised andunpolarised lightfalls normallyon a polarisingsheet.On rotatingthe polarisingsheet about the direction of the incident beam, the transmitted

intensity variesbya factor 4. Find the ratioofthe intensities

and Iq respectively of the polarised and unpolarised components in the incident beam. Nextthe axis ofpolarising sheet is fixed at an angle of 45® with the direction when the

Advance Illustrations Videos at www.phvsicsgalaxv.com

AgeGroup - Advance Illustrations Section - OPTICS

"Topic-Wave Optics

Illustrations - 22 In-depth Illustrations Videos

[Wave Optics

~

"""

"T" """agr

Discussion Question Q6-1 Why does an aeroplane flying at a great altitude not Q6-11 Some motor cars have additional yellow headlights, case a shadow on the earth?

Why?

Q6-2 When sun rays pass through a small hole in the foliage Q6-12 Why does acolumn ofsmoke rising above the roofof atthetopofahightree,theyproduceaneIIipticalspotoflight houses seem blue against the dark background of the on the ground. Explain why. When will the spot be a circle? surrounding objects, and yellow or even reddish against the background of a bright sky?

Q6-3 If a mirror reversesright and left,whydoesn't it reverse

up and down?

Q6-13 What are primary colours ?Why are they socalled ?

Q6-4 Is itpossible to photograph a virtual image?

Q6-14 Abeam ofwhite light passing through ahollowprism gives no spectrum. Is this true of false ?

Q6-5 Is is possible for a given lensto act as a converging lens

inone medium, and asa diverging lens inanother?

Q6-15 Ordinary paper becomes transparent when it isoiled. Explain whey.

Q6-6 A cameralensis marked//1.8. Whatis the meaning of

this mark?

Q6-16 Whey does the moon, purely white during the day, have a yellowish hue after sunset ?

Q6-7 If these are scratches on the lens of a camera, they do

notappearonaphotographtaken with the camera. Explain. Do

Q6-17 What is the best position of the eye for viewiing an

the scratches affect the photograph at all ?

object through a microscope?

Q6-8 The sun seems to rise before itactuallyrises and itseems

Q6-18 What is the function ofthe circular stop at the focal

tosetlong after it actually sets. Explain why.

plane ofthe objective of a telescope?

Q6-9 Does the focal length ofalens depend on the medium in Q6-19 Why isthe objective ofatelescope oflarge focal length which the lensis immersed? Is it possible for a lensto act as a

andlargeaperture ?

conversing lens in one medium and a diverging lens in another

medium?

Q6-20 Can the optical length between two points even be less than the geometrical path between these points ?

Q6-10 Explainwhy the use ofgogglesenablesan underwater' swimmter to see clearly under the surface ofa lake.

*

^

^

Wave Optics

:392

ConceptualMCQs Single Option Correct 6-1 A parallelmonochromatic beamoflightis incidentnormally on a narrow slit. A diffraction pattern is formed on a screen placed perpendicular to the direction ofincident beam. At the first minima of the diffraction pattern, the phase difference between the rays coming from the edges of the slit is : (A) 0

(B)f

(C) TT

(D) 2sz

6-2 In Fraunhofer diffraction experiment, L is the distance between screen and the obstacle, b is the size of obstacle and "k is wavelength ofincident light, the general condition for the applicability ofFraunhofer diffraction is: (A)

— »1 LX

(B)

^ =1

(Q

^«i Lk

P)

^.1

Lk

Lk

6-3 In a Fraunhofer diffraction experiment at a single slit using a light of wavelength 400 nm, the first minimum is formed at an angle of30°. The direction 0 of the first secondary maximum is given by:

• -1^2

(A) sin .|(Q sin

-1

(B) sin-M^ (D) sin

(A) tan ' Ij

(B) tan-' ! -

(Q tan -1

P) tan"' I-

4^

6-7 Which ofthe following cannot be polarized ? (A) Ultraviolet rays P) Ultrasonic waves (Q X-rays P) Radiowaves

6-8 An unpolarised beam of intensity 7^ falls on a polariod. The intensity ofthe emergent light is :

(A) ^

P) ^0

(Q f

p)' Zero

6-9 Which ofthe following is a dichroic crystal ? (A) Quartz P) Tourmaline (Q Mica p) Selenite

6-10 An unpolarised beam ofintensity /qis incident ona pair ofNicol's prisms making an angle of60° with each other. The intensity oflight emerging from the pair is:

(A) 7, '

P) 7^2

(Q V4

P) 7^/8

6-11 When anunpolarized light ofintensity 7^ isincident on a polarizing sheet, the intensity ofthe light which does not get transmitted is:

6-4 If we observe the single slit Fraunhofer diffraction with wavelength k and slit width b, the width of the central maxima is 20. On decreasing the slit width for the same k: (A) 0 increases P) 0 remains unchanged (C) 0 decreases p) 0 increases or decreases depending on the intensity of light 6-5 In diffraction from a single slit, the angular width of the central maxima does not depend on :

(A) P) (Q P)

Xof light used Width ofslit Distance of slits from screen Ratio ofX. and slit width

(A)

P)

(Q Zero

P) ^

6-12 An optically active compound: (A) Rotates the plane polarised light P) Changing the direction ofpolarised light

(Q Do not allow plane polarised light to pass through P) None of the above

6-13 A 20 cm length ofa certain solution causes right handed rotation of38°. A30 cm length ofanother solution causes left handed rotation of24°. The optical rotation caused by 30 cm length ofa mixture of the above solutions in the volume ratio 1 :2 is:

6-6 The critical angle of a certain medium is sin~' polarizing angle ofthe medium is:

(A) P) (Q P)

Left handed rotation of 14° Right handed rotation of 14° Left handed rotation of3° Right handed rotation of 3°

_Wave Optics 393

6-14 Specific rotation ofsugar solution is 0.5 deg m^/kg. 200 kgm-3 of impure sugar solution is taken in a smaple polarimeter tube oflength 20cm andoptical rotation isfound tobe 19®. The percentage ofpurity ofsugar is : (A) 20% (Q 95%

(B) 80% p) 89%

-

6-15 Two polaroids are placed in the path ofunpolarized beam

CA)f ^I

bd

(Q

2d

^ ^ Id' Ad

Id

6-20 Ratio ofintensities between a point.4 and that ofcentral

fringe is 0.853. Ifwe consider slits are ofequal width then path

of intensity such that no light is emitted from the second Polaroid. If a thirdpolaroid whose polarization axismakes an

difference between two waves at point Awill be:

angle 0 with the polarization axis offirst polaroid, is placed

(A) I

(B) T

(Q V

(D)X

between these polaroids, then the intensity oflight emerging from the last polaroidwill be: (A)

— |sin^20 cos'^0

(Q

2

(B)

^|sin^20 4

p) L ens'* 0

6-21 Two waves of same intensity produce interference. If intensityatmaximum is4/,then intensity atthe minimum will be: (A) 0 (Q 3/

p) 2/ p) 4/

6-16 Two light waves arrives ata certain point on a screen.

The waves have the same wavelength. At the arrival point,

6-22 Intensity ofcentral bright fringe due to interference of

their amplitudes and phase differences are: (I) 2aQ, 6aQand7irad (n) 3oq, 5c7gand7iand (HI) 9aQ, la^, and3^ rad (IV) 2aQ, 2aQ and0.

two identical coherent monochromatic sources isI. Ifoneofthe

The pair/s which has greatest intensity is/are: (A) I • P) H (C) II, m P) I,IV

source is switched off, then intensity ofcentral bright fringe becomes:

(A) I

P)/

6-17 Two beams oflight having intensities/and 4/interfere to producea fringepatternonthe screen. Phasedifference between

(B)|

6-23 In an experiment to demonstrate interference offlight using

7U thebeams is —at point Aandtc at point B. Thedifference in Young's slits, separation of two slits is doubled. In order to maintain same spacing of fringes, distance D of screen from intensities ofresulting light at points AandB is :

(A) 3/ (Q 5/

P)4/ p)6/

slits must be changed to : (A) D

(B) f

6-18 A plane electromagnetic wave offrequency Wq falls normally on the surface of a mirror approaching with a

(Q 2D

relativisitic velocity v. Then frequency of the reflected wave

willbe ^given p=— j: (A)

IzP

U+P

6-24 In Young's double slit experiment, 12 fringes are obtained tobe formed in a certain segment ofthe screen when light of wavelength 6000A is used. Ifwavelength oflight is changed to 1+ P

Wn

screen is given by:

(l + P)wo (Q

(1-p)

4000A, number offringes observed in the same segment ofthe

(D)

(1-P)

(A) 18

p) 24

(Q 30

p) 36

(1+P) Wo

6-19 White light isused to illuminate two slits inYoung's double

6-25 In Young's double slit experiment, interference pattern is found to have an intensity ratio between bright and dark fringes

slit experiment. Separation betweenthe slits is b and the screen

as 9. Then amplitude ratio will be:

isata distance d(» b) from the slits. Then wavelengths missing

(A) 1 (Q 9

at a point on the screen directly in front ofone ofthe slit are:

P) 2 P) 3

•

Wave Optics

^3^

6-26 IfA is the amplitude ofthe waves coming from a point

TL

source at distance r then which of the following is correct:

(A) cos 0= -^

(B) cos 0 - —

(A) Accr^ {Q Acct^

(Q sec 0 - cos 0 =

p) sec 0 - cos 0 =

(B) ^ cc r' (D)^ccr'

4?.

6-27 If^ istheamplitude ofthewave coming from alinesource at a distance r, then :

(A)^ocr

(B)v4xH'2

{Q A

(P) Ax

6-33 In ayoung double slitexperiment, D equals thedistance ofscreen and is the separation between the slit. Thedistance ofthenearest pointtothecentral maximum where theintensity is same as that due to a single slit, is equal to :

^-28 Phase difference between two coherent light waves (A) ^DX having same amplitude A is Itz/S. If these waves superpose each other at a point, then resultant amplitude at thepoint of

DX

2DX

DX

P)

superposition will be:

(A) 2A

(B) 0

(Q A

(D) A^

6-29 Ratio ofamplitudes ofthewaves coming from two slits

6-34 Inthegiven figure-6.109, ifa parallel beam ofwhite light is incident on the plane of the slits then the distance of the white spot onthescreen from Ois [Assume dv>wv.physicsgalaxy.coni 6-1 Aprism having angle A° (which isvery small) isplaced in front of a point source 5 at a small distance d. A screen is

placedat a large distanceDas shownin the figure-6.113. Find the fringe width of interference pattern given that source is emitting light of wavelength Xand refractive index of prism

of100 cm from theslit. Itisfound thatthe9^ bright fringe isat a distance of 7.5 mm from the second dark fringe from the center ofthefringe pattern on same side. Findthewavelength of the light used. Ans. [5000 A]

IS p.

H—H •D-

Figure 6.113

I

6-6 Light ofwavelength 520 nmpassing through a double slit, produces interference pattern of relative intensity versus deflection angle 9 as shown in the figure-6.115. Find the separation

between the slits.

Ans. [ 7—• , ] ^ iy-\)Ad '

6-2 A thin glass plateof thickness t and refractive index p is inserted between screen & one of the slits in a Young's experiment. If the intensity at the center ofthe screen is /, what was the intensity at the same point prior to the introductionof

0(degrees) Figure 6.115

Ans. [1.99 X 10"^ mm]

the sheet.

6-7 The distance between two slits in a YDSE apparatus is

Ans. [/„ = / sec^ \

3mm. The distance ofthe screen from the slits is Im. Microwaves

of wavelength I mm are incident on the plane of the slits 6-3 A young's double-slit arrangement produces interference fringes for sodium light(X = 5890A) thatare0.20® apart. What is the angular fringe separation if the entire arrangement is immersed in water(refractive indexofwater is4/3). Ans. [0.15°]

normally. Find the distance of the first maxima on the screen from the central maxima. Ans. [35.35cm app., 5]

6-8 One radio transmitter.^ operating at 60.0 MHz is 10.0m

from another similar transmitter Bthatis180° outofphase with 6-4 Two coherent narrow slitsemitting lightofwavelength X transmitter How far must an observer move from transmitter in the same phase are placed parallel to each other at a small Atowards transmitter B along the lineconnecting AandB to separation of 2X. The light is calculated on a screen 5 which is reach the nearest point wherethe two beams are in phase ?

placed a distance (D » X) from the slit 5", as shown in figure-6.114. Find the distance x suchthat the intensity at P is equal to the intensity at point O.

Ans. [1.25m]

6-9 Thecentral fringe oftheinterference pattern produced by the light ofwavelength 6000 Ais found toshift totheposition of4^^ bright fringe aftera glass sheet ofrefractive index 1.5 is introduced. Find the thickness of the glass sheet. Ans. [4.8 pm]

'D-

Figure 6.114

Ans. [-JiD]

6-5 In Young's doubled slit experiment the slits are 0.5 mm apart and the interference is observed on a screen at a distance

6-10 A board source oflightofwavelength 680nm illuminates normally two glass plates 120 mm long that meet at one end and are separated by a wire 0.048 mm in diameter at the other

end. Find the number ofbright fringes formedover the 120mm distance. Ans. [141]

I40'2 \

:

"

.

6-11 Two microwavecoherent point sources emitting waves

^

^ "l

'Waye^Optlcs.1

ofrefraction index \iisfilled: (wavelength oflight inairisXdue

ofwavelength Xare placed at5X distance apart. The interference to the source, assume l»d,D » d). isbeing observed on a flat non-reflecting surface along a line

passing through one source, in a direction perpendicular to the line joining the second source (refer figure-6.116) Considering as4mm, calculate the positions ofmaxima and draw shape ofinterference pattern. Take initial phase difference

H-/s

between the two sources to be zero. Observation surface Screen

Figure 6.118

(a) Between the screen and the slits.

Figure 6.116

(b) Between the slits & the source 5. In this case, find the Ans. [48

minimum distance between the pointsonthe screen wherethe intensity ishalfthemaximum intensity onthescreen.

, 21, 4r,-|o, m.m; (((©)))]

6-12 In a YDSE with visible monochromatic light two thin

transparent sheets are used in front of the slits

Ans. [(a) Acj) =

"I / dJ X '

2 2d '•

and S'j.

p, =1.6 and Pj = ^•4- Ifboth sheets have thickness t, the central 6-15 In aYDSE, aparallel beam oflight ofwavelength 6000 A maximum is observed at a distance of5 mm from center O. Now is incident onslits at angle ofincidence 30°. A&Bare two thin the sheets are replaced by two sheets of same material of transparent films, each ofrefractive index 1.5. Thickness of^ is refractive index

t =

h+h

but having thickness

suchthat

. Now central maximum is observed at a distance of

2

20.4 pm. Light coming through A&Bhave intensities I&41 respectively on the screen. Intensity at point O which is symmetric relative to the slits is 31. The central maxima is above O.

8mm from center O on the same side as before. Find the

thickness/,(inpm) [Given: i/= 1mm.D= 1m].

A

ll

^ 30^ y

0.1mm

1

/15

Im

(u,, / -> n, tj)

Figure 6.119 Figure 6.117 Ans. [33]

(a) What istheHfiaximum thickness of5to do so ?Assuming thickness of B to be that found in part (a) answer the following parts:

6-13 Inatwo slitexperiment with monochromatic light, fringes are obtained ona screen placedat somedistance from the slits. If the screen is moved by 5 x m towards the slits, the change in fringe width is 3 x 10""^ If the distance between the slits is 10"^ m, calculate the wavelengthof the light used.

(b) Findfiinge width, maximum intensity&minimum intensity

Ans. [6000 A]

(d) Intensityat 5 cm on either sideof O.

6-14 A sourceS is kept directlybehindthe slit

in a double

slitapparatus. What will bethephase difference atP, ifa liquid

on screen.

(c) Distance ofnearestminimafrom O. Ans. [(a) t, = 120 nm; (b) [J - 6 mm; (d) I (at 5 cm above 0) = 9/]

= 91,

- /; (c) p/6 = 1 mm

[VVave^Optics

403]

6-16 A central portion with a width ofd= 0.5 mm is cut out of a convergent lens having a focal length of 20 cm. Both halves

same distance from theslitbyplacing a convex lensat a distance

of0.3 mfrom the slit. The images are found tobe separated by

are tightly fitted against each other and a point source of 0.7 X10~^ m. Calculate the wavelength ofthelight used. monochromatic light(A = 2500A) isplaced in front ofthe lens Ans. [5550 A] at a distance of 10cm. Findthe maximum possible numberof interference bands that can be observed on the screen. Ans. [5]

6-21 Two identical monochromatic lightsources A andB of

intensity 10"'^ W/m^ produce wavelength oflight 4000>/3 A.

A glass ofthickness 3 mm is placed in thepath ofthe-ray as 6-17 A plastic film with index ofrefraction 1.80 isputonthe shown in the figure-6.121. The glasshas a variablerefractive surface ofa carwindow toincrease thereflectivity and thereby index n=l +-Jx wherex (in mm) is distance ofplate from left to keep the interior of the car cooler. The window glass has to right. Calculate resultant intensityat focal point F ofthe

index ofrefraction 1.60.

lens.

(a) What minimum thickness isrequired, iflight ofwavelength 600 nm in air reflected from the two sides of the film is to

interfere constructively?

(b) Itisfound tobedifficult tomanufacture andinstall coatings as thin as calculated in part (a). What is the next greatest thickness for which there will also be constructive interference? Ans. [(a) 8.33 x lO""^ m; (b) 2.5 x lO"'] Figure 6.121"

6-18 Two coherent monochromatic sources AandB emit light

Ans. [4 X io-'5 W/m^]

ofwavelength X.The distance between A and B is d= 4X. \D

6-22 Two fine slits are placedvery close to each other and theyare illuminated bya strong beam of light ofwavelength 5900 A. Fringes are obtained at a distance of 0.3 m from the slits. The fringe width is found to be 5.9 x 10"^ m. Calculate the distance between the slits. Ans. [3 X 10"^ m]

Figure 6.120

(a) If a lightdetector is moved alonga line CDparalleltoAB, what is the maximum number ofminima observed ?

(b) If the detector is moved along a line BEperpendicular to AB and passing through B, what is the number of maxima observed ?

6-23 Biprism fringes are produced with the help of sodium light (A = 5893 A). Theangle ofthe biprism is 179.5? and its refractive index 1.5. If the distance between the slit and the

biprism is 0.4mandthe distance between thebiprism andthe screen is 0.6 m, find the distance between successive bright fringes. Also calculate the maximum width ofthe slit for which the fringes will still be sharp. Ans. [3.376 x 10^ m, 112 ^m]

Ans. [(a) 8, (b) 4]

6-24 A glass plate 1.2 x 10~^ m thick isplaced in thepath of 6-19 A beam of lightconsisting oftwo wavelengths, 6500 A

and 5200 Ais used to obtain interference fringes in aYoung's double slit experiment. Findthedistance ofthethirdfringe on thescreen from thecentral maximum for thewavelength 6500A. What is the least distance from the central maximum at which

one of the interfering beams in a biprismarrangement using monochromatic lightofwavelength 6000 A. If the central band shifts by a distance equal to the width of the bands, find the refractive index ofglass. Ans. [1.5]

the bright fringes due to both wavelengths coincide? The distance between the slits is 2 mm and the distance between

the plane ofthe slits and the screen is 120 cm. Ans. [1.17 mm, 1.56 mm]

6-25 A monochromatic light of intensity 7 is incident on a glass plate. Another identical glass plate is kept close to it. Each glass plate reflects 25 per cent of the light incident on it and transmits the rest, find the ratio of the maximum and

6-20 In an experiment using Fresnel biprism, fringes ofwidth 0.185 X 10"^ m are observed at a distance of 1 m from the slit. Images are of the coherent sources are then produced at the

minimum intensities in the interference patternformed bythe twobeam obtainedafter reflection from eachplate. Ans. [49 : 1]

Wave Optics j

1404

6-26 Aplane light wave ofwavelength X=700 nmfalls normally on the base of a biprism made of glass (p = 1.560) width refracting angle0 = 5°.Aplateisplaced behind thebiprism and thespace between themisfilled witha liquidofrefractive index (p'= 1.500). Find thefiinge width onascreen behind thebiprism.

6-31 Two pointcoherent sources ared= 4X apart. Two light detectors are moved - one along the line parallel to the line

joiningthe sources andtheotheralong thelineperpendicular to this line and passingthrough one of the sources. Find the maximum number ofmaxima registered by each detector. Ans. [9, 9]

6-32 In aYoung's arrangement, the distance between the slits is 1.5 mm and the distance ofthe screen from the slits is 2m. At

what minimum distance from the central fringe will red and

violet brightfringes coincide iftheslitis illuminated withwhite light?The wavelength range in white light is from 420 nm to 690 nm.

Ans. [12.88 mm] Figure 6.122 Ans. [66.8 pm]

6-27 In a biprism experiment, the angular width of fringes was 0 = 2'. calculate the separation between the slits if the wavelength of the incident light was X= 700 nm. Ans. [1.2 mm]

6-33 A double-slit apparatus is immersed in a liquid of refractiveindex 1.33.It has slit separationof 1mm, and distance between the plane of slits and screen is 1.33 m. The slits are illuminatedbya parallelbeamof light whosewavelength in air is 6300 A.

(a) Calculate the fringe-width, (b) one of the slits of the apparatus is covered by thin glass sheet of refractive index 1.53. Find the smallest thickness ofthe sheet to bring adjacent

6-28 In a biprism experiment, calculatethe intensity oflights on the screen at a point 5p/6 away from the central fringe in termsofthe intensity ofthe central fnnge,which \sIq. Here P is

minimum on the axis.

the fringe width.

6-34 In a Young's double slit experiment, the separation

Ans. [0.633 mm, 1.58 mm]

between the slits is 2 x 10~^m whereas the distance of screen

Ans. [3/^/4]

from the slits is 2.5 m. A light of wavelengths in the range of

6-29 Twoparallel beams of light P and Q (separation d and 2000 A - 8000 A is allowed to fall on the slits. Find the mutuallycoherent)are incidentnormallyon a prismas shown, wavelength in the visible region that will be present on the ifthe intensitiesofthe upper and the lowerbeams, immediately screen at 10"^ m from the central maximum. Also find the after transmission from the face AC, are 4/and /respectively, wavelength that will be present at that point of the screen in find the resultant intensity at the focusofthe lens which brings the infrared as well as in the ultraviolet region. Ans. [8000 A, 4000 A 2666.7 A ... 8000 A (infrared), 2666.7 A

them into focus.

(ultraviolet)] p

6-35 A narrow slit ofwidth 0.025 mm is illuminatedby a parallel

\

beam of light. The diffraction pattern is observed through a telescope. It is found that to reach the first minimum, the telescope has toberotated through 1°24'from thedirection of the direct ray. Calculatethe wavelength of light used.

d

, i Q

Figure 6.123

Ans. [6100 A] Ans. [91]

6-30 Twopoint sources are d=nk apart. A screen is held at right angles to the line joining the two sources at a distance D from the nearer source. Calculate the distance of the point on the screen where the first bright fringe is observed. Assume D»d. 2D(D + nX) Ans. [J—^ ^1

6-36 Angular width of central maximum in the Fraunhofer diffractioh pattern ofa slit is measured.The slit is illuminated by another wavelength, the angular width decreases by 30%. Calculate the wavelength of this light. The same decrease in angular width ofcentral maximum isobtained when theoriginal apparatus is immersed in a liquid. Findthe refractive indexof the liquid. Ans. [4200 A, p = 1.43]

Optics

405 i

6-37 In a Young's double slit set up the wavelength oflight

6-41 In a Young experiment the light source is at distance /j = 20pm and /j=40pmfrom the slits. The light ofwavelength

used is 546 nm. the distance ofscreen from slits is Im. The slit separation is 0.3 mm.

X=5000A is incident on slits separated atadistance 10 pm. A

(a) Comparethe intensityat a pointP distant 10mm fromthe

screen is placed at a distance D = 2m awayfrom the slits as

central fringe where theintensity is1^.

shown in figure-6.125. Find

(b) Findthe numberof brightfringes between P and thecentral fringe.

.Ans. [(a) 3.0 X10-6 m; (5) 3 q x lO"' 7^]

6-38 In a double slit pattern (X. = 6000A), the firstorderand tenth order maxima fall at 12.50 mm and 14.75 mm from a

particularreference point. If Xis changedto 5500A, findthe position of zero order and tenth order fringes, other arrangements remaining the same. Ans. [ 12.25 mm, 14.55 mm]

Figure 6.125

6-39 An interference is observed due to two coherent sources

(a) The values of0 relative to the central line where maxima appear on the screen ?

placed at originand ^2 placed at (0, 3X, 0). Here Xis the wavelength of the sources. A detector D is moved along the (b) Howmany maximawillappearon the screen? positive x-axis. Find x-coordinates on thex-axis (excluding (c) What should be minimum thickness ofa slab ofrefractive X= 0 andx = 00) wheremaximumintensityis observed. index 1.5beplaced onthe pathofoneoftheraysothatminima occurs at C?

Ans. [1.25X, 4?.]

(b) 40; (c) 5000A] 6-40 AYoung double slitapparatus isimmersed ina liquid of Ans. [(a) sin"' refractive index p,. Theslitplane touches theliquid surface. A parallel beam ofmonochromatic light ofwavelength X(inair) in 6-42 A screen is placed 50 cm from a single slit, which is

incident normally on the slits.

illuminated with 6000 A light. Ifdistance between the first and third minima in the diffraction pattern is 3.0 mm,what is the width ofthe slit? Ans. [ 0.2 mm ]

6-43 In a double slit experimentthe distancebetween the slits is 5.0 mm and the slits are 1.0 m from the screen. Two interference

patterns can be seen on the screen one due to light with

wavelength 480nm, andtheother duetolight with wavelength 600 nm. What is the separation on the screenbetween the third Figure 6.124

(a) Find the fringe width

orderbright fringes of the two interference patterns? Ans. [0.072 mm]

(b) If one oftheslits (say 5'2) is covered by a transparent slab ofrefractive index ^2 and thickness (as shown, find the new 6-44 Ina Young's double slitexperiment using monochromatic position ofcentral maxima.

light the fringe pattern shifts by a certain distance on the screen

(c) Now the other slit 5, is also covered by a slab of same thickness and refractive index pj asshown infigure-6.124 due

when a mica sheet ofrefractive index 1.6 and thickness 1.964

microns is introduced in the path of one of the interfering

to which the central maxima recovers its position find the value

waves. The mica sheet is then removed and the distance

between the slits and screen is doubled. It is found that the

ofp3.

(d) Find theratio ofintensities at Ointhethree conditions (a), (b) and (c),

distance between successive maxima(or minima) now is the same as observed fringe shift upon the introduction of the

mica sheet. Calculate the wavelength of the monochromatic (b)

(c)

(d)

light used in the experiment. Ans. [589 nm]

• Wave Optics j

i406

6-45 Two very narrow slits arespaced 1.80 pmapart and are reflection theintensity decreases by90% andoneach refraction placed 35.0 cm from a screen. What isthe distance between the theintensity decreases by10%, find theratio oftheintensities first and second dark lines of the interference pattern when the slits are illuminated with coherent light of ^ = 550 nm? (Hint: The angle 0 is not small)

ofmaximum to minimum in reflected pattern.

Ans. [12.6 cm]

6-46 In a Young's double slit set up the wavelength of light used is 546 nm. The distance of screen from slits is 1 m. The slit separationis 0.3 mm.

(a) Compare the intensity ata point P distant 10 mm from the

Figure 6.126 Ans. [361]

centralfringewherethe intensity is/q.

6-51 A convergent lenswitha focal length of/= 10cmis cut

(b) Find the number ofbright fHnges between P and the central

into two halves that are then moved apart to a distance of d= 0.5 mm (a double lens).Findthe flingewidthon screenat a distance of 60 cm behind the lens if a point source of

fringe. Ans. [(a) 3.0 X lO"^ /(,; (b) 5]

6-47 Interference pattern with Young's double slits 1.5 mm

apart are formed on ascreen ata distance 1.5 mfrom theplane ofslits. In the path ofthe beamof one ofthe slits, a transparent film of lO-micron thickness and of refractive index 1.6 is

interposed while in thepath ofthebeam from the other slita transparent film of15 micron thickness and the refractive index

monochromatic light {X = 5000 A)isplaced infront ofthelens at a distance ofa = 15 cm from it. Ans. [0.1 mm]

6-52 In theYoung's double slit experiment a pointsource of X= 5000 Aisplaced slightly offthecentral axis asshown inthe figure-6.I27.

1.2is interposed. Find the displacement of the fringepattern. Ans. [3 mm]

6-48 A glass plate (« = 1.53) that is 0.485 pm thick and surrounded byairisilluminated bya beam ofwhite lightnormal

T

1 . 10mm T Inim *

tP

5mm

to the plate.

(a) What wavelengths (inair)within thelimits ofthevisible spectrum {X = 400 to 700 nm) are intensified inthereflected

m—nr—

-2m-

beam?

Figure 6.127

(b) What wavelengths within the visible.spectrum are Ans. [(a) 424 nm, 594 nm; (b) 495 nm]

(a) Find the nature and order ofthe interference atthe point P. (b) Findthenature andorder ofthe' interference at O. (c) Where should we place a film ofrefractive index p= 1.5

6-49 An oil film covers the surface of a small pond. The

and what should be its thickness so that maxima ofzero order

refractiveindex of the oil is greater than that of water.At one

is obtained at O.

point on thefilm, thefilm hasthesmallest nonzero thickness

Ans. [(a) 70"' order maxima; (b) 20"' order maxima;

for which there will be destructive interference in the reflected

(c) t - 20 pm, in front of 5,]

intensified in the transmitted light?

light when infrared radiation with wavelength 800nmisincident normal to the film. When this film is viewed at normal incidence

6-53 A convex lens offocallength/= 25 cmwascut alongthe

at this same point, for what visible wavelengths, if any, will

diameter into two identical halves and was again set leaving a

there be constructive interference? (Visible light has

thin gap of width a = 0.10 cm. A narrow slit, emitting

wavelengths between 4000A and 7000A) Ans. [5330A]

6-50 A ray of light is incidenton the left vertical face of the

glass slab. Ifthe incident light hasan intensity / andoneach

monochromatic lightofwavelength X= 0.60pmwasplaced at the focus of the lens. A screen was placed behind the lens at a distance b = 50 cm. Find the width ofthe fringes on the screen and the possible number of fringes. Ans. [1.5 X 10^ m, 13]

jWave Optics

407 !

6-54 Light ofwavelength X—500 nm falls on two narrow slits placed a distance = 50 x 10^ cm apart, at an angle (j) = 30°

6-58 Inthe YDSE the monochromatic source ofwavelength X ^

relative to the slits as shown in figure-6.12S. On the lower slit a is placed atadistance —from the central axis (as shown in the transparent slab ofthickness 0.1 mm and refractive index —is

figure-6.129), where is the separation between the two slits

and Sy

placed. The interference pattern is observed on a distance Z) = 2m from the slits. Then calculate:

Imm

i..

-"77???777/Z"

^

Mirror

^

H—Scm-

D. = 2m

-190cm

Figure 6.129

(a) Find the position ofthe central maxima (b) Find the order ofinterference formed at O.

(c) Now S is placed on centre dotted line. Find the minimum thickness of the film ofrefractive index p.= 1.5 to be placed -D-

Figure 6.128

(a) Position ofthe central maxima? (b) The order ofmaxima at point Cofscreen? (c) How many fringes will pass C, ifwe remove the transparent slab from the lower slit?

Ans. [(a) At 0 = 30° below C; (b) 50; (c) 100]

6-55 A double-slit arrangementproducesinterference fringes forsodium light(A, = 5890A) thatare 0.20° apart. Whatwillbe the angular fringe separation if the entire arrangement is 4' immersed in water

p =

3

infront ofS2 sothatintensity at Obecomes —th ofthemaximum intensity. TakeA, = 6000A;t/=6mm Ans. [(a) 4 mm above 0; (b) 20; (c) 2000 A]

6-59 Fringes are producedbya Fresnel'sbiprism in focalplane ofa reading microscope which is 100 cm from the slit. A lens inserted between the biprism and the eyepiece gives two images of the slit in two positions of the lens. In one case the twoimages ofthe slit are 0.45 mm apart in the otho-case2.90 mm

apart. If sodium light of wavelength 5893 A used, find the width ofthe interference fringes. If the distance between the slit and biprism is 10 cm and refractive index ofthe material of the biprism is 1.5, calculate the angle in degrees which the inclined faces ofthe biprism makes with its base.

Ans. [0.15°] Ans. [2.91 or 1.73 mm]

6-56 In a double-slit pattern(A, = 6000A), the zero-order and tenth-order maxima fall at 12.34 mm and 14.73 mm from a

particularreference point. If A, is changed to 5000A, findthe position of the zero order and tenth order fringes, other arrangements remaining the same. Ans. [Zero-order at the same position but tenth-order maxima at 14.53 mm]

6-60 In a biprism experiment withsodium light(A. = 5893A), the micrometer reading is 2.32 mm when the eyepiece is placed at a distance of 100 cm from the source. Ifthe distance between

two virtual sources is 2 cm, find the new reading ofmicrometer when the eyepiece is moved such that 20 fringes cross the field ofview. Ans. [2°]

6-57 The distance between a slit and a biprism ofacute angle 2° is 10 cm. Find the fringe width and the width of the entire band when observation is made on a screen at a distance of

90 cm from the biprism. The refractive index ofthe material of

thebiprism is 1.5 andA, = 5890A. Ans. [0.168 mm, 31.5 mm]

6-61 In two slit experiment with monochrmoatic light, fringes are obtained on screen placed at some distance from the slits. If the screen is moved by 5 x m towards the slits, the

change in fringe-width is 3 x 10"^ m. If the distance between the slits is 10"^m, calculate the wavelength of the light used. Ans. [6000 A]

Wave Optics]

408

6-62 In a Young's double slit experiment, the separation between slits is 2 x 10"^ m whereas the distance ofscreen from

by another sheet of thickness = 1.25 pm as shown in figure-6.131.Both sheets are made of same material

the slits is 2.5 m. A light of wavelengths in the range of 2000 - 8000 A is allowed to fall on the slits. Find the wavelength

inthevisible region thatwill bepresent onthescreen at 10"^ m from the central maxima. Also find the wavelength that will be

S, L.

present at that point of screen in the infrared as well as in the ultraviolet region. Ans. [2000 A (ultraviolet)]

6-63 A double slit apparatusis immersedin a liquidofrefractive index 1.33.It has slit separation of 1mm, and distance between the plane of slits and screen is 1.33m. The slit is illuminated by

-M. \*-

Figure 6.131

aparallel beam oflight whose wavelength inair is6300 A. (a) Calculate the fringe-width

(b) One of the slits ofthe apparatus is covered by a thin glass sheet of refractive index 1.53.Find the smallest thickness of

the sheet to bring the adjacent minimum ofthe axis. Ans. [(a) 0.63 mm; (b) 1.575 pitn]

6-64 The Young'sdouble slit experiment is done in a medium

ofrefractive index 4/3. Alight ofbOOOAnm wavelength isfalling on the slits having 0.45 mm separation. The lower slit ^2 is coveredbya thin glass sheetof thickness 10.4 pm and refractive index 1.5. The interference pattern is observed on a screen placed 1.5 m from the slits as shown in figure-6.130 "

havingrefractiveindex p = 1.40.Waterin filledin spacebetween diaphragm and screen. A monochromatic light beam of

wavelength A= 5000Ais incident normally on thediaphragm. Assuming intensity of beam to be uniform and slits of equal width, calculate ratio ofintensity at C to maximum intensity of interference pattern obtained on the screen, where C is foot of

perpendicular bisector of 5, 52. Refractive index of water

Ans. [|] 6-66 In a Young's experiment, the upper slit is covered by a thin glass plate of refractive index 1.4 while the lower slit is covered by another glass plate, having the same thickness as the first one but having refractive index 1.7. Interference pattern

is observed using lightofwavelength 5400 A. It is found that

5

,

Screen

the point P on the screenwherethe central maximum (n = 0) fall beforethe glass plates were inserted now has (3/4) the original intensity. It is fiarther observed that what used to be the fifth maximum earlier,lies belowthe pointP while the sixth maximum lies above P. Calculate the thickness of the glass plate. (Absorption of light by glass plate may be neglected).

Figure 6.130

(a) Find the location of the central maximum (bright fringe with zero path difference) on they-axis.

Ans. [9.3 X 10"^ m]

(b) Find the light intensity at point O relative to the maximum fringe intensity.

6-67 In Young's experiment, the source is red light of

(c) Now, if 6000A nm lightis replaced bywhitelightofrange 4000 to 7000A, find thewavelength ofthe light that form maxima exactly at point O. [All wavelengths in this problem are for the given medium of refractive index 4/3. Ignore dispersion] Ads. [(a) 4.33 mm; (b) 0.75

(c) 1300A, 650A, 443.34A, 260A]

6-65 A screen is at a distance £) = 80 cm from a diaphragm

having twonarrow slits 5"] and

whichare 2 mm apart. Slit5,

is covered by a transparent sheet of thickness

= 2.5 pm and

wavelength 7x10"^ m. When a thin glass plate ofrefractive index 1.5 at wavelength is put in the path of one of the

interfering beams,the central bright fringe shifts by 10~^m to the position previouslyoccupiedbythe 5^^ bright fringe. Find the thickness ofthe plate. When the source is now changed to

green lightofwavelength 5 X10"^m, the central fringe shifts to a position initiallyoccupied bythe 6^^ bright fringedueto red light. Find the refractive index of glass for green light. Also estimate the change in fringe width due to the change in wavelength. Ans. [1.6, 5.7 x IQ-^ m]

jAtomic Physics

409

ANSWER & SOLUTIONS CONCEPTUAL MCQS Single Option Correct

1 4 7 10 13 16 19 22 25 28 31

(D) (B) (A) (A) (A) (B) (A) (C) (C) (C) (A)

2

(D) (A) (D)

5 8

11 14

17 20 23 26

.29

3 6 9

(B) (D) (D) (C) (A) (B) (B)

12 15 18 21 24 27 30

(C) (B) (A)

(C) (D) (A) (B) (A) (B) (B)

'Be

IIL

Be

^Be

z

"Be = ^

(iii) As

cc —— we use n

E

Be

= 4

NUMERICAL MCQS Single Option Correct 1 4

(C) (B)

7 10 13 16

(A) (B) (B) (D)

19 22 25 28 31 34 37 40 43 46 49

(D) (A) (B) (A) (C) (B) (D) (A) (D) (A) (A)

(1)^

(A) (D) (B) (B) (C) (B) (D) (C) (C) (C) (D) (D) (B) (B)(A) CD)

2 5 8

11 14 17 20

23 26 29 32 35 38

41 44 47

(D) (C) 9 (D) 12 (C) 15 (B) 18 (B) 21 (B) 24 (D) 27 (C) 30 (B) 33 (C) 36 (B) 39 (D) 42 (A) 45 . (C) 48 (C) 3 6

(iv) E — 3.4eVisfor« = 2stateandweuse 1

V

cc —

"

n

V

''2=2

E^ cc —- we use

(v) As

n

±u_ n,;

H

ADVANCE MCQs One or More Option Correct 1 1 4 7 10 13 16 19 22 25 28

(D) (All) (A, C, D) (A. D) (A, C, D) (A) (B) (A, B, C) (D) (B) (A, B, D)

2 2 5 8 11

14 •

17 20

(D) (B, C)

3

(All) (A, C) (B. C) (A, B, C) (B, C, D) (C, D)

6

3

9

12 15 18 21

(A, C, D) (A, C) (A, D)

23

26 29

24 27 30

(C) (A, D) (A. D) (B, C) (A, C) (B) (B, D) (A, B) (A) (A, C) (B, D)

Solutions ofERA CTICE EXERCISE 1.1

"irli] x(3)' «i/=3

(vQ As

TE^

(vii)As

=> (i)

As

KE. =

2r„

TE =-

KZe^ 2r^

=-l

KE^ =~TE^

XE=+3.4''eV'

oc

Solutions ofPRACTICEEXERCISE 1.2 (i) Minimumwavelength is correspondingto transition from

'lO

/•io = 100 X1.06=106A

00 —> 1 hence

1 ^ 12431 „ (ii) As

cc —

we use

^

13.6

Atomic PhySfcSi

andmaximumwavelength is fortransition2 12431 10.2

1

A

When six wavelength are emitted that means electrons are excited tow= 4and byabsorption of10.2 eV ifelectrons excite from to «2 ~ 4 we use 1__J_

10.2 = 13.6 Z2

^max

2

16

V"i

4

y

for Z= 1 this is not possible

for Z = 2 we get n^ = 2 (ii) We use

X=

^ =4113.2A

for Z = 3 onward this is not possible

Thus Z= 2andtransition is from n^ = 2Xon^ = A. Minimum wavelength (maximum energy) emission is for transition 4

(iii) (a) We use

.

12431

'^min

Maximum wavelength emission isfor transition 4-)•3

(b) From« = 3 totalspectral linesare

We use

. =77^^ 12.75x4 =243.74A

= 114.26

• 13.6(3)2

1

=3

A£: =13.622 [i-i Z2=-P?|^=36 5x13.6

12431

^max 0.66x4 =4708.71 A. Solutions of PRACTICEEXERCISE1.3

(i) Frequency if e~ revolution in fn =

Z=6

A:£'in;i=l is A:£',=+13.6Z2 = 489.6eV

X=t{^^=25.39A 13.6x36

and

orbit is

47i^K^Z^e^m n'h'

for transition from «to « -1 orbit radiation frequency is given by

A£ (v) We use =>

12431

12431

10.2

2.55Zi2 Z=2

v =

Energies offirstfour levels ofX are

£:j=_13.6x4=-54.4eV £:^=_3.4x4= -13.6eV £'3=-1.51 x4= -6.04eV £^=-0.85 x4= -3.4eV Iffor thisatom is+E^ = 54.4 eV (vi) Given that and

E^-Ej = \0.2 +17.0 =27.2 eV E^-E^= 4.25 +5.95=10.2 eV E3-£i=27.2-10.2=17eV

(«-l) 2T^K^Z^e'm h'- •

1

1

2Ti^K^Z^e^m

v =

n'

Zn-\

rp-{n-\f

for large n wecan use2«-1 = 2« and « -1 = « Asi^K^Z'^e^m v =

-^fn-

3?.3

n'h

(ii) First lineofBalmer series ofH-atom is

'Kh

U

9j

rn^+ntfi

first line ofBalmer series for T-atom is

1.89Z^ = 17eV Z=3

4

9J

rrig+mj-

1 J_

Now we use

272 =13.6Z2| 4-72 Xj.

36(Wg 5RmH

« = 6

12431

(vii) Incident photon energy.£= ^213 ~ 10.2 eV

)

36(We +mj-) SRmj

36

(m^ + nifj )m-p - {m^ + mj)mu

5R

mffmj-

IAtomic Physics

"

(h) For inelastic collision H-atom must at least exite from n = 1 36

to « = 2 byobsorbing 10.2eV energy. In givensituation initial momentum is zerothusmaximum possible energyloss can be

5R

36

9.109x10"^'x2xl.67xl0"^'^

5x10967800

3X(1.67x10"")

2X^mv^ so we use wv2 = 10.2eV

=2.387 A

10.2x1.6x10"'^

1.67X10-"

(ill) Photon

(v) (a) We use mvr = -r—

for « = 1

2;c

h

Wavelength of incident photon is "12431 1=

Photon momentum is

v =

2nmr

= 2266.28 A

5.4852

mv

and

m( h ^_ _

P=-

^

= evB

X

r \2Tzmr} ^

By conservation of energy we use

AE= i m^vf+

-(1)

2neB

By conservation ofmomentum we use

...(2) and

h

j^=W^.V2C0S

Yih

^

(b) Usingmvr= — and

...(3)

= evB weget

nh r =

2-K.eB

Squaring adding (2) and (3) we get

h

(c)

heB

nh

2jcot

V27cw^k

2;ce5

1 -2 2 ^^

r

1 2Wg

2 2 h^

— myf= —

Vis maximum when n = 1

_

From equation-(l) we use .rr

heB

.,2

AE= -z—V, -

2m„

2mJX

2

wl"

A£' + -

2m.V = 14.2 m/s

Vn = m

•Li

+ 1 me

By angular momentum conservation we use

from equation-(3) we use

mvb = mv^

...(1)

by energy conservation we use cos 9 =

Xwy^V2

0 = 88.9°

= 0.0178

1 2 :rwv/+ 1 2^ -rm\p= 2

2

^

1

ZZ'e'

An e.

Sabtitutingvalueofv^from (1) to (2) we get

..-.(2)

Atomic PhysioS|

;4i2

1

n. 1

( v6 1 ,

4mv2 fl-4^ for

Sol. 5 (A) In Rydberg's formula the wave numberis directly

t.2

1

proportional to Z^ and Balmer series of hydrogen atom is corresponding to the transition of electron from any level to

-ZZ'e

n = 2. Shortestwave length for the spectrumof hydrogenlike

r.2

1

ZZ'e

atom for transition to « = 4 level will be inversely proportional to t? and ifit matches with that ofthe Balmer series ofhydrogen

4n Gq

atom then we use

6 = 0 we get 1 All'

2ZZ'e^ =0

mv

Z2= -- x42 = 4

(vii) Usingforce on massm is conservative field F =-

dr

Z=2

- mb^r

for circular orbit ofparticle mv

mb^r =

...(1)

r

nh

and

mvr =

-(2)

2%

Sol;7 (A)' It has already beendiscussed in theorythat by the wavenature ofelectron and itsDeBroglie wavelength weanalyze for formation of stationary waves in an orbit the orbit circumference shouldbe an integralmultipleofthe wavelength so for consistancy of quantization postulate option (A) is

m(br)r=^ nh r =

Sol. 6 (B) Balmer series in thehydrogen atom haswavelength of most of its lines lying in the visible region of the electromagnetic specturm and someare in infraredregion.

correct.

2%mb

Solutions ofCONCEPTUAL MCQSSingle Option Correct

Sol. 8 (D) Transition A is correspondingto infinity to « = 1 whichis the last line (serieslimit) ofLyman series,transitionB

Sol. 1 (D) As energyin

Orbitof hydrogen atom is givenas

is from « = 5 to « = 2 which is third line of Balmer series and transition C is from « = 5 to « = 3 which, is second line of

2 k^2'7'2_4 2n K Z e m

Paschen series.

Z

E_=-

Sol.9 (A) In lower energylevels kinetic energyis higherand

hence option (D) is correct.

Sol. 2 (D) As radiusof

potential energy and total energy is lesser hence option (A) is orbitofhydrogen atomis givenas h'

r.. =

"

n' X

An^Ke^m

Z

correct.

Sol. 10 (A) In Bohr'sModel the kineticenergyofelectron in orbit is given as

hence option (D) is correct.

1

Sol. 3 (C) As alreadystudiedthat due to differencein masses of nuclei we consider the effect of reduced mass of electron

E=-LKZe^

energies ofthe energylevels slightly varyin the twothat cause

2

given as

Sol. 11 (B) InBohr's model thespeed ofthe electron inn"^orbit is given as

1 KZe^ •

hence option (B) is correct.

r„

hence option (A) is correct.

Sol. 4 (B) The total energy of electron in hydrogen atom is

.

2

1 KZe]

and total energy is given as

which is different in hydrogen and deuterium because of which the difference in the spectrum.

2

r

2nKe' v.. =

hence option (B) is correct.

h

fAtomic Physics

41^3.]

Sol. 12 (C) Weuse -J_il

47160 ^0

Sol. 22 (C) In case ofelectron we have studied that during collision itcan transfer almost its complete energy to the particle it is colliding hence E^ is the minimum required energy for ionization ofhydrogen atom. In case ofhydrogen and helium

—

ion as helium is more massive than hydrogen it transfers less

v = m

Sol. 13 (A) In

energyduring inelastic collision compared to the case when hydrogen is colliding hence option (C) is correct.

orbit maximum number ofelectrons can be

hencethe possible electrons in energylevels 1, 2, 3 and 4 are 2,8,18 and 32 ofwhich the sum is 2 + 8 +18 + 32 = 60. Hence option (A) is correct.

Sol. 23 (A) The wavelength emitted isinversely proportional to the energy radiated hence option (A) is correct.

Sol. 24 (A) To excite the hydrogen atom the kinetic energy of neutron must be more than 20.4 eV sothatin case ofperfectly

Sol. 14 (D) Because of the energy level differences in the hydrogen atom which explained thattheenergy level difference inelastic collision the energy loss which is half of the total decreases as we go away from nucleus which explained the kinetic energy (10.2 eV) may exite the hydrogen atom soincase hydrogen specturm which other models bythat time couldn't if kientic energyof neutron is less than 20.4 eV then collision must beperfectly elastic hence option (A) is correct. explain hence option (D) is correct. Sol. 15 (D) A quantum number givesthe state of an electron

Sol. 25 (C) As during de-excitation it is emitting six

principle no two electrons can have same set of quantum

emitted energies have less, equal or morethan the excitation energiesthen only option (C) can be correct.

inextranuclear partofanatom soaccording toPauli's exclusion wavelengths that means the atom is excited to «2 =4 and the numbers, hence option (D) is correct.

Sol. 26 (B) Due to the effect of mass of nucleus we consider Sol. 16 (B) From the figure we can see that AE^ =AE^ +AE^ reduce mass of electron in analyzing the energies of energy

hence option (B) is correct as Af = hc/X.

levels of the atoms and due to this the binding energy of deuterium is more thanthatofhydrogen inground state hence

Sol. 17 (D) At room temperature all atoms of hydrogen gas option (B) is correct. Bohr's model is valid for all one electron are inground state inwhich only radiation ofultraviolet region systems (hydrogen like) and some of the lines of Balmer series can be absorbed to excite atoms from « = 1 to higher energy lie in infra red region thats why options (A) and (C) are not levels sovisible light cannot excite any atom in ground state correct. hence option (D) is correct.

Sol. 27 (B) Balmer series is corresponding the de-excitation

Sol. 18 (A) The angular momentum ofelectron in hydrogen

ofelectrons inhydrogen atom to «= 2state hence option (B) is

atom is quantized andaccording toBohr's second postulate for

correct.

energy level it is given as nh/ln so for two successive orbits option (A) is correct.

Sol.28 (C) ByusingRydberg's formula we can calculate the values of wavelengths emitted in above four cases and

Sol. 19 (A) Hydrogen atoms atroom temperature are inground considering same value ofZ for hydrogen and deuterium and state and can only absorb radiation of ultraviolet region to will not consider the effect ofnucleus asin given options we excite from k = 1 to higher energy levels due to which in need tocheck for approximate relations, option (C) iscorrect. absorption specturm of hydrogen gas only Lyman series is obtained hence option (A) is correct.

Sol. 29 (B) As thetotalenergy of electron in hydrogen atom is given as

Sol. 20 (C) a-particle scattering occurs due to Coulomb repulsionof nucleuson incidenta-particle hence option(C) is correct.

^ "

1 KZe^ 2

r„

hence option (B) is correct.

Sol.21 (B) Inany given orbit angular momentum ofelectrons

Sol. 30 (B) 12.leV isthe difference inenergies of« = 1and

are same as given by Bohr's second postulate and energies of electron in anyhydrogen like ion is directly proportional to 2?hence option (B) is correct.

« = 3 forhydrogen atom sothe electron isexcited to « = 3 state hence theangular momentum will become 3/7/271 frran hlln hence theincrease inangular momaitura isgiven inoption (B).

Atomjc Physic^,

|414

Sol. 31 (A) Inhydrogen atom themagnetic moment duetothe Sol.6 (C) First excitation energyofthis atom is given as motion of electron in

orbit is given as , _ enh "

=>

eV=RhC, ,

= =^RhC

,

4

2'J

47cm

Thus ionization energy ofthis atom is hence option (A) is correct. AV

RhC=^ SolutionsofNUMERICAL MCQSSingle Options Correct

Sol. 1 (C) We use X=

Sol. 7 (A) As we have E oc —

=1218.72A

Sol. 2 (A) We use ^ , V nh nhv frL^-z—xrx— =—^ 2nr

(1/ir

Hli

(3/3)^

-^3Z./ ^\H ^

2%

for Bohr's model we have

V=

1ML

Sol. 8 (B) First line ofLymenseries ke^

1243 \ =^=1218.7A

X 27t

10.2

frl

nh

ke^

An'

nh

X 2%=

last time of Lymen series

ke'2ji

.

^

. ,

k^

ke'n"-

12431

13.6

hence option (B) is correct.

x=0

Sol. 9 (D) velocity of Sol. 3 (D) Pot first line of Balmer series

orbit is inverselyproportional to n

hence the excited'state orbit is « = 3 and radius of

orbit is

directlyproportionalto squareof n hence option(D)is correct.

X

9). 36

...(1) h

for second line ofBalmer series

Sol. 10 (B) We use X= — =

^'

3R

X

16

16

...(2)

From equaion-(l) & (2) V

P

,

h

=>

yl2mE •

i

h

yjlmiVq)

Thus forhighervalueof mand q; Xwill besmallerand for an — 'a' particle; both'm' and are higher.hence lesser is the wavelength.

'36

X

As, we known (penetrating power) oc Energy cc

T6

X'-^X X-^^X Sol. 4 (B) We know \K\-L-'

=Y=\=c-' r' L cT

From above; penetrating power of an a-particle is more than that ofa proton, hence option (B) is correct. Sol. 11 (B) Time periodofrevolution is directlyproportional to

Sol. 12 (C) Radius of

orbit is proportional ton^ hence the

radius of first excited state (« = 2) will be four times that ofthe ground state hence its area will be 16times that of ground state.

Sol. 5 (D) Allthetransitions given inoptions (A), (B] and(C) give photons of energy more thanbytransition from « = 4 -> 3. Hence none of the transitions given in options (A), (B) and (C) give out infrared radiation.

hence option (B) is correct.

Sol.

13 (B) Weuse ^

Solving we get « = 5.

/g ^2 , 400

R

[Atomic Physics _ _

415

Sol. 14 (C) By the incident radiation electron will excite to

« = 3from which during de-excitation number ofspectral lines will be ^€2= 3lines.

Sol. 20 (D) We use 47^ = 1.8922 =>

Z=5

Sol. 15 (B) Speed ofelectron isinversely proportional ton so for excitation from «= 1to« = 2 thevelocitywill reduce tohalf.

H-atom

Sol. 21 (B) Casc-I

Sol. 16 (D) After removal of one electron it becomes one

electron system from which energy required to remove the second electron will be 13.6(2)^ =54.4eV sototal energyrequired

H-atom

to remove both electrons will be 54.4 + 24.6 = 79.0 eV.

Sol. 17 (B) The H like atom is in the third excited state i.e. n = 4.

=> Energy corresponding to this wave length 12431x51

In the first case K.E. ofH-atom increases due to recoil whereas in the second case K.E. decreases due to recoil.

^

E^>E,

_

62000 This isthe

Case-Il

10.2 eV

Sol. 22 (A) Linear momentum => wv a — n

- E-^ for H and E^ - E^ for He^ and angular momentum

we get 2=2 for 4->2 radiation

mvran

Hence the atom is Helium ion.

=> product oflinearmomentum andangular momentum a

Sol. 18 (B) Let u be the speed of neutron beforecollision

Sol. 23 (C) For a Bohr atom velocity is proportional to Z/« hence option (C) is correct.

O

m

4m

At end of the deformation phase (when the kinetic energy of

Sol. 24 (D) We use the concept of reduced mass so we have

(neutron+ He"^) system is least)

Ke

m

—T' =

^

...(1)

or r

o—r'"")— m ^' 4m

Where

is velocity of centre ofmass. From conservation of

momentum mu

u

m + Am

5

The loss of kinetic energy

and

m ,

.

nh

...(2) 2,2

2a

n^h

By(l)and(2)

n^h x2

In^mKe^

r = (0.529A)x2«2 For first excited state w = 2

It K is the kinetic energy of neutron then the maximum loss in K.E. ofsystem is

=>

r = 8x0.529A

=>

r =4.232 A

•jif=12.75x4 =51eV or

Sol. 25 (B) We use, kinetic energy=^m

^=ll^=63.75eV

1

Sol. 19 (D) Forequilibrium ofoil drop 3.2x 10-Mx ]0 = (Nx 1.6 X10-19) X

2 •

4

Ke^ . ~ Ar

1200

6x10" N=10

m-co^r^

K=

Ke'

4x(0.529^x2)

\3.6eV

. 4

= 3.4eV

Atomic Physicst

Sol. 26 (C) We use i^=R 1 _

Forcompletely inelastic collision both come torest after collision and net energy of 4£ + E = 10.5 eV is lost. But electron in ground state ofH-atom canaccept only an energy of 10.2 eV.

1__1_

Li

Hencethe collision maybe inelastic butit canneverbeperfectly

-1

inelastic.

XR

n^= {n^-\)XR

Sol. 34 (B) Momentum of photon is given as

XR

„2 =

^

XR-\

—

hv

E

c

c

I3.6_iM|xl.6xlO''^

XR n

h

X

XR'\

^

Sol. 27 (C) Energyofelectron in hydrogen atom in inversely proportional to

stateis

hence option(C) is correct.

Sol. 28 (A) Toget six spectral lineselectron mustbeexcitedto n = 4 fromw= 1forwhichphoton energyneededis 12.75 eV of which wavelength is given as

:=>

3x10^

p = 6.45 X10"^'kgm/s.

Sol. 35 (D) As the electron returns to ground state after emittingsix differentwavelengths in emissionspectrum, there must be a difference of'2' between the ground and excited states. Also the ground state should be « = 2.

Hence n, = 4, «2~2. Sol. 36 (B) Longest wavelength corresponding to the transition from « = 2 to « = 1 for which we use

Sol.29 (C) Themaximum energy difference isfor«= 2 to« = 1 hence opiton (C) is correct. .1

X=

GrnMp Sol. 30 (B) We use r'

3RZ'

...(1)

= r

Sol.37 (D) For2"''lineofBalmerseires inhydrogen specturm nh

and

mvr'

-(2)

2jc nh

v =

27rwr

For Li2+

which is satisfied by only (D).

from (1) we have

GmMp

( nh V

r

Sol. 38 (B) We use force on electron

t2jcmrJ

dU

F= —r— -Kr= dr

n'^h' r =

Ait^Gm^Mp

rh

orbit is proportional to r? hence

of

mw=

~ 2tz

sate in Hydrogen is same as energy

nh r =

state in Li"^.

=> For 3

271

1^1

option (C) is correct.

Sol. 32 (D) Energy of

r

m\?- = KPand

Sol. 31 (C) Radius of

mv^

1 transition in H would it give same energy as the

3 X 3 -> 1 X 3 transition in Li"^.

Energy of

orbit is £.=

Sol. 33 (C) Total momentum of system is O hence maximum possible energylosscan be 10.5 eV outofwhich 10.2eV can be absorbed hence option (C) is correct.

In-ylKm Kr^

Knh "

. Kr

+

2

2

nh IK 271 \ w

jAtomic jPhysics

,417j

Sol. 39 (D) We use

Also for both w = 1 we have

_ E2-E^

_ £3 -£•]

h

h

£3 - £*2

'

•

Hence option (A) is correct.

V2-V, = --^=V3 As

Sol.47 (D) FirstlineofLyman serieswavelength is

^

V3 =V2(l-i)=(^J-ljv|

_ 12431 , 10.2 ^

Hence the only incorrect option is v =

first time of Paschenseries wavelength is Sol. 40 (A) For an electronwith kinetic energy£ we use X=

12431

0.66

h

yjlmE

h. = 0-66 _ 7

and for a photon we have 1

A

X2

10.2

108

_

Sol.48 (C) A£ocZ2 ' Thus it is maximum for Li^ and minimum for H.

^

/zc V2w£

Rat.o=---^

2mc^ Sol. 49 (A) As radius of

orbit is inverselyproportionalto

the mass ofelectron for mesoriic atom we use reduced mass of

Sol. 41 (B) Maximum energyofradiation incident onH-sample = ££maxOfsl^ctron +i3.6eV =51eV thisenergycorresponds

electron as

nip(lOlntg)

to the transition w = 4 -> « = lin Helium

P =

For electrons ofthe He to get excited to « = 4 As

.

12431

,

ntp+lOlm^

w^ =1840mg

=243 A

1840x207 1840 + 207

Sol. 42 (A) For lower value of n the energy and hence the frequency is higher. Hence (A)

p=186Wg Thus radius of first orbit of mesonic atom is

Sol. 43 (D) In the process of collision, the electron (because of very low mass in comparison to hydrogen atom) can loose upto 10 eV. 10 eV is sufficient to ionise the hydrogen atom or excite to higher excited state from first excited state.

r=

0.53x10"'® —

„

=2.85xl0-'^m

186

ADVANCEMCQs One or More Option Correct

Hence, the collision can be elastic,inelasticor perfectlyinelastic. Sol. 1 (All) The potential energy in ground state is-27.2 eV Sol. 44 (A) Potential energy= 2 Total energy =>

=2x-21.76x10-'9 --43.52 xiO-'^j.

Sol. 45 (C) The photon'swith energiesequal to that required for upward transition A

X, A

B and A ^ C would be

absorbed, hence only lines 1, 2 and 3 will be present in

andthat in first excited stateis- 6.8 eVsothe potential energy offirstexcited state is 20.4eV higherthan that of groundstate so option (A) is correct. The kinetic energy of electronin first excited state is 3.4 eV so total energy becomes 23.8 eV hence options(B) and (C) are correct. In groundstate totalenergywill be the kinetic energy which is 13.6 eV so option (D) is also correct.

.

.

obsorption spectrum.

Sol. 46 (A) Doublyionisedpositivelylithium ion is a hydrogen

Sol. 2 (B, C) In higher energy level Kinetic energy decreases hence speed or angular speed also decreases.

like atom so we use

r

nh

Sol. 3 (A, D) Time period of electron in an orbit is directly proportional to the cube ofthe quantum number hence options

2n

(A) and (D) are correct.

L = mvr= —

Atomic Physics;

Sol. 4 (A, C, D) Wavelenths emitted in the two cases are ^32

^1=

2T(K2e' V=

12431

rh

and U-,= 32 \.89eV

^21

2n'^K^Z^e^m E =

10.2 _ 27

X 32 a =

~ 5

^21

n'h'

Hence option (A) and (C) are correct.

Momentum ratio of photons is X/ b =

Sol. 9 (B, C) Any orbit the angular momentum ofelectron is an integral multiple of/i/2;rso for any transition it can change by a factor integral multiple of/i/2;rhence options(B) and (C)

5_

32

27 ^21

are correct.

Energy ratio of photons is hc/

5_ _\_

•32 c =

21 ~ a

hc/ '•21

Hence options (A), (C) and (D) are correct.

Sol.5 (All) Energyofelectroninn'^orbitis m.

E —

Momentum of electron in

orbit is

nh P = mV= — " 2nr

Sol. 10 (A, C, D) In radiation A incident beam passes from which some of the wavelengths will be absorbed by the hydrogengas hence option(A) is correct.Due to absorption of ultraviolet radiation electrons get excited and during reverse transition differentspectral series ofhydrogen atom are emitted in radiation B hence option (C) and (D) are correct. Sol. 11 (B, C) Whenever a photon is emitted in Bahner series, it will be corresponding to the transition of electron to « = 2 state from where it will further transit to ground state and emit a line corresponding to the transition of n = 2 to w= 1 in Lyman series ofwhich the wavelength can be given asX,= 12431/10.2eV « 122nm hence options (B) and (C) are correct.

where r is given as n'^h'

2nKZe^m^ from above relations all options (A), (B), (C) and (D) are correct.

Sol.6 (A, D) WeuseAE'j ^ AEj + AE^ hv^ = h\^ +hv2 ^

V3 = V,+V2 C

^

^

Sol. 12 (A, C) It is already been studied in spectral series of hydrogen atom that some of the lines of Balmer,series are also lying outsidethe visibleregion hence options (A) and (C) are correct.

Sol. 13 (A) AsalreadystudiedinbasictheorythatbycoIIision of a physical particle any fraction of maximum possibleenergy loss can be absorbed by the atom for excitation so here only option (A) is correct. For inelastic collision(by excitation) is possibleonlywhen the maximumpossibleenergyloss is more than the energyrequired for excitationofelectronto first excited state to n = 2 level.

^1 ^2 Hence option (A) and (D) are conect.

Sol. 14 (A, B, C) As we know

r„cc«2

Sol.7 (A,D) EnergyofHe^ ionis = 4 times thatofhydrogen atom for a given orbit so all transitions where square of ^2and

«j is amultiple of4for He"*" ion, itwill match with some transition

A., oc

and

/oc —

ofhydrogen atom hence option (A) and (D) are correct. Sol. 8 (A, C) From Bohar Model we have -2.2

R =

r„« .2 — =}t\

A„n 4. . =n A1

J n

-7 =

n'h'

n

2Ti^KZe^m

Thus options (A), (B) and (C) are correct.

J\

1^

— n

[Atomic: Ph^ics

419:

Sol. 15 (B) Among above options only option (B) is correct because for due to slightly heavier mass of nucleus the bindingenergyis more if we consider the mass of nucleus in

calculation of total energy of electron bytaking the reduced mass oforbiting electron.

he

he and

Sol. 16 (B) Due toquantum nature, photon is absorbed only if its energy is eqaual to any ofthe transition energies other ^=>

1 —

wise it will not beabsorbed hence option (B) iscorrect.

he ^ A,2

Xq

^3

1+ 1 1 +

=

^0

^2

^3

henceoptions (A)and (B)are correct.

Sol. 17 (B, C,D) With the standard data for kinetic energy, potential energyand angular momentum we can see that opitions Sol. 22 (D) From the given information we use Rydberg's (B), (C) and (D) are correct.

formula for hydrogen atom andthehydrogen likeatom 'JT as

Sol. 18 (B, D) We use the given potential energy for dU F=

,1 1 1 = 13.6Z2| —

13.6

calculation of force electron as

Fromthegiven options onlyoption (D)is correct.

3 Ke^

dr ~ 2

Sol. 23 (A, C, D) Binding energy of hydrogen like atom is given as 2

4

Z^ £=-13.6-T-=-122.4eV

nh

Using Bohr's II Postulatewvr= — we get

n

2%

For m

( nh ^

3 Ke^

\2nmr

2

roc

Excitation energy ofthishydrogen likeatom for excitation from

27c

« = 1 to « = 2 is given as

J_

A£i2= 10.2

rP' and

« = 1 we get Z= 3 so option (A) is correct.

10.2 X9 = 91.8 eV

Thus option (C) is correct.

rocw

hence option (B) and (D) are correct as energy is inverssely When an electron of 125eV collides withthisatom, 122.4 eV energy is absorbed by the atom for ionization of electron and

proportional toP. Sol. 19 (A, B, C) For a hydrogen atom in second orbit the

total energy of electron is- 3.4eV hence options (A), (B) and (C) are correct.

Sol. 20 (C, D) As we know that radius of velocity ofelectron in

orbit

orbit

orbit r^ ocl/Z,

cc Z andenergyofelectron in

cc Z^ sooptions (C) and (D) are correct.

Sol. 21 (A, B) Using the Rydberg's Formula for the given transition

remaining energy 2.6eV will getconverged tokinetic energy of the ejected electron hence option (D) is correct. Sol. 24 (A) Hydrogen atoms at room temperature are all in ground state so these can absorb radiations onlyof Lyman series aselectrons can excite from «= 1tohigher energy levels so in absorption spectrum onlyLyman Seriesis observed.

Sol. 25 (B) On changing the reference potential energy in ground state to be zero, this will not affect the variation of energy with the energy levels as well as the difference in

energies of any energy levels so total energy increases with 1

1

higher energy levels hence statement given in option (B) is incorrect.

X.n

solving we get n = 4. Thus electron can cometo groundstate byemitting two orthreephotons such thatthe sum ofenergies Sol. 26 (A, C) As discussed in the basic theory theaffect of of each transition remain same so we can use

he

he ^ he 7^2

massof nucleus is accountedby consideringthe reducedmass of electron in the orbitalmotion. So considering the effect of reduced mass in the expressions ofradius and energy we can

state that option (A) and (C) are correct.

- Mqmic^Ph^cs

[•^.20

Sol.27 (A,C) For inelasticcollisions the maximum possible Sodifferenceof energyshould be 12.0eV (approx) energyloss in a collision should be more than the minimum Hence Wj=l a^td energy required for excitation of any of the colliding atoms which in this case is 10,2 eV for hydrogen atom. As the

(-13.6)eV

(-1.51)eV

hydrogen atomandneutron areofsame mass, maximum kinetic Forlongest wavelength, energy difference should beminimum. energy loss can be the half of initial kinetic energy of the Soinvisible portion ofhydrogen atom, minimum energy emitted colliding particle-hence in abovecase initial kinetic energy is in transition IV. must be more than or equal to 20.4 eV for inelastic collision hence options (A) and {C) are correct Sol. 28 (A,B,D) ForBalmersCTies,

Hi=2, »2=3A, (lower) (higher)

In transition (VI),PhotonofBalmer series is absorbed.

Sol. 29 (A, D) The hydrogen atom is in « = 5 state.

•Thus maximum number ofpossible photons = ^€2 =10. To emit photon in ultra violet region, it must jump to « = 1, becauseonlyLyman series lies in UVregion. Once itjumps to n = 1 photon, it reaches toits groundstateandno morephotons canbe emitted. Soonlyonephotonin u. v. range can beemitted. If H-atom emits a photon and then anotherphotonofBalmer series, option D will be correct.

Intransidonll

E2=-3A eV,£^=-0.85 eV M=Z55eV

SoL30 (B, D) Anyfraction ofincidentelectronenergycan be absorbed by the atom hence minimum energy required for excitation is 10.2 eV hence option (B) and (D) are correct.

AE=-^ ATT

A,

=>

=> ^

^ —

AE

l=486nin.

Wavelength ofradiation = 103 nm = 1030 A 12400 AE=

1030^

AE = 12.0 eV

fPhoto Electric Effect & Matter Waves

421}

ANSWER & SOLUTIONS CONCEPTUAL MCQS Single Option Correct 12431

Threshold wavelength X,= ^ 1 (D) 4 (D) 7 (D) 10 (A) 13 (B) 16 (C) 19 (A) 22 (D) 25 (C) 28 (B) 31 (C) 34 (A) 37 (B) 40 (B) 43 (D)

2 5 8

11 14 17 20 23

26

29 32 35 38

41

(A) (B)

6

(A) (D)

12

3

9

(A) (D) (C) (C) (A) (C) (D) (D) (D) (A)

15 18 21 24

27 30 33 36 39

42

(D) (A) (B) (A) (C) (A) (D)

2.30

he

(ii) We use

;tx3xl0^

nr =6 + 3 X 10-^9

(C) (C) (B)

(C) (D) (B) (D)

=5404.78 A

3310x10"'® /jx3xlO^

and

...(1)

^

=(l)+0.97x l0-'9

5000x10

...(2)

solving (I) and (2) we get /i X 3 X 10'^

I

1

3310

5000

= 2.03xl0"i9

NUMERICAL MCQS Single Option Correct 1 4 7 10 13 16 19 22 25 28

(B) (B) (D) (C) (D) (A) (C) (B) (B) (B)

31 34 37 40 43 46 49 52 55

(C) (A) (B) (B) (D) (B) (C) (D) (B)

58

(B)

h = 2

5 8

11 14 17

20 23 26 29

32 35 38

41 44 47 50 53 56

59

(A) (B) (B) (C) (B) (D) (A) (A) (C) (A). (A) (C) (A) (B) (A) (D) (A) (B) . (A) (C)

3 6

9 12 15

18 21 24 27

30 33 36 39 42 45 48 51

54 57

(B) (B) (D) (C) (C) (A) (D) (B) (B) (A) (D) (A) (C) (A) (B) (A) (B) (A) (C)

2.03x10"'^ 3.06x10'"'

= 6.634 X 10-34 j-s

(m)KE^=lOAey Incident photon energyis £•=({»+KE = 1.7+10.4 = 12.1 eV

Incident wavelength X=

12431

~ 1027.35A

Energygap 12.1 eV is corresponding to « = 3 to « = 1energy level transition in hydrogen atom.

(iv) ©-

ADVANCE MCQs One or.More Option Correct /

For circular motion ofelectrons around the ion we use 1 4 7 10 13 16 19 22 25

(B,D) (A. C) (All) (B, C) (B, D) (A, B, C) (A) (A, B, C) (A, D)

28

(A, D)

2

5 8

11 14 17 20 23

26

(B) (A, C) (B) (A, B, C)

3

(B)

6

(B. C) (A, D) . (B, C) (A. C) (All) (C) (C) (A, B, C)

9 12

(B) (A, C) (C) (A, B) (B. D)

15 18 21 24

27

1

mv

4n' =0

d .2

—

^ a

A

831

^

Sti Eq dhc

^-~2 e +87ceo(()cf (v) (a) To ionize H-atomkinetic energyofphotoelectronsmust

Solutions ofPRACTICE EXERCISE 2,1

beatleast 13.6 eV 12431

(i) Incident photon energy £"=

= 1.7+13.6

12431

4800

= 2.59 eV

=-e-(()=2.59-2.30 = 0.29eV

• 12431 X= 15.3

= 812.48 A

Photo Electric Effect & Matter Waves}

•422

(b) To excite H-atom from w= 1 to « = 2 kinetic energy of photoelectrons must be atleast 10.2 eV

If Vis the retarding potential we use e(K-0.6) = 0.9eV =>

V=l.5V

12431

= 1.7 + 10.2

(b) Xshould be such that initial maximum kinetic energy of 12431

photoelectrons should be

^ ^ = e(l-0.6) = 0.4eV

=1044.62 A

12431

(c) To emit visible light H-atom must be excited atleast from « = 1 to « = 3 so that in Balmer series it can emit visible light thus kinetic energy ofphotoelectrons must be atleast 12.09 eV (ill)

12431

= 1.7+12.09

Workfunction

6.63xl0"^'*x6xl0''^ 1.6x10"'^

/7V^^^ =

12431

901.45 A

/7V,^ = 2.486 eV /7V = 2.486 + 3 = 5.486 eV

We use

(iv) Initially we use

5.486x1.6x10

/7V = ([)+13.6eV

...(1)

V =

= 1.33x1015 Hz

6.63x10

later we use

h

-19

,-34

12431 _5vi =(j,+ i_J!£ieV=(l)+ 10.2eV

6J ^

1218

-(2)

^

^max=2.5eV

(iv) We use

12431

using (1)-(2), gives

= (l)+2.5eV

1980 hv -=3.4

(|)=3.78eV (j)

Threshold frequency

h

3.4x6x1.6x10"^^ V =

3.78x1.6x10-19 ,-34

6.63x10

V = 9.122x101'! Hz

n-34

6.63x10"

= 4.923 X 1015Hz 12431

from (1) we get (l)=/iv-13.6eV

(v) We use

=>

(1)=((6.63 X10-34 X4 923 x iQiS) 1.6x lO"'^-13.6 eV

=>

^=6.80eV

WhenXchangedto 4272 A, we use

6402

({)= 1.402eV

12431 — = 1.402

Solutions ofPRACTICE EXERCISE 2.2

(i) Incident photon energy E=

^m. = l-508eV

12431 = 2.198 eV

5896

Maximum kinetic energy of photoelectrons KE Work function

= (t)+0.54eV

KE„ K„ =

= 0.36 eV

= 1.508 eV

(1)=£-KE T max

(|)=2.108-0.36= 1.748 eV

Threshold frequency

(j)

1.748x1.6x10 -19

v=- =

6.63x10

he

(>i) We use

— =(l)+1.85eV

and

— =(l)+0.82eV

A,i

-34

he

v = 4.218xl0i'!Hz 12431

(it) (a) Incident photon Energy,£ =

2300

= 5.4 eV

Using, (1)-(2) gives

Maximum kinetic energy ofphotoelectrons is = 5.4-4.5 = 0.9 eV max

...(1)

=>

1

1

X,

X,

he

= 1.03eV

...(2)

^Photo Electric Effect & Matter Waves

h =

423!

1.03x1.6x10"^^

3000x4300x10"'°

3x10®^

1300

(ill) Electron momentum initial

h

6.63x10"^^

Pi =

/i = 5.451 X 10-34J-s

x-lO

10"

;^, = 6.63x 10-24 j_s (vii) No of photons incident per second on plate are PX W=

he

5x4x10

-7

6.63x10"^"

:t2 ~ 0.5x10-'°

6.63xlO"^'*x3xlO^

;7i= 13.26 X10-24 j-s

Ne

Photo current

h

Find

ICinetic energy ofelectron

/=

10^

5x4x10"^ xl.6xl0"'^ /=

E =

2m.

6.63x10"^'^ x3xl0^xl0®

Energy added to electron is

/=1.6xl0-^A

AE^

(viii) (a) Work function of metal is (t)=

2m„

he

6210A

he We use

[(13.26r-(6.63r]xl0,^8

hE =

2x9.109x10"^'xl.6xl0"'^

he

4140A

Pi-Pi

+ leV

6210A

=452.4 eV

Ix 1.6x10"'^ x4140x 6210x10"'° 5 3x10^x2070

=>

h=

=>

/? = 6.624x 10-34 j.s

(iv) Force on space vehicle is F

100

F'=- = -r-:j=3.33xio-7N 3x10®

(b) We use

—w 2

= 1 eV ^

3.33x10

Acceleration is

a

m

Ixl.6xl0"'^x2

-7

=

50

a = 6.66 X 10-^m/s^

v =

9.109x10"^' =>

(v) As 30%lightis incident andreflected bymirror, force due

v = 5.927xl05m/s

to reflection of light on mirror is Solutions ofPRA CTICE EXERCISE 2,3 F=2x

0.3F

0.6F

e

e

(1) In first Bohr orbit electron speed is

To support the weight ofmirror, we use

nh v =

27cwr

0.6F

deBrogli wavelength is h X= —

mge _ 20x10 ^xl0x3xl0 P =

mv

0.6 ~

0.6

X-Itv

P= lO^watt

(ii) At temperature T,deBrogli wavelength ofany moleculeis given as h

mv —mV H

X=

...(1)

By conservation of energy

4x400 ^ "^hTh

¥ e

yl3mkT

"He

(vi) By conservation of momentum

8

2x300 ~V3

]-m^-^lsE=\mv^+hf

...(2)

Photo Electric Effect ^Matter Waves^

424

Solutionsof CONCEPTUAL MCQSSingle Option Correct A£ = ^/¥

AE=hf- —w(v+v')

mc

hf 2v iSE^hf--^ —(v=V)

Sol. 1 (D) As studiedphtoelectric current depends uponthe intensityof incident light,work flinciton of the emitterand the condition of its emission is that wavelength of incident light has to be less than the threshold wavelength ofthe metal hence option (D) is correct.

Sol. 2 (A) As work ftinction ofP is more than that ofy4 the threshold frequency ofB is morehenceoption (A) is correct.

M = /2/ 1-2:

Sol.3 (D) Photoelectric effect startswhen incident wavelength When atom as at rest, we can use

is less than that of the threshold wavelength of the metal. For

AE = hfQ

the given thresholdwavelength only ultraviolet wavelengths are less than this hence option (D) is correct.

¥o = ¥

I— Sol. 4 (D) Energy of a photon alwaysremain same as in any optical phenomenon frequency of light does not change.

/=/oM+Sol. 5 (B) de-Broglie waves have wavelength inversely

proportional tothe momentum of theparticlehenceoption (B)

(vii) Initial momentum of each particle is h •:

is NOT true for de-Broglie waves.

-

h "

Pj = —/ and P2 = —J A,2

vi =

h , ^ h and V2 = ——J mXi /wA,2

Velocityof center of mass is gives as

/M1+W2

2wl^X]

X2

Momentum of particles in frame ofcenter ofmass is h

= Kvi-Vcm) = 2

and

he = ^"(^2 _

It

1

^2

2

_L---L^1

^1^2

deBrogli wavelength of particles in frame of their centre of mass is

h

2^1X2

^ ^Vxirtl and

h

2X,X lA,2

•2c

•^X^ -1- X

X. =

Sol. 7 (D) Photons having same momentum will have same de-Broglie wavelengths but photonshaving samewavelengths will have same magnitude ofmomentum but their directions may be different hence option (D) is corerct. Sol. 8 (A) For same speed de-Broglie wavelength is inversely proportional to the mass of the particle hence option (A) is

)

+ A. 2c f

correct.

correct.

>1

=2

-

Sol. 6 (A) For same speed de-Broglie wavelength is inversely proportional to the mass of the particle hence option (A) is

Sol. 9 (B) As already discussed in theory that the electrons when ejected in photoelectric effect may loose their energy during emission process in collisions so the emitted electrons have their kinetic energy distributed from zero to {hv hence option (B) is correct.

Sol. 10 (A) Photoelectric emission starts only when the frequency ofincident light is more than the threshold frequency hence option (A) is correct. Sol. 11 (D) Photoelectric effect explains the quantum nature

of light particles called photons hence option (D) is correct. Sol. 12 (A) Once photoelectric emission starts at frequency above threshold frequency the current is directly proportional to the photon flux in incident light which is proportional to the intensity of light hence option (A) is correct.

fPhbto EJectric_Effect & Matter Waves

J

So!. 13 (B) Duetocollisionselectronsmaylooseenergyduring the ejection process in photoelectric effect so the ejected electrons have their energy distributed between zero to the maximum possible kinetic energy of ejected electrons hence option (B) is correct.

Sol. 14 (A) Both momentum and energy of a photon are inversely proportional to the wavelength of radiation hence option (A) is correct.

425j

Sol. 24 (C) Atsamepotentialdifference,kineticenergygained by both electronand proton will be same but momentum will be

more that in.photon hence its de-Broglie wavelength will be less hence option (C) is correct.

Sol. 25 (C) Kinetic energy of photoelectrons will increase if energyofincidentlight is increased henceoption(C) is correct.

Sol. 26 (A) Based onexperimental analysis studied in theory option (A) is correct.

Sol. 15 (C) Fromtherelativisticrelation£'^=/j^c^+WQVonly for a photon rest mass m^ iszero thus theabove relation isonly

Sol. 27 (C) As frequency of incident light is kept constant

valid for a photon hence option (C) is correct.

maximum kinetic energy of ejectedphotoelectrons will remain constantand intensityis increasedwhichmeans photon fluxis

Sol. 16 (C) No. of photons emitted per second by a bulb of power P„^,.

increased due to which more number of electrons are emitted

N=

n

=

EIl he

Eh: he

As

and n, =

^

—

he

n_ > n.

hence option (C) is correct.

Sol. 28 (B) Due to increasein both frequency and photon flux by a factor of two, light intensity increasesby four.times. Due to doubling thefrequency themaximum kineticenergyincreases bya factor morethan twoand bydoubling the photon flux the photo current increases by a factor of two hence option (B) is correct.

Sol. 17 (D) As already studied that in the processof ejection photoelectrons loose thier kinetic energy in one or more Sol. 29 (C) As kinetic energy of photoelectrons is related to wavelengthof the incident light by a relation given as collisions with other electrons hence option (D) is correct. Sol. 18 (A) Among all the above options the photoelectric hence in this case option (C) is correct. current is directly proportional to the inciden light intensity Sol. 30 (B) At stopping potentialthe emitted photoelectrons hence option (A) is correct. are not able to reach the collecter electrode and repelled back Sol. 19 (A) As X-rays haveenergy (and frequenty) more than and absorbed by the emitter metal again hence option (B) is that ofultraviolet light option(A) is correct.

Sol. 20 (C) As we have studied that the photoelectric effect equation obtained in experiment is

correct.

Sol. 31 (C) The relation of frequency of light and stopping potential.is given as

eVQ = hv-^ hv.

hence option (C) is correct.

Sol. 21 (D) As during ejection process an electron may loose energy in collision process the ejected electron can have kinetic energy less than the maximum possible energy {h\ - ([>) hence option (D) is correct.

Sol. 22 (D) As maximum photocurrentin experimentdepends upon light intensity hence option (D) is NOT true. Sol.23 (C) Asenergyofredlightislessthanthatofblueand

yellow light, no photoelectric emission will occur hence option (C) is correct.

hence option (C) is correct.

' •

Sol. 32 (D) As the distance of light source is doubled from

the photocellthe light intensitywhich is inverselypropcational to the distance will become one fourth hence option (D) is correct.

Sol. 33 (C) By changing the distance between source and emitter metal the photon flux changes and stopping potential depends upon frequence hence there will be no change in stopping potential in this case hence option (C) is correct. Sol. 34 (A) As frequency ofradiation is kept fixed there will be no change in maximum kinetic energy of the emitted photoelectrons hence option (A) is correct.

Photo Electric Effect & Matter-Wavesi

U26

Sol. 35 (D) As photon flux and intensity of incident light is inversely proportional to the square of the distance from the light sourcefrom metal and saturationphotocurrent is directly proportional to the photon flux, option (D) is correct.

Sol. 2 (A) We have hv = eV

eV_ v

h

1.6xl0~*^xl0xl0^

Sol. 36 (D) •The relation of frequency of light and stopping potential is given as

v =

6.6x10

-34

v = 2.4x IQiSHz

eVQ = h\-^ hence option (D) is correct.

=

Sol. 3 (B) no. of photoelectron emitted per second are ,-31.5xlO-^ffx(IO-")

Sol. 37 (B) Due to downward direction of electric field the ejected electrons will experience upward acceleration due to this field and because of this the total kinetic energy ofejected

N=

0.48

\240(nm)(eV) xe(V / 5) 400{nm)

xlO-6

electrons will increase.

=> Photo current = ne = 0.48 ^lA

Spl. 38 (D) In photoelectric effect the incident photon is absorbed by the free electrons ofthe metal in conduction band and these are emitted ifincident light energy is more than the work function hence option (D) is correct.

Sol. 4 (B) Change in momentum due to photon =

F=rate ofchange ofmomentum

F=n-^ =ma

Sol. 39 (B) The maximum kinetic energies of the ejected electrons from the two photons will be 1 — 0.5 = 0.5eV and 2.5-0.5 = 2.OeV hence option (B) is correct.

Sol. 40 (B) As the aperture ofthe lens is reduced to one fourth so photon flux coming out from lens will also become one fourth so current will also reduced to one foruth, hence option (B) is correct.

Sol. 41 (A) The relation in stopping potential and wavelength for a photoelectric effect experiment is given as eV^ = hclX- -(j) using above relation we can see that option (A) is correct.

h

nh a

=

Xm

12431

Sol. 5 (B) We use

ij)=

eV

6000

=>

(t>=2.07eV

=> =>

(t)=2.07x 1.6x10-19 (l)=3.315xlO-i^J

Sol. 6 (B) Number of photons per second are PX N=

Sol. 42 (D) In photoelectric effect electrons are ejected only when the photon energy is more than work function ofmetal and if photon energy is less than work function then electron will not be ejected no matter how much is the intensity. This fact supports the quantum nature of light hence option (A) is correct.

Sol. 43 (D) As studied in the theory of Eienstien's photoelectric effect experiemnt, option (D) is correct.

he -9

100x0.05x555x10 N=

-34

6.63x10""^ x3xl0'

=>

A^=1.39xl0-'9

Sol. 7 (D) The electron ejected with maximum speed are stopped by electric field E = 4N/C after travelling a distance = 1m. Thus we have

Solutions ofNUMERICAL MCQS Single Options Correct

The energy ofincident photon =

Sol. 1 (B) We use he

\240{nm)eV -1.9eV 400(«w)

= 6.21 eV

From equation of photo electric effect we use

e7^ = 1.2eV F, = 1.2V Thus the cesium ball can be charged to a maximum potential of UV

=>

(k = 6.21-4=2.21eV

jPhotq^Electric Effect &^Matter Waves

427!

Sol. 8 (B) The radiation pressure depend on the intensity of Thenumber ofphotons incidents per second on theplate light used and noton itswavelength and frequency. = number ofelectrons emitted per second Also, the radiation pressuredepends on thenature ofthe surfece on whichlight is falling. Hence(B).

Sql. 9 (D) Theptoential ofanode withrespect to thecathode must be -5V to repell fastest photoelectrons.

Sol. 10 (C) The maximum K.E. ofejected photoelectron is

= 1^ hv

The momentum of each photon and electron respectively are

and ~j2m{hv-^) where m is the mass of the electron. Hence the net force exerted on the metal plate is

Ifthefrequency ofphoton isdoubled, maximum kinetic energy of photon electron becomes

Sol. 15 (C) Power percent area ofwave is

P=^,ElC Numberof photonsfalling per unit area are

PA,

(KE)r {KE)r nu *

Ef)-0 eIcX -^-0

>2

^

Photo current oc

Eq ^^0

he

=

Sol. 16 (A) By conservation ofmomentum we use

intensity of beam ^ /rv

h

If intensityand frequency both are doubled, the photocurrent remains same.

Sol. 11 (C) Number of photons per second enering human eye are

-

r,= 4 ^ mX Sol.17 (D) Number ofphotons strikingper secondare

PX

=>

N=

=>

10"'®x660xl0"^ TT T 6.63x10"^^ x3xlO^

Area Ahereisthearea perpendicular tothedirection ofintensity 10^

or direction ofenergy flow.

Consider a ringofradius x andwidth dxonthe disc. Intensity I

#=3.31xl(r»

on the ring due to source is /=

Sol. 12 (C) At temperature T, average kinetic energy due to

r Aizr

thermal motion is given as 1

3

2

2

-m\P-= -kT

Momentum is given as

P = mv= ^3mkT de-Broglei wavelength is given as

T^sinG

h

X=

y/3mkT

FcosG

Sol. 13 (D) Energyofaphoton==£';f=0.25£= 4^ 4A,

dA^= dA cos 0 dA =2wcdx

AX~X

So percentage change in wavelength = —— x 100% = 300%.

Sol. 14 (B) The net force on the plate is due to incidence of photons + due to emission of electrons.

Gives, dA^ = i^-Kxdx)QOS^ A photon willexertforce F asshown, onlytheFcos0 component willremain andPsin 0 will cancel outasweintegrate onthering Nh F=

Photo Electric Effect & Matter Waves]

428

[as N photons strike per second and this leads to less of

Nh

3^

=4eV 4eVo he

momentum per second.] as only Fcos0 component of force remains

IdA, \

h \

~

he

Sol. 21 (D) We use •j/kv^= (27iriccos0) -hC\

k=i

h

\

,1 .

,

he

,

4he

,

l"" =(3X74)-^=ir-'^

and

4

v'> J-V

[As r= V4^^+?] Solving we get

F-

Sol. 22 (B) Potential ofthe sphere at any time 20c

Ofi + Qf

V{t)= AneoR y r. =V+

Sol. 18 (A) We have

>.,=4100 A >2=4960 A ^3= 6200 A = 3eV

41.0

r[e x t

made halfthe work function ofthe metal. Hence photo-emission shall stop altogether.

12431 2

PX

An&QRhc

Sol. 23 (A) Under the given condition energy of photon is

12431

^11 =

Here we have Qt =

-nXPet

= 2.5eV

4960

KE„ 12431

Sol. 24 (B) Stopping potential =

= 2eV EnergyofK^lineisA^2i = 25.31-3.56= 21.75KeV

=> EnergyofKp line is Afijj =25.31 - 0.530 =24.78 KeV (ii) For X = 1A we use

=> EnergyofL^ line is 12431

= 12.431 kV

Photon energy

he

6.63x10"^" x3xl0^

SE-y,

Thus

h Vi- =

io-'« = I.99x 10-'5J

= 3.56- 0.530= 3.03 KeV

V,^oc

AE'31 h

AE.

=5.249 X 10'8 Hz

= 5.98xl0i8H2

^ =7.312 xlO'^Hz

h

X-rays

i«6

Solutionsof CONCEPTUAL MCQSSingle Option Correct

Sol. 1 (B) In Coolidge tube high speed electrons areallowed

Sol. 12 (B) The energy of X-ray photons is directly

proportional tothe radiation frequency with which the photons penetrate a physical substance hence option (B) is correct.

to incident on the targetanodemetalfor production of X-rays

Sol. 13 (B) A good target for X-ray production must be able towithstand against highenergy electron which looses energy Sol. 2 (C) In X-ray production the electric work done on toheat upthetarget soit should bea heavy element with high

hence option (B) is correct.

electrons transformed to their kinetic energy and then this

energy isradiated inform ofX-rays during interaction ofthese high speed electrons with thefield ofnucleus oftarget anode atoms. Hence option(C) is most appropriatehere.

melting point henceoption(B) is correct.

Sol. 14 (B) Continuous X-ray spectrum is produced due to loss of electron energies when these interact in the positive field ofthe targetnuclie at different approach distances hence option (B) is correct.

Sol. 3 (D) If the incident energy allowed to emitX series of X-rays then line frequency will remain same but the minimum Sol. 15 (C) X-rays are electrically neutral so these are not wavelength ofthecontinuous spectrum will get doubled hence deflected byelectricfield hence option(C) is correct. option (D) is correct.

Sol. 4 (B) The minimum wavelength ofX-rays emitted from an X-ray tube is given as 12431

X = —= eV

V

A

hence option (B) is correct.

Sol. 16 (D) As the incident photon supplies some energy to the electron its energy decreases due to which its frequency decreases andwavelength increases, henceoption (D)iscorrect.

Sol.17 (A) X-rays areveryhigh energy electromagnetic rays which can damage the targetandreceiver alsoovera periodof time due to regular reflection and absorption thats whythese are not used in RADAR hence option (A) is correct.

Sol. 5 (C) Hard X-rays have higher energy and lower wavelength so the potential difference across the tubeneed to be increased hence option (C) is correct.

Sol. 6 (A) Molybdenum hashigh melting point duetohigher valueof itsmassnumbCT so it canwithstand againsthigh ehergy electrons so this is preferredhence option (A) is correct. Sol.7 (B) Asminimum wavelength ofthe continuous spectrum isinversely proportional totheapplied voltage inthetubehence option (B) is correct.

Sol. 18 (D) X-rays areelectromagnetic radiation which travel at speed of light henceoption (D) is correct. Sol.19 (C) X-rayphotons haveenergymorethan ultraviolet radiation but less than gamma radiation hence option (C) is correct.

Sol. 20 (C) X-rays are highly energetic electromagnetic radiation which will pass through the muscular organs and tissues ofthe body and no shadow is obtained hence option (C) is correct.

^ 8 (C) MosleVs X., I j law I is • given • C a (Z-a) /v \ hence as Vv= Sol.

Sol. 21 (A)^ The energy in the of hydrogen . ^ differences , • ,shells.mv

option (C) IS correct.

spectrum region hence option (A) is correct.

Sol.9 (B) From thespectrum ofcontinuous X-rays theemitted

Sol. 22 (D) As X-rays are electrically neutral and does not produce anymagnetic phenomenon, theseare not deflected by

^

radiation has wavelength above a minimum wavelength and extend upto infinitevalue hence option(B) is correct. Sol. 10 (A) line wavelength depends upon the atomic charge which is same in all isotopeshence option (A) is correct.

are so less that the emitted photon energies do not fall in X-ray

magnetic field hence option (D) is correct.

Sol.23 (C) Whitetermisusedto denote the mixtureofseveral wavelengths in the continuesspectrumof X-rays hence option (C) is most appropriate.

Sol. 11 (B) Sodium has less work function than copper so the kinetic energy of ejected electrons in case of sodium will be Sol. 24 (A) Wavelength ofX-rays is inverselydependentupon morehencethestopping potential willbemoreforsodium hence the applied voltage in the tubehence among the options given (A) is correct. option (B) is correct.

;X-rays

Sol. 25 (C) wavelength of

4371

line has energy less than line hence line is morethan henceoption(C) is correct.

Sol. 26 (C) For higher energy levels in an atom screennig constant cannot be precisely determined so the above relation

hC ° =>

—

30000e = 30000 e

mainly gives good approximated results for lower values ofn[ By using the formula of energy of electrons according to Bohr's and «2 henceoption (C) is mostappropriate here.

model and considering shielding effect 30000 = (13.6)

Sol. 27 (A) As X-rays wavelength is ofthe order ofinteratomic spacing, good level of diffraction pattern can be obtained by

the lattice atoms forthe studyofstructuralanalysis hence option

1.100,/^ =^=54 17 ^

(A) is correct.

Z=55

Sol. 28 (A) As applied voltage across the tube increases, the cut offwavelength decreases at hus the difference between and increases hence option (A) is correct.

Sol. 29 (D) From the figure it is clear the cut off wavelength of A is less than that of5 so applied voltage for A is more than that of5 and the characteristic line wavelength ofB is less than that ofA that means the energy level difference ofelement B is more than that ofX for « = 1 and n = 2 thus atomic charge ofB is more than that of^ hence option (D) is correct.

Sol. 6 (C) Energy gap between X and Z, is 12431 AE-=59.19keV

Thus option (C) is most appropriate. Sol. 7 (C) We use

=2&(6.25 X101^+3.125 xIOi«)x 1.6 xiQ-i? P = 300 watt.

Sol. 8 (C) We use Solutions ofNUMERICAL MCQS Single Options Correct

E, =Ey -E^.

Sol. 1 (B) To knock out the innermost electron the voltage must be more than 40kV then only a vacancy will be created in Kshell and Kseries X-rays will be emitted.

1 1 =12431 f y-2 „ 12420 E, =—r—=6215.5eV 2

Sol. 2 (Q The minimum wavelength is

E, =6.21 KeV.

12431

X =

40000

= 031A

Thus wavelengths lesser than 0.31 Amust not be present.

Sol. 9 (B) We use Z2-I

Sol. 3 (D) Minimum wavelength is 12431 X =

80000

= 0.155A

Z, =(Z,-l)jTJ- +I

As

so characteristic X-rays will be there Z,= 50xJ—+ 1 4X

Sol. 4 (C) We use

Z2 = 26 vZ,-i = Xx

Sol. 10 (C) We use

56 , as Vcharges to V/2wavelength X will get doubled.

Sol. 5 (B) WeuseX- =

he

hC

'I™ 20000e'^"^

Given that 4 (X^-X^„) = (X„-X'^) 3Xa = 4X„i-X' nun

nun

Sol. 11 (C) We use

lOOOOe -=i?(Z-l)2

1-i 4

X-raySf

1438

_

=>

+ 1

Z=

200 fJ

=> Z=

=>

3x0.76x10"^® X10967800

Sol.17 (Q KnockoutanelectronfromAlevel,energysupplied mustbeenough to driveit to anunfilled higherlevel ofnegligible

Sol. 12 (C) Energy of photon is given by mc^ now the maximum energy of photon is equal to the maximum energyof electron = eV

hence, mc^= eV m

=

energy (n -> oo)

=> Energyrequired=B.E. ofA level = 69.5keV =?> Acceleration pot^tial = 69.5 kV Sol. 18 (B) Power ofelectron beam is

1^ =20x103x10x10-3

1.6xl0"'^xl8xl0^ 2 = 3.2 X10-32 kg (3x10^)

=>

I

Sol. 19 (D) On increasing the applied voltage across the tube the electrons which knock out the orbiting electrons will left over with extra energy and these may knock out more orbitin electrons in the target which leads to emission of more characteristic X-rays hence option (D) is correct.

ne

1 V=

6.25x10^^x1.6x10"'^

1^= 10^ volts he

12431 »

X.= = 0.12A. mm -j7 qY = —f^A V Sol. 14 (B) We use for

Sol. 20 (D) Using ^ =^(z-l)'

line

— =R{Z-lf

Fora particle;

1-i i-i

^2

9

r-=fi(Z-l)2

^ --X?.i

4 8

1

Wo

Wi

«j = 2, «2 ~ 1

FormetaU;

=>

1

1

...(1)

4

for ifp line we use

(2)

F=200fV

^.-,a.= ^x200=lw.

Sol. 13 (B) We use

ill

Z=27

+ 1

Z=41.

eV

78

3 \33~ 3

z, =26

...P)

Formetal B;

675R=72 (Z2 - 1

Therefore, 4 elements lie between A and B.

27,

27

Sol. 21 (B) Using Moseiys law for both cobalt and impurity = 0.27 A

47 =K(z-i)

Sol. 15 (B) Maximum photon energy is

t -^(2-1)

12431

E=^rT:r =37.67 keV. 0.33

t

aIi:

Sol. 16 (A) We use

Z.-l Z. -1

Z =40.

Sol. 22 (D) Minimum wavelength emitted is 12431

66000

(Z-1)-

Thus option (D) is correct. 3xl.lxlO'xl.8xlO"'°

= 0.1883A

jX^rays^

^

"

^

Sol.23 (C) Power dissipated in target anode is P =1^x0.01

=>

P=150A^^x lOx 10-3x0.99

=>

P = 1485 watt

1485 =>

.

'

.A39!

Sol. 5 (A) As analyzed by DeBroglie option (A) is the only correct option as given in question.

Sol. 6 (A,B,C) Shortest wavelength ofX-rays emitted isgiven

^

P= —=355.26 cal/sec.

as

• -

.

12431

Sol. 24 (A) Noof electrons through the table per second are ,-3

Hence option (A) is correct. Astheincident energy20 KeV is

3.2x10 N=

,-I9

= 2 X 1016

morethenthe ^ergy ofZ, shell X-raymaybeemitted which willhaveenergy 19.9 KeV orless hence option (B) and(C)are

1.6x10

Sol. 25 (A) Penetration power of X-rays depend upon the photon energies thus the photon energy is more for lesser wavelength radiation.

correct.

Sol.7 (A, C,' D) The energyrequired to knock out K shell electron (from « = 1 to infinity) is more than-that of Z, shell X-ray photons is directly electron (from « = 2 toinfinity) hence option (A)iscorrect. With

Sol. 26 (C) The energy, of propotional to (Z- 1)^. The energy ratio of two

photons

obtained in X-ray from two metal targets ofatomic numbers Zj and Z2 is |

the energy levels involved intransition ofL^^,

and

X-rays

it is clear that option(C) and (D) are correct. Sol. 8 (B) The accelerating potential difference for the

electrons toproduce minimum wavelength 66.3pm s given as ADVANCE MCQs One or More Option Correct

12431

Z=—^ volts«18.75 kV X

Sol. 1 (A, C) With thefrequency spectrum ofelectromagnetic waves, the rays in the short wavelength neighbourhood: of X-rays are y-rays and in the long wavelength neighbourhood

TheDeBroglie wavelength oftheelctrons reaching theanode is given as

of X-Rays are ultraviolet rays hece options (A) and (C) are

- •, h

correct.

sllmeV

Sol. 2 (A, C) K^, Xp and

« 8.9 pm

lines are corresponding to the Hence only option (B) is correct.

electron transition from n = 2 to k = 1,«=3 to « = 1 and « = 3 to n = 2 respectively. Thus we can use By energy levels we can also use

ho

. .

he ^he

Sol. 3 (A,C,D) Ifthe potentialdifference is same,theelectrons

Sol.9 (A, D) As the potentialdifference appliedto the tube increases, the energyof ejected X-ray photons increases due towhich X-ray intensity increases and the minimum wavelength emitted which is inversely proportional to theapplied voltage decreases. Hence options (A) and (D) are correct.

Sol. 10 (A) Cutoffwavelength depends upon theaccelerating

will gainsame kinetic energy hence willhavesame speed (but potential difference so it will remain same and characteristic lines are dependent upon the energygap between the energy levels of the elment so these maychange. Hence onlyoption

may not be in same direction) so due to same magnitude of momentum they will have same DeBroglie wavelengths. As theseelectrons havesameenergy theyproduce X-rays ofsame

(A) is correct.

minimum wavelength asshortest wavelength ofX-ra}^ depends upon the accelerating voltage only. Sol. 4 (A, B, C) When a high speed electron strikes a metal

surface then, no amountofthe energydissipates initially. When electronpassesin the fieldof nucleus, X-Rays are emitted and for any passing by electron its frill nergy can be convereted into X-Rayphoton or partially also partial energy can also be convered intothe X-Ray photon energyhenceoptions (A),(B) and (C) are correct.

Sol.11 (B, D) Asweincrease the applied potential difference, the electronenergyincreaseswhich increasesthe frequency of the photons emitted hence option (B) is correct. With increase in filament current more electrons per second are incident on the targetwhich increases the numberof photons emitted per unit time hence intensity of X-rays increases but no effectwill

bethereonthecutoffwavelength duetothishence option (D) is correct.

[440.

:

X-raysj

:•

SoL 12 (A, B, C) As value of Z increases the difference of energies of energy levels also increase so wavelength corresponding to each characteristic line decreases hence option (A) is correct. As applied voltage across the tube increases, cut off wavelength decreases hence option (B) is correct. If power of cathode is increased which is done by increase in current or increase in number.ofelectrons per unit

increases due to which the emitted photon energy increases

and cut off wavelength decreases. Dueto increase in overall energies ofthe emitted photons X-ray intensity alsoincreases hence options (A) and (D) are correct.

Sol. 15 (All) X-ray is an electromagnetic radiation which carries energy andmomentum, when it isincident ona substance

time incidenton the target,then duetothis moreX-rayphotons and absorbedor reflected, it supplies energy and momentum to

are emitted per unit timefrom the targetwhich increases the intensity ofX-rays emitted henceoption (C) is correct. Sol.13 (C, D) Forthe different characteristic X-rays givenin options we know that the transitions for these X-rays are correspondingto the energy levels as given below K.. n —2 to W=1

it hence all given options are correct.

Sol.16 (A,B) Asstudied inbasictheoryofcontinuous X-rays are produced dueto highkinetic energy of electrons because

ofapplied potential difference across thetube isconverted into radiation due to interaction with the ncuclear charge and

characteristic X-rays are produced due to excitation of lower

n= 3

to

n= I

shell electrons in the atom which creates a vacancy into which

« = 4

to

n=\

electrons ofhighershelltransitsand emitscharacteristic lines

K

n= 3

to

n= 2

hence options (A) and (B) are correct.

M_

k= 4

to

n= 3

For the above Imes we know that correct energy and wavelength relations are-

- E{KJ

ADVANCE MCQs One or More Option Correct

4

16

3

4x3.14x{l.2x10"^^)^

p„= 2.3xl0"kg/m3. Mass defect • Am= 1.007825+1.008665 -2.014102 = 0.002388 amu

=>

Binding energy A£" = 0.002388x931.5 =2.224 MeV

(v)

Mass defect

Am =3(1.007825)+4(1.008665) -7.016005 =0.04213 amu

Nuclear Physics and Radioactivity| => Bindingenergy

(iii)

A£' = 0.04213 x 931.5MeV = 39.231 MeV

Activityof sample is A=}^

.

Binding energy per nuclus

6.023x10

0.693 A =

39 231

= 5.604 MeV

28x86400x365

23

90

.4= 5.106 X10'^dps (vi)

Radius of

nuclei are

(iv)

•r = r,{Ay''

We useA - Aq2 .4= (lmCi) 2-1/3-3

= 1.2 X10-15 X(12)1/3

.4=0.877mCi

= 2.747 X lO-i^m

Electrical potential energy is

(v)

For substance A we use

m=mp) '/^

v2

u=

1 (6^)^ 471 €o (2r)

m, = 10-2x(2)-

^=—xl0"^ =6.25x10"^ kg.

,-19 \2 9xl0'x36x(1.6xl0"'^)

u=

2x2.747x10

=>

f/=9.435xl0^eV

=>

(7= 9.435 MeV

(vii)

Mass defect

-15

16

For substance B we use.

m«=10-2x(2)-i6/8 mR=--x 10-2 = 2.5x10-3kg. s 4

Am =20(1.007825)+36(1.008665) -55.934939

(vi)

(a) Half life = 10s

= 0.533501

=> Binding energy

1

=496.95 MeV.

=>

(a) We use

A,

(vii)

m=

137,

CS-> ^g^/CS+

55 '

0.01 X10-3 =10-3(2)-'/1390 / = 1590x

N=Np).-tIT

log(lO)

1590x2

log(2)

0.301

^23

6.023x10-

137 .

r = 10564.78yr.

r = 1590x

A=Ap)-'^

(c) We use

log

A = il mCi)2-3/30=0.89mCi

100 99

A=Ap)-"^ '

(viii) Using

log(2)

6=15(2),-//5730

5730xlog|^^^

1590x0.004365 0.301

= 23.05 years t =

(ii)

Weuse^ =^o(2) 2700 = 4750 (2)-3/^ T=

5xlog(2) (4700 log .2700J 5x0.301

=6.135 mim 0.2453

log(2)

/ = 7575.40 years

(is)

0.693

(a) X= —-— = 0.231 sec

(b) Using =>

T=

^

W=3.913 X10^1 atoms

(b) WeuseO.99 x 10-3= jq-s (2)-'/1590

t =

14.43 s

(a) Nuclear reaction for p decay is

(b) We use

=>

T 0.693

0)) Further amount willreduce tohalf afternexthalflife= 1Os

Solutions ofPRACTICE EXERCISE 4.2 (i)

Mean life = t" ^

=0.533501 x 931.5MeV

,

W=7Vo(2)-'//' 1000 = 8000(2) r = 9s

(c) .4 = >77=0.231 X1000= 231 dps.

ajid R^ioactivity Solutions ofPRACTICE EXERCISE 4.3 (i)

Explained in article 4.4.3.

(ii)

213 d;

and

^

;

';44^;

Solutions ofPRACTICEEXERCISE 4.4

(i)

(a) Using charge and mass conservation equations are 83

.

0.1%ofmassofumingis0.001gm

Thus energy released is

^ 0.001 xiO"^

. 20871,^^

,^-27 ^ 931.5 XlO^x 1.6 X10-19

81 Tl + a

2]2p„ +e- +v

1.66x10 ' £"= 8.978 X lO'Oj

=>

(b) Decayconstant of a and p decay are 0.693 X =

1

(ii)

7 X— 20

Power output =

Total released energy Time

Total energy

0.693

13

2x10^x6.023x10^^

1

20

235

and

After 1pm amount of£/ will reduce to halfi.e. 0.5 gm. and production rate of77and are in ratio of7/13 hence the quantity

X 185x106x1.6x10-19

= 1.517x10^4;

1.517x10^'*

after 1 pn are '"77-0-5 X— =0.175gm

(iii) No offissions required toproduce 1ODD J/3 energy are ^20

13

/«/'o= 0.5x — =o.325gm.

200x10^x1.6x10"'^ ^

(iii)

Explained in article4.4.3

(iv)

If in time at total atoms inside the body are

(iv)

A'=3.125 X loi^fissions/s Mass defect of reaction is Am =2 x 2.0141-4.0026

=>

"Am = 0.0256 amu

Energy released per fission is A£:=0.0256x931.5MeV Ifafter time/

N, = -^ ^

2

Total activeatoms inside bodyat time t are

=>

A£=23.846MeV

„

•^= Ifafter time t

^

we use

- 200x10^86400

1 777x2 23.846x10^x1.6x10"'^

iV=9.058 X 10^4atoms

2

Mass of deuterium needed per dayis

^ =7V-o(2)- £'i=4.0324MeV

equal top-emission rate only. Thus y-emission rate will drop to Number ofD atoms required per dayare halfafterhalflifeofp-dacay i.e.270days. (iv)

Explained in article4.4.3

N=

10^x86400 x2

4.0324x10^ xl.6xl0"'^x0.5 iV=5.356 X 10^6^jQjj^g

Nuclear Physic£"and Radi^^iyit^ timerequired toattain thischarge is

Mass of D atoms required per day is 5.356x10 m =

6.023x10

^

26 23

x2

Q ro x0.4xl.6xl0~'^

m = 1778.5 gm.

2x0.01 t =

(vl)

Givenreactionis

iH+

9x10^x5x10'" X0.4x1.6x10 -19

/ = 6.94xl(r's

2^e +AE (ii)

Released energy is

A£:=4(7.0)-4(l.l)

Nuclearreactionis

f^Th^ f^Ra+ \He +energy

A£:=23.6MeV

mass defect of reaction is

(vii)

Am=230.033131-226.025406-4.002603

Fission reaction is

f^U +

56^-5a +3o «+energy

mass defect of reaction is Am =235.043925 -91.8973

-140.9139-2(1.008665)

=>

Am = 0.005122amu

Energy released is 5=0.005122 x931.5 MeV 5=4.771 MeV

Kinetic energy of a-particle is

Am=0.215395 amu

Energy released

=0.215395 X931.5 MeV

t ^ J

A£:=200.64 MeV

=>

230

5^ =3.267 MeV

(viii) Reaction is 41

20 Ca

(ill)

1

20^a+ ort

Energy y-photon is

5^=^^=0.09945MeV

mass defect of reaction is Am =40.962278 -39.962591

mass defect of reaction is

-1.008665

Am=239.052158-235.043925-4.002603 Am= 0.00563 amu

Am =-0.008978 amu

=:>

Asrnass defect isnegative, thisreaction requires energy for its Energy released is completion and amount ofenergy required is 5 = 0.008978x931.5 MeV

Mass defect of this reaction is Am = 1.007825 + 7.016004-2(4.002603)

=>

5 = 5.24434 MeV

Kinetic energy distributed in and \He is approximately E^^E-E^ =5.24434-0.09945

5 = 8.363 MeV

(ix)

5 = 0.00563 x 931.5 MeV

Am = 0.018623 amu

= 5.1449 MeV

Thus kinetic energy of a-particle is

=I

Asmass defect ispositive, thisreaction releases energy soit is

=111 >

5 = 17.347 MeV

Now we use

im„v^ =5.0588MeV

Solutions ofPRACTICE EXERCISE 4.5 (i)

Charge required for 2Vpotential canbegiven as v =

1

2 =

47C Sn

R

2x0.01

^

9x10^

C

=>

(iv)

2x5.0588x10^x1.6x10"'^ 4.002603x1.66x10"^'^

v = 1.56x lO'm/s.

Mass defect of reaction is Am= 12.018613-12.0000-2(0.00055)

^

Am = 0.017513amu

Nuclear Physics and Radioactivity

445

Total energy released =>

Kinetic energy of

£ = 0.017513 X 931.5 MeV

Sol. 5 (D) By conservation ofcharge and mass for the above reaction wehavechargeofx is2 andmassis4 hence option (D)

£= 16.3133 MeV

is correct.

and p-particle willbeapproximately e,=e-e^

=>

.

=16.3133-4.43MeV

=>

=11.88MeV

As mass of

is very small almost whole of this energy is

carried by bata particle. (v)

4 xE^. zr = 4 x5.3MeV 'ai

206

206

=0.1029 MeV 4

Er. =

mean density of nuclear matter almost remain constant hence option (D) is correct.

Sol. 7 (D) By conservation ofcharge and mass for the above reaction wehavecharge ofZis 0 andmass is 1hence option (D) is correct.

Energy ofdayafter nuclei in two cases ofa-emission are £. =

Sol. 6 (D) We have studied that for all stablenuclei average

4

x^a2 = —>^4.5 MeV

206

Sol. 8 (B) As we have studied that in fusion binding energy per nucleon released is more than that compared to a fission reaction hence option (A) is correct. Sol.9 (A) Chemical properties ofan element isdependent upon the valance electrons in an atom hence option (A) is correct.

206

Er. =0.0873 MeV

Thus total energy released in first case reaction is E^==E^ T Z?! +£aj =5.3+0.1029MeV

Sol. 10 (D) Critical mass is the amount offissionable material which can selfsustain the chain reaction in it, this can be reduced by putting a surrounding shield fi^om which the neutrons can

get reflected and reused in the fission processhence option (D) is correct.

= 5.4029 MeV

In second case when y-photon is released, we use

Et=E^^ +E^^ +E^ =>

E=E^-E^ -£

=5.40-4.50-0.09

°2

£.^=0.81 MeV Solutions ofCONCEPTUAL MCQS Single Option Correct

Sol. 11 (D) When in a nuclear reactor Cadmium rods are inserted, these absorb neutrons and slow down the rate of

reaction and when pulled out, rate of reaction increasesso by these rods the rate at which energy is being produced by the reactor can be controlled hence option (D) is correct. Sol. 12 (A) A P -partcle carries -1 charge which is emitted when inside a nucleus a neutron transformsinto a proton.

Sol. 1 (D) If in above decay x alpha and y beta particles are emitted then we use 4x = 238-222=16 =>

x=A

and 2x-y-l => ^ y=\ Hence option (D) is correct.

Sol. 13 (D) Tostart fusion nuclei must have very high kinetic energy to overcome the Coulomb repulsion between the nuclei

and this high kineticenergyis obtainedbythermal energydue to high temperature only hence option (D) is correct. Sol. 14 (B) As studied in basic theory mass defect is refered to the difference in masses ofthe nucleons ofa nucleus and the

Sol. 2 (A) Out ofthe above curie is the largest unit of acitivity

hence option (A) is correct as 1 Ci = 3.7 x io'° Bq and lRu=10^Bq. Sol. 3 (A) Among the given radiations gamma rays are of highest energy hence option (A) is correct.

Sol. 4 (B) Due to each alpha emission Zreduces by 2 and for beta minus emission Z ioCTeases by 1 and for beta plus emission Z decreases by 1 hence for the given sequency option (B) is correct.

sum of masses of its independent nucleons which produces energy when the nucleus is produced by anhilation ofnucleons hence option (B) is most appropriate here. Sol. 15 (C) By basic definition ofPackingfi-action option (C) is correct.

Sol. 16 (A) Fusion ofelements occur at veryhightempearature and pressure as at high temperature and pressure the closest distance of approach of nuclei during collision decreases to a level where fusion can start by appearance ofnuclear attraction hence option (A) is correct.

|446

^

Nuclear Physics and Radioactivityj

_

Sol. 17 (C) As explained in article 4.6.1 of basic theory of

Sol. 29 (A) By conservation of charge and mass we can see

fission ofuranium nuclei option (C) is correct.

that Xis an electron hence option (A) is correct.

Sol. 18 (B) As already studied that majority of space in an Sol. 30 (C) By conservation of charge and mass we can see atomis emptythatswhyin goldfoil experiment mostofincident that option (C) is correct. alpha particles pass undeviated and very few gets deflected. Due to very small size of nucleus very very few particles get •SoL31 (D) Halflifeofaradioactiveelementisanuclearproperty reflected hence option (B) is correct. which doesnot dependupon external factorshence option(D) Sol. 19 (B) C-13 and C-14arenot availablein abundance so it is not possibleto fusethese elementnuclei with other whereas in option (D), the given reaction is a fission reaction hence option.(B) is correct.

Sol. 20 (C) Carbon dating is a process in which age of a specimen is estimatedbyanalyzingthe amountofC-14present in it as compared to C-12 hence option (C) is correct. Sol. 21 (A) In radioactive elements fi"om their neclei protons are never emitted hence option (A) is correct. Protons can only transform into neutrons within the nuclei by emission of emission.

Sol. 22 (C) As studied in theory that beta minus decay occurs due to transformation of a neutron into a proton inside the nucleus hence option (C) is correct. Sol. 23 (D) By conservation ofcharge and mass in the above reaction option (D) is the correct. Sol. 24 (A) If in above decay Xalpha and

beta particles are

emitted then we use 4x = 200-168=32 =>

is most appropriate here.

Sol. 32 (D) From the given information we have

=>

ln(2)a_y= l/Xy > Xy

hence Xis more active than Fhence option (D) is correct. Sol. 33 (B) By conservation of charge and mass we can see thatoption (B) is correct.

Sol. 34 (A) In above reaction nuclear charge decreases by 1 hence it occurs due to an electron capture among the given options hence option (A) is correct.

Sol.35 (D) After200minutesthenumberofXwillreducetol/ 16 and that ofY will reduce to 1/4 hence option (D) is correct.

Sol. 36 (A) p"-particle is a fast moving electron emitted fi-om the nucleus due to transformation ofone neutron into a proton hence option (A) is correct.

Sol. 37 (B) By conservation of charge and mass we can see thatoption (B) is correct.

x = 8

and 2x-y = l0 => y =6 Hence option (A) is correct.

Sol. 38 (A) In a nuclear reactor moderator is used to slow

Sol.25 (D) TheactivityofelementXistherateforformation of Y and the activity decreases exponentially according to radioactive decay law hence option (D) is correct.

Sol. 39 (Q In path 3 it showsthat the fired particle is attracted by the nucleus which is not possible as alijha particle is always repelled by the nucleus.

Sol. 26 (C) By conservation of charge and mass we can see

Sol. 40 (D) As studied in theory the radioactive property of

that charge ofx is 2 and mass is 4 hence it is an alpha particle so option (C) is correct.

an element is a nuclear process and not associated with any

Sol. 27 (B) By conservation of charge and mass we can see that option (B) is correct. Sol. 28 (B) According to Pauli's neutrino hypothesis an electron emission is accomplished by emission ofan antineutrino and a positron emission is accomplished by emission of a neutrino hence option (B) is correct.

down the neutrons to a level where fission can start hence

option (A) is correct.

physical and chemical properties of the element hence option (D) is correct.

Sol. 41 (B) By conservation of charge and mass we can see thatoption (B) is correct. Sol. 42 (B) By conservation of charge and mass we can see thatoption (B) is correct.

[Nuclear Physics and Radioactivity

447]

Sol. 43 (A) For the same energy alpha particles have largest Sol. 4 (B) Rate ofdecayof^ keeps on decreasing continuously in size due to which these cannot penetrate much in physical because concentration of^ decreases with time => (A) isfalse. substances and have least penetration power and gamma Initial rate ofproduct!on ofB is decay iszero.

radiation being electromagnetic radiation have maximum penetration depth hence option (A) is correct.

With time, as the number of5 atom increase, the rate of its production decrease and itsrate ofdecay increases. Thus the

Sol.44 (B) Duetoemission ofbetaparticle mass number does not change hence option (B) is correct.

(B) is the correct choice

number of nuclei of5 will first increaseand then decrease. ^

Sol. 45 (B) Fast neutrons in a nuclear reactor are sloweddown

by use ofmoderator which are heavy water and lad shielding henceoption (A) is mostappropriate.

Sol. 46 (D) When a nucleus ruptures into two parts their velocitiesare in inverse ratio oftheir masses as linear momentum of the system remain conserved thus the mass ratio of the two parts will be 1:2 hence from the relation offermi radius their size

The initial activity of5 iszero whereas initial activity of^ is A-iA'q => (C)isfalse.

Ast\met->co:N^ =Q,Np = OandN^=-N^^ => (D)isfelse Sol. 5 (D) Using #=7^0(2)-'"® wehave ^0

att = 30,

N=~

andat/ = 40,

N=-—r

Nr.

16

ratio willbe 1:1}'^ hence option (D)is correct.

hence option (D) is correct.

Sol. 47 (C) From thecurve ofvariation ofbinding energyper nucleon with mass number it is clear that the two elements with

Sol. 6 (B) Using 7V=A^o(2)-'/^

mass number more than 100 when fuse the binding energy per nucleon oftheresulting element islower than thefusing elements hence energy must be supplied for this purpose hence option

at / = 1 day

0.97V,(2)-"^

(C) is INCORRECT.

Sol. 48 (C) Energy isreleased ina process when thebinding energy per nucleon of the product is less than that of initially

log(2) at / = 2 days

reacting elementshence option(C) is correct.

SolutionsofNUMERICAL MCQSSingle Options Correct

log

(l)M

^gC2)

log(2)

Sol. 1 (B) ^QmzAp =AQe''^'=AQe ^ 100 =>

N

t=Tln~ Ad

=>

Sol. 2 (C) We use

81

W=0.8IiVo

Sol. 7 (A) Using 10.81 = l(h:+l 1(1-x)

N=NJ2r"'

=>

N.

10.81 = U-x

=>

x=0.19

=>

l-x=0.81

? = 15 days.

Sol. 8 (C) We use Sol. 3 (B) No ofatoras left after 10 years are

® ^

N=N,{2r'^

,v„

forA

4

forR

Thus (3/4)^ of sample decay in 10 years thus probability of decay is 75%. Nb

4

Nuclear Physics and Radloactivityj

i448

Sol. 13 (D) After 4hours the sample will contain 30 gm of isotope B and 12.5 gmofisotope^.

10 ml/sec

Mass ofisotope Adecayed to^' is37.5 gm which will have a niass after four hours

Sol. 9 (C) ICQ ml_,

w'=37.5x Y5Q=36.5gm

2ml/sec

solution

Thevolume ofliquid in beaker atany instant oftime t is

Total mass = 30+12.5 + 36.5 = 79 gm.

K=100 + 8r 37.5

The volume of liquidejected in t seconds is 2/

Sol. 14 (B) No. ofa-particles emitted = "[50"

Number of active atoms being taken out is N -dN=Y2dt

^

2N

^M

• dt

100 + 8/

V

Sol. 15 (C) Energyreleasedis £=234{8.5)-236(7.6) £ = 1989-1793.6

multiplying both sides with disintegration constant. ^ 2di -XdN=Xl^—

or

£= 195.4MeV

, 2dt -dA=A.—

Sol. 16 (D) Ifxamount isremaining then after three halftubes where A is activity ofthe solution. Thetime taken for 10 ml activity will be solution to come out is 5 second. dA

ff

1 C

-2t -2^

-X 8

.

J ^ ~ J 100 +8/

100

=>

,,

C

x(a:) = —

10

x = 10cm^

1/4

A=AJj

Sol.17 (A) Afterfirsthalf hrs N=Nq—

Thusrequired activity of the ejected solution is

for /= ^to/= 1^N= ^^0^

a/4'

^ ^0 ^0

-'f

=^2

for /= 1-| to/=2hrs. Sol. 10 (B) Weuse2l=^o(2r''^ .

[for both andB

+yy^ =2+4=6/,^2= 1/6 hrs.

/=12hrs.

Sol. 11 (C) After four halflives

ofthe sample will left

undecayed.

Sol. 18 (C) Energy releasedis £=4(7.1)-2(1.15)

Sol. 12 (C) At 300 K temperature most probable speed of

=>

£=23.8 MeV

thermal neutrons is 2kT v =

Sol. 19 (C) We use77 = - [a - (a - XN^)

m

Thus

1 2 £^= -mv^kT

1.38x10"^^ X300 1.6x10"'^ £2^=0.026

£„=

exampl&4.24 100= [200-(200) 100 = 200(1-6-0 e

-

2

/=ln(2)

taken from

jNucjear''Physics and Radioactivity

449;

Sol. 20 (B) Rest mass energy ofan electron is £:=0.00055x931.5MeV =>

£:=0.512MeV

A=A^9. dN.

Sol. 21 (B) for

~=XN^-2XN^

Sol. 27 (D) We use

A=Aq e-"^

to be maximum,

_y= x

dN2

Noofatoms disintegrated during

=0

dt

is

AA''=(.x->')x

m.=2XN. or -^=2. ' 2 N2 Sol. 28 (A)

X2

Sol. 22 (C) We use

decay rate of a decay

N=NJ2r'^ iVn

^2 decay rateof |3 decay Xj probability of a decay X2 ~ probability of p decay

,-2/r

y -^o(2)

log 2

T= 2-

75

log 6

.?± _ 100 =_ 3

7'= 0.7736 hrs

25

r=46.41 min.

Sol. 23 (D) Tfafter 5 halflives ~ for thesample will remain

100

Sol. 29 (B) ^2^^^

isotopes similarly goTh^^"^

and 9QTh^^® are isotopes.

hence it is

Sol. 30 (A) Using N=NQ(2y'^

16 32

Sol. 24 (D)

= 0.5gm

n=XN

=>

X= —

Sol. 31 (D) Noof fissions per second required are

^ 'N 0.69

/ = 6 hrs.

3.2x10^

0.69A^

^1/2 . X

200x10^x1.6x10"^^ =^>

Sol. 25 (B) A

^ B

C

dB

Let bbe the number ofnuclei ofB-jp =x-Xb or y=x-Xb =>

A^=10''

Sol. 32 (D) Energy of Yphoton = Diffemece in energies ofa particles= 0.4MeV.

7J)=x-y

Activity of

B = -Xb=y-x.

S0I.33 (B)

+

i

Sol.26 (C) Aftert = 9yer

2a +

A -A,(2)-''^ 3

T=9

=>

log(2) Iog(3)

A=A,i2y^^

a.tt= 18yrs. 18-

log(2) _ , log(2)

log (A A

+v

log(3)

Given.

^-8=224

&

Z-5=89

^ = 237,Z=94

Sol. 34 (D) Using A^=AL'"^'we have

N=NQe~^ N=

Nr.

(2.71)

2 =0.13W,.

Nuclear^hysics jnd Radioactivity^ Sol.44 (Q Inthemixture,initiallyA''=W|+^2 Atanytime(/):A(0 =A^i(0 + ^2W

Sol. 35 (B) 100= 800e-'^(^^®'^ e-360>. = i/8

=> Decay rate at time (r) (obtained bydifferentiating above)

-360A.=>^| =-Xn8

Hence (C) X=

^ _M =

^1/2

or

X

Inl

360

120

Inl

(/«2/120)

Sol.45 (C) g^Ra+«(2He^) +m(LiP^.

= 120min

206 ->

82

Pb

where n andmarerespectivelythe required number ofa andp

particles. Clearly from the options given, only (C) satisfy the

rj/2 = 2hrs.

above reaction.

A=AQe

Sol. 36 (C)

after one year:

Inl^

Sol. 46 (A) Since there is no change in atomic number and . mass number due to the emissionor absorptionofa y-rayphoton

A=

((,yO). Hence (A).

^

10

Sol. 47 (D) Energyreleasedbylkgflielis

X 10

£:=

Afterfurther 9 years(i.e. after 10years): °

10

Ix0.001x(3xl0)^

.

— =2.5 XlO^kwh

1000x3600

Sol. 48 (D) Energy released is £ = 8(7.06)-7(5.60)MeV £ = 56.48-39.20 MeV

Sol. 37 (B) We use

m= mo(2) 1=256(2)-^'2.5

=>

t = 100 hrs

E = 17.28 MeV

Sol. 49 (B) Fora-decay : ^A^-

^ _jeO + 22He'»

>x-2 0

Forp'-decay:

>x+iB^ + -iP

Hence the unknown nuclide is

For p"^-decay: ^A^

>x-\ B^ + ^ip''

Sol.39 (C) Aftera verylongtimeat equilibrium

For k -capture: there will benochange inthenumber ofprotons. Hence, onlycase in which no ofprotons increases is p~-decay

Sol. 38 (B) jLi^ + on' -

Hence (B).

f=> Rate ofdecay=rateofformation = lO'^dps.

501.50 (C) RateofformationofFisthedecayrateofZwhich is given as

Sol. 40 (D) Using

= A^„(2)~"^ we have A^= 10^(2)-^^'° =

IX

' dt

^0^

IfA^q are no. ofnuclei ofXat/= 0. 501.51 (C) Energyreleased inanuclear reaction isproportional

Sol. 41 (C) Nuclear Reaction are

to

so if C will reduce to halfthe energy released become one

fourthand will reducedby(3/4)*'' ofinitial volume Sol. 52 (D) From given graph we have

SoL42 (C) Explained in example 4.24 =>

Sol.43 (C) In above reaction four hydrogen nuclei are fusing toproduce a helium nucleus soit is a fusion reaction.

^ = ^(-1/^0 +2,5) ^ = g2.5 . g-0.1/

=>

^=12e-*'".

[Nuclear Physics and Radioactivity

15JJ

Sol. 53 (C) Activity of an element is

As per given condition

A=XN

for same number ofmolecules

X^N^(2r=2X^N,(2T^ -« = l-2«

A

« = 1.

A,

Sol, 61 (D) Reaction is

Sol. 54 (A) We use

N+^He^^^O+\P

A=AJi2)-"Sol. 62 (C) We use

=>

^q = 5 X16= 80dps

N=\(2y" so after time t we have

Sol. 55 (D) Explained in article 4.4.2

-r =JV„{2)-'°

Sol. 56 (C) We use ViV. K2

KE_^

log 5 t = 2—-—r =4.64hrs.

4n er

log 2

,-19x2

_ 9xl0^x2x92x(1.6xl0~"') 5x10^x1.6x10"'^

Sol. 63 (D) By radioactive decay law

r = 5.29xlO-i''ra r = 5.29x 10"'^ cm

and

N2=NQe-'^'

=i =g-9X/

A

Sol. 57 (D) We use

N, N,

=0.2Nq

...(1)

=O.SN^^

...(2)

9X

and dividingequation-(2) by (1) we get

In 2

Sol. 64 (A)

1620"^810

-4

=>

/'= 8 XlO^years

In 2

A^o — =N^{e)-Kst

Sol. 58 (B) By conservation ofmomentum we use

A^:A2 = 1:2

=>

21n(2)

r^:r2 = l: 2"^

t =

ln2

ln2 + •

1620

Sol. 59 (D) Total no of decays,

AN=N^-N{t)

=>

=A^oO-^~'0

=>

= ^•o(l-2"'^'"0

=>

810

f = 1080 years.

Sol. 65 (C) Mass defect of reaction is Am =3(2.014102)-4.002603-1.008665 -1.007825 Am = 0.023213 amu

This quantity is maximum for option (D)

Energy released

Sol. 60 (A) Given that

=>

£ = 0.23213x931.5 MeV

2A,^. After« half lives ofyf their

£=21.623 MeV

Total supplyby 10"*® deutrons will get exhausted in time

activities are

A.^xM^r as

T =A ^

and

2.

A^= X^^{l)-2n

21.623x10^ xl.6xl0~'^xl0'"' t =

3x10

/= 1.153x101^5.

16

Nuclear Physics _and Radioactivity]

^52

Sol. 66 (D) Cu nucleus is heavier than Zn so it will decay through beta decay.

quantity before time and ^2 are

and 2Nq

Sol. 67 (C) Number ofnucieiof//^ will be maximum when its decay rate becomes equal to its production rate.

and

Sol. 68 (C) de-Broglie wavelengths ratio will be inversely proportionalto the momentumofthe twoparticleswhich will be same as parent particle was at rest so has initial zero momentum.

2^V„=

2A^e-^ = Z4,

Sol. 69 (A) Numberofnuclei will becomemaximumwhen the production rate of nuclei will be equal to their decayrate hence we use a = 'kNthus option (A) is correct. Sol. 70 (C) We use

^ r- A t - t = ^2 h b 2 In

N=NJ2r"^ ^0

,-l/T

-T =W

Sol. 75 (C) We use

=>

/ = 3x 1.37x10^

=>

r=4.llxioVs.

N=N^2'"'^ 63.5

N= Nq2 12-7

Sol.71 (B) ForA^jatomsoftheradioactiveelementbetaactivity is dps hence we useWj = XNy Sol. 72 (D) Power ofSun is

^5= 1.4x10^x471(1.5x10")^

^0

N= — 32

There total number ofnuclei decayed will 1^® ^31 ^ro•

energy released by Sun per day is

^suWday =Ps>'86400 mass lost by sun per day is

Sol.76 (D) Toachievekineticenergyofelementsoftheorder of potential energy at fusion reparation we use 2A:7'=7.7x 10-1"

•'sun/day m

=

=>

r=

1.4x10^ x4x3.14x(1.5xI0")^x86400 m

=

,8\2 (3x10®)

w= 8.79xl0"kg Sol. 73 (A) Energyreleasedby2kgofuraniumis 2000x6.023x10

^=2.78xl0^k.

2x1.38x10""

Sol. 77 (D) Initial nuclei attime/j are N,=A,x attime t^ = N-A^ Total nuclei decayed in time

are

N^-N={A,-A^)x.

,23

E=

7.7x10"'''

x185x10^x1.6x10->9j

235

Power output is

Sol.78 (A)

dN

_ £ _ 2000x6.023x10^^x185x10^x1.6x10"'^ t ~

235x30x86400

dN d^N for -jp to be minimum; "^i" ~^

= 5.853 xlO^w. d^N

Sol. 74 (C) Given that present quantity are

A

dN

—- =2t-\-r =-2t-X(f-XN) = 0 dp(it • ^

^A

2/o - XtQ N= X-

fNudear Physics an^ Radioactivity

4531

Sol. 79 (D) If day-1nuclei decaythen are use Decayed Day-1 Day-2 Day-3

Day-9 Day-10

t:

N,

= 2«7V^ = 2'W^

2N^ 4N^

Hence options (B) and (C) are correct.

= 22

Sol. 7 (A) In y-decay no effect is there on atomic and mass numbers hence option (A) is correct.

= 2'N^ =2'N^

256

512N^

Sol. 8 As^ and5 areisotopes, thesewillhavesameatomic number and belong to same element. B and C are isobars so

1023iV^. Amount leftafter 9'^ dayis Total Decayed

these will have same mass nutfibers but different atomic numbers

sothese will be different elements. Byaradioactive process of

512

alpha or beta decay atomic number always changes so B can

xIOO 1023

, change to C by any of these processes hence option (D) is

= 50%.

correct.

ADVANCEMCQs One or More Option Correct

Sol. 1 (All) In a-decay nuclear charge decreases bya factor of 2 and mass decreases by a factor of 4 and in P-decay no effect is there on the mass of nucleus but charge increases or decreases bya factor of 1and in caseofy-decay no changewill be there on charge or mass. Hence all the above given otions are correct.

Sol. 9 (A, D) As already discussed in basic theory that in a-emission the energy ofallthe emitted a-particles aresameso these will have almost thesame speeds butincase ofp-emission due to emission of neutrino the energy is distributed among betaparticleandneutrino soemitted p-particles havedifferent speeds hence options (A) and (D) are correct.

Sol.10.(A,B,D) Inoption (A)the reaciton given is thefission reaction of bya slowneutronhenceoption (A)is correct.

Sol. 2 (C, D) Onlyiny-decay or K-capture process nocharge is emitted from the atom hence options (C)and (D)are correct.

Option (B) iscorrect asHatoms inordinary water may capture neutrons, while Deuterium in heavy water does not capture neutrons. Option (C) is not correct as Cadmium rods decreases

Sol. 3 (B, C) As wehavestudied in basictheorythat a neutron decays radioactivelyto a proton but a proton can transform to a neutron only within the nucleus and cannot decay freely hence options (B) and (C) are correct. Sol. 4 (A, B) From the basic definitonofhalflife and mean

lifetimeoptions (A) and (B) are correct. If in decay equation we put t = mean life = l/A, then we get A^== 0.37

hence

option(C) is not correctsimilarlybysubstituting r=3T we can

the reactorpower byabsorbing neutrons. Option (D)is correct as for

slow neutrons causes effective fission whereas in

fast neutrons causes effective fission:

Sol, 11 (A, C) Weuse decay equation

In a given timet the amount ofradioactive substance decayed is given as

AN={N^-N^^e-^') In time r - 1/X we have

check that option (D) is also not correct.' AN=NJ 1--

501.5 (B,C) Assumingptobe^,p®thenucIearreactionwith atomic numbers and mass numbers ofthe elements are written as

AA^=A^„(l-0.37) = 7/„(0.63) AN

>o(0-63)

No

N.

X 100 = 63%

hence option (A) is correct.

hence options (B) and (C) are correct.

501.6 (B,C) UsingradioactivedecayequationA^=A'Qe"^'we Half life time of the substance is given as have

At / = 0 number of nuclei are

=

= 4 year

thus option (C) is correct. and at time t number of

nuclei left are

Thus number of nuclei decayed in time t are {N. -

=

Probabilitythat a radioactive nuclei doesnot decayin t = 0 to

Sol. 12 (A, D) Wehave studiedthat the stabilityofa nucleus is due to the mass defect in formation of a nucleus so the rest mass of a nucleus is less than the sum of rest masses of its

nucleons considered independently henceoption(A) iscorrect.

Nuclear Physics and Radioactivityl

454

In case ofnuclear fission heavy elements disintegrate in smaller fragments by inducing the reaction through a neutron and due to the fragmentation of nucleus energy is released hence option (D) is correct.

Sol. 20 (B, C, D) According to decay law for activity we use

A=Aq{2)-^^^''^ 800=.4^ {2)-'^^^''^

at t- 200 days

...(1)

8000

at /=300days ^ Sol. 13 (A, B, D) According to the decay equation we have N= hence option (A) and (D) are correct and according to radioactive decay law option (B) is correct. Sol. 14 (A, B) The parameter represented by y-axis in the figure growswith time and becomesconstant after a finite/long time interval.Accordingto radioactivedecaylaw the number of nuclei of a radioactive element left undecayed decreases exponentially with time so the total nuclei decayed can be represented like this curve hence option (B) is correct. If in a reactor a radionuclide is produced at a constant rate then these start decayingjust after their production starts and after some time equilibrium comes when the decay rate becomes equal to the production rate then the number ofsuch nuclei can also be represented by this curve, hence option (A) is correct.

...(2)

Dividing-( 1) by (2) we get

^ _ 2l/7'(300-200) T=200days from equation-( 1) we have

8000(2)-' => = 16000 dps at / = 400 days we have

^ = 16000(2)-^°^®® 16000

=>

A=—-—=4000dps

hence options (B), (C) and (D) are correct. SoL21 (A,Q Given, X=0.173

Sol. 15 (B, C, D) By radioactive decay law we can find that options (B) and (C) are correct. The disintegration energy of the nucleus is distributed among a-particle and the

T

daughter nucleus hence option (D) is correct.

Also

Sol. 16 (C, D) As Ca nuclues is more stable than Ne nucleus

for

1/2

=

In 2 X

0.693

=

,

— 4

0.173"

No-N =Noe-^' 0.173

rest mass of-former is less than twice the rest mass of the

latter hence option (C) is correct. As Ne nucleus is a stable

year:

N, N.-N= —=0.37N..

element its rest mass is less than the sum of rest masses of

its nucleons hence option (D) is correct. Sol. 17 (B,D) From the figure it can be seen clearly that aiergy is released in a nuclear reaction when the B/A of elements

produced is more than that ofthe reactants or the nuclei present before the reaction. Iftwo nuclei in the range 51< < 100 will fuse then they will produce an element with mass number above 100 and less than 200 which has moveB/A thus en^gy is released hence option (B) is correct. Similarly a nucleus in the range 200

=>

N^>Ng

Activity curves are Activity

Activity of B decreases rapidly than that ofA.

455!

IGeometrical Optics

ANSWER & SOLUTIONS Solutions ofPRACTICE EXERCISE 1.1

CONCEPTUAL MCQS Single Option Correct

(i) (A)

1

2

(C)

,

3

(B)

4

(C)

5

(A)

6

(D)

7

(D) -

8

(B)

9

(A)

10

(C)

11

(C)

12

(D)

13

(D)

14

(A)

15

(A)

16

(B)

17

(C)

18

(B)

19

(A)

20

(A)

21

(A) (B)

22

(D)

23

(B)

24

25

(B)

26

(D)

27

(C)

28

(C)

29

(C)

30

(D)

31

(D)

32

(D)

33

(B)

34

(A)

35

(A)

36

(A)

37

(A).

38

(A)

39

(B)

40

(C)

41

(B)

42

(D)

43

(A)

By Similarly in trignel ABC and DEC

In the triangles IsCDE and taCABby similarity we have

J__'2± 0.3 ~ X =>

NUMERICAL MCQS Single Option Correct

/=0.6m

(ii) Ifthe mirror center ofcurvature is considered on its left and object is placed on the right side of it, here for mirror

1

(A)

2

(C)

3

(B)

4

CB)

5

(B)

6

CD)

7

(C)

8

(C)

9

(D)

10

(C)

11

(B)

12

(A)

13

(A)

14

(B)

15

(C)

16

(B)

^ 17

(A)

18

(D) (D)

formula we use

/=-20 cm,

m = (- v/w) = —

and

or V= - (w/4) Using the mirror formula,

19

(B)

20

(C)

21

22

(D)

23

(B)

24

(C)

25

(A)

26

(B)

27

(D)

28

(C)

29

(D)

30

(A)

31

(A)

32

(A)

33

(A)

34

(A)

35

(B)

36

(D)

^

37

(B) ,

38

(C)

39

(A)

Solving we get u = 60cm.

40

(C)

41

(B)

42

(B)

.

.

?

1

1

1

— = —+ —, we have

/

w V

• 20

43

(B)

44

CC)

45

(B)

(iii)

46

(A)

47

(A)

48

(C)

mirrorsMjand

49

(B)

50

(B)

51

(A)

In A/f5Cwehave

M H

The figure below shows the incident angles on the

and in the trianglewe have

2/ + 2(6O-O + 0 = 18O° ADVANCE MCQs One or More Option Correct 120^ + 0 = 180°

4 7

2

(A, D)

3

(A, D)

5

(A, C)

6

(A. C)

(All)

8

CA. C)

9

(A. C)

(B, C, D)

1

.

(A. B)

10

(A, D)

11

CB. C)

12

CB, C, D)

13

(A, B, D)

14

CA, G)

15

(A, D)

16

CC, D)

17

CA, D)

18

(A, B, D)

19

CB, C)

20

CB. C)

21

(A. B)

22

CB. C)

23

CB. C)

24

(All)

25

(A, C)

26

CB. C)

27

(B, C, D)

28

CB, C, D)

29

CB, C)

30

CA)

31

CB, C)

=>

0=60°

'^^•^77777Z^7/7777P7h777^W^777:^,M, 'N,

Geomelrical Opticsl

•456

0v) Figure below shows the multiple images formed by the combination of each pair of mirrors listed below. As studied earlier that two mirrors placed at an angle 60° produces 5 images, all ofwhich lie on a circle, here we consider 5 images formed by each pair of images as shown. Combination

Images

of mirrors AB and^C

AB and BC AC and BC

1,2,3,4,5 r,2',3',4',5' r',2",3",4",5"

mirror gets reversed hence we use along the surface of mirror image velocity components are

\

= 5i m/s

\ = ~3jm/s

&

Due to motion of mirror theimage velocity in the direction normal to mirror the image velocity is increased by twice the velocity ofmirror thus in direction normal to the mirror image velocity is given as V- = 2v '7

=>

'"z

—V

Oz

\ = 2{-3k)-(-nk)=+5k

Thus final elocity ofimage is given as

V,- = 5i-3j + 5k m/s (vii)

Different angles of incidence and reflection at mirrors

M, and

are as shownin Figure.

3' B C 3"

a = 90°-9

360 ^ For the mirror AB and BC -r- = 6 60

Thus by three pair ofmirrors total 15 images will be produced but three images out ofthese locatedat (1,1'), (5,1") and (5',5") are coinciding so total 15-3=12 images are produced by the given configuration ofmirrors.

In triangle AABC we have 0 + 0 + 0 = 180° =>

(v) Figure shows the field of viewIPQ of the image of light source B from the line along which the man is walking. In this figure fi^om triangle IPQ and IMNwe use PQ

3L

0=60°

(viii) As we have studied that due to motin of object the image velocity component normal to mirror gets reversed and due to mirror motion the velocity ofimage in direction normal to mirror gets doubled in same direction so the final image velocity we will have here will be given as =>

v. = -2-5=-7 m/s

Thus fmal speed ofimage is 7 m/s and direction is towards left.

(Lx) We know that along the direction parallel to the mirror which is Xand z directions in this case the velocitycomponents of the image will remain same as that ofobject and along the normal to the mirror which isj direction here the image velocity is equal and opposite to the object velocity and added to twice the mirror velocity which becomes—(— 4) + 2 (+ 5) = 14m/s thus the final velocity ofimage is given as

1^ =-2i+14j +4yt (vf) Due to object motion, image velocity components parallelto mirrorremainsameand the componentnormal to the

(x) The light ray retraces its path only when it falls on the mirror normally so the light ray in its third reflection falls on

[Geometrical Optics ,45-:

mirror

normal to itas shown in the ray diagram shown in (^/= 60 cm

figure. Thusfrom triangletiOEC,wehave

1

12 cm

75° +0 + 90° = 180°

+x

=>

0 = 15° •u - 84 cm

h-v= 35 cm-H

(iii) Bycoordinate convention for mirrorformula weuse M= - 25cm,/= - 20cm By mirror formula we have v =

«/ u-f

-25X-20 =-100 cm

-5

Thus magnification in this case is Solutions ofPRACTICE EXERCISE 5.2

(i)

V-100 m = ~ ~ = - —-7- =-4 u -25

Considering object placed to the left ofmirror then by Image velocity along theprincipal axis isgiven as

coordinate convention for mirror formula, weuse

/(along PA) =

V,

v = + 35 cmand/= + 60 cm Using mirror formula, we have U =

yf

35x60

v-f

-25

^

ffP' X Vo(aIong714)

^/(along PA)

^

And image velocity normal toprincipal axis isgiven as

= -84 cm

^/(normal to PA)~ Magnification

40-v/2 cm

w~84~12

^o(nonnalto/34)

^

%oniialtoPA)'=^^^ =^0^/2 Cm

(iv)

Here

Sizeofimage = w x sizeofobject

=^ X12=5cm

M=-40cm v = + 80cm

Figure shows the position ofobject and image with respect to convex mirror. •/= 60 cm

2K

1

hn - 12 cm

+ x +x

•40cm-

•80cm-

•u = 84 cm-

H-v = 35 cm-H

(ii) As object is real and image produced is inverted that means image isalso real. Inthis case the magnification produced is given as V

5

u

-12

2

Focal lengthofmirrorisgiven bymirrorformula as

^ u+v

/=

uv

-40x80

-40x80

u+v

-40 + 80

40

' /=-80cm

Thus focal length ofmirror is 80 cm. V

Imagedistance is v = -12(2.5) = -30cm

uv

using mirror formula

-12X-30

-42

60

7

8.6cm

(v) Using magnification formula, ifwe consider both object and image arelocated ontheleft oftheconcave mirror, we use V u

size of image size of object

V

5

-12

2

v=-30cm

Geometrical Optics!

{458

(ii)

From mirror formula we have -12X-30

MV

/=

w+ v

Apparent depth due torefraction byparallel sides glass

slabs is given as = -8.6cm

h

-42

Ih

aPP

velocity magnification along theprincipal axis ofthemirror is

^2

given as np- so the velocity of imageis given as 25

v2

Image velocity =\^—j ^Object Velocity

=22.61 cm

app

Image velocity

15

happ =— + TT =16.67 + 6 1.5 2.5

xl.2 = 7.5cm/s

(iii) In the figure /j is the image formed by the glass slab refraction and I^ isthe image of /, formed by the reflection by asobject ismoving toward themirror, image must bemoving in fi-om center ofcurvatute.

the convex mirror and image offormed by the refraction through glass slab is ly

(vQ

First we consider shift by glass slab in air which is given as

opposite direction as it is inverted so it mustbe moving away

Consideringfirst Reflection H= + 25 cm&/=+20cm

Shifl=?| 1-^1 =6^ 2

using mirror formula V /, =

= 3 cm

25x20

«/ u-f

= 100 cm = 40 cm

ConsideringsecondReflection at convexmirror M=+ 50&/= + 30cm •~..^-12cm'-~

=3cm,_---*::7^ 1:^

50x30 = + 75 cm

0

h

E

;—H

h

h

shift/= 3 cm u = 2 6 cm -33cm-

Now for reflection at ctmvex mirror we use mirror formula

1 1=1 V u

From figure wecansee thatfinal imageis produced at 75cm behind M2 and it isvirtual.

f

Hereweuse/=+20cm and M=-30cmthenweget

Solutions ofPRACTICE EXERCISE 5.3

I J_

J_

V

20

-30

v=+12cm

(!)

The system now comprises of two slabs of different

materials so the net shift of point of convergence is given as

1-+

Shift = /,

+ '2

Thusthe position of is as shown in figure. Nowagain dueto refraction through glass slab a shift of 3cm is included so

1'

image I^ is obtained at adistance of12 - 3=9cm from the pole

^2.

ofmirror hence the distance between object and final image is

.

33+ 9 =42 cm.

r

21

= 6 1— 3

+

r, ii 4 1—

= 4 cm.

2

As the slabs are denser than surrounding the direction of the shift is in the direction ofthe incident rays which is to the right. Therefore,the rays will finally converge to a point at a distance

equalto (14- 6 - 4) + 4 = 8 cmto the right ofthe second slab surface.

(iv)

Speed ©flight in glass is given as C v =

The raydiagramin figure-xxx below shows the path oflight ray in the glass slab.

'Geometrical Optics

459!

rf=8feet

=>

width of canal is 2d=\6 feet.

(vO

Lateral displacement ofhightdue to a glass slab isgiven

as

t, = Im

t sin(/ - r) A = cosr

/[sin i cos r - cosi sin r] A = COST-

By Snell's law we use

I.sin30°-psinl5'' => p = 2cosl5° Time taken by light to cross the slab

AC

/,/cosl5' C/p

/ =

1

/=

X

2cosl5'

cosl5°

3x10^ For small 6 we use sin 0 = 9 and sin r=r and both cosine terms

to be almost equal to unity so we have

/= jxlO-%

A=/(y-r)

...(1)

(v) Figurebelow shows the raydiagram of imageproduced at 5ft mark for boththe marks - One at 4 inch byrefraction and

In the abovefigure-xxxshown,by Snell's law we have

other at 17ft by reflection.

=>

sin 0 = p sin r 0 = pr r =

+ d

Thus from equation-(l) we have

vertical shaft

A = /| tJ--| =

/0(p-l)

066ft

(vii) Figure shows the container filled with water upto a height Xso that when observed from top, it appears to be half filled.

By using Snell's law we have

1 Xsin 0 = —sin (90- (p) (21-X) 4 Ix

—

yBe + d 2 =>

—X-

3

21cm

32^

%ll024 +9d^ = 4-j26 +d'^

On squaring both sides, we get . 1024+9£^= 16(36+ =>

1024 +9£^ = 576+ 16£^ 7^=44S

=>

(P-= "Tp =64

,

448

The apparent depth of container is such that it should be equal to the empty length of container for it to appear halffilled, so we use

-

=21-x

Geometffcal Optics]

1460

(ix)

3x

The situation is shown in figure

— + x =21 •

4 Ix

x = \2cm

(viii) The situation is shown in figure-xxx. As per given condition we use i + r=90°

=>

r=(90-0 From figure, Similarly,

By Snell's law we have V''-

sin /

sin i

sin i

sin r

sin(90 - i)

cos i

...(1) ...(2)

A5 = /i tan a AB'=h'tanQ

Now/2tana-A'tan0= constant

Differentiating both sides, we get

p = tan i

i = tan~^ p

da

...(1)

h sec^ a-jz - h' sec^ 0 = 0 aO

From ABCF, BF cos r =

h

t

^

EC ~ BC

BC^

h'

cos^alt/0j cos^ 0 hcos^

t

t

cosr

cos(90°-0

t

•••(2)

sin/

da ••.(3)

cos

a

By snell's law, we have sin I

N

1^^

sma

...(4)

sin 0 = p sm a

Differentiating both sides, we get

sfeC'--'-)

' i J I 1 •

1

\

cos 0 = p COS a

V

da

t

! \

!

#

(da^ _ COS0

in.

...(5)

^dQj pcosa Substituting the value of (daldQ) from equation .(5) in equation-(3), we get

From ABCE,

/icos^0 sin (/ -r) = BC =

h'= BC

cos^a

COS0 Hcosa

hcos^B

...(3)

sin(/ - r)

^— X

...(6)

pcos a

From eqs. (2) and (3), we get sin0

From equation-(4), sin a=

/

X

sin/

sin(/-r)

sin/

sin(2/-90)

sin{/-(90-/)} sin

-cos 2/

/(- cos 2/)

t(2sin^ /-1)

smi

sin I

cos a

=

113/2

X-

•-1

1+ p^ x

=

VaV)_

_ y-i)

cos^ a =

1-

sin^0

(p^-sin^0)^^'

iGeometrical Optics

4611

Substituting thevalue ofcos^ a in equation-(6), we have

H^Acos^G -sin^ 0)3/2

h'=

refraction at water surface the final image isproduced at an apparentdepthh/\i = 30/(4/3) = 22.5cmbelowthe watersurface

orata height 80-22.5 =47.5cm from thepole ofmirror. In second case when thedepth ofwater is40cma^r reflection

(x) Thesituation is shown in figure. Thelength ofshadow of poleABat the bottom of river is given as

from mirror, the reflected ra)« from will produce an image at

From LAPQ, we have

from the water surface and produce image 7^ as shown in in

Xj

tan60° =-v/s meter

...(2)

itsfocal point ata distance 50cm from the mirror pole asshown infigure. But before reaching thefocus, light rays gets refracted figure.

From dtRPM, we have

Xj =/'Mtan r = 2 tan r

•••(3) Water

surface

T C

s u

ly^

p" ^ i/?

-J '

M

1^—*2—•

From Snell's law, at point

•

^*1-

For refraction atwater surface object will be taken as 7j which

we have

sin 60

4

smr

3

3'

isata distance 10cm from thesurface and for this theimage produced will beata distance /i/p= 10/(4/3) = 7.5cm above the water surface or at a height50 +7.5 = 57.5cm from thepoleof

3^

miiTor.

smr= — sin 60 = — 4 8

=>

'

(xii) Figure shows theraydiagram oftheimajge formation as described in thegiven condition. Byusing Snell's law, wehave

cos r= yj{l-sm^ r) =

8

sin 0 = 1 sin ())

Wesubstitute the value of r in equation-(3) X2 = 2 tan r

-'

;(373)/8 V(37)/8

.43

•

"

=sin(j) => ^=53°

...(3) 1

(xO The focal length ofthe concavemirror is lOO/l-SOcm. In &st case when the depth of water is 80 cm the situation'is

J7l y /fc

h'

i

4cm

shown in fi^e. Asthe rays are incident normally on themirrb

from infinity, the image /j is produced by mirror at its focal . »

point i.e. at 50cmfrom the pole of mirror.

^

—

1cm

nV

si

—

-

Here we have

tan ^ = —

...

E u

n

...

rjT7~

-.V"-

-

Solutions ofPRACTICEEXERCISE 5.4 W\vo>—

(i)

We use refraction formula for which we have

The light rays from this image 7, will incident on the water surface at a height of30cm and gets refracted out in air. Due.to

4

v =

oo

j14i= "J; P2==

3

and 7? =+30 cm

Geometrical Optlcsl

1462

Using these values in refraction formula, we have

Here BO = DC so from triangle OBC, we have 2r+8=180

M2

Ml

1^2-Ml

V

u

R

4/3

3/2-4/3

CO

M

30

^

90—

s i n r = sm

=cos —

21

...(1)

2

From figure,total deviationof light ray is given as

1

3u ~~ 180

=>

r = (180-9)/2

Taking sine on both sides of above equation, we get

3/2

4

=>

M=-240 cm

a = ZEBC+ZEDB

=>

a = (/ - r) + (/ - r) = 2(i - r) a

Thus objectis locatedat a distance240 cm fromthe poleof the

ct

0

refracting surface to the left ofit.

(ii)

viewing the bubble.

(O-a)

/ = 90-

Figure shows the situation in which the observer is

Taking sine on both sides of above equation, we get

(8-a)"

90-

sin I = sin

(8-a) sin I =

...(2)

cos

By Snell's law we have 2.5cm sin;

sinr

We use refraction formula

Ml i i V

Substituting the value of sin i and sin r from eqs. (1) and (2),

M2-M1

we get

cos(8-a)/2

R

u

M=

For refraction formula we use

/?= + 5cm;M = + 3cm;p.,= 1.5 and Substituting the values in the formula, we get 3

5

i _ 1_J_ - A V~ 2

10 ~ 10

10 V = — = 2.5 cm 4

Thus the image of the bubble will be formed at a distance of 2.5 cm behind the surface P ofthe sphere.

(ili)

(G-a) cos

2

= }I cos 2

1-1-5

1

or

COS0/2

The ray diagram of the situation is shown in figure.

(iv) Since, the distance of all the points lying on the sphere is constant from the centre, all the angles of refraction are

same. Here we consider 5, is the deviation of the light ray at first refraction, 82 is the deviation of the transmitted ray through partially polished surface and 83 is the deviation of the light ray emerging out of the sphere at final refraction. Figure-xxx showsthe ray diagram of the given situation then according to the given condition, we have from figure 1

(60- r) + (60 - r) = - [(60- r) + (60 - r) + (188 - 2r)]

Here we consider the total deviation of incident ray and emergent ray is a.

5, = 60°-r

V^60°\

|Geometrica[ Optics

-4831

=>

120-2r= j [300-4/-]

=>

360-6/- = 300-4/-

(vO The image formation is shown in figure.The mark considered as object is at point O. First refraction takes place

from the inner surface and the image is produced atOy Now this image acts as the objectfor the second outer surface and secondrefraction takes place from this outer surfaceand final

60=2/=>

r=30°

image isformed at Oj.

Now using Snell's law, we have l.sin 60° = (i.sin/'

S 2-

=>

(v)

—

=

1 ux — ^2

Figureshowsthe refraction ofa parallelbeamon a solid

glass sphere atnear normal incidence. Atfirst refraction image /, is produced whichacts as an object for second refraction at other surface ofsphere at which afterrefraction final image

For the first refraction at inner surface for refraction formula we

is produced.

use,

u=-2R; R =-R; iij= 1 and ^ M2-M1 V,

Wi

R

Substituting the values, we get 1

2R

+iL ^ M-1 V,1

-R

Using refraction formula at first surface we have

1 -+, (M-1)

ii Ml

M1-M2

"i

R

Hereweusep,= l;

-_E5_ (M-1) For the refraction at the second surface, we use

1

(2/2-1)

...(2)

u^=~

R+-

2p/2

(4m-1)/2 (2)1-1)

(2m-1) Now using the refraction fonnula, we have

R = -R and

""cm-i)

(M-1)

For the second surface,

(M-1)

(1-p)

U2

-R

ii

M2 ~Mi

"2

R

m(2|4-1)^J_ ^ 1-M (4n-l)/2

A-

V2

-2R

1

1 m(m-i) _ (1-m) —

ii(2-.|i)

~R

m(m-i) _ (1-1^) '2 -^(2-m) Simplifying, we get

2R

For the refraction at second surface, we use, = ii2 = l; R = -2R and

M2

V2

(2m-1)

2iiR V, =

Substituting the valuesin refraction formula weget

V2

R

...(1)

«i = '» and R = +R

^j = p;

_2R

Vl

R

On solving we get v- = -

1-M 2R

m(2m-1) (4p-l)/!

2/2(4^1-1)

{3n-l)

So the distance of the final image O2 from O is given by equation-(l). The shift of objectis given by

Shift =3/?-^^^"*^ ^^=R (3p-l)

v., =

2(m-1) Shift = R

2(4p-l)'

3--

3p-l

9p-3-8n + 2) (3p-l)

(3m-1) •

Geometrical Optics!

U64

(vii) We consider refractive index of glass to be and after first refraction, image is produced at v then by refraction

We apply refraction formula ^2

M'l

1^2-^1

V

u

R

formula, we use ^•2

1^1

HereweuseM =-10cm; i?=-15cm; Pi=1.5 and

1^2 "1^1

Substituting the values, we have

R

1=90°

Thus,it willhappen at grazingincidence.

^2 _ sin i

(iii) We draw a tangent at any point (x, y) onthetrajectory which makes anangle 0with optical normal parallel toy-axis as

Pi

shown in figure.

sinr

Byusing Snell's law attheinitial point and atthegeneral point

^ _sin45° .1

of the trajectory of light, we have 1. • Sin 90°= p sin 8

sin r 1

smr =

^

^/2^/2

2

=>

. 0^ = — I = , I— sm

li

r = 30°.

...(I)

yjl+y

It shows that the refractive ray thus'becoraes parallel to AD Slope of tangent is inside theblock. Soparallel rayisincident onspherical surface dy

^=tan(90-e)

CD. Now for refraction formula we have

M=co; ii=0.4m; Pi=V2 and ^2=1-514 li2._lh _ P2-1^1 ' V

«

R

dy =>

...(2)

— = cot 0 dx

Fromequation-(1)and (2) weget,

'

= y\/2

Substituting values we get dx

_ LSM-^ V

00

" = iff-h

0.4

0 y-

Solving, we get V= 6.06m =>

0E=6.%m .

=>

0

2^ =x => y=~

Solutions ofPRACTICEE^RCISE 5.5 (i)

(iv) Figure shows thesituation in which from thetopcircular area light will come out in air. The light rays falling on the circumference ofthecircle make anangle equal tocritical angle

Here^p^= 1.66 and ^p^= 1.33 .

14,=

,

a\^g

=

..p,,

1.66

so all the light rays falling on the water surface out of this circulararea will internallyreflected into water.

1.33

sin(^ + 8^)/2

1.66

sin^/2

1.33

We know that

/i

""72 + 5. ^

/

sin

sin36°

'72 + 5^ sin

/

1.33

7A

_ I.66sin36°

~

2

1.66

1.33.

From the figure we have

= 1.248x0.5878 = 0.7336

tan 0^= T

72 + 6.

=47° j r

^

r= 6_ = 22°22'.

h

tan 0,

-

air

water !

Geometrical Opticsj

1466 Area of circlethrough which light comesout from water is

A= tzi^ = ^ nh^ tan^ 0^ wherecriticalangle0^;. is givenas

Squaring and adding equation-(l) and (2), we get ji^ = 1+ sin^ 0

=>

.==sin"' 1-

|j,= Vl +sin^ 0

(b) Maximum value of0 canbe90° for which 02 = 63 = 45° Thus

(v) Figure shows the raydiagram of the lightray incident on the givenprism.Herefromthe givenconditions, we have

=

(c) Ifangle 0isslightly increased, 02 will increase and ©3 will decreasesolight ray emergesout in air and it will no longerbe grazing emergence.

As at point B the light xoyAB is reflectedwe have

Ifangle 0 isslightly decreased, ©2 will decrease and©3 will

Zi = Zr=45°

increase beyound critical angle and it point Q total internal reflection oflight will take place.

A

D J

(vii) For glass critical angle is given as .-1

0 = sm

rA

^1^

\ \ N

.V

0 = sin

-1

= 42°

As incident light is normal to the face, this light incident on other face of prism at 60°which is more than critical angle so light will be internallyreflectedas shownin figure

For internal reflection ofthe light ray at face EF^we use sin />sin 9^> — > —

.4 = 60°

sin 45'

l^n,;n = 1-414 r'mfn

When theprismimmersed in water, then pathofthelightrayis shown in figure-xxx /

A

/

'6^

Water

1 6o\

D J

ER

y/vlZXtl S

Thus angle between incident emergent ray is

yi

/I

6=60°

(viii) p.sin 0 = 1 Xsin 90°

As the incidentray is normalto the prism, it passesimdeviated throughwater and glass at firstrefraction but at point B after refraction light rayentersin water again at angleof refraction r

=:>

sin 0 = —

tan0 =

=

e

which is given by Snell's law as sin 45° = .-xnd

Li= —:

^

sinr

dx

sin 45° sin r

When_v = (i,then

Zr = 48° 35'

(vi)

^

=2

(a) Using Snell's law, we have

sin 0 = p sin Oj p sin ©3 = sin 90° p cos ©2 = 1

—= l-e^^

...(1)

x = 2t/ln2 2d]n2

...(2)

= Vl +e

= >/r+4 =^/5

.-x/2d

f^ometrica] Optics (ix)

467

Forminimumdeviation ofa lightraythrougha trihedral

By Snell's law

prism we use

p sin ^ ~ sin e •

•

and p sin 2/1 = sin r for small angles, we use

sin

K 2

e = pi4

siny4/2

The refi-active index p is a function of A,. Thus a change in wavelength Xcorresponds to a change in refi"active index of

and

r=2p^ = 6.5

Now we have

the material and it also causes change in 5. Therefore, 6 is a functionof p. A differential changein refractiveindexAp.causes

a differential change A6 in 5. If the rate of change of p with

=>

respect to 5 be (d^dS), then we use

=>

3.25-^=125 vl=2"

^

P=Y

13

(xi) . From the given conditionwe use the deviationangle of light ray which passes through the prism is given as 5=1,+ ^2--4 3(F = 60°+ 1,-30'' => i2= 0 or Tj = 0 => r^=A=30°

^ + 5 sm

A6 siny4/2

^ + 5 COS

A^.= -

Now by using Snell's law we have

A5

sini, _ sin60°

sin^/2

sin/;

2sin^/2 A5 =

1/2

sin 30°

(xii) In the given situation as incidence angle is 0° for the prismif light rayincidenton the otherinsidefeceof prismat an

Ap

angle r then we use

1-sin'

r=A = 30°

2

Now by using Snell's law we have 2sin^/2 AS =

sini2

Ap

7(l-p^ sin^ A/2)

sinr2

sini2

2sin60/2[1.520-1.515] AS =

1A =

V{l-1.555sin^(60/2)}

sin i2= 1.5 sin30°

2sin30x(0.005) A5 =

sin 30^

sin i2=1.5x -^=0.75

V{I-l.555sin^30}

1*2 = sin-'(0.75) =48.6=

A5 = 0.5®approx

(x) Using the given conditions the deviation angle oflight after passing through the prism is

The deviation angle of light passing through the prism is given as

5=(ii + i2)-'4 8=^(p-l)=-

...(1)

=>

5=(0+48.6)-30

=>

5= 18.6°

(xiii) Suppose 0 is the angle of incident ray at any point {x,y) on trajectory of light in the medium as shown in the figure. The slope ofthe trajectory at this point is given as dy dx

= tan(90°-e)

dy — = cot 0 dx

.GeometricaT Opticsj

1468

AtA:= 0,y= 0 aiidso'c = 0, .6 -> 90^

(c) At^ = 1,fromequation of trajectoryye have .1/4

A90^-e /01- •

Thus coordinates ofthe upper surface, where ray intersect are t:

(4,1)

90°

(a) By Snell's law at initial point and the general point we

(d) At upper interface, we have p sin€ = l ,

have

p sin 0 ^ 1 sin 90° 1 sin 0 = —

• r 1 sin e = — = 7 = 1 H i

... (i) =>

•

e=90°

Thusthe emergent raywillbe grazingoutfrorn the uppersurfece

cose=^l_sin2

„ ll_J_

ofthe medium".

' "'

Applying lens formula for the given case with • w= -15cm and/=+10cm

Slope of tangent is given as

1

COS0 cot 0 = —: sin0

^

:1. .

1

V -15 ~ 10 • ... 1 V

1

1

3-2

10

15

30

v=+30cm

cot 0 =

1

6=

20cm

(b) From equation (i) we have p sin 0 = 1

=e>

^

Solutions ofPRACTICE EXERCISE 5.6 (i)

cot

"

30cm

IScm

p^ sin^ 0 = 1,

diameter

AB- —:

= 1.5 cm

1

^

= —T~ = cosec^ 0

r = radius ofthe illuminated part •

sin^ 0

= 1 + cot^

UsingsimilartriangleSPQl-rr AABI, wehave L5

{ky^'^ + 1) = 1+

dx

30

1

So, Area ofilluminated part =

=Wi =^ cm^ (u)

ax

As lens power is 5D, its focal length is 11

dy 2IA

/= — = —^0.2m=20cm

= k^'^dx

y

On integrating the above relation, we get

4yl/4=f^l/2^+C

Given condition is

v = 4w

As image is virtual, we use

N

v=-4x

figure. Aswe know toproduce image on object reflected light

by using lens formula, we have

The convex lens and the convex mirror are shown in the

rays retrace the path of incident rays so here for the ray to retrace its path it should be incident normally on the convex mirror,or the rays shouldpass through the center of curvature ofthe mirror.

4x

X

2

^ ;

From the diagram we see that for the lens

' 20

1_

4x ~ 20

=>

^

x = 15cm

(ill) ' For final imageat infinityobject must at focal point of lens i.e., 20 cm. lOcm^

If object is considered to be placed at a distance x from lens

tO 00

then for lens formula we use

u=-x\ /=-i-20cm and v = -t- lO-P 15 = -i-20cm From the lens formula, we get

-25 cm-

As actual object is kept at 25 cm so 5 cm is the shift due to glass slab, given as

V

J

^

Shift =/

/

u

1

1_

25 -X ~ 20

V

=> 5=.r

X = 100 cm

(vi) Figure showthe imageformation when light beamfalls firston the converging lens.In the absence of lens the rays

1.5

t -15 cm.

will be focussed at P" at a distance 20 cm fi^om lens Lj. This point F acts asvirtual source for lens L^. Itsimage isformed at

Civ) ..

P.

Thur for lens formula atlens L^, we use u = + 12cmand/=-30cm

Parallel

.•

substituting in

0,

0.

1-1^1 V u

f

FJF,

.d= 5 mm H 10cm10 cm

fy = 20 cm

we have

i-JV

Bysimilarity ofi^5/and CDI, wehave

^

12

30

v=20cm

Thus therays converge toa point P, 20cm from L^.

AB _ CD

Secondcasewhen parallel beamfalls first on diverginglens is shown in figure.

^

^ _A

20 ~ 10 d

=>

-

d,-— =2.5 mm. ' 2 (a)

^

GeometriMl Opticsj

acts as object for the concave mirror and it is located at a distance 10 cm from the mirror which is the center ofcurvature

of the mirror so again image

is produced at this point only

after refraction as shown in ray diagram.

Further refraction takes place through the lens and we can see

that image Z2 is at 2/point of the lens and its final image is (b)

When a parallel beamfalls on the diverging lensZ,^, the rays

produced at the original object whichis ofsamesizebutinverted as shown.

divergeand appear to come fromF. For lens formula to be used

at converging lensZ,,, we have

(ix) The x-axis is the principal axis for first lens & PQ is the principal axis for second lens. First lens produces image at its focal point at a distance f on x-axis.

m= -(30 +8)=-38 cm and /=+20cm

J

1_

20

38

12-10

9

380

380

Now for second lens, we use lens formula as 380

v=

]_ 1 l_ V -If ~ f

=42.2 cm

Thus the point P in this case is located at a distance 42.2 cm

fromLensZ,,. (vii)

[ _J_ f

Focal length of an equicanvex lens is given as

1_ '^f

v=2/ R

/=

2(11-1)

For lenses ^4 & 5, we use

A

/\

R.

R.

2(H^-1)

2(h5-1)

0

..

Magnification for the second lens is given as

= 1 + 0.7= 1.7

hi

(viii) \{AB is the object placed at a distance of20 cm from the convex lens L as shown in figure then we can calculate the position ofthe final image after first refraction through the lens and then reflection from the concave mirror then again

, V . +2/ • • -

Here negative sign indicates that the image will be formed above the principal axis PQ for the second lens.

Thus final coordinatesofimage are {5f, 2d)

refraction from the lens.

(x) The convex lenses and the plane mirror setup arrangment is shown in figure.To produce final image on object itself, all the reflected rays are to retrace their path or the rays should be incident normally on frieplane mirror.

(♦—«, = 20cm—

V, = 20cm

»)♦—

10cm

As object is at If point, its image will be produced at same

2f= 20cm from die lensas shown in figure-xxx. ThisimageZj

IGeometrical Optics

471]

For light rays tofell normally onmirror theimage Ijproduced by first lens must be located at the focal point of the second lens which is at a distance 10cmfrom it.Fromthe diagram ifwe

when image is real its distance is taken positive and in second case when image is virtual its distance is taken negative so the two magnifications are taken as

use lensformula forlensL^, wehave v = 30-10=+20cm,

m, =

/=+10cm,«=-x From the lens equation we get

h.

+Vi

K

-u.

h

-v-,

16

—

-tin V

u

/

As magnifications are same, we use

J

^

"

1

1_

20

~ 10 16

=4>

(xi)

x==20cm

6V| = 1bvj

For lens formula, we have 11 cm

1

.V

1

w

1

1 -16

...(2)

/

1 =

—

1

/

-6

-V-,

111

^

...(1)

by lens formula we have in the two cases

/=+6cm =>

6

By solving above equations we get, /=llcm Here positive sign indicates that the lens is a convex (or

v^ll ~ 6 i _ i_± V~ 6

1r

converging) lens. 66

(xiiO Wefirst find the image producedby mirror using mirror

66 V =

—

...(3)

/

cm.

formula as

V' u 5 mm

f

We use u = - 12.5cm and/= - 10cm, we get

•II cm

66/5 cm •

/= 6 cm

i.

^

1

vi

-12.5

-10

V, = —50 cm

Magnification by lens is given as

^'1

V 66 6 — = —TT = 7 u 5x11 5

Longitudinal magnification is 36

m,= n^= — 36

Image size =—x 5 mm

V-• u

(-30) (-12.5)

Hereimage formed by the mirror is at a distance of50 cm from the mirror to the left ofit. It is inverted and four times enlarged. Image formed by mirror acts as an object for lens. It is located at a distance of 25.0 cm to the left of lens for the light rays

incident on it from right so using the lais formula with w=-25 cm and/= + 16.7cm, we have

••25

i_i 36

V

Image size = — mm ~12 mm 1

(xii) As per the condition given in the question, magnifications for the two are same in magnitude. Iffirst case

u

/

1

I

-25

+16.7

V2=+50.3cm

Geometrical Optics!

1472

Magnification by lens is given as V

my= - = ^

u

x = d+

50.3 -25

(/i+/2-^)

=-2.012

/i/2+^(/i-^) (/l+/2-^)

overal magnification by the system is

w = Wj X/M2 = 8.048 Thus, the final image is produced at a distance 25.3 cm to the right of the mirror, virtual,upright enlarged8.048 times. Positions ofthe intermediate and final images are shown in figure.

The ^'-co-ordinate of the focus F can be obtained by using magnification by second lens as

_

/2 «

ifi+fi-d)

Forsecond lensthe image produced at F,.by first lens acts as an object which is located at a distance A in jv direction from the principal axis of the second lens. So the deistance offinal image at point P from principal axis of second lens can be given as

•25.0cm-

_A B'

/2A A, =ot.A= ——-—— Ux+fi-d)

M—12.5 cm——12.5 cm-

-25.3 cm-

The y-coordinateof point P thus can be written as A- A, (xiv)

If the second lens is not present, the parallel incident

yp=^-\ = -}

(/i + /2-^)

rays will focus at Fj, at a distance/j from the firstlens on its principal axis. These rays now intercepted by second lens, and fmallly focusat some point P which is obtained bylens formula for second lens. Thus for second lens, we use

and/=+/2

(xv) For refraction through lens, we use M=.+ 20cm,/= -15 cm From lens formula, we have

By lens formula, By lens formula we have

i_l =1 V u

i_l V

u

1

^

/

1

1

1

V (f\-d)

.1

V 20 ~ -15

^ 1

f

v=-60cm

Magnification by lens is given as

fi

V m

=

Solving, we get

fiifi-d)

=>

V

lmagesize= ~ Objectsize =

v =

W+/2-^)

~ u

-60

=-3.6cm

The image formed by lens acts as an object for mirror. It is located (3.6 - 0.6) cm below the principal axis of the mirror and

0.6cmliesabove principal axis. Letimage ofCj5j is C'5', and that of Cj.^, is C!^'. Now we use mirror formula for reflection at concave mirror as

i+i.i V u

The Xcoordinateof the final image at point P is x = d+v

J

f

I_

V -30 ~ +30 v=-15cm

fcsometrical Optic^

473S

Magnificationbymirror is given as C'B',

V

3.0

1 = J__±

4-3

J_

V ~ 15 20

60

60

u

V= + 60 cm

=>

C'B'

n

3.0

30

Thus radius ofcurvature ofthemirror isgiven as -fi = v-5 = 60-5 = 55 cm

CB'=~\.5cm

Solutions ofPRACTICE EXERCISE 5.7

Also we have for same magnification C'A!

V

0.6

u

(i) If object is placed at focus of lens, light rays become parallel and fall normal onplanemirror. So, rays retrace their path.'

C'A!

15 =

0.6

—

(ii)

30

=>

C'A' =-0.3 cm

We use

A'B' = C'B'+ C'A'

=>

A'B' = 1.5 + 0.3 = 1.8 cm

Focal length ofthin concavo convex lens is given as 30x60

fr-

Then 1.5 cmofimage liesabove principal axisand0.3 cmlies belowprincipal axis

(fi-l)(i?,-/?2)

1/2x30

^2" 120 cm focal lengthofmirror is givenas . 30 /a/= Y =15cm.

Equivelent focal length oflens mirror combination isgiven as

fe, "///az "120''i5 ^ A =12cm Thus for image tobe produced on object, it should be kept at

30cm

a distnce

(xvO Figure shows optical setup asdescribed in thequestion. K

5cm

P

(iii)

= 24 cm.'

(a) For thecombination oftwo lenses and aplane mirror,

the equivalent focal length is given as

A,

1 —

feeq

fix

+ —

—

fll

fu

Focal length of the piano concave lens is f*

R

« = 20cm

A.= Pi-1

Hereconvexmirror is separatedat a distanceof 5 cm from the lens. An object O is placed at a distance 20 cm from the

30

1;5-1

= 60 cm

Focal length of the piano convex lens is

lens. The ra)^after refraction through lens form animage at C ^

sothat the light rays fallnormallyonthe mirror andreflected

R

30

Pi-1

1.25-1

= 120 cm

rays will retrace the patli ofincident rays to produce image on Focal length ofplane mirror isf^= oo the object position.

'

For refraction at lense we use

.•

M=-20cm and/=+15cm By lens formula, we have

Equivalent focal length ofthe combination is given as 1

feq

2U—+— 60 120

^

60

= - 60 cm

Asequivalent focal length is negative, thusnatureofequivalent

V « ~/

mirror diverging or convex.

_ Geg^Qtrica] Opticsj

!474

(b) For the equivalent mirror, we can apply the mirror formula

/, =25cm _^ = 20cm

to find the final position of the image. Thus we use

'1 i=V

u

ft

1' 20cm

_1_

1 J_

v"^-15

^1•

60 v= + 12cmi

Magnification oflens

=

Magnification of lens

=

Vi

100 20

Thusmagnification bythecombination is m

(iv)

=

V_

_4

u~

-15 ~ 5

If/^^ is the focal length ofcombination, we use M= -12.5 cm

For real image

V2

220/9

"2

110

2

2

10

Total magnification = 5 x - =

v = mw = + 50 cm

(vi)

1 V

1

12.5

feq

_1

1_

=2cm.

1 1 , i11'11.1'

~ 10

=>

Indisplacement method experimeit, as explained inarticle

5.18.2, the size ofobject is given as

feq

u

J_ J_ 50

^ -5

(vii) Suppose at a height h of the liquid in the beaker, an incident rayfalling from source totheedge ofthedisc incident atcritical angle soit gets refracted atgrazing manner asshown in figure. Any other lightrayfalling onwater surface will be totally internally reflected since theangle ofincidence would begreater than critical angle. This inthefigure shown, we use

:1h11 1 1i'1!, (

/^ =10cni

Equivelent powerofthe lens system is

•^^5" 0.1 optical power of each lens is

^.,.h=50;..(1)

tan 0^= 7" ^

(v) Figure-xxx shows the raydiagram for image formation. For lens we use Mj = - 20cmand/j = + 25cm

^

Vi ~ 25 20 ~ 100

h

100

Vj =-100 cm

So the lensZj forms avirtual image,5 jat100 cm behind the lens. For lens1-2, this image acts as areal object for which we use

-

M2=—(100+ 10)= —110cm, and_^=+20cm

1 _J

^

1

11-2

220

V2 =+-^ =+24.4cm 24.4 cm from lensZ-2

^

T"—~

9

^ ~ 20 110 ~ 220 ~ 220

Thus final image ^2^2

-

By Snell's law we have .^13 ji 5

sin0,., = — =T

-(2)

real and is formed at a distance cos

...(3)

[Geometrical Optics 3/5 _

~ 6 1

=>

3

h~ A =>

/eq =6cm.

...(a)

With respect to air, focal length oflens I3is given as

^ = 1.33 cm

/3 =

(viii) Figure below shows the given situation and for solving the above problem.

=>

R

8

^i-l

4/3-1

.^=24cm.

With respect to air, focal length oflens L, is given as R

/4=:

2(p-l) 2(1.5-1) -.hi

In figure we consider whole system may be considered to

consist oftwo equivalent lens combinations separated by a fix

8

^

f4=-6.67cm.

Equivalent focal length oflenses ^,3 and £4 isgiven as

distance in air.

First combination isontheleft end oftheoptical setup (0 double convex lens (glass) Lj

1

1

/eq2

24

1.2

(ii) plano-concave lens (water)

The aboye two lenses form one group of lens, say of focal

length/^^j. Second combination is ontherightendofthe optical setup (iii) piano convexlens (water)

120

4=-77 cm

Lens combinations of which focal lengths are given by equations-(a) and (b) are separated by air distance 3 cm so combined focal length F of the system is given by

(iv) double concave lens (glass)

The above two lenses form one group of lens, say of focal

The above two groups are separated byparallel plane water slab of thickness 4 cm. The apparent thickness of water slab

will be4/(1 =4/(4/3) = 3cm. Thus theabove two groups can be consideredto be separated by air of thickness 3 cm.

to air is given as

2(1.5-1) ...(i)

Now with respect toair, focal length"oflens Z,2 isgiven as R

fi - ^-1

4/3-1

=> ^=-12cm. ...(ii) Equivalent focal length ofthis combinationis given as

J_-i J_

F

1

1

^

feq\

feql

feqxfeql

13

120

+

6x(120/13)

120-78 + 39

F=8.9 cm.

(ix) The plane mirror forms avirtual image ata distance {a+ b) behind it.Given thatthereTs noparallex between the images formed bymirror and lens and soboth theimages are formed at

R

y;=4cm. -

V

1

6x120

Bylens makers formula, the focal-length oflens I,with respect

fx =

...(b)

a distance {a + b) behind mirror. The distance of the image from lens v={a+b)+b={a + 2b). Herethe lateralmagnificaticai produced bythe lens is 2 because

it is given that thetransverse length of final image formed by lens is twice that of image formed by mirror and the mirror forms an imageof samelengthof the object. As the distance ofthe objectfrom lens is a, hence distance of image from lens should be equal to 2a.

Geometrical Optics

=> =>

{a+2b) = 2a b = al2

•• .(1)

The silvered lens may be regarded as a combination of

fea

fl

fe,

30

fu

30 7.5

equiconvex lens anda concave mirror incontact. Thus for an equiconvex lens, focal length is given as R

=>

/i = 2(11-1)

4 =5cm

Using mirror formula, we have

/, =40cm

M = ~ 40 cm

f=-S cm eq

For concave mirror, focal length is given as R

-40 X+5

eq

f = — •//n 2 =-20 cm

V =

+35

U-feeq

The combined focal lengthof the lensmirror system is given 40

as

v=

cm

1-A J_-J_ J-

F~

^

fm ~ 2o"'20

F'=-10cm

(xi)

Fortwo thin lenses keptin contact the equivalent focal

length of the combination is given as

For the combination for mirror formula, we have M= -a, v = + 2a and

7^=-10 cm.

1 -i,

V

u

1

are given as

, R

F

/i =

1

1

i-a)

-10

...(1)

F~

Whenlenses areimmerged inwater, theirindividual focal lengths

Using the mirror formula in this case,

i

1=— —

iiL-i

•+•

2a

R

a=5cm

/.=

i) = a/2 = 5/2 = 2.5 cm

and

(x)

and

i^_l

The optical setup is shown in figure-xxx. For a piano

convex lens the focal length ofthe lens is given as

1

_1_

1^1-M2

R

R:=A^-A)

>v2(Hi -fc) R=30

1-1

.2

J

= 15 cm

4

10

3 2(1.7-1.5)

, 40/7 cm

(xil

= 33.3 cm

If the distanceof the lens from the object be L when a

real image is formed on the screen. Then by lens formula we have 1

1

I

100-X

-L

i? = 15 cm •40cm-

7.5 cm

If/^ is equivalent focal length oflens-mirror combination then we use

• 23

On solving, we get

L=(50± 10>/2)cm

f^ometricai,6ptics

4771

Now, aslens isexecuting SHM and areal image isformed after If image

is obtained when the lens passes through two positions at same distance fi"om the mean position and having separated

istheemerging angle for blue light, "we use Snell's law at

other surface of prism, weget sine

bya timegap of on fourth ofthe period ofSHM;

smr-,

Thus phase difference between thetwo positions ofreal image

sm h

sin27.4° %

formation must be —. As the two position are symmetrically

located about the mean position, phase.difference ofany of

thesepositions fi-om origin must be j. IfAis the amplitude of SHM then we have

•

e=50.6°

Deviation angle for red light when itpasses through prism is 6. = 65® + 50.6°-60® = 55.6°

...(i)

For red light, by Snell's law we have

J .

sines® 1.6S = smn

=>

'

' 10V2cm =Asin'— ^ •4

=>

•

-4 =20 cm

=>

r,"=33.3® -

=>

r2=.4-r,=26.7®

If

Velocity of lens at mean positionin this case is

-

. .

'

is theemerging angle for red light, we useSnell's law at

othersurface of prism, weget

v^ =A(o=aJ~ ..

-

•

sme

\m

1.6S=sin26.7'

Required impulseto attain this speedis J-mvQ=A^Km =8kgm/s NOTE : Students can alsothink that realimagesareobtained on screen at two extreme positions oflens in SHM wh eh time

gap becomes halfofthetimeperiod !!!

• ^

6^=47.8°

Deviation angle for redlightwhen itpasses through prism is =

As we know that for achromatic combination we use

(iii)

If theangles ofcrown andflint prisms betaken asA and

^'then thedispersive power ofprism materials isgiven by

D=D'

=i'

. co =

• (i.42'^1.35)4°

-

A'=

^

5^-5^ =2;8®

Solutions ofPRACTICEEXERCISE 5.8 (i)

+

=> 5^ =65® +47.8®-60° =52.8® . ...(ii) From Equations-(i) and(ii) we getangular dispersion as

:

^—^=140

1:9-1.7

P-I.-

'

Net deviation oflight for achromatic combination ofprism is Angular dispersion produced bytheprism is given by given as

,

=>

. 0i)

-.(m-1) (pA ^« = (1.40-l)4°~(r.8-l)1.4® . 5^^= 0.48-,, ;

So the angular disperrions produced bycrown and flint prisms respectively will be(p- 1)(q^ and (p'- 1) to'^'for achromatic combination

(p-l)oi4 = (p'-

For blue light, by Snell's law, we have ^ p=

^

sin/,'I

=>

-

" =0.50 Totaldeviationproduced is (6-5% given as 5-5'=(p-1>4-(p'-1>4'

sines'' 1.68= . , smq. -

=>

:.

r, - 32.6°

0.517xQ..03, ,

.4 .(p'.-l)®'^ 0.621x0.05

Sinn

. >

'4'

•

Other refi-action angle oflightray at second surface ofprism is given as

r2=^-n,=27.4''

r =0.517^-0.6ll2'Solving equations-( l^and(2), we get

_

^ = 4.8®and^'=2.4®

...(1)

' ^

•

:..(2) "

.

Geometrical Qpticsi

(iv)

The deviation produced by thecrown prism is

sin49°12'

0.7570

^

6=(p-lM

and that produced bythe flint glass prism is

=>

r,=29°39'

and other refraction angle is given as

The prisms are placed with their angles inverted with respect

r2 = 60°-29°39'

to each other.The deviationsare also in the oppositedirections. Thus, the total deviation of light ray is

r2=30°2r

BySnelPs law atother surface ofprism, wehave sine

(a) If the net deviation for the meanray is zero,

= 1.53 smr2

sine

-

^

(b) Theangular dispersion produced bythecrown prism is:

=1 53

sin30°21'

=>

sin e = sin 30°2rx 1.53 =0.7732

=>

e = sin'-'(0.7732) = 50°38'

and that by the flint prism is.

(c) Angular width = angle ofemergence for red - angle of Thenet angular dispersion is,(p^,- (p'^,-p^M =(1.523-1.514) x5°-(1.632-1.613)x4.2° =-0.0348°.

emergence for violet = 50°38'-49°12' = 1°26'

The angular dispersion is 0.0348°.

(v)

Thecondition for thecombination to beachromatic is

(vii)

For achromatism oftwo lenses incontact, thecondition

is

given as 03i

(p^-p^-)^ = (/«;-/ic')^'

CO,

(1.523-1.515) X20°= (1.664- 1.644)a' 0.008 a-

^ fi

X20° = 8°

0.02

CO,

(vi) (a) Asit is given thatinstrument is settogive minimum deviation for red light, therefractive index ofthematerial for

The dispersive power of crown glass is given by Fr If

red light is given as

I

f A+ d, sm

...(1)

y;

"—

\^R

P,,+P;j

60 + 8. sm

1.55-1.53 sinSO-^

sinyl/2

sm

60 +5^) =1.514x^=0.7570

For the equiconvexlens, its focal length is given as R

2

fr 2(Py-l) feo+s.

^ I =sin-'(0.7570)=4$°12 5^= 2(49°120-6O = 38°24'

and

'/=49°12'

sin49°12'

smri

sm/1

0.54 _

2(1.54-1) ~2x0.54 ~

If the combination has a focal length F, then we use

(b) While considering the refraction for violet colour through the prism,we useat first surface of prism sin/i

0.54

^

= 1.53

J_ -i_ J_

^ ~/i''/2

f2~Ffv

[Geometrical Optics

4791

J

^

1_

(ix)

_2_

/2 " 1.5 0.5

1.5

(a) Condition for achromatic combination isgiven as Or

From equation-(l), we use

Oo

—+— =0

fx

0)2

0.5

0.037

0.75

h fi

0.5 0.75

(0,

\ fx j

X 0.037

(-30)

0)2=0.055

(0.18)=0.27

(+20) (viii) Condition for minimum chromatic aberration is that the

distance between lenses = mean of the focal lengths of the

(b) Equivalent focal length of lens doublet can begiven as J

component lenses 0.06 + 0.04

^

1__J_

F~

=0.05 m

^

20 30

F=+60cm.

So for thedesired eye-piece thelenses must beplaced co-axially 0.05 m apart. Thefirst lens(0.06 m focal length) willbethefield

Solutions ofPRACTICE EXERCISE 5.9

lens and the other the eye-lens.

A=/, +/2- Jc = 0.06 + 0.04 ^ 0.05 = 0.05 m fjy

(!)

We aregiven thatthefocal length ofthelens is/= 10cm

and distance between the screen and lens is v = 500 cm.

0.06x0.04

Using the lens formula, we have

i-1 =1

Magnifying power for normal vision D

V H

0.25

/

1 =>

~ F~ 0.048

—

1

The distance of the first principal plane fi*om the field lens towards the eye-lens is

V f

_1_ _J

^

1__

M~ 500 10

X ^ 0.05 a=—71 =7777 x0.06 = 0.06m A

I

=

U

500

500

0.05

»=-—cm

The first principal focusis at a distanceof0.048 m fromthe first principal plane towards the object. Withreference to the field lens, it is at a distanceof(0.06- 0.048)= 0.012m on the other side. Hence it is a negative eye-piece.

Nowthe linear magnification is given by Size of Image

v

Size of Object ~ u

A=/;+7^-x=0.06 + 0.04-0.05 = 0.05 m /1/2 F=

A

0.05

= 0.048m

Magnifying power for normal vision

a: ^

u

500

x 2 = 98 cm

=> Sizeoftheimage = 98cmx98cm

= 5.2 P

Thedistance ofthefirstprincipal planefrom thefirstlens (field lens) towards the eye-lens is given by a= —fx =

500x49

Suppose the illuminating power ofthe source beP, then

D _ 0.28 F ~ 0.048

V

=> Image size = —x Objectsize =

0.06x0.04

0.05

Illumination ofslide = 2x2

Illumination ofpicutre = 98x98

x0.06 = 0.06m

The first principal focus is at a distance 0.048 m from the first principal plane towards the object. With reference to the field lens it is at a distance of(0.06 - 0.048) = 0.012 m on the other side. Hence it is a negative eye-piece.

Illumination of slide

P

98x98

Illumination of picture

2x2

P

(ii)

= 2401.

For the objective, we use

w^=-0.31an arid^ = 0.3cm

G eom etricai "0 ptics j

When the distance is increasedby 2 cm, then new value of v^

Using lens formula, we have

will become

1_

56-68

fo

V '= — + 2 = —. • ® 6 . . 6

I L_^_L v„ -0.31 " 0.3

^ =>

now magnificationby objective will be

v^= + 9.3cm

(68/6) •

62

For the eyepiece, we use

Vg=-25cm and^=+5cm

As magnification'by eyepiece will remain same, total

Using lens formula, we have

magnification will nowbe

My.=MQ'xM^=

' _i

i_ _ r_ .

-25 u, ~ 5 ^ "w^=—25/6cm=~4.166^

^)

h^agnifying power of telescope is given as vo I -

M= ^

0.3U

"o V fey

Using lens formula for objective, we hpe

' ••

lL(i+21\=m

D

1+ -

'"X-J

- Vq

1.1

Vg = - 25cmandf^= 5cm ^

Using lens formula for eyepiece, we have _i_ fe

'1

+0.25

•

1.

_ _

.-0.25 -u^ ~ +0.025

For the eyepiece, we use

r

!_

-5

=> Vq=-0.2632 m Nowusing lens formula for eyepiece, wehave

5

Separation between lenses = VQ + Mg= 9.3 +4.166- 13.466cm; (Hi)

62

•

1

'

^4 + 40 = 44

Mg=0;0227m Tube length ofthetelescope is= 0.2632 + 0.0227 = 0.2859m

1

r

6

-25

5

25

magnifying power =

25

(v)

Vq

0.2632

•" o.0227 " ^^

^en thetelescope is focussed ona distant object and

the final imageis alsoat infinity, the distance between lenses Magnification of eyepiece is = 50+5 = 55 cm

V,

M = —

^

25

=

Ue

When it is turned oh a nearer object,the image formedby the

= 6

(25/6)

objective will notbeatitsfocus, but a little away fi-om it. the eye-piece has to be shifted exactly by the same distance •through, which the image is shifted from focus of objective

The magnification ofmicroscope isgiven as

becausethe fmalimageis at the sarnedistance.withougt change in accomodation of eye. Figurebelow shows the ray diagram

50 = 6 XM. 50

Afn = ~~ = Magnificationof objective The magnifying power ofobjective is given by . * r •

^

VO ' "o

Q

;T>*^

VO —Jo f Jo

of this situation.:

p'

p -

^offi~ ^0 "fo

-

•

Using lens formula for the objective, we have

Vo=/o-fSO - +1."l = 56

- ^

. I Vo

1 I_ -10 " +0.5

[Geometrical Optics

4811

50

M

Vq= +—m=+52.63 cm 95

«0

The distance bywhich the eye-piece is to be shifted is, 5 = 52.63-50 ^

28

29

3

Mo

4

112

•

(^0)2 =-^(^1

5 =2.63 cm

When the object is at a finite but large distance then the Magnifying power of telescope is given as

If in the first arrangement, the distance of the object from eyepiece bex. Thenby lens formula for eyepiece, wehave 111

a

...(3)

75

u.

For the case of focussing the telescope in both cases, using

/ ^ 75 485 (Vo),=20-= —cm.

lens formula for eye-piece, we have

J_

_ J_

cc -Ug

^

112 and

+5

485

(Vo)2=TT'

,^= —=31cm

and

^=90 cm 111 = ~r and u =3cm -u^ +3 ®

CO

...(1)

.3

Then

Similarly in second case

Vn=90 + 3 = 93cm "^0

J +93

(yoh 4

«0

1_ ~«o

1 +90

1 90

^ Given that

Mn

4

...(2)

93

90x93

m^=30 x 93= 2790cm=27.9m 93

Magnification = — = — =31. M.

3

Geometrical Opticsj

r482

Solutions ofCONCEPTUAL MCQSSingle Option Correct Sol. 1 (A) In first case, the apparent depth of the liquid = {b-a) Real depth = p(6 - d) In second case, apparent depth = {d-c) ^ • Real depth = |a{i/-c) The difference in depth ofliquid = \x{d-c)-\3i{b-d) Here we use - d)

fromexperimentaldata, the difference is equal to {d- b) Hence

ix{d- c)-\i(b~d) = d-b d-b

{a + d-c-b)

'71

Sol. 2 (C)

Vo = v, = ca4

+v]-2vq COS120' Fromthe similartriangle PQ

=M\M -

andAZY", 3L

xxrT

Sol.3 (B) IfOis the centre and ^ is the point source for the From the similar triangle, MSand P/g light ray incidentat any pointP on surface ofsphere to angle

have maximum angle ofincidence/, then situation ofmaximum i is shown in figure-xxx. It happens when OA and AP are perpendicular, for which we have

Yy ~~L Then

^ PS~9L

RP-i-QS = RS-PQ = 9L-3L^6L 6L Time = —

• -1 • -1 ^ r• = sm'—=sm'-

u

Sol. 5 (A) Here, P0 is the length of reflected light from the mirror. From the similar triangles, PMA and MXS' PA

^

3H

XS' ' H

Then, PA=3(XS') = 3H From the similar triangles QNA andXNS' 2H

Sol.4 (C) Whenthe manMj is in theregion RP andSQ, then man M, seesthe imageof M2.

XS'

2H

jGeometrical Optics 483-

QA=H PQ==PA-QA^ 3H-H= 2H

Sol. 9 (A) Using refraction formula for spherical surface taking '5' as object, if its image to be at oo, we have 00

{-~2R)

-R

^=2 Sol. 10 (C) If we draw an incident rayalong thetop side of rectangular strip,which happens to be parallel tothe principal axis. After reflection this ray passes through focus. Thus image 77777777/P^.

ofall points on the top surface ofthe strip O,, Oy.. .etc lie on this reflected ray at locations 7,, ly ...etc in between focus andcentre ofcurvature. Thus theimage ofthis strip isa triangle as shown in figure:

Sol. 6 (D) Focallengthoflens does notdepend ontheaperture of lens. It depends on the refractive index of lens and that of surrounding and the radius of curvature of its surfaces. For intensity of image, we have I oc (aperture diameter)^ blocked

oc

—I and/pccr^2

Sol. 11 (C) If we consider two point objects O, and

as

shown in figure-xxx on the two sides ofthe center ofcurvature

Cofthe surface. Figure also shows the incident rays from O, and Oj at point P and as the other medium is denser both

Hence,

incident rays bend towards normal and thus thecorresponding refracted rays will bediverging hence inboth cases theimage

7.blocked

Final image intensity

produceddue to refraction will alwsybe virtual.

^original Allocked

normal

7=/-^ V A 4 =^4

« = 4/3

Sol.7 (D) Ifmirror is turned,aboutan axis perpendicular to

plane ofmirror, then there will benochange inincident angle andreflected angle soangle between incident &. reflected rays after rotation will be same as before.

1and2 is Cj andbetween 1and3 is Cj

Sol. 12 (D) Slab only shifts the image of point P by some distance which will remain constant and does not depend upon the location of slab so the final image formed by slab has a

mu sin • c,- — ^2 and, sm . c-= — 1^3 Then

fixed separation from'O' and will not move.

From TIR atsecond interface 90 - Cj > c^. taking sine ofboth

Sol. 13 (D) Weusev = c/«. For air, we consider the index of refraction =1. The SnelTslaw is given bythe equation :

Sol. 8 (B) Let the critical angle of interface between media

sides, we get

cos Cj > sinc^ or

11-1^1 1^1

sin0, = «sin02 •••(!) We see that for the normal component of velocity to be constant,

Ml

V, COS0, =

COS02

C COS0,= - COS0.

...(2)

Geometrical Optics^

1484

0

Qcomes at lowest position at time

Air

travelling a distance

rj%

K.

Kj

downwards. In time —, which is also the time period of 2

P, the block P comeback to original position,-so we use 2mg

Thedistance between Q and its imageis

Glass

Wemultiply Equations-(l) and(2),obtaining sin0,'1 ""'""'I COS0, = ""••^2 sin0T cos02 sin20,'1 = sin "^1 20.

The solution 0j =0^does not satisfy Equation-(l) and must be rejected. The other solution is 20 =180°-20,

K

Amg

^2-

K

Sol. 17 (C) When the object moves from infinity tocentre of curvature, the distance between object and image reduces from infinity to zero. When As the object moves from centre of curvature-to focus, the distance between object and image increases from zero to infinity. When the object moves from

focus topole, thedistance between object and its image reduces from infinity to zero. Hence the distance between object and its imageshall be 40 cm three times.

02 = 90-01 Sol. 18 (B) In thesituation given in questiqn, wecanseethat

Then Bquation-(l) gives

afterthird reflectionthe reflectedray becomesparallel to mirror

sin 0, = « COS01,

M2 after which no more reflections will take place. Thus light

tan 0 ,=«

raycanreflect maximum threetimes in this case.

= tan~'«.

SoI.14 (A) Theimage7/ofobject'0'formedbyplanemirror moves towards left. Here I^ acts areal object for concave mirror. As 7 moves towards left, its image formed by concave mirror whether real or virtual will alsway move towards right.

Sol. 19 (A) Focal length isminimum incase I,therefore power is maximum. In cases given in options (C) and (D) the focal length and power remain same.

Sol. 20 (A) Ifdistanceofimageisvfromlens,weuse f

2 2 1111 Sol. 15 (A) To see head, thelight rayfrom head after reflection - = — + -r + ^ + ^ f f f shouldcome to his eyes a shown in figure. In this situation for V^i R. not to seethe top ofhis headthe pointfrom which theselight Herewe usedthat ray of light passes through lens thrice and rays aregetting reflected, we should create ahole atthispoint. reflected twice from the two spherical surfaces of lens Thusthis point is located 167cm from the bottom layer. ^13 cm

13(11 V-

By lens maker'sformula, we have 167 cm

_i_ j_ R-)

1 V

Sol. 16 (B) Both blocks loose contact immediatelyafter the release as the springs are different and net forces on the two blocksare alsonot equal.The time periodofoscillationsof the

V-

two blocks is given as

7- = 2jiJ—,7^= InJ— 'Uk 2 \K Tq =2T,

3p-l

7

Sol. 21 (A) As we knowin directionnormal to mirror we use

^Image

^Object

[GeornetriCcd Optics

485j

No effect of Vcos 0 as its in the plane of mirror =>

V = 2VsmB

Sol. 22 (D) Fornoraytoemergeoutofside/7?, weuse

Sol. 27 (C) The acceleration of the elevator does not affect the apparent depth. Sol. 28 (C) For first case, we use

^>26.

i_+J_ _ J_

A

V]

sin — > sin 2

. A

sm —

...(1)

/

and for second case, we use

^/3

> —

2

...(2)

—+ ~

2

V2

7

U2

A > 120®

/

Sol. 23 (B) As AB is common and O^B = BJ and triangles AOj BA and ABAl are congruentso V] V2 Bysymmetry is normal to to O2 and AO^AB = ZBAI But — = —, thus we use

/-«2

w,

/ _ / «!-/ /-«2 Wi +W,

and

Z5^C=90®

Sol. 29 (C) In Pa biconcave lens of p = 1.33 is formed. In we have a combination of two convex and one plao-concave lenses but overall the converging power is more so both cases

Sol.24 (B) The line joiningOand/crossesprincipalaxisof the mirror at centre ofcurvature. In a convex mirror, for a real

are convergent.

object image is always virtual.

Sol. 30 (D) Foriraageofcenterofmassofyf andP, wehave WjVi

Hence the given situation can be achieved only by using a concave mirror with object placed a distance 'o' above centre of curvature C on principal axis as shown.

(-0

/W| + W2 W2V2 V

(+0

=

W] + W2

Sol. 25 (B) Figure shows the image formation in focal plane. In AOAB, we use

miVi

mlvl

(w,+W2)

(mi+mjY

2

2

v„J =

tan0 = 2

2

2

2

W| V, + /W2 Vj + m2

Areaoc /

Sol. 31 (D) For the image of mango when it is located at a distance x fi-om the water surface, apprent distance of mango appear to tortoise is, app

d^xapp dt'

Sol. 26 (D) The optic axis ofa lens does not change even on cutting it so in this case as magnification is 2, image ofPwill be produced 1 cm below optic axis of lens which is 1.5 cm below lineAT.

r*'

d^x

^ dt^

^app = ^ Thus the acceleration offalling mango with respect to tortoise is given as relative

Geometrical Optlcsj

486

Sol.32 (D) Maximum separation will be 4A when A is the

for total internal reflection at top face weuse

amplitude which is given as

02>9c

4© = v

COS0, = sin 0,= — Vm

...(2)

P2

A

Eliminationof 0, between (1) and (2), we get

r"

Therefore the maximum separation is 4v

sin 0, =

.

1

sinr

p

-1.

^1

Sol. 35 (A) Aray4B is incident onmirror OMj at angle a and is reflected alongBCsuffering a deviation 6j = ZFBC

Sol. 33 (B) BySnell'slaw, wehave sin i

1^2

C

In triangles A4 CD and ACEF

Theray5C falls onmirror DMj at anangle ofincidence p andis reflected along CD suffering another deviation

X

+4h'

5;= ZGCD

...(i)

2h —x

Thetotaldeviation is 5 = 5j+ §2

4{2h-xf+}?

It is clear from the diagram that In A6CG, wehave

' 8, =180^-2aand52= 180*^-2p

2h-x sin r =

4i2h-xf+h^

4h^ +x^

...Cii)

=>

5=6i + 52 = 360°-2(a+P)

Now, in triangle OBC, we can use We also have by similarity oftriangles a +p=0 h

2h—x

• h

...(iii)

Solving equation-(i), (ii) and (iii), we get

mirror.

Sol.34 (A) Consider Points on which bySnell'slaw, wehave

and angle of incidence on top face is

02= 9O''-0^

5=360^-20

Which is independent of the angle of incidence a at the first

fj

PjSin 0j = PjSin02

=>

•••(!)

Sol.36 (A) Let 0 be the angle between the twomirrors OMj and OM2. The incident ray AB is parallelto mirror OM2 and strikesthe mirror GM, at an angleofincidence equalto a. From figure we have

ZMjBA = ZOBC= M^OM^ = 0

;Geometrical Optics

4871

Focal plane of lens

As the lens oscillates. The image shifts on jc-axis in between F c

and P.

Thus distancebetweentwo extremeposition of the oscillating

Similarly for reflection atmirror OMj, we have

image is Ax = PF=

ZM^CD = ZBCO= ZMpM^ = 0

COS0

~f=/(sec0 - 1)

^

Now in triangle OBC, 30 = 180®, therefore 0 ^ 60°. Sol. 41 (B) We use magnification

Sol. 37 (A) There is no effect on the spot. Rotating the glass slab will shift a ray parallel to axis. The direction ofa ray before and after the glass slide is unaffected. All parallel rays are focussed at the focal point of the lens according to ray optics, and there is no effect on the focussed spot. Sol. 38 (A) The slab does not object for minimum deviation by prism, thus for minimum deviation, we use

T] =

=30° as shown in figure

w

/ w- /

/ distance of object from focus

x'

Sol. 42 (D) As we know that a light ray after successive reflections from two mutually perpendicular mirrors reverses its direction. Thus as the finally reflected ray is having same slope as that of the given incident light ray so only option (D) can be the correct answer for a given set of mirrors. Sol. 43 (A) Ifthe angle ofemergence from the first prism is 'e' calculating step by step by Snell's law, we get 2

e = sin"'

=> sm/= V2sin30° or i = 45°

Then for net deviation to be double, the incident ray on side A' B' ofsecond prism should make angles / or e with normal.

Thus minimum deviation - 2/-A = 90® - 60° = 30®

sin(= V2sin30° or / = 45® Sol. 39 (B) As mirror is moving parallel to itself, it does not affect the position of image so length of spot on ground will remain same. This can also be seen bysimilar triangles in figure.

Thus the angle between the given situation be 2e or i + e.

Solutions ofNUMERICAL MCQS Single Options Correct Sol. 1 (A) We know that for a prism deviation angle is given

Sol. 40 (C) When the lens is tilted by 0, the image is formed at the intersection point P of focal plane oflens in tilted position andx-axis.

as

,

-

b=i+e-A =>

=>

45° = / + e-60°

"

/+ e =105°

...(1)

Geometrical Optics]

Sol. 5 (B) From the figure in triangle QRS, we have

As per given condition, we have (2)

i-e=20°

9O-; + 0 + 2r=18O°

Solving. (1) and (2), we get

'9O + /-0^

i=60°30'ande = 42®30' r =

Now

^

sin;

sine

sinrj

sin(60°-r,)

p = 180-(90-r)-2;

sin / (sin 60° cosrj - cos60° sin /-j)

/7= 180-90 +

= sin e sin

=>

^90+1-0

-2/

0.8870(0.866 cos r,-0.5 sinTj) = 0.6756 sinr.

Solving we get tan r, =

0.8870x0.866 90° - 0\

1.1191

rj=34°28' sin/j

0.8870

sinn

0.5659

90° - j

= 1.567

777777777777777777ZC

P

\ R 90°-r

18O-4/ + 9O+./-0

Sol. 2 (Q

I

. ^1^

-3m-

5 „

•

f'270-3/-e'l

f 3-*

3-x

X

J

X

6-2x

Distancebetween images 7, and I2is given as 7,72 72 = 6-: 6-2x=4m a: = 1 m

m=;?-0=18O-(9O-z)-2z r27Q-3i-0'

I

2

-0 =90-z

Substituting 0 = 20°, we get Sol. 3 (B) After projection of the particle its velocity component in vertical direction is given as

Uy = yl2 sin 45° =1m/s

z=30°

S0I.6 (D) In the figure iWis the length ofimagehe seesin

In vertical direction if we consider i'jas the displacement of the mirror. From the similar triangles EAB and EMN, we have particleand Sjthe displacement ofmirrorin timet thenweuse

5, =(1) (0.5)-

A

=0.5~^gP-

rr— :^0.3m

1.5 m

and

0.8 m

Vertical distance ofparticle from mirror is given as 52 = 0.5m j= Thus, distance between particle and its image is As = 2s = lm.

•

i

N

AB _ 0.3 MN ~ I As AflV= 2.45, we have

Sol. 4 (B) As we know that along the normal the velocity of image is double the velocity of mirror and opposite to the velocity ofimage and along the surface ofmirror image velocity components are same as that ofobject, so velocity ofimage is

given as (3z -h 47-i-11^).

I=0.6m

Sol. 7 (C) Here, PQ is the size of spot formed on the wall. From the similar triangles AP5'0 and AMS'N, we have

489f

Using refraction formula, we have Q wall

M2-K1 V

u

R

u

w/////y/'N

/ scuj''

3v

R

=>

-R v=+2R

Thus theimage ofchild's nose will appear ata distance 3^ from the center of the bowl. The ray diagram is shown in below figure.

PQ MN

=>

d

Pe = 2x8-I6cm.

Sol. 8 (C) Using refraction formula for air-glass interface, we use

'u=-x; /? =+10cm; jXj^l and ii2 =3/2 By refraction formula, we use

Sol. 11 (B) Final image coincides with the object when the V

u

}h_ 00

•-

image produced by lens is formed at centre of curvature of

1

1.5-1

-X

10

x =

mirror oritselfonthe pole ofmirror. So there are two possible conditions here.

-20 cm

For the lens, we use u = -15cm and /= +10cm so in lens

As the second surface is flat, rays must become parallel after formula, we use

first refraction only asfi-om flat surface rays will not suffer any deviation when fallsnormally.

Vi

M

/

Sol. 9 (D) Using lensformula wecansee thatfirst image after refraction through lens is obtained 10cmbehind the lens from

J_

which when light rays are reflected from plane mirror next image will be ogtained at a distance 20cm behind theplane mirror

Vj

which will act as an objectfor the next refractionat lens sowe

-15

10

Vj=4-30cm. Thus image produced must be at the centre of curvature ofthe

use

M= + 30cm and/=+10cm Using lens formula, we have

V

u

mirror as it is not possible at the pole because the distance of pole to lens is only 10cm. /?=20cm

i_i

^

1_

/

i_J_ = J_ V 30 ~ 10 £ _ J_ J

/=10cm "

Sol. 12 (A) Equivalent focal length of the system of three

-

lenses in contact is given as

^

V ~ 10 30 ~ 30 V =4-7.5 cm

Thusfinal imageis obtained 7.5cmto the right of lens.

11 i

1 1

fa " fw

Wherefocal length of glass lenses is given as R

Sol.10 (C) Here nose of the boy is the object & fish is

fr = p-l

observer. So for refraction formula, we use

u = +R; 102 = 4/3; H, = l and R = ~R

fc

7? =5 cm

= 2R

...(1)

Geometrical Optics,

'490

P eq ' = 2P,L + P. m'

Focal Length of water lens is given as

P^•=2(^ [ R

3R

R

= —- = 7.5cm

fw= ^ 2(^-1)

2

Given that,

Nowfrom equation-(1),we have

-L= |J_

2_

_1_

I 15

15

+

fe,

Equivalent optical power of the given system is given as

R

f =28 cm

J eq

R

So we use

28 =

...(1)

2(11-1)

When curved surfece is silvered, we use R f ,' = — =10

P = —xlOO =6.67D ®

R

...(2)

2p

15

From equation-(l) and (2) we get,

Sol. 13 (A) Equivalent power of the system is given as

14

P eq =2(P ^ w +P) + P m

Where P^ and P^ are the respective powers ofthe water and glass lenses and P^ isthe power ofthe mirror. These are given Sol. 15 (C) For the minimumangle ofdeviation producedby the prism,/ = e and Tj = ;25

as

L

-60

=>

Tj +

=Athen, r=45° (since =90°),

Using Snell's law for the 1" interface, wehave 1. sin / = p •sin r

-1

• •

in Ir = sin

and P = —, where we use =>

1

r 1

f

and

1 ^ -20,

1

P

2 V2

1= 60°,

Then

6^^^ = (/- r)+ (e- r)= (60 - 45) + (60 - 45),

=>

Sn.„=30° mm

2

= —

"

^

- /— X —^

= — cm"'

L

20

Equivalent power of the systemis now given as P eq =2(P *• w +P) g' + P ni', P

eq

= 2

180

60J

Sol.16 (B) Graph between (5 - /) for the prism is shown in figure. Here i = Angle of incidence when it is undergoes 20

minimum deviation, we have deviation angle 5=i+e-A

13

P

=>

= — 90

.

90

7 = —

=>

44° =42° +62° ^=60° 5

cm

13 44

Sol. 14 (B) When plane surface is silvered,-we use eq

L

p-r R J

Then in case when curved surface is silvered, we use

38 .

iV

^

1

j

1

1

1

42

i

/

y 1 1 1 1 1 1 1 1

62

For the condition ofminimum deviation

.

IGeometrical Optics

4911

f-^ + f2=A

_M__1 =lizi

(As /•j = r2= r)

r=3d°

2R

5„^-20--r)

oo

R

g = 2^1 - 2 g=2

38 = 2(/-30°) /=49°

Sol. 19 (B) Apparent shift in the object 0 dut to three slabes

5,, ^2 and

.4

with respectto the medium of p = —is given as ^

\

1 ^

Shift = 45

3/2

.

V2 sin 0, = 1sin 90

=>

1 sin0^= ^;0^=45'

Then, we have

1 ^

3/2

4/3;

4/3;

. ' 4/3,

( 1— 4^ +54 f 1— 8"!

I 3J

Using snell's law at the second surface, we have

+ 54

1

Sol. 17 (A) Ray diagram ofthe grazing emergence is shown in figure.

Herewe have r + B^=A => r + 0^ =6O

1

+ 24 1-

\

I 9)

Shifl:=5 + (-8) + 6 = 3 cm Object distance w= 150cm Thus image will be formed on the object itselfas light rays fall on mirror normally as the object appears to be located at its center of curvature.

Sol. 20 (C) As, we know

r=^ - 0^ = 15°

dv

du

dt

dt

j_ _ We use

f

u

1

Grazing

V

-1^1

-12 " 20

emergence

V

v = -30cm/s

Using snell's law at the U' surface, we have

1sin /= yflsinr and

sin i= ^/2sinl5 1 = sm

dv

^30"'

It

,20,

X 4 = -9 cm/s

i.e., 9 cm/s away from the mirror.

-1

Sol. 21 (D) Acceleration of block AB =

'^rng 3w +w

V3-1

(As sin 15°=^)

acceleration ofblock CD -

2w + m

3

Acceleration ofimage in mirror

Sol. 18 (D) For refraction formula at spherical surface, we

= 2 acceleration ofmirror

use

v = 2R\ W= co; Pi=l and Using refraction formula, we have 1^2 V

^i_ 1^2-1^1 u

R

= 2-

-3g

-3

= Yg 4g

Acceleration ofimage in mirror CD=2-\ — I=—

Geometrical Optics]

|492

Acceleration of the two imagew.r.t. eachother is given as 4g

^rd

Applying mirror formula to5 and A, wehave 1

IZi

3

2 }

1

-(1)

/

6

Sol. 22 (D) Thesituation described in thequestion is shown in the ray diagram below 0/2

A

W-Wfl

A'

H '

21.-1 2U+-

r--'

^2-1-21

1

2l L + —

L +

2J

—

•...(2)/

2

Solvingequation-(l) and (2) we get a~L. We are given that ZA®£)=30° =>

'

Sol. 25 (A) As shown in figure, we use

..

ZS£D = 120°

As BCDE is a cyclicquadrilateral, we have ZBCD=m°

g PC' _ 1 P " OC " |x

The line CE will be angle bisector of ZBCD S£' = crtan30°=-p V3

For small angles, we can use

^ BE _ _2_ AB ~ al2~ yl3

We use

sm/ =

P=pa=|^ (6°) =«°

V?

Using snell's law, we have l.sin/ = «sinr 2

1

n

=

=?• Bendingangle= p-a'=2°.

Sol. 23 (B) Incidence angle on face5C is Sol. 26 (B) This is a case oftotal internal reflection, we use

1=90-0

i = ^ = 90- 0 > 0^(forlightnotto cross BQ 6/5

4

1

=>

0>e„(=sin-'—)

cos 0> sin 0^ ==-j — < sin 0

=>

'

0 < cos"^ — =37°

As at point B angle of incidence of light ray is maximum, we use

Sol. 24 (C) If the length of rod be 'a\ The magnitude of transversemagnificationof ends^ and 5 is 2 each.The image of5 is virtual and that of. Required width ofregion AB is

3/2,^2=1.1^3 = 1

(9 4^ V=[^+Tj-l =10cm i.e. the final image formed at 10 cm from side ^5 of the slab.

Thus at the objectitselfas shown in figuare. Herethe image is virtual.

45

—— =tan53°= -

!

•

15

5

25

^=^,-^2=^-4-^™ Sol. 30 (A) This case is the refering to the setup of displacement method experiment as product of the two magnificaitons is unity. This is shown in figure-xxx below.

Geometrical Optfcsl

494

Sol. 32 (A) In the figure shown O't

-hx-

t -to-r

InAOO'P& A/TP using similarity, we have Using lens formula, we have

J_ +i -i y\x X

f

00'

OP

ir

IP

also inAOO'C & A/TCusing similarity, we have 00'

f=

...(1)

OC

...(2)

~1F~JC

Ti + 1

By equation-(l) and (2)

So we get the given ratio is

^.

OP

OC

IF ~ IC Sol. 31 (A) The image coincides with object O if the ray starting from O is incident normally on mirror as shown in figure-(a) Here the image is located at

OP

^

IP

OC ~ IC

Sol. 33 (A) In the figure shown belowthe line'CM'is normal to the mirror passing through the end point Byray diagram

4 16 . = 4x- = -

it can be seen when the insect is located to the left of'O' all its

In second case the image will coincides with object O if ray starting from O is incident normally on mirror as shown in figure-(b)

reflected rays will be towards right of'O' so it cannot see its imagebecause rays are not reachingit and whenthe insect is to the right of' O' its reflectedrays will be on both sides of the insect that means the insect will be in the field of view of its

own image so it can see its image.

40+16/3

So it will be able to see its image till it reaches the point '5' of the mirror from point 'O', ift is the time for which it will be in field ofview ofits own image, we use 3 X 100

2x/ =

cos 60"^

/ = 300 seconds.

(b)

(a)

Thus here we use

136

,,

4

136 3 h'= —- X — =34cm 3 4

[Geometricai Optics Sol. 34 (A) By using mirror formula, we have

Sol. 37 (B) From the given vectors for the direction of light rays we can use

V

-10

10

/I

For the first medium sin 0, =

• V7T7

v=+5cm

Thus magnification by mirror is

and for the second medium sin 0^= 1

By Snell's law, we have PjSin 0j = pjSin Oj 1

So the image revolves in circle of radius —cm. Image of a radius is erect so particle will revolve in the same direction as the particle. The image will complete one revolution in the same time 2s sovelocityof image will be given as

Sol. 38 (C) Using mirror formula, we have

i i-1

27C 1 7C v = a)r.= — X — = — cm/s= 1.57 cm/s. ^

V u

f

^

Sol. 35 (B) Ifcable ofan elevator breaks then elevator will'be in free fall at acceleration g.

i+-i V =>

-90

1

-30

v=-45cm

For velocity of imagewe usevelocity magnification alongthe principal axis as

/////////////////////

dv

du

dt

dt

45"

•^14=1 cm/sec

Sol. 39 (A) In first case, we use

u = "12 mis

1 +i - J_ P Mirror

and floor after time / is '

5= «sm

f

9

and

And particle starts moving vertically up at speed u sin 0 which will be constant with respect to elevator as both elevator and particle both are in free fall. Thus separation between particle

9

0)

=>

In the second case, we use

)J= yl2.sm45°-~ =^

1.^1

m

q+x

1-

p'

f

Thus separationbetween the particle and it's imageis given as 1 1 s =2s= — H— = Im. 2 2

Sol. 36 (D) Using mirror formula, we have

i

J_ _

1

V 10 ~ -20

q+ x and

^

= w.

m^=

'

q + x

/

From equations-(l) and (2), we have

m^ —m^=xlf

20

|v|= —cm in front ofmirror

f= m-, —m.

(2)

Geometrical Opti

496

Sol.40 (C) Intheraydiagramshowninfigure.wecanseethat at refraction at point by Snell's law, we have sin 60*^

sinr]

Sol. 43 (B) By lens makers formula, the focal length of the lens is given as

\

= V3

1

sm r, = -

l_

fj__0

^(1.5-1) 1^20 20,

0.5x2

20

1

~ 20

/=20cm

Since^2 = r, Now considering refraction at point Q, by Snell's law, we have

Thus the light rays will converge at 20 cm from the optical centre of the lens.

sin /o sin i2

Sol.44 (C) Ifprmtsaresimilar,totallightenergyrequiredfor the exposure will be same, so we use

73

A'l ~ hh

Where /j & aretheintensities reception onprint area. If andP2 are thelightintensities when thesource is at a distance rj and r2, we have

A

^

„2 '1

-5..

„2 h

PrP^

As we have

t = t '2 ^2 M

^2 = 5 sec

Puttingr2=30® weget = 60® The angle between reflected and refracted ray at point Q is

Sol. 45 (B) Maximum Deviation is forgrazing incidence

given as

a=180°-(r^ + /2) =>

a= I80°-(30® + 60®) = 90®. 1 = 90

Sol. 41 (B) For multiple layersof parallel sided media, apparent depth is given as

h =iL+i2_ ^1

The deviation is maximum when ? = 90® or e = 90® that is at

P2

grazing incidence or grazing emergence. 36

5

3

7 ~ 5/3 p2

1=90®

Let

/•j = Csin"' (l/p] = 36 li2

7

15

r,=sin-i (2/3) = 42®

7

r2=^-r, = 60®-42®=18'' By Snell's law

tt3=- = 1.4

sinri sing

Sol. 42 (B) As object is located at center, all the incident ra>^ will be normal to both the inner and outer glass surfaces ofthe

spherical shell and will not suffer any deviation, so final image as seen from outside will be produced at the center only.

sme = u sinr2= 1.5sin 18® sine=0.463 e=28®

Deviation =5_

[G^^etrica|^Optics_^

497

=i+e-A

= 90° + 28°-60°

10

=58°

l2

R

00

Sol. 46 (A) Asthesun isvery far away, the incident rays are

J

considered parallel.

2R ~ 10

Normal

1_

i?=5cm

Normal

Forimage tobeobtained onobject, lightrays on mirror smust fall normally toretrace thepathofincident rays after reflection. So by refraction formula, we have 3/2

I

00

~d

fl

[-d

Forrefraction at first surface ofsph'ere weuserefraction formula as

1

-

d "

1 10

10 cm u

1 =4>

— +

00

=>

R

V

1.5

1.5-1

Vj

+8

Sol. 49 (B) By lens formula, we have

Vj=24cm

V w

...[18(A)]

/

.

The rays will converge at/j which will actas an object for the second surface, so using refraction formula for the second

J

surface, we have

20 ~ +30

u

^

1.5

-8

=>

V

R

1

1-1.5

V2

-8

1—

j± 1 V

1 ^ J__J I_ V 20 30 " 60

^

V2=4cm

Sol.47 (A) Bylensmakers formula power ofa lensisgiven as

L-i R\

If

is the lens power in air and

20cm

Rj

l«-30cm-H h

•60cm •

is the lens power when

submerged in the liquid we can use

=>

'

•'

• i)=60cm

•

As triangles OPC & ABC are similar ..Hq

•y = -5

-P,

-100/100

Hz.

^

60 ~ 10

.

1

=>

j:= —=0.3cm

'

5-

Sol. 50 (B) By lens makers formula, we have Sol. 48 (C) By lens .makers formula, we can find the radius of curvature of the lens surface as

r-'fi-i

Geometrical "Optics!

Differentiating both sides, we get -llOcm-

\

I

\

1_

_1

1_

V.^1

^2

V-^1

^2J

dn

lOOcm

\*—25cm

From O to / intensity increases and then decreases as it is inverselyproportional to the.areaofimage on screen. Sohere at X=90 cm «fe 110cm the intensityis same. Radiusatx=200cm is equal to the radius of lens.

M-1 but

-df=

;— /

=:.

C^=-G3f

=>

/.-/s=-®/

SoI.2.(A,P) For the given lens /=.20cmandpg=

weuse

3-1 A-Ji? ~ 20 2

i?=20cm

=-0.4x 10 = 4mm.

Now M= 30 cm. Iff willchangeto 30 cm,imagewillbe at infinity that is possible in two ways. Case-t :Ifanother concavelens of60 cm is placed in contact,

Sol. 51 (A) ForobjectOj, bylens formula, we have J_ J_ __1_

•0)

v"^20"/

then equivalent focal length becomes - -

J L _L = i/' " 20 " 60 " 30

for object Oj, we have

i_J_- 1V 40""/

..g) =>

- " • -

/'=30cm' . 9

(ii) Ifthis lens is immersed in aliquid ofrefractive index -g, "02

Q^

/

H

20

•

«

40

then its focal length becomes

—H

_v

Y3 ,9^

_1_ From equation-(l) and (2) we get

2

/'

8

_2

1_

20 ~ 30

/=80/3 .

ADVANCE MCQs One or More Option Correct

Sol.1 (B, C, D) Bylens formula, wehave'

1

f 1

I

•

.. • ,

Hence options (A) and (D) are correct.

^

V (-25)

20

1

1

1

1

V

20

25

100

v= 100 cm

• .

Sol.3 (A, B) For given conditionthe incident light rays after refraction from lens should fall at mirror normally.

1-1 _1 V II

/'=30cm

fGeometrical Optics 499i

SO

f^ = d+2f2

=>

^ = 1/,1-21/21

Sol. 7 (All) Figure-xxx explains how deviation angle varies with the incidence angle. With this figure we can analyze that all given options are correct.

Asthe lightbeam is reflected as it is, thebeam diameter willbe same as it isretracing it's origianal path.

5

5=7c-2c

Ifwhole arrangement is submerged in water then focal length of .lens would change as it depends upon surrounding medium also but that of mirror,

\^=jt-2i

will remain same.

Sol. 4 (A,D) First three images on Af, are formed atdistances 5cm, 35cra and45cm. First three images on

/ = J0,

for 0< / < 0_.

are formed at

/=90

. -I 8 =sin («sin i)- i

distances I5cm, 25cm and 55cm.

Sol. 8 (A, C) For convex mirror always [/« | < 1for any real

h—20 cm-

object

•

. i

1/.

•♦-15cm -•n

•

25cm 15cm

5cm

55cm

1

•//nil

As we know that Kimage

4/,3

1^* ^ 5cm

^*~35cm —45cm

' obie«

=> IF. I image |l 1

sin 45®

=5,

In above figure, we have 7t

...(1)

'i +'2=3" Bysnell's lawat firstsurface, wehave

g>1.414

Therefore forrefractive index greater than r.414{which isthe case for Cand D)total internal reflection takes place.

SoL 17 (A, D) Ifthe angle between the two plane mirror be 0, the light ray incident atangle iand angle'a on miffdf Mj &M2

l.sin30° = ^sini,

respectively as shown in figure below. /, =sin '

2\i

...(2)

By snell's law atsecond surface, we have ^ sin »2 -1 /., = sin '' ~

sin

-1

+ sin-1

...(3)

2pj" 3

77777777777777777^^777777/. Mx \

Sol.14 (A,C) Bylensformula, wehave

i_i^i V u

f

Initiallyasmagnification is0.5,wehave

-u

and the deviation of light ray due to

is

= 180°- 2a

Total deviation 6= +Sj=360® - 2(i +a)

u

But e + (90-0 + (90-a) = 180®

2

As we use from figure

1 _ 1 _1 «/2

Thedeviation oflight raydue toM, is6j- 180® - 2/

i+ a = 0

=>

5 = 360®-20.

f

m=3/

When displaced by 20 cm, real image ofequal size is formed.

Sol. 18 (A, B, D) Ifa parallel beam oflight incident on lens then for refraction at first surface oflens, wehave

[Ge^etrical. Optics

V

R

00

Then for refraction at second surface of lens, we have

v'

V

•••(ii)

(-/?)

From (i) & (ii), we get

Source

,-.-±

/=

'7=;^[2^2-(^,+^3)] 1

2

V'

^3i^

\^2-

Sol. 21 (A, B) If = 120 cm then by refraction at curved surface, by refraction formula, we have

(^1+^3)

^-1 2

1

IfM^>

2v

v'=+ve (converging lens)

2

^

60

-120 V= op

^1+^3 2

^1+1^3

v'=-ve (diverging lens) v'=00 (neither diverging nor converging)

\h =

Sol, 19 (B, C) As already studied for emergent ray to exist, refracting angle at surface AC in side the prism must be less than critical angle and maximum value ofangle r in figure can

also becritical angle sofor thistohappen A< 20^ A

'

'

Thus parallel light incidentnormallyon mirror and henceimage will form on objectitself,hence option(A) is correct.

If d^= 240cmthenusingrefraction formula wegetv =360cm soif£^2=360 cm,thenplane mirror form image atsame place and again final image formon objectitselfhence option(B) is also correct.

Sol. 22 (B, C) Distance offish as observed by bird is given as 0.8

+ 6 = 6.6 m

And distance of bird as observed by fish is given as

6

If light incident at angle/. For emergent ray to exist, we must

^4

4

3 =6.8x-=8.8m

Sol.23 (B,C) From figurewe can seethat

have

r>{A-Q^)

AB =

HT

170

= 85 cm

=> sin /> n sin{A - 0^) . sin I >

sin(^-0c)

10 cm

— sin I

Sol. 20 (B, C) If P is the power of the source. Then energy

160 cm

emitted from water surface

-2/'(l-cose^) Fraction of light

/=

(I - cos0^)

and h£BT we have EG = GT = BD =

= 80 cm.

Geometrical Optics!

1502

Sol. 24 (All) Ifthemirrors OMand 07/make anangle 0 with each other as shown in figure, then we have .

TVC-

ZBi:C=240-180 = 60°

In A/4fiC,wehave

x

=

At x= *2 2 . %

=>

S

y = — sin ^ = — •^71 3 71

If lightrayis incidence from rightthanslope would be45° -1

-1

tan 45° = Vi = y-i, '^TTTTTTTTTTTTTTT^TTTTTTTTTTTTTTT?^

1=

2 cos Tlx

2 COS Tlx

x =

24.+2(0-(!))+60° = 180° 2(|)+28-2t|>+60° = 180°

2

360°

Number ofimages = I

. 71X2

y = — sm -^71

0=60°

3

71

r2 V3 So points are 11 | — I andj 1y'"

-1 =5,

Hence all option are correct.

Sol. 25 (A, C) Let us assume that incident ray is AB and Reflected ray is BC

Sol.26 (B,C) Sincefirstreflection should be at planernirror then imageof AB would be .4'R'which will be equal distance from mirror.

Nowimage^'5'will behave likeobject andimage will beformed by convex mirror

/= + 60cmforconvex mirror

"'a- - 60 cm (distance ofobject from mirror) R = I20cni

(1,0)

Point ofincidence is B

Let as calculate the slope of tangent at point B

40cm

d Cy) = —d (2 . 7CX^ =—XjtC0S7LC=2 2 — —Sin COSTCC dx^'

dx\%

)

71

Since ED and MN are _L to each other

If slope of If

w' = -90 cm (distanceof objectfrom mirror)

is mso(slope ofMV) x (slope of^D) = —1 "A

W X 2 C0S7IX=-1 -I m =

uf

2 cos Tlx

ZABM=45'' ZMBC=45°

and

uf u-f

w = tan 135°

'5 u-f

-60x60 -60-60

-60x60

-90x60

-90x60

-90-60

-150

size ofimage = 36 -30 = 6 cm size of object=30 cm 6

-1

tan 135° =

2 COSTCC

cosTlx = 2

-120

1

overall magnification ^ ^ ~ "5 •

= 30 cm

= 36 cm

[Geometrical Optips

:5d3l

Sol. 27. (B, C, D) Giventhat angleofincidence= 60^ When= we can see that Q=,0,=^,Zi = Zr . UsingSneil'sIawwehave • ,

|a, sin / = ^2sin/" If

. i=r Angle of incidence

^=1

In general forsmallangledprism, wehave

1^2

=>

Before

kQ = l

5=(p-lM

i.e. kAf.

In experiment there are two positions ofthe lens for which sharp images are obtained on the screen for which the product of magnification is unity, thus we have

ave Optics^

505!

ANSWER & SOLUTIONS The ratio of maximumto minimumintensityis givenas

CONCEPTUAL MCQS Single Option Correct 1 4 7 10 13 16

^max

(D) (A) (B) (C) (D) (D)

2 5

8

11 14 17

19

(A) „

20

22 25 28 31 34

(C) (B) (C) (C) (D)

23 26 29 32 35

(C) (G) (A) (A) (C) (B) (C) (C) (B) (B) (B) (A)

3 6

9 12

15 18 21

24 27 30 33

(B) (C) (B) (A) (A) (C) (A) (A) (B) (B) •(C)

^2 )

•^min

(''i

a+2)^ _ (1-2)2

^min

(ii) Figure showsthe YDSEsetupas describedin question. At a distanc x from screen centre path different between the

waves from 6", and $2 is given as dx D

As itis given that at x = dH, path difference is given as 2>X

NUMERICAL MCQS Single Option Correct

because third bright fringe is located in front of one slit on 1 (B) 4 (D) 7 (C) 10 (D) 13 (B) 16 (C) 19 (D) 22 (D) 25 • (B) 28 (D) 31 (A) 34 (D) 37 (B) 40 (D) 43 (D) 46 (C) 49 (B)

32

(A) (C) (B) (B) . . (A) (B) (D) (C) (C) (C) (A)

33

35

(B)

36

38

(D) (B) (C) (C)

39

2

5 8

11 14 17 20 .

23

26 29

41 44

47

3 6 9 12 15 18 21 24 27 30

42 45

48

(C) (C) (B) (A) (B) (C) (B) (A) (C). (B) (C) (A) (B)

1

(A, C, D)

2

(A. C)

5

7 10

(A, D) (B, C)

8 11

(B, D) (B. C) (C) (All)

3

6 9

(B, C, D) (C) (A, C)

Solutions ofPRACTICEEXERCISE 6.1

0)

2D

d= -Md = V6x6xl0"'xl V36xl0"' m d= 0.6VlO mm Sc

D-

(D) (A) (D)

ADVANCE MCQs One or More Option Correct

4

screen. Thus we use

The intensity of the wave is proportional to tHe area of

theslit.Thuswecan usetheintensities /, and from theslits

/.

DX

ZL

^2

nl

4

= —

A= 0

P ^DXIld^

^=2 = y=

the amplitudes ofthe waves, then we have

j

x

2

(iii) (a) Just above point O the direct waves from source and reflected waves have a phase difference tcdue to reflection from mirro so these wave interefare destructively so therewill be dark fringe at O, therefore intensityoflight at O will be zero, (b) From the figure, we can see that the distance of first maximum(first brightfringe) from0 willbe locatedat half fringe width above, so we have

on screen are in ratio

If Oj and

A= 3X

4d

Wave Opticsl

^506

(iv) We know that intensities due to slits is directly proportional to the area ofslits or proportional to the width of

Thus we use 1

slits as these are of equal lengths hence we use

tan 0 =

=

yf24

h 4 Ratios of maximum to minimum intensity on screen is given as L

3+ 2

L

3-2

10

Thus, central bright fringe is observed at 2.04 cm abovepoint Q on side CD of the vessel. (b) The central bright fringe will be observedat point Q if the path differencecreated bythe liquid of thickness / = 10 cm^ is

=25

^min

equal to A,, sothat the net path difference at Qbecomes zero.

(v) (a) Figure shows the calculation of path difference on screen at a general point R which is located at a distance,^2

Figure shows the situation after vessel is filled with the liquid.

from the center Q.

' 1

Xe/

D. = 2m

0/ y}

i

1

1

\*-D2—lOcm-H B

The path difference in waves reaching at Q after the slit plane is given as()a,- l)r, thus to obtain central bright fringe at we

D

h-£>2= lOcm-H

have

In above figure angles a and 0 are given as 1

(p-l)r=A,

^2 •

tan a = — and tan 0 = —

As slit separation is small compared to Dj andD2 in figure, we can use tath difference between 55, and SS2 is A, = SS^ - SS2 A, = i^sin a=(0.8mm) A, =0.16mm

...(I)

If at point R on the screen, central bright fringe is observed then at this point the path differen in waves before slit plane and after slit plane must cancel each other.

Path difference between S2R and iS'jR would be A2 = S2R-S^R => Aj^tfsina

...(2)

Central bright fringe will be observed when net path difference is zero, thus we use

=> => =>

(p-1)0 DO) =0.16 p-I =0.0016 p = 1.0016

(vO (a) Phase difference in the two waves-reaching at C from the two slits must be zero for central bright fringe to be obtained at C. Here the phase difference is given as 25X

27C

Ait)= Y^sincj)-. Where first term in above expression is the phase difference due to the path difference d sin f which is there before the slit plane due to inclined incidence of the light beam as shown in figure. The second term here is the phase difference due to the path difference introduced by the. thin mica film. Thus.for central maxima at C, we use A4) = 0

i/sincp / =

A2-Aj=0 if sin 0=0.16 0.16 sin 0 =

0.8

1

tan 0 =

2xI0-^xsin30°

= 5x 10"^

[Wave Optics

1

Hence refractive index ofmica slab = 1.6

507:

yg, sotheoptical path oflight waves from source isgiven as X, —iS'|P+(jj.| —

The optical path oflight waves from source $2 Xi —S2P + (^2—1^2

The path difference between the two waves reaching atP is Ax = :c2 - Xj

Screen

(b) Ablack lineis formed at theposition where dark fringe is formedfor boththe wavelength. Thus distanceof the first dark fringefrom centerbright fringeis (277-1)XD y=

...0)

2d

For blackline, dueto both wavelengths, we have (2«i - 1)X|Z) (2/72 ~ 1)^2^ .

Here Xj and

2d

M

arethe wavelengths present in the lightbeam in

the water.

(2bi-1) _ X2

^

(2^2 ~1)

X., ' X

^

_ ^20 andX,-^

X, = ^

where

As weknow the physical path difference from slitsto a point on screen is given as

(2/7i -1)

7

(3/72-1)

5

^0

AX=— +(p2-l)/2-(lii-I)/, For zero order maxima, Ax=0

.

For first line we choose minimum values ofn,and n^, taken as /7j =4and/72 = 3. Hence distance of the first black line from center is

(2X4-1)4000 X10"'® X40 X10"^ X3 y-

=>

2x2xl0"^x4

1)^!-(M-2-1)'2]

7'„= 5 Forfindthecondition for zeroorder maxima atthe centre O,we use

>7 =2.1 X10^m=210pm.

(vii) Path difference due to insertion ofa thin film at screen centre is given as

>'0 = 0 0=

^

-1)^1-(M2-1)^2]

(^i-1)/, = (^2-1)'2-

A = /(n-l) => =>

A= 1.5 X 10-^x0.5 A = 7.5xio-7ni

(ix)

In this setup of YDSE, we can use d=2mm = 2x 10~^m

Phase different between the two waves at screen centre is given as 271

D = Um

. Region of

intensity 11

2ji

=> =>

•(})= o -4 x7.5x 10"' ^ 8x^1^ (() = 37C

Thus resulting intensity at C is given as

4 = 4/qCos2(()>/2) =0

1 max

(viii) Figureshows the twowaves from the twoslitsreaching point P on screen. If this point is the position of zero order maxima and the distance ofP from the centre O ofthe screen is

Ih

5 cm——5 cm

nocm

Wave Optics|

;£08 . ,

(b) If thegap between L, and

Fringewidth in YDSE orsimilarsetup isgiven as

is reduced, d will decrease.

Hence, the fring width will increase so the distance OA will 6xl0~^xl.2

XD

P=

=>

increase.

2xl0"^/«

Screen

p = 3.6xl0-4m

=> P=0.36mm Width of interference pattern on screen is given as ...{1)

AB = OA-OB

5 mm

Here by similar triangles, we use OB

0.1

D = 1.0 m

=>

0.30m

0.15m

110 ~ 10

.30m

05 = 1.10cm

And we also have

(xi) OA

0.1

115 ~ 5

=> A= (ti-l)/ = (1.5-l)r=0.5/ Due to this slab, 5 red fringes have been shifted upwards so

0^=2.3 cm.

Now from equation-( 1), we have ^

(a) Path difference dueto insertion of the glass sla.b,is

given as

we use

= 2.3-1.1

=>.

= 1.2 cm

0.5/ =(5)(7xl0^m) •

=>

(x)

/ = 7xlCr^

(a) For the lens formula, we use

w= -0.15ra and/= + 0.10m By using lens formula, we have

(b) Let p'be the refractive indexof glassfor greenlight then we have

A' = (H'-l)f

i_l_l V u

1

1

(-0.15)

(0.10)

i-l+lV =»

u

Now shift is equal to that of 6 fringes of red light so we use A' = 6X,'red

f

f

t(p-l) = 61'red

v=+0.3m

=>

In this case lateral magnification by the leiis is =-2

Thus by the two lenses, two images 5, and

n' = 1.6

of S will be

which implies that

formedat 0.3 m fromthe lens as shownin the figure. Image5, isduetopart 1which willbeformed at 0.5mmabove itsoptical

axis(w= - 2).Similarly the image

7x10"^

_

=0.6

(c) Inpart (a), shiftingbf5 bright fringes was equal tolO'^m,

-0.15

u

(6)> = 1.30 - 0.30 = 1.0 m = 103 mm; Light wavelength is X.=500nm=5 x lO^mm.

XD

P= We can use r green

'"green

Pred

^red

Thus fringe width is given as

XD P=

X'green-

(5xlO"^)(100

1

P^een Veen = Pr Hred

,-7

= (0.2x 10-3)

5x10

.-7

7x10

mm= —mm.

d

(1.5)

As it is given that the point A is at the third maxima, we use Q^=3p=3(l/3)mm OA=\nm.

Pgreen-0-143xl(r-^m .^P =Pgreen- Pred= (0.143-0.2) XlO-^m Ap=-5.71 X10"^

;Wave Optics

509!

(xii) As the given refractive indices of the glass plates are = 1.4and Pj = 1-7and if / bethe thickness ofeachglassplate then the path difference at the screen center O, due to insertion ofglass plates will be

A= (^2-^l)/=(1.7-1.4)^ = 0.3r

...(1)

• d sin

As 5*^ maxima (earlier) liesbelow pointOand 6*^ minimalies above point O, this path difference must lie betwen 5X and 5.5X. This situation is shown in figure.

T 6

As sin 0 < 1 therefore

Minima

•th

2n-l

2x1.0

(2«-l)

< 1 or n < 2.5,

^ So, n can be either 1 or 2 For n = 1, we have Minima

sin• 0,1 = -7 4

Due to the path difference A, the phase difference at O will be 2k .

(2«-I)0.5

4

0

5

{2n~\)X sin 0 =

tane. =^

2k

When « = 2, we have

(|t= [1071+—Aj

-(3) sin 0., = — ^

In above equation Aj is considered less than X/2. Intensity at point O is given —so we use the intensity at

4

tan 0., =

or

a point where the phase difference between the two waves is 9 is given as yy

^(9)=7„^cos,21 1

^^max -^max 3

—

=

cos

4 •

/

2

219

y

y

y

S

'

\

>

'

y'

V

s

s

'

0

—

\

...(4)

—

.I2

[In this case, net path difference Ax= Ax, - Atj]

From equation-(3) and (4), we have X

•

^^=6

;

• A=5X.+- = —X=0.3r 6

6

Solving, we get

31X

(31)(5400xl0-'®)

'"6(0.3)"

1.8

r = 9.3 X10~^m = 9.3|im. [In this case, A.v = Av, + Av,]

(xiii) For the setup given values are - ^ = 0.5 mm, d=\.0 mm andZ)=lra

=> j' = £)tan 0 = tan 0 (Z)= Im) So, the position ofminima will be:

(a) When the incident beam falls normally the path difference

1 >*1 =tan 0, =-y^ m=0.26m

between the two waves and S^Pis, A= S^P-S^P«dsinQ. For minimum intensity, we use i^sin0 = (2«-l) —,77=1,2,3,...

and

3 _V2 =tan02= -^m= 1.13m

And as minima can be on either side of centre O. Therefore

Wave Optics]

i51'^ _ •

there will be four minimas at position ± 0.26 m and ± 1.13 m on

X

c^sin ~ Y

the screen.

(b) When a = 30°, path difference between the rays before

- X

So, there is already a path difference of X between the rays. Central maximum is defined as a point where net path difference is zero. So we use

=>

1 s

n/Is

y, =

—

d

m=0.26m

the central maximum are

e = a = 30° tan 0 =

'

Therefore, y-coordinates ofthe first minima on either side of

tfsin a = f/sin 0

=>

tan 0, —

=4>

•a.=a,

(Q-5) (2)(1.0)

sin 0, = ^

reaching5, and^j is Aj = t/sin a = (1.0) sin30°= 0.5mm= X.

y, =0.26mandy2= 1.13m.

2^

Note : In this problem the relation sin 0 a tan 0 « 0 is not valid as 0 is large

= — D

1

(xiv) (a) For first minima, we have

^0= 0.58m Ax, = Ax2 Above P Ax2 > Ax, and Below pointP Ax, > Axj

^/sin 0 =X

At point/',

d^sinO

Central 2 maximum ^=0

P

-r

y = 0.58m

=?>

6500x10"'® —7::^—=251 nm

=

sin 15°

=> • =2.5 pm. (b) If X' is the required wavelength, then for first order maximum

t/sin0=—X' dsinQ X' =

1.5 251 IxsinlS'

X' = 1.5

]^ow, let P, and P^ be the minimas on either side of central maxima. Then, for ^2 ''

=>

X'=4300A

Solutions ofPRACTICE EXERCISE 6.2

•

X

a,-a, + -

X

,

X

A-—A, + — - X+ — —

2

12

2

3X 2

3X

dsm ^2~~2 sm

'3X (3)(0.5) 3 0., = 2 2d ~ (2)(1.0) ~ 4

tan 0, =

3

(i) Incident ray AB is partly reflected as ray 1 from the uppersurfaceand partlyreflectedas ray 2 fromthe lowersurfece

of the layer of thickness / and refractive index p, = 1.8 as shown in the figure. Path difference between the two ra>^ would be given as

A= 2pjf =>

A = 2(1.8>=3.6/

yi

r-

D B

m= 1.13m

T I

Similarly for pointP,, weuse

i

X

A,-A." 2

=

X XX - =X- - = -

12

2

2

-

jWave" Optics

511,

Ray 1isreflected from a denser medium, therefore, itundergoes a phase change of k, whereas the ray 2 gets reflected from a rarer medium, therefore, there is no changein phaseof ray 2. Hence, phase difference between rays 1and 2 would be (j) = jr.

'min

2>/6

= 1224A

2

Therefore, condition ofconstructive interference would be

(Hi) Path difference in the two waves constituting the reflectedlight at normal incidenceis given as

n-—

A =

2

« = 1,2,3...

where

A=2p/-3.6/=

Fordestructive interference when the filmappearsdark weuse

Least values of t is corresponding to «= 1 so we get •

X

X

A=2iU-j=(2N-l)-

X- ' 2x3.6

_ 6480

^min~ 72

A

/„:_=906A (ii)

By using Snell's law, we have 1 sin i = p.sin /•

=>

2ut=NX

1 sin45= V2 sinr

=>

Nxsmxio

2/

r=30°

Path difference in air between two reflected rays can be calculated from the figure as Ar = 2p/ cosr+

0 — 2

=> Where we use

p = 1.318

(iv) As indicated in the figure only one of the reflected ray suffers a phaseinversion. As alreadydiscussed we ignore the

=nX 2

2p/ cos r ~

p = 0.263 JV W=l,2,3,

As per the given conditionon refractiveindex 1.2< p < 1.5 Weget refractiveindex in the given range at //= 5

For constructive interference, we use 2u/cosr

2x1.1x10"^

{2n + l)X

reflections from top surface of top plate and bottom surface of bottom plate due to plate thickness. At the thin end of the wedge, wherethe thicknessis negligible, the tworays interfere

2

i2n + \)X 4pcosr

destructively. This region is dark in the reflected light. The condition for destructiveinterference in the reflected light is 2t=Nk

whereA'^=0,1,2,...

The change in thickness between adjacent dark fringes is At= X/2. The horizontal spacing betweenfringes is given as

r = 30=

d= 1/12 cm= 8.3 x 10^m. From figurewe can see bysimilarity in triangles that DfL = Add, thus we have

^ XL (5.5x10"^ m)(Q.2m) 2d =>

For minimum thickness we use « = 0, so we get X

'mm • 4^cosr

for minimum thickness

16.6xl0-^m

D = 6.6 X 10"^ m

(v) For strong reflection ofa colour, its wavelength reflected fromtop and bottomsurfeceoffilm must interfereconstructively fo which the path difference in the two waves must be an integral multiple ofthe wavelength.

Wave.Optics

"5i z

(b) Intensity oflightreaching onthescreen directly from the

source is say /j = /q and intensity of light reaching on the screen after reflecting from the.mirror is

X/2 + 2U/

36%of/o= 0.36/q 1

1

As shown in figure, the path difference in the two waves

^h

/

\ 2

1—

F-ilj 1^2

^min

reflected from the film is given as

0.36

,0.6

2\kt _ 2x1.5x3x10 -7 N

N

9000i N

=>

,0.6 16

A= 2^=M, X=

0.6

for

(c) Initially path difference P between two waves reaching from 5 and 5" is A = 2/) + 2

1,2,3...

>,=9000A,4500A,3000A....

White light approximately ranges from 4000A to 8000A sohere

Here Vl is the path differenceadded in reflected light due to reflection from the mirror which is a denser medium.

light of wavelength 4500A will be strongly reflected bythis film.

(vO

Accordingto the given conditionin question,Intensity h

T

of ray AB, 7, = —, and Intensity of ray .4' 5' is given

T -s

v/z/z/A////////////////. h

corresponding to 20% reflection at each surface as

vzzzzzzi^zzzzzzzzzzzzzzzz. h±x'

i ,5'

16/„

i •$>

Initial

^2 125

Final

For maximum intensityat P, weuse forconstructive interference A = 2/i + — = 2

...(1)

Now, if the reflecting surfaceis displacedbya distances away fromthe sourcethen newpath difference will befor maximum intensity at P will be given as A = 2/i+2x+ — = (?ri-l)X

2h+2x =

...(3)

n + \— 2

Forlight raysy45 and/4'5', afterinterference maximumand minimum intensities are given as

Solving equation-(l) and (3), we get X 2

=

6000 j A = 3000A 2

Solutions ofPRACTICE EXERCISE 6.3 and ''min

(vii) (a) Shape of the interferencefringeswill be concentric circles with center at point P. This is because at a point on a circle with center at P on the screen, the path difference in the

light waves coming from source and from its image remain constant.

(i) This is a case of Fraunhofer diffraction by a narrow rectangular slit in which condition for minima, we use, sin 0 = nX

WhereK= 1,2,3, => 0.025 X 10-3 sin r24' = 1 x ^

=>

=>

•

^

0.025 xKT^x 0.0244

X=6100xio-'°m = 6100A.

•Wave Optics

(ii) First order minima on either side of the centralmaxima are at an angularseparation 0 from thescreen center, whichis given as

the second Polaroid is having 20% intensityofthe light entering this second polaroid,-sowe use

i>=63.4°

6sin 0 =

For small 0, we use sin 0 = 0, so we have X

Where b is the slit width. Thehalfangular spread of central maxima (upto firstorderminima) is given as ^

^

2mm

tan0«0 = -

(ii) Suppose angle between first and second nicol's prisms is0. Then the angle between second and third nicol's prisms becomes 90° - 0. If/^ isthe intensity ofthe incident light on thefirst prism, then intensity ofemerging light from this will be Iq

.

.

.

intensity of light emergingfrom secondand third

= 0.002m

Im

Thus we have

nicol's prisms be .-9

6 =

(iii)

0.002,

- 0.3 mm

The condition of minimadueto singleslit diffraction is

(iv)

The angularspreadof centralbrightspoton the screen

Thus according to the given condition'

is given as

^ cos2 0 sin2 0= 77

•

2

i.227. •

-

sin^ 0

X(9o°-e)

. . />sin 0 = ^ , , ?. = 0.05xsin3a° >.=0.05 X 0.5 = 0.025m=2.5cm

=> =>

cos 0

respectively.

600x10

0.002,.

cos^0 and

sin 0 = —rz—

•

16

(sin 0 cos 0),2_1

b

For small 0, we use sin 0 = 0, so we have

(2 sin 0 cos 0)^= —

1.22X

Theradius ofthecentral bright spotonscreen canbegiven as

=> '

sin2(20)=-

^ 1.22X

r = dQ = dx

1

=5"

sin 20 =

^-IO

=>

r= 20x10"^

m

=>

10x10"^ ^

r= 1.4396 xlO^m

(iii)

Solutions ofPRACTICEEXERCISE 6,4 (i)

0=22.5°'

Theintensity ofthelightafterpassing through thefirst

polarizer is1=cos^ 0|. This light ispolarized inthe direction of the axis ofthe first sheet, and so its axis makes the angle (a) Using MalusTaw for the secondPolaroid, we have 02 —0] with the axis of the second sheet. Consequently, the /,//2 = cos2 (j) = cos^ 30° =(0.866)2 =0.75 intensity of the light after passingh through the second

Thus three-fourth of the light striking the second Polaroid is

transmitted. As we know that thefirst Polaroid allows only half the intensity ofunpolarized lightthroughit, thuswe havetotal intensity which passes through the second polaroid is

•

20=45°

"=>

• \_ 3 _ 3

polarizer is

/' =7C0s2 (02 - 0i) =Tq C0s2 0j C0s2 (02-0,)

•' •

2^4 ~ 8 thus (3/8)"' of the original light is transmitted out from the second polaroid. '

(b) Nowthe final intensity is 10percent ofthe intensity ofthe lightentering thefirstpolaroid thusthelightcoming outfrom

/Axis

Wave Qptics}

5^:¥;

(iv) Is shouldbe remembered that the transmitted intensity of unpolarised light will forall orientation ofpolariser

yx 1 3^ sin0 sm0'=yx= -x—

sheet whereas the intensity ofpolarised light varies from zero

3 3 = -xsin30®=-

to Jp. Thus intensity ofemerging light from polarising sheet will be:

)'=sin-'[^ / mac =—+L 2 "

and

Sol. 4 (A) The first diffraction minima of light waves of wavelength Xdiffracted bya single,longnarrowslit ofwidth b

According to given condition;

•^max

^^min

is given by 6sin0=X.

2

"

= 'i-

2

X

sin 0 = T

/=^ p 2

0

When b is decreased for same wavelength, sin 0 increases, hence 0 increases. Thus, width ofcentral maxima will increase.

3 '0

2

Sol. 5 (C) For a slit of width 6, light of wavelength X, when light falls on the slit, the diffraction patterns is obtained. The

For 0=45®,

2

P

first diffraction minimum occurs at the angles given by •X

sin 0 = T b

I, ^21,12 _ 51, Solutions ofCONCEPTUAL MCQS Single Option Correct

Sol. 1 (D) The phase difference(^) betweenthe waveletsfrom 27C

Fromthe equation, it isclearthat widthofthe centraldiffraction maximum is inverselyproportional to the width ofthe slit. One increasing the width size b, the angle 0 at which the intensity first becomes zero decreases, resulting in a narrower central band and ifthe slit width is made smaller, the angle 0 increases,

giving a wider central band.

the top edgeand the bottomedge ofthe slit is (}) = 'irid sin 6) K

where d is the slit width. The first minima of the diffraction

Sol. 6 (C) Brewster's angle is given as

pattern occurs at

1

tan I = sin0 =

d'

X 27ir, %

So

. Q

sm 0c

cot i = sin = 2tc tan i - -

Sol. 2 (C) The generalcondition forFrounhofer diffraction is

^«i. LX

Sol. 7 (B) Ultrasonic waves cannot be polarized as these are longitudinal waves.'

Sol. 3 (B) For first diffraction minimum a sin 0 —X • a

X

—

sin0

For first secondary maximum, we use asm a

3X -

Sol. 8 (A) As explained in article 6.8.8, ifan unpolarised light is converted into plane polarised light by passing through a Polaroid its intensity becomes half. Sol. 9 (B) Some crystals such as tourmaline and sheets of iodosulphate of quinine have the property of strongly

absorbing the light with vibrations perpendicular to a specific

|Wa^e Optics

,5i5l

direction (called transmission axis) transmitting the light with The sugar smaple dissolved in a ofwater is200 kg in which vibrations parallel to it. This selective absorption oflight is 190 kgispuresugar. Therefore, purity isgiven as called dichroism. 190

Sol. 10 (C) According to Malus' law, we use

X100 = 95%

200

/=/q cos^ 0=/q(cos^ 60°)

Sol. 15 (A) Ifinitialintensityofincidentlightis/Qtheintensity

oflight after transmission fi-om first polarid will be y and intensity oflight emitted fi"om

Sol. 11 (A) Accordingto Malus'law, we use

isgiven by Malus' law as

/ = /q cos^ 8

Itensityofpolarized light=^ Intensityofuntransmitted1ight ~ ~^ ^^

'

2

cos^0

Intensity oflight transmitted from last polaroid will be given by Malus' law as

P2=IiCos^ (9O°-0)

Sol. 12 (A) Optical rotation or optical activity is therotation oflinearly polarized light asittravels through certain materials.

7^., = —cos^0sin^0

It occurs in solutions of chiral molecules, solids with rotated crystalplanes and spin polarizedgasesof atomsor molecules.

P2~

Sol. 13 (D) For liquid A, we have

P2= •~sin^28.

2

2

h

(2 sin 0 cos 8)^

Lj =20 cm, 0] =38; concentration = C, Specific rotation is given as •

Sol. 16 (D) The resulting amplitude in the given four cases are la^, 2(2^ andAoq. Thus in case I andIVintensitywillbe

38°

maximumand equal.

iiQ ~ 20xCi Similarly, for liquidB, we have

Sol. 17 (B) Resulting intensity at points

1-2= 30cm, 02 =- 24°, concentration = C2

=>

Specific rotation is given as e,

(-24°)

L2C2 30XC2 The mixture has 1partofliquid.4 and2partsofliquid5. Ci':C2'=1:2 e={a,Cj'+a2C2')/

=>

C 3

(-24°) ^ 3OXC2

X—i- + -

^

2C, 2 3

=> 7^ = 5/ Similarlyresulting intensity atpoint5 is given as

/g =/, +/2 2-^/1/2 cos 02 x30

0 = 19^-16° = 3°

Thus, the optical rotation of mixture is + 3° in right hand direction.

Sol.14 (C) The strength of solution is given by C=-—

/X5

0 = 19°,/-20cm=0.20m S = 0.5 deg m^kg~' 19 0.20x0.5

=>

cos 7C

/g=/

Difference in lightintensities atpoints AandB is given as 7^-/g = 5/-/=4/ Sol. 18 (C) Frequencyofelectromagneticwavesgoingtowards .

Where the symbols have their usual meanings.

C =

=>

the approachingmirror is given byDoppler's effectas

8

Here,

/^ =/]+ + 2.^/j/2 cos8j

/,=/,+/,+2V7;}7cosf

a.,=

38° 20xq

is given as

c + v

w'= ——Wq

Frequency ofwavesreflected fi-om mirror and movingtowards source is also given by Doppler's effect as -3

w" =

•w c-v

fsie'

^ c

w"=

,,

^ c+.v

^

X

,

^'aye.Opficsl

•W„ "0 =

intensitywill be due to constructive interference of the light

I c+ u

[ r-y, 1^0

duetothesetwosourceswhichwillbe4/'andthisisgivenasI,

.

w"=

thus we have

,

r-L

TWo

4

cl 1-II

Sol. 23 (C) InYDSE, weknowfringewidth is givenas

c}

(1+ P w^ =

*^0

l-P

I--I

.

Wf, where p= ~

^

c

^ .

....

^

,

..

Sol. 19 (A) Path difference in the two lights from the silts at a

«i^i

. ih £ • d

2d

• ,

For missing wavelength, light waves should destructively interfere atthispoint soweshould have path difference between waves to be A =

2 '

3X2 2

b' 2d

or

^

2

and

2d

5X3

b'

2

2d

12x600

locations ofbright and dark fringes are given as

^

2 '

2

_

—V

j min

'

hp=—

^

^

max i 2 - mm -^max _ (^1+^2) _ 9

7^4

2

i

2

^ ^ _,

1

1 2 and i72

2 , —~r —2 a2 1

1

-

•.

.

2d

3X2 _ b'

and

^

Sol. 25 (B) Aswe know maximum andminimum amplitudes at j

Xi

Id

Sol.24 (A) «,X,, kJUI. \ i \ l Herewe IICIC W«5 use uas/«,/V, = — M2^

point in front of one of the slits is given as

^

d

=> D'=2D.

Sol. 26 (B) For a point source oflight we use or

or

and as weuse A

.4 oc —

y-^ =0.853

.

• • .

•*max

^

/ =0 853/ =0 853x4/ R

max

Sol. 27 (B) For aline source oflight, we use

The intensity at point A is given as

4 =/+/o +2/cos(l)=2/(l +cos(l)) =0.853 x4/ . jt . ^=4 ,

^

Thus corresponding to the above phase difference the path X

r

^ ^

•

.4 oc

•

Sol. 28 (C) Asweknow atthepoint ofsuperposition, resulting amplitude is given as

difference is TT.

,

Ai^= 'JA^ +A^ +2A -AcosQ

Sol.21 (A) It is already discussed that intensityat maximala isis ^ ^ ^ Ia^+A^+2A^ cosf—^ lual four times the individual intensity ifboth waves areof equal

^ V

V3 J

intensities. In this situation the intensity at minima is zero.

=> A,= .lA^ +A^+2A^(-l Sol. 22 (C) If each of the sources is producing an intensity/' at the point of intereference then at the central maximum,

_;

^

+ A^-

=A.

.

aye. Optics517i

Sol. 29 (B) In aYDSE setup the light intensityon screen due to any slit is directly proportional to its slit width, so we use _

/o=47oCOs2^

As Intensity X slit width

('•' .

A

Given in the question)

271

1

^=.4=2

2n

•We know that,

— Ar , ''

Sol. 30 (B) Since there is no shift in centr^m^imathe.pat^i difference introduced by the two sheets should be equal and

'A

In the YDSE setup, wehave•

nullify each other, so we have

A

where p, and ^2 are refraction indices ofthetwo sheets =>

t2

(Hi-1)

(1-5-1)

0.5

(1.25-1)

0.25

D

[From the above relation]

3

= 2 • y=

Zd

Sol. 31 (C) Increase in optical path isgiven as A =pr-/=(p~l)/.,

. (v-

Sol. 32 (B) and

PR = d PO = dstcQ-.

and

CO= PO cot 2Q = d sec0 cos20

path difference between thetwo rays is,

As: = CO -+ P0 = (dsec 0+ dsec 0cos 20) path difference between the tWo rays is, Aj) =7t((me is reflected, while another isdirect) Therefore, condition.for constructive interference should be '/////////////////////y/z/xA

Sol.34 (D) Ifaparallelbeamofwhitelightisincidentonthe plane, then there is.only one point where all light falls on the screen, which in the central maxima. Hence, position ofthe central maxima is at bisector ofthetwo slits, which is at I

6

distance above the point "0\

Sol. 35 (A) Path difference in air atpoint O, is given as Ax =[{5, - O- /) +tn^ _ (5-^ o)]t => • .Ax=[S^0-Sp)nY(n,~n2)t], ' =>

Ax=(«3-H2)?

Phase difference, ^

27C

Path difference inair

h X

X 3X'Ax = ~,'.— ... 2 2

sec 0'(L-F'c6s 20) ='— •

Solutions ofNUMERICAL MCQSSingle Options Correct

Sol. 1 (B) In YDSE setup angular width offringes on screen is given as

^

—^1 (2 cos^ 0) =— '2

0-^

, cos 0 y

X

d

' 6x10-^x180x7

cos 0 —

4d

0.1x22

Sol. 33 (C) In such type ofquestion, we should obtain phase difference (A(j)) at thegiven point •" ^Atp -

/= 4/nCOS^ 0 2 ;

= 0.344 mm.

Sol. 2 (A) In YDSE setup fringe width is given as XD

5893x10-^x200 0.08

= 0.1473 cm.

1

•

Wave Optibsj

15m

Sol. 3 (C) Difflaction Limit ofresolutionofthe telescope is

- =tan60°= ^/3

given as

V

a

c

= X

a

3x10^

\.22lx d =

v= >/3 X10®m/s.

a

1.22x5x10"'x8xl0^16 =>

Sol. 8 (B) For the path difference due toplate iscompensated bythe physical path difference, we use

£/=1.95xl0"m.

xd

(1^-1)'=^

Sol. 4 (D) In single slit difflaction pattern, angular separation offirst darkfringe from central bright fringe is given as

\

xd

•

h sin 0 = X,

For small angle we can use sin 0 = 0

p=l +

Separation distance offirst dark fringe from central maxima is

0.5x0.2 10x0.05

= 1 + 0.2=1.2

given as x —D^

Sol. 9 (B) Weuse, from the figure

Uk

600x10"^ x2

b

a-3 1x10*

x =

h = r+o

= 12x lo^m

So, distance between thefirst darkfringes oneither side ofthe central bright fringe s = 2x

5 = 2x12x10-^111

=>

5 = 24x 10^m=2.4nim.

Sol.5 (C) Width of central maxima is given as 2DX

'

'

Moreover, If slit width is reduced to half and wavelength is changed to 6000A, newwidthof centralmaximais givenas

,

2D

/^= r+24 jj + r=90°

..•0)

From Eqs. (i) and (ii), we get

/, + (i,-24)=90°

6000

^^ "(^j/2) ^4000"^^

Sol. 10 (D) ByBrewster'sIaw, wehave p=tan Qp =3. p=tan 54.74°

Sol. 6 (C) "After first polaroid theintensity reduces tohalf. If =>

p=1.414

1q is the initial incident intensity then after next four polaroids, For an equilateral parism, we use prism angle 60° and the angle intensitywill become

7=:^.CDs'* (60°)

ofminimum deviation is given bytherelation . r^ + 5„ sin

-

2

sm||

2 (4)^ 1=

512

Sol. 7 (C) From Brewster's law,wehave

"p = tani^

1.414 =

. f60° 2

fWave Optics

1.414x1

Sol. 14 (A) Ifshift offringes isx then we use

. (60'> + 5, sin

-

2

D =

sin

_i_

("60^ + 5. ^

_

. f60°+5^^

-ys = sm

•Jl

I,

a: =

(1.5-l)xlO-3xlOO

=> ^

5;25

=0.2cm.

2 Sol. IS (B) Angular width ofcentral maxima in diffraction

60°+ 5.

pattern is given as

45° =

=>

20=30°

Where we have

5_=30°

=>

Sol. 11 (B) Distance offifth bright fringe is given as

sin 30° =7b

XD . 5x6.5x10"'xl 10

sin30°

b =12xlO-7ni

Distance ofthird dark fringe isgiven as

Sol. 16 (C) For angular positions ofcentral minima, we use

^ ^ ^ 5x6.3xl0~'xl =16.25xi0-»m X3d-(2«-1)--2

^

d

6000x10"'° = 12000 xlo-JOm 1/2

b =~

-32.5xl0-4m

-3

6sin 9 = «A,

2x10

-*^56-^30® 1-63 mm.

=>

b-~=nX

Sol. 12 (A) As ratio ofslit widths =Ratio ofintensities => b = nX— X

a-x

=>

D =

nX

9

=> "J" - T

^2

4 D=

a\

3

a2

2

Maximum and Minimum amplitude are given as

(3-2)' ~ 1

^

_ =>

c+v

V'

X

AX= ~xX c

Sol. 13 (B) In case ofmoving light source, apparent frequency will be

=2.0m.

V^ M c

^ ^max = (gl +g2)^ ^ (3 +2)' _ 25 (fll-a2)'

— 1x6000x10

Sol. 17 (B) ByDoppIer's effect in light we use

'^max ~'3j+Ct2'~3+2 = 5 ''mi„ = 3-2 = l

•^min

3xl0"'x4xl0,-3

AX AX

1.2x10^

12 10^

X100=0.4i.e.,0.4%.

Sol. 18 (C) Diffiaction minima on either side ofcentral maxima

-v

c

is given as —

V

c+ v

X

c

X'

6 sin 0 = nX =>

X' =

Xc

5500 xc

c + v

c+0.5c

AX = (5500-3667)A=1833A

b ~=nX

= 3661k

r.

bx

D= — =

2xl0"*x5xl0"' 1x5000x10"

^10

=2.0m

Wave OpficsJ

;520

Sol. 19 (D) Dopplershift is given as

Xv

Totalchangeis 6X = 2AX = 2 c• cjk v =

2X

Timeperiodofrevolution isgivenas Substituting values, we get T=

i2x2jt

^

^

dX

T=>

47i/?X

^5^.

«X=(p.-l)f

(n-l)/

4x3.14x6.95x10^x0.59x10 3xl0^x8xl0"'^x86400

days

Sol.24 (A) Forfirst diffraction minima at angle0, wecanuse f^(sin0-sin0Q)=fiX,.'_, "i.

(1.5-l)x2xlO-'

=

T =

r= 24.8 days a 25 days

Sol.20 (D) Shiftinno.offi-ingesisgivenby

X

2X

X —

c

..

^ „

-2nRx

X

r= 25 days

=> w=

2iiR

TK—fringes = 2 5000x10'^® ^

Here 0^ isthe angle atwhich central maxima is obtained. , Sol. 21 (B) When source is fixed and observer is moving towards it,byDoppler's effect theapparent frequency isgiven as

,

For one side ofcentral maxima, we use

sin0j = sin0o+

.

c + a

v'=

sin 0j=O.5+^^ =0.55

•V

When source is moving towards observer at rest, we'use c

v"=

.

V

c + a

1+ ^

0, = 33.37^

Sol. 25 (B) Forfirst diffraction min.rfsin6 = X ,

v = c

c-a

c-a

=>

and if angle is small, sin 0 = 0

1-^

:

7

did=X

Angular width offirst dark fringe from central maxima is -1

v" = c

1+^

1+^

1-^

0-^ ® d

c

Full angular width ofcentral maxima is 2a\ =^>

Av = v-v=

•

C

2a • ' h

•

^

Av 0.5x1000 = 250 ms a = X— =

=>

a = 900 km/hr

Also we can use

'

.

is given as

•

'

>

.••

'

..."

-•

0.3 X10-3x0 = 6000 xlQ-io

.

0 = 2xio-3rad.

w

AX _ y_

••

•

-

--

w

?^'=6000 x 0.7 = 4200A.

Sol. 26 (C) BySnell'slawwehave

Sol. 23 (C) By Doppler's effect, we have X

21^ X

w' •,Xf=X —

i sin 0 = «X

sin/

'

ix-

AX—

— sinr

c sm r =

^

2X'

'

w-

Sol. 22 (D) Insingle slitdiffraction pattern, n^.order minima

=>

-

c

•

.

sin I _ 0.788 = ' ".-4.33---

0.6

,

Change in wavelengthfor two edges= ±AX

cos

r= Vl-sin^ r - -*/l-(0.6)^ '-0.8 -

lyVave Optics 521

For constructive interference on reflection 2W (n-l)tZ) d~d

2^^cos/•=(2« +l) j _ (2n +l)^

,

(2;n-l)x0.6 '

(M-l)r

=

4ncosr ~ 4x1.33x0.8 '

X= 0.5892xl(^6m=5892A.

=> r=0.14(22+l)^m

Sol. 27 (C) For diffraction minima due to asingle slit ofwidth b, we use 6 sin 0 = nX

^ _

nX _ 2x6000x10-^0 ,

Sol.30 (B) Fringe width in YDSE setup is givenas P=^ Difference infringe widths isgiven as

sin 8

=> b=

(I.6-l)xl.964xl0'^ 2

P1-P2-

sin 30°

2x6000x10"^®

, ,

,

=4x6000x10-'®

d

3x10-^x10"^

,

,

5x10^ => i> = 24x l0-^ni = 24x lO-^cm.

Sol. 28 (D) Fringe width inYDSE setup isgiven as

Sol. 31 (A) Fringe width on screen isgiven as „ XD

P- ^ -

p=¥

5000xl0-^°xl

5x10^

=O.OOIm=Imm

Shift in fringe pattern xdue to insertion ofglass plate is given

Angular fringe width is givenas

as

xd.,.

D- d

Angular widths oftwo wavelengths are given as (p-l)Dr 0

= —

(l.S-Dxl.SxlO-^xl

x =

e - —

5x10"

-1.5x 10"^m=1.5mm=I.5p

01 01

Xi

ft.

5890x^ =6479A.

Thus after insertion, at screen center there will be adark fringe so intensity will be zero.

Sol. 32 (A) Path difference at a distance x from the central maximaon screenis givenas

^

xd

A=-^ =

10"^x2xl0~^

—

8OOOA

Sol. 29 (Q Shift in fringe pattern due toinsertion ofa sheet in

For bright line at this position, we use

front of slits is given as

8000 = njX, If «, = 2,weget

xd

•

Xj=4000A Thus second bright line for visible region ispresent.

X =

Sol. 33 (C) In aliquid the wavelength oflight changes to

When distance between screen and slit is doubled, thanfringe width becomes

. _ ^air i" Pi

2XD

P=

As per the given condition, we have P=x

In YDSE setup thefringe width is given as

-

_ Wave Optics}

522

Whenthe setupis submerged in a liquidfringe widthchanges

Sol. 38 (D) Diffraction minima oneither sideofcentral maxima

to

is given as 6 sin 0 = kX

_ KirD d

M

„

^

6300x10"^" xl.33 P= 1.33x10"'

nX

6328x10"^® xl

a

0.2x10^

sin 0 = — = —T^TT^—

=> sm0=3164x 10^rad=0.003164radian

= 0.63 mm.

Angular widthofcentral maxima isgivenas

Sol. 34 (D) Onscreen distinguished fringes areseen untilthe bright fringe ofonewavelength overlaps with thedark fringe of

20 = 2xO.OO3164x

180®

=0.36®.

7C

other so we use

Sol. 39 (B) Location of diffraction minima in single slit nkj

n+

1

diffraction pattern is given as 6 sin 0 =

"4j X, n

fr —=nX

A-i 1 => X= —7:

nD

1

5895-5890

5

2n

5890

5890

2xl0"'x6xl0 ^

.

1x2

^ «=^ =589

Sol.40 (D) Angularspreadofcentralmaximain a singleslit diffraction patternwith slitwidthb andforlightofwavelength

Sol.35 (B) Formaximum reflection ofa specific wavelength,

Xon either side is given by

we use

0=i = irad

2pf=(2n+l)| X (2« + l)

(2k + 1)x5320x10

-10

Sol.41 (B) Fringe width in YDSEsetupis givenas

2x5x10"^ xlO"

XD

p=1.33

5x10"^ x2 10

/

-3

= 10~^m=lmm.

Sol. 36 (A) ByBrewster'slaw, weuse

p^tanip

Sol. 42 (D) ByDoppler's effectin light shift in wavelengthis

tani^=1.54

=>

given as

i^ =tan-'1.54 =57®

At polarizing angle we use

AX=-X c

r +r^ =90® => r=90®-ip =>

cAX v =

=90®-57=33®

Sol.37 (B) Fordiffraction minimabya singleslit ofwidth6,

3x10^^x1.9x10.-10 =4>

v =

,-10

= 100kms '.

5700x10

we use

b sin 0 = nX,

Sol.43 (D) Fringewidth in YDSEsetupis givenas

asin0 X=

«

AX)

X=1.0x 10-5xsin30® X.= 1.0 X 10-5

1

As in both setups fringe widths are equal, we use XjDj

X=0.5x 10-5=5x10^ cm

X=500A

P=

d,

X,2jD2

fWave Optics'

A

^2

2 2

Sol. 49 (B) Fringe width, 03=—GC>. a

Sol. 44 (C) Distance of bright fringe in aYDSE pattern from central maxima is given as

wavelength is decreased from 600 nm to 400 nm, ^ fringe width will also decrease by a factor of - or - orthe

x = nk-T={n-\-\)V^

u

"

.

nuber offringes inthe same segment will increase by afactor

=> «> K=5andx=/jX^ a

segment =12x-2 =ig

^

Concept: Since moc^, and ifms'fiapparatus is immersed inaliquid of

^

=5x12000x10-'° x2^

. refractive index p, the wavelength landthus the fringe width

i^-2x io-3uj=2mm.

will decrease ptimes.

Sol.45 (A) FringewidthsPandp'inaYDSEsetupisgivenas ^VANCEMCQs One orMore Option Correct P'=~^

=> R--3'=

(-^CjD) Glass slab introduces extra path differenced

fringe pattern will get shifted towards covered slit. Due to

~-Q ^ d

reduced intensity ofone ofthe slits, intensity ofbright fringe decreases and that of dark fnnge increases. Fringe width is

"

given as

KD-D")

''"IFpT

-p=^ d

, 6000 X10"^® X5X10"^

Hence fringe width willremain constant. = l'^'m=lmm.

Sol. 2 (B,D) For sustained interference the phase difference

Sol. 46 (C) First polariser, polarises the light and hence remain conststant between the interfering waves, ifit intensity ofli^t reduces by 50%. varies with time then the resulting intensity at the point of interference will change with time so the interfering waves Sol. 47 (C) The concentration ofsolution is given as ' coherent, >^dth change in amplitude and intensity of C=

'

two waves, only theresulting intensity will beaffected but

it does not change, hence option (B) and (D) are not necessary for sustained interference.

e

9.9

^ iC ~ 2x0.075 => 5=66"

As analyzed in article 6.4.5 options (B), (C) and (D) are correct.

Sol.48 (D) Theangularwidthofinterferencefringes inYDSE L"dS setup for the two wavelengths are given as

^

a —j • o A ~ tt Sin 0

9, ^ ^ = 02

_

«

and 9,^ =^d

A.2

02

fr«q"™cy ofthe waves is/= iO'Hz and wavelength is

given as '

- 3x10^

=30Qm

Case-1: If0 = 9O°, wehave A = rf

Thus phase differencebetween waves is

^=5890xf|=6479A. A

-Wave Optics

;524

Sol. 7 (A, D) When light is reflected from theboundary ofa 2k

A