TB Engineering Physics - I | Pages-444 | Code-801 | Edition- 21st | Concepts + Theorems/Derivations + Solved Numericals + Practice Exercises | Text Book

Table of contents :
Engineering Physics I (B.Tech)
Dedication
Preface
Syllabus
Brief Contents
Detailed Contents 1
Detailed Contents 2
Detailed Contents 3
Detailed Contents 4
Detailed Contents 5
Detailed Contents 6
Detailed Contents 7
Detailed Contents 8
Detailed Contents 9
Detailed Contents 10
Unit-I
Unit-I Relativistic Mechanics
Unit-II
Unit-II Electromagnetics
Unit-III
Section–A : Interference
Section–B : Diffraction
Unit-IV
Section–A : Polarisation
Section–B : Laser
Unit-V
Section–A : Fibre Optics
Section–B : Holography
Appendix-A
Appendix-B
Examination Paper 1
Examination Paper 2
Examination Paper 3
Examination Paper 4
Examination Paper 5
Examination Paper 6

Citation preview

E ngineering

P hysics -I for (B.E. / B.Tech. / B. Arch. students of first semester of all Engineering Colleges affiliated to Mahamaya Technical University (MTU), Noida) (As per revised syllabus, w.e.f. 2012-13)

By Dr. S.K. Gupta Associate Professor & Head,

Department of Physics D.N. P.G. College, Gulaothi C.C.S. University, Meerut (U.P.)

Sankalp Gupta (B.Tech.)

KRISHNA Prakashan Media (P) Ltd. KRISHNA HOUSE, 11, Shivaji Road, Meerut-250 001 (U.P.), India

Jai Shri Radhey Shyam

Dedicated to

Lord

Krishna Author & Publishers

Preface The interest shown for the immensely popular title ''Engineering Physics-I'' by professors and engineering undergraduates has encouraged us to update the book in accordance to the syllabus framed first time by Mahamaya Technical University (MTU) Noida, which came into operation with effect from 2011. We feel highly delighted in presenting the thoroughly redesigned and revised new edition of Engineering Physics-I based on the syllabus recently framed (w.e.f. 2012-13) for B.E./B.Tech./B.Arch students of first semester of all Engineering Colleges affiliated to Mahamaya Technical University (MTU), Noida. This year, for the first semester of B.Tech course MTU, Noida has introduced new advanced topics, such as recapitulation of vector algebra, anti-reflection and high reflection coatings, polarization by reflection and refraction, sheet polarizer etc. All these topics are incorporated in a very simple and explanatory way, in this present edition. In view of the explosive developments in various areas of technologies in this fast growing century, the knowledge of physics is necessary for an engineering student, if he has to cope up with multi-disciplinary technologies. The revised edition is rewritten in its entirely improved form and the illustrations are redrawn. In the light of meaningful comments and suggestions received from the teaching faculties, colleagues, friends and students from various corners of the country some worthwhile additions, subtractions and modifications have been made in numerous articles of different sections of existing units to provide extensive coverage of the field. Moreover, present edition has been rewritten retaining the original theme of emphasis on concepts and updated in the light of recently revised and framed syllabus by the Mahamaya Technical University Noida along with University examination question papers of previous years. The revision of this market leading text-book maintains its classic strengths, contemporary approach, flexible chapter construction, simplicity, lucidity and clear exposition. In addition, authors have further enhanced the book's traditional strength and conceptual understanding. We trust this revised edition, in present form, should satisfy both average and brilliant students. Our debts are enormous to esteemed colleagues, professors, friends and students from all part of the country who readily interacted with us for improving the quality of the text. We hope they would continue to help us in making further improvements. We owe a special debt to Prof. D.S. Srivastava, Head, Dept. of (vii)

Physics, Mangalayatan University, Beswan, Aligarh and former professor, Z.H. College of Engg. & Tech., AMU, Aligarh. We thank him for going through the entire matter and making invaluable suggestions and corrections for the enhancement of the ability and utility of the book. We would like to express our sense of appreciation to our family members as without their unconditional support, active encouragement, complete co-operation and honest sacrifices, the book could not be finalized. We would like to thank Dr. R.K. Pandey, Professor & Head. Dep't of Physics, VIET, Dadri, Dr. K.P. Singh and Dr. S.K. Tomar, Dep't of Physics, ITM Aligarh for their help during the preparation of this revised edition. We unreservedly acknowledge our sincere gratitude to Sh. S.K. Rastogi, Managing Director, Er. Sugam Rastogi, Executive Director, Krishna Prakashan Media (P). Ltd., Meerut and their subordinate staff for publishing, composing and printing the present edition in the recorded short time. Any genuine and authentic suggestions for rectifying shortcomings and for the substantial improvement of the book shall also be hearty welcome.

–Subodh Kumar Gupta –Er. Sankalp Gupta 98/6, Shastri Nagar Meerut (U.P.) Ph. 0121-4003868

(viii)

Syllabus E ngineering

P hysics -I

for (B.E. / B.Tech. / B. Arch. students of first semester of all Engineering Colleges affiliated to Mahamaya Technical University, (MTU) Noida. (As per revised syllabus, w.e.f. 2012-13)

L-T-P 3-0-2

Unit- I : Relativistics Mechanics Inertial & non-inertial frames, Galilean transformation equations, Michelson-Morley experiment, Einstein ’s postulates, Lorentz transformation equations, Length contraction & time dilation, Addition of velocities, Variation of mass with velocity, Mass energy equivalence. Unit – II: Electromagnetics Recapitulation of vector product, Gradient, Divergence & Curl, Statement and explanation of Gauss divergence & Stokes theorems, useful vector identities. Maxwell ’s equations (Integral and differential forms), Equation of continuity, Transverse nature of EM waves, EM - wave equation and its propagation characteristics in free-space, Poynting vector. Unit- III: Interference Spatial and temporal coherence, Interference in thin films of uniform thickness and in wedge-shaped film (qualitative), Newton’s rings, Anti reflection and high reflection coatings (qualitative), Interference filters(qualitative). Diffraction: Single and N- slit diffraction, Grating spectra, Rayleigh’s criterion of resolution, Resolving power of grating. Unit-IV: Polarization Polarization by reflection and refraction, Double refraction, Nicol prism, Sheet polarizer, Production and analysis of plane, circularly and elliptically polarized lights, Basic concepts of optical activity, Polarimeter (Half shade). Laser: Spontaneous and stimulated emission of radiation, Einstein’s coefficients, Construction and working of Ruby, He-Ne and semiconductor lasers, Important laser applications(qualitative). Unit–V: Fiber Optics and Holography Fundamental ideas about optical fibers, Types of fibers, Acceptance angle and cone, Numerical aperture, Propagation mechanism and communication in optical fiber, Attenuation, Signal loss in optical fiber and dispersion. Basic principles of holography, Construction of hologram and wave reconstruction, Applications of holography (qualitative).

(ix)

Brief Contents

Unit-I Relativistic Mechanics.............................................................................(01-82) Unit-II Electromagnetics.....................................................................................(83-148) Unit-III Interference and Diffraction...........................................................(149-266) Section–A : Interference Section–B : Diffraction

Unit-IV Polarisation and Laser ......................................................................(267-352) Section–A : Polarisation Section–B : Laser

Unit-V Fibre Optics & Holography................................................................(353-412) Section–A : Fibre Optics Section–B : Holography

$ $

Appendices....................................................................................................(413-416) Examination Paper ..........................................................................................(01- 08) (x)

Detailed Contents Unit-I Relativistic Mechanics.............................................................................(01-82) Introduction

03

Frame of Reference

04

Galilean Transformation Equations or Galilean Transformation for the Position, Velocity and Acceleration of a Particle

05

Failure of Galilean Relativity

07

Michelson-Morley Experiment and Its Outcome Background (The ether hypothesis) Experiment

09 09

10

Explanation and Interpretation of the Negative Result Einstein Novel and Revolutionary Idea

12

13

Significance of Negative Results

13

Einstein’s Postulates of Special Theory of Relativity

14

Postulate I: The Principle of Equivalence (or Relativity)

14

Postulate II: The Principle of Constancy of the Speed of Light Lorentz Transformation Equations

15

Inverse Lorentz Transformation Equations (Lorentz-Fitzgerold) Length Contraction Proper Length

14

16 24

25

Time Dilation (or Apparent Retardation of Clocks) Proper and Non-proper Time

31

33

Experimental Verification of Time Dilation: An Example of Real Effect Concept of Simultaneity: Relative Character of Time

33

(Relativistic) Addition of Velocities or Velocity Addition Theorem Consistency of Einstein Second Postulate

33

43 (xi)

42

Variation of Mass with Velocity Derivation

52

52

Experimental Verification

54

Relativistic Momentum and Force

55

Mass-Energy Equivalence (Einstein’s Mass-Energy Relation) Limiting Value of Relativistic Kinetic Energy Experimental Verification

60

61

62

Relativistic Relation between Energy and Momentum

62

Relativistic Relation between Kinetic Energy and Momentum Massless Particles

$ $ $ $ $

64

Exercise

75

Question Bank

75

Short Answer Type Questions

77

Unsolved Numerical Problems Answers

63

78

81

Unit-II Electromagnetics.....................................................................................(83-148) Introduction

85

Recapitulation of Vector Product Triple Product

86

87

Scalar and Vector Fields

89

The Gradient of a Scalar Field

89

The Divergence of a Vector Field

91

Line Integral Surface Integral and Volume Integral Curl of a Vector Field

93

94

Gauss- Divergence Theorem or Divergence Theorem Stoke's Theorem

98

Useful Vector Identities

100

Electrodynamics before Maxwell

101

Concept of Displacement Current

102

Maxwell’s Equations

97

103

Physical Significance of Maxwell's Equations

104

Derivations of Maxwell's Equations in Differential Forms

(xii)

106

→

→ →

→

Maxwell's First Equation, div D = ρ or ∇. D = ρ or div E = →

ρ ε0

→ →

Maxwell's Second Equation, div B = 0 or ∇. B = 0

106 107

Maxwell's Third Equation or Faraday's Law of Electromagnetic Induction →

→

Maxwell's Fourth Equation or Curl H = J +

→

→ → ∂D ∂ E or curl B = µ0  J + E0  ∂t ∂t   

Maxwell's Equations in Integral Form Equation of Continuity

108

→

109

111

113

A Note on Continuity Equation Electromagnetic Waves

114 116

Poynting Theorem and Poynting Vector

116

Electromagnetic Waves in Free Space

119

Solution of Plane Electromagnetic Wave (Transverse Nature of E.M. Waves)

121

Electric and Magnetic Vectors are Mutually Perpendicular to Each Other and also to the Direction of Propagation of Wave Vector

122

Energy Flow in Plane Electromagnetic Wave (Poynting Vector) in free space Energy Density in Plane Electromagnetic Wave in Free Space Nature of Electromagnetic Waves

125

126

Propagation of Plane Electromagnetic Wave in Free Space Momentum of Electromagnetic Waves

144

Radiation Pressure and Energy Density

145

$ $ $ $ $

Exercise

127

146

Question Bank

146

Short Answer Type Questions

147

Unsolved Numerical Problems Answers

124

148

148

Unit-III Interference and Diffraction...........................................................(149-266) Section–A : Interference .................................................................151-198 Introduction

151

Interference of Light

151

Coherent Sources

152

No Interference by Two Independent Light Sources (xiii)

152

Incoherent Sources

152

Spatial and Temporal Coherence Temporal Coherence

152

153

Purity of a Spectral Line

154

Spatial Coherence

155

Non-realization of Strictly Monochromatic Light in Practice Interference by Two Independent Laser Beams

157

157

Types of Interference (Formation of Coherent Sources in Practice) Methods of Obtaining Coherent Sources Change of Phase on Reflection

158

159 160

Interference in Thin Films

160

Interference Due to Reflected Light

160

Condition of Maximum and Minimum Intensities Interference Due to Transmitted Light

162

162

Condition for Maximum and Minimum Intensities Colours in Thin Films

163

164

Colours in Thick Film and Blackness of an Excessively Thin Film in Reflected White Light Interference Colour Pattern on the Surface of a Soap Bubble Wedge Shaped Film Fringe Width

165

171 172

Determination of Wedge Angle θ

173

Necessity of an Extended Source

174

Newton’s Rings

164

174

Experimental Arrangement

175

Theory (or Explanation of the Formation of Circular Newton's Ring) Diameters of Bright and Dark Rings For Bright Rings

176

For Dark Rings

177

175

176

Newton's Rings by Transmitted Light

178

The Rings Observed in Reflected Light are Exactly Complementary to those Seen in Transmitted Light The Rings in Transmitted Light are Much Poorer in Contrast Than Those in Reflected Light Determination of Wavelength of Sodium Light Using Newton's Ring Determination of Refractive Index of a Liquid Determination of Optical Flatness

180

181

(xiv)

179

179 179

Newton's Rings Formed by two Curved Surfaces

181

Effect of Introduction of Liquid between the Plate and Lens on Newton's Rings

182

Effect of Increasing the Distance between Lens and Plate or Lifting up the Lens from the Flat Surface Effect of Placing the Lens on Silver Glass Plate or Mirror

183

183

Newton's Rings are Circular but Air-wedge Fringes and Straight

183

The Effect of Placing the Concave Surface of the Plano-Concave Lens Towards the Plane Glass Plate 183 Effects of Using a Lens of Small Radius of Curvature Effect of not Using Monochromatic Light

183 184

Newton's Rings with Bright Centre due to Reflected Light Newton's Ring with White Light

184

184

Anti-reflection Coatings or Non-reflecting Films

194

High-Reflection Coatings or High Reflection Films

197

Interference Filters

197

Section–B: Diffraction ....................................................................199-266 Introduction (Diffraction of Light)

199

Distinction between Fresnel and Fraunhofer Diffractions

200

Resultant of n Simple Harmonic Motions of Equal Amplitude and Period and Phases Increasing in Arithmetic Progression

201

Fraunhofer Diffraction at a Single Slit (Single Slit Diffraction) Explanation

202

203

Positions of Maxima and Minima Secondary Maxima

204

204

Spread of Central Diffraction Maximum Effect of Making Slit Narrower

206 206

Distinction between Single-slit Diffraction Pattern and Double Slit Interference Pattern Fraunhofer’s Diffraction at a Double Slit (Double Slit Diffraction) Effect of Increasing the Slit Width ‘a’

216

219

Effect of Increasing the Slit Separation ‘b’

220

Effect of Increasing the Wavelength ‘λ’

220

Missing Orders in Double Slit Fraunhofer Diffraction Pattern

220

A Plane Transmission Diffraction Grating (N-Slits Diffraction)

222

Theory of Grating

223

Secondary Maxima

225

Angular Half-Width or Width of the Principal Maximum Effect of Closeness of Rulings Effect of Width of the Ruled Surface

226

227 227

Advantage of Increasing the Number of Rulings on a Grating (xv)

227

207

Formation of Multiple Spectra with Grating

228

Condition for Missing Order or Absent Spectra with a Diffraction Grating Maximum Number of Orders with a Diffraction Grating

228

229

Determination of Wavelength of Light with a Plane Transmission Diffraction Grating Measurement of θ

230

231

Dispersive Power of a Plane Diffraction Grating Linear Dispersive Power of a Grating

232 233

Chief Characteristics of Grating Spectra

233

Difference between Interference and Diffraction

247

Resolving Power of an Optical Instrument

247

The Rayleigh Criterion of Resolution

247

Resolving Power of a Diffraction Grating

249

Difference between Dispersive Power and Resolving Power of a Grating

251

Section-A

$ $ $ $ $

Exercise

257

Question Bank

257

Short Answer Type Questions

259

Unsolved Numerical Problems Answers

260

261

Section-B

$ $ $ $

Question Bank

262

Short Answer Type Questions

263

Unsolved Numerical Problems Answers

264

266

Unit-IV Polarisation and Laser ......................................................................(267-352) Section–A : Polarisation...................................................................269-322 Introduction

269

Polarisation of Light

269

Mechanical Experiment to Demonstrate Polarisation of a Wave Transverse Nature of Light Waves

270

Methods of Production of Plane Polarised Light Plane Polarised Light by Reflection Brewster’s Law

272 272

273 (xvi)

269

Production and Detection of Plane Polarised Light by Reflection Law of Malus

274

Plane Polarised Light by Refraction (Pile of Plates) Doubly Refracting Crystals

280

Geometry of Calcite Crystal

281

Ordinary and Extraordinary Rays

279

281

Phenomenon of Double Refraction (In a Calcite Crystal) Polarisation by Double Refraction Nicol Prism

273

282

283

285

Huygen’s Theory of Double Refraction

287

Optic axis inclined to the refracting face and lies in the plane of incidence

287

Optic axis is parallel to the refracting surface and lies in the plane of incidencc

288

Optic axis is perpendicular to the refracting surface and lies in the plane of incidence Optic axis is parallel to the refracting surface and perpendicular to the plane of incidence

289 290

Mathematical Treatment for the Production of Plane, Elliptically and Circularly Polarised Lights Superposition of Two Linearly Polarised Waves with Their Optical Vectors Parallel Superposition of two linearly polarised waves with their optical vectors mutually perpendicular Special Cases

293

Plane Polarised Light

293

Elliptically Polarised Light

293

Circularly Polarised Light

293

Phase Retardation Plates

295

Distinction between Quarter Wave Plate and Half Wave Plate Sheet Polariser

303

1. Wire Grid Polariser 2. J-Sheet Polariser

303 304

3. H-Sheet-Polariser 4. K-Sheet Polariser

304 305

Applications of Sheet Polarisers

305

Production of Plane, Circularly and Elliptically Polarised Lights Plane Polarised Light

306

306

Circularly Polarised Light

306

Elliptically polarised light

307

Detection of Plane, Circularly and Elliptically Polarised Lights Plane Polarised Light

296

307 (xvii)

307

290 291 292

Circularly Polarised Light or Distinction between Circularly and Unpolarised Lights

307

Elliptically Polarised Light or Distinction between Elliptically and Partially Plane Polarised Lights Conversion of Elliptically Polarised Light into Circularly Polarised Light Analysis of Polarisation in a given Beam of Light Basic Concept of Optical Activity

313 313

314

Polarimeters

314

Laurent’s Half Shade Polarimeter Apparatus

309

309

Experimental Demonstration of Rotatory Polarisation Specific Rotation

308

315

315

Action of Laurent's half shade plate

315

Determination of Specific Rotation of Sugar Solution Biquartz Polarimeter

316

317

Determination of Specific Rotation

317

Strength of Sugar Solution

318

Section–B: Laser............................................................................323-352 Introduction

323

Spontaneous and Stimulated (Induced) Emission of Radiation Absorption of Radiation

324

324

Spontaneously Emission of Radiation

325

Stimulated or Induced Emission of Radiation

325

Difference between Spontaneous and Stimulated Emissions

325

Spontaneous Emission and Einstein’s Coefficient of Spontaneous Emission Stimulated Emission and Einstein’s Coefficient of Stimulated Emission Transition Probabilities

326 326

327

Einstein's Relation between Spontaneous and Stimulated Emissions or Transition Probabilities Condition Necessary to Achieve Laser Action Various Pumping Methods

329

Optical Pumping

329

Electrical Pumping

329

Inelastic atom–atom Collisions

329

Chemical Pumping

330

Metastable States

330

Principle of Laser Action

328

330

[Active System, Population Inversion (Inverted Population and Pumping) ] Active System and Cavity Resonator

330

Population Inversion (or Inverted Population)

331 (xviii)

330

327

Types of Lasers

332

Production of Laser

332

Construction and Working Principle of Ruby Laser

333

Construction and Working of Helium-Neon Laser (Gas Laser) Characteristic (or Properties) of Laser Beam Difference between Ordinary Light and Laser Radiation Applications or Uses of Laser Radiations Semiconductor Laser

334

335 336

336

339

Galium-Arsenide Laser

339

Comparison with Other Lasers

340

Section-A

$ $ $ $ $

Exercise

345

Question Bank

345

Short Answer Type Questions

347

Unsolved Numerical Problems

348

Answers

349

Section-B

$ $

Question Bank

350

Short Answer Type Questions

351

Unit-V Fibre Optics & Holography................................................................(353-412) Section–A : Fibre Optics .................................................................355-398 Introduction

355

Fundamental Ideas about Optical Fibres History

356

356

Optical Fibre

357

Fibre Fabrication

358

Principle of Operation of a Fibre Types of Fibres

359

359

Step Index Multimode Fibres (MMF)

359

Step Index Single Mode or Monomode Fibres (SMF) Graded Index Multimode Fibres (GRIN)

361

362

Acceptance Angle and Acceptance Cone of a Fibre Numerical Aperture

363

365

Numerical Aperture for Graded Index Fibres

366

Number of Modes and Cut-off Parameters of Fibres (xix)

366

Propagation Mechanism in Optical Fibres 369 Communication in Optical Fibre 370 Attenuation and Signal Losses in Optical Fibres 377 Absorption Losses 377 Rayleigh Scattering Losses 378 Waveguide Scattering Losses 379 Bending Losses 379 Signal Dispersion 382 Modal Dispersion or Intermodal Dispersion 382 Dispersion in Graded Index Fibre 384 Intramodal or Chromatic Dispersion 386 Material Dispersion 386 Waveguide Dispersion 388 Total Dispersion and Maximum Transmission Rates 389 Optical Cables 395 Advantages of Optical Fibre Over a Coaxial Cable 395 Disadvantages of Optical Fibres 397 Applications of Optical Fibres 398

Section–B: Holography....................................................................399-412 Introduction 399 Basic Principle of Holography 400 Construction and Reconstruction of Image on Hologram Mathematical Treatment of Holography 402 Requirements in Making Hologram 404 Main Characteristics or Features of a Hologram 404 Orthoscopic and Pseudoscopic Images 405 Types of Holograms 406 Applications of Holography 406

401

Section-A

$ $ $ $ $

Exercise

408

Question Bank

408

Short Answer Type Questions

409

Unsolved Numerical Problems

410

Answers

411

Section-B

$ $ $ $

Question Bank Short Answer Type Questions

412 412

Appendices..................................................................................................................(413-416) Examination Paper .......................................................................................................(01- 08)

(xx)

Unit

I

R elativistic M echanics

Relativistic Mechanics ........................................01–74 I nertial and N on-inertial F rames M ichelson- M orley E xperiment E instein's P ostulates L orentz T ransformation E quations L ength C ontraction T ime D ilation A ddition of V elocities V ariation of M ass with V elocity M ass-energy E quivalence

Exercise ...........................................................75–82 Q uestion B ank S hort A nswer T ype Q uestions U nsolved N umerical P roblems A nswers to Unsolved Numerical Problems

3

U nit

I

R elativistic M echanics

Introduction

T

h e age old classical mechanics is entirely based on Newton’s laws of motion and gravitation. Initially it was thought that Newton’s second law of motion is universally applicable at all speeds. But new

experimental evidences in the beginning of twentieth century revealed that Newton’s second law is valid only for the objects moving at low speeds and fails when applied to the objects moving with high velocities comparable with the velocity of light. The failure of classical theory to explain many new experimental facts led to the development of remarkable special theory of relativity by Albert Einstein in 1905. The new concept of relativity revolutionized the age old concepts of absolute mass, space and time that gave rise the consideration of absolute motion. Einstein proposed the idea that the motion through an ether filling empty space is a meaningless concept; only motion relative to material bodies has physical significance. According to new concept, everything in the universe is relative, nothing is absolute, all rest and motions are relative, position is relative, length is relative, time is relative etc. Einstein, in his new theory, extended and generalized Newtonian mechanics as well. He correctly predicted that the old classical mechanics is the limiting case of the new theory of relativity. The new theory, special theory of relativity, touches all branches of modern physics and can be applied to certain areas, such as high energy physics, quantum theory, atomic theory and many other branches of science and technology.

4

Theory of relativity deals with the way in which observers who are in motion relative to one another describe physical phenomena. The theory of relativity is studied in two parts– special theory and general theory. The special theory of relativity deals with observers which are either in relative motion in a straight line at constant speed or are at rest. The general theory of relativity applies to the cases of arbitrary relative motion or accelerated systems with respect to one another and is still under developed stage. In this chapter, we shall limit ourselves to the special theory of relativity.

Frame of Reference In order to specify the location of a point object or an event in space, we require a co–ordinate system. We choose a point of origin and directions for three axes. Therefore, the location of a point object or an event is expressed in terms of three real numbers, called coordinates of that point object with respect to the origin. For complete information about an event, we must not only know about its true location or position, but also its correct time of occurrence. Thus, we need another co–ordinate axis that of time. Such a co–ordinate system with respect to which we measure the position of a point object of an event is called a frame of reference. The motion of a point is completely described if we express the three point co–ordinates as function of time. A point is at rest relatively to our frame of reference if these three functions are constant. The frames of reference are of two types: 1. Inertial or unaccelerated frames of reference. 2. Non-inertial or accelerated frames of reference.

1. Inertial Frames of Reference or Inertial Frames The frames of reference in which Newton’s law of inertia and other laws of Newtonian mechanics hold good are called inertial frames of reference or simply inertial frames. It is also defined as the frames with respect to which an unaccelerated body appears unaccelerated, that is, is at rest or moving with constant linear velocity. Hence, unaccelerated frames are inertial frames. Inertial frame is also defined as the frame in which a body at rest or moving with uniform velocity and not under the influence of any force, remains at rest or moving with the same uniform velocity. All other frames of reference moving with a constant velocity with respect to it are also referred to as inertial frames of reference. It is experimentally proved that all inertial reference frames are equivalent for the measurement of physical phenomena. Observers in different frames of reference may get different numerical values for measured physical quantities but the relation among the measured quantities, that is, the law of physics will be the same for all observers. For our ordinary purposes, the earth serves well enough as an inertial frame of reference and hence, also the interior of a vehicle, such as train, car etc. moving smoothly and with a constant velocity on its surface.

2. Non-inertial Frames The frames of reference with respect to which an unaccelerated body appears accelerated are called non-inertial frames or in other words the accelerated frames are called non-inertial. In non-inertial frames Newton’s law does not hold. The simplest example of a non-inertial frame is provided by a rotating merry-go-round. The fictitious force can only be experienced by the observer in the inertial frame.

5

To study the motion of a body we should always prefer inertial frames because all the mechanical laws stated with reference to inertial frames preserve the same form whatever may be the relative motion between the observers. We never choose non-inertial frames because these involve different forces and thus, is very complicated to formulate the laws, common to all observers.

Galilean Transformation Equations or Galilean Transformation for the Position, Velocity and Acceleration of a Particle If an event be observed by two observers simultaneously in two different inertial frames of reference, it will naturally have a set of observations (co–ordinates such as the three mutually perpendicular X, Y and Z axes) in one, different from that in the other. The equations relating the two sets of observations of an event in two different frames of reference are called

Y S

transformations equations. In classical or Newtonian mechanics where the speed of the observer or object is very small compared to the speed of light these relevant transformation

O

v (x, y, z, t)

Y' S' x

P (x, y, z, t)

vt x

X O'

X'

equations are called Galilean transformation equations or Galilean transformation.

Z'

Z

Case (i) : S′ frame is moving with constants velocity v

Fig. 1

along positive direction of X-axis: In order to find the Galilean transformation equations, let us consider two inertial frames of reference S and S′ as shown in Fig. 1. Suppose S′ is moving along the positive x direction with a constant velocity v relative to S which is at rest. Let ( x, y, z, t) be the space and time co–ordinates of an event occurring at P for an observer on the frame S. Let the space and time co–ordinates of the same event P for an observer on the frame S′ be ( x′ , y′ , z′ , t′ ). For simplicity, let us choose our axis such that the X and X′ axis are parallel to v, and Y ′ and Z′ axis are parallel to Y and Z axes respectively. Let the time be counted from the instant when the origins O and O′ of the two frames momentarily coincide. Then after a time t, the frame S′ is separated from S by a distance vt in the direction of X-axis as shown in Fig. 1. Therefore, the co–ordinates of O′ in frame S are x = vt, y = 0, z = 0. At t = 0 let both observers in the frames S and S′ record the coordinates of an event at P in space, then the co–ordinates recorded by both are x, y and z at t = 0. After time t, let they again record the coordinates of the event at P. Suppose the co–ordinates recorded by an observer in frame S is x, y, z and t, and that by an observer in frame S′ is x′ , y′ , z′ and t′. As there is no relative motion along Y and Z axes, the co–ordinates in the two frames of reference along Y and Z–axes are same, that is, y = y ′ and z = z′. From Fig. 1, we have

x ′ = x − vt ,

y′ = y ,

z′ = z

and

t′ = t

...(1)

and similarly,

x = x ′ + vt ,

y = y′ ,

z = z′

and

t = t′

...(2)

The equations (1) and (2) are known as Galilean transformation equations for space and time or for position of a particle. For velocity transformation, we have from eqn. (1) d x ′ = d x − vdt,

dy ′ = dy,

dz ′ = dz and

d t′ = d t

(∵ v is constant)

6

Thus, ∴

d x′ d x dt = −v , dt′ dt′ d t′ v x′ = v x − v,

but

dt = dt ′,

dy ′ dy and = d t ′ dt

∴

d x′ d x dt = −v d t ′ dt dt

dz′ dz = dt ′ dt

Hence, Galilean transformation equations for velocity of particles are v′x = v x − v, v′y = v y and v′z = v z

...(3)

For acceleration transformation equations, differentiating eqn. (3) with respect to time, we get dv y′ dv y d vx ′ d vx dv ′ dv and z = z = − 0, because v is constant, = dt′ dt dt ′ dt dt ′ dt ∴

α ′x = α x , α ′y = α y and α ′z = α z

...(4)

These are Galilean acceleration transformation equations. Case (ii): The frame S′ is moving along a straight line relative to S along any direction : Y S Y′ S′ v Let us consider another situation in which frame S′ is (x, y, z, t) moving with respect to the frame S with constant velocity P → (x′,y′,z′,t′) v along any direction (as shown in Fig. 2) such that y′ x′ vxt → ...(1) v = v x i + v y j + vz k O′ X′ vyt

→

where v x , v y and vz are the components of velocity v along

O

X

X, Y and Z axis respectively. Let at t = 0 the origin O of vz t frame S concides with the origin O′ of frame S′. Suppose the t w o o b s e r v e r s s i t u a t e d a t O a n d O′ o b s e r v i n g simult aneously an event occurred at P. Let the Z′ Z co–ordinates of P relative to the observers at O and O′ be Fig. 2 ( x, y, z, t) and ( x ′ , y ′ , z ′ , t ′ ) respectively. After a time t, the frame S′ is separated from S by a distance v x t, v y t and vz t along X , Y and Z-axis respectively. Then the measurements of event at P taken by the observers may be related (Fig. 2) as, x′ = x − v x t ,

y′ = y − v y t ,

z′ = z − v z t

and t′ = t

...(2)

The equations (2) are the Galilean transformation equations relating the observations of two observers when one of the two frames is moving along any direction. Now let us consider the Galilean transformations of the velocity of the particle. For this differentiating equations (2), we get d x′ = dx − v x dt ,

dy′ = dy − v y dt,

where v x , v y and vz are constant velocity components. Equation (3) may be rewritten as,

d z′ = d z − vz dt

and

d t′ = d t

...(3)

7

d x′ d x = − vx , dt dt But ∴

dy′ dy = −vy dt dt

and

dz′ dz = − vz dt d t

uy′ = uy − v y

and

uz′ = uz − v z

d t = d t′ ux′ = ux − v x ,

...(4)

where ux , u y and uz are the components of velocity of the particle in the frame S and ux′ , u y′ and uz′ that in the frame S′ along X , Y and Z axis respectively. In terms of unit vectors i, j and k, the equation (4) can be represented as →

→

→

u′ = u − v

→

(∵ u′ = iux′ + ju y′ + kuz′ )

...(5)

Equation (5) represents the Galilean transformation equations for velocity of the particle. Obviously the inverse transformation gives u = u′ + v. This clearly indicates that the velocity is not invariant under Galilean transformation. Galilean transformation equations for the acceleration of the particle may be obtained by differentiating equation (5) with respect to t, that is, d u′ d u du′ d u or = = dt′ dt dt dt or

(∵ d t′ = d t)

α′ = α

where α = d u/d t is the acceleration of the particle in frame S and α ′ = d u′ / dt ′ that in frame S′. Thus, the acceleration observed by the observers in different inertial frames of reference is same. It means that Newton’s second law is valid in every inertial frame of reference or it is invariant under Galilean transformations. From the above discussions it is proved that the basic laws of physics do not change under Galilean transformations.

Failure of Galilean Relativity Galilean relativity not only fails to explain the actual results of Michelson-Morley experiment but also violate both the postulates of special theory of relativity. According to first postulate, the laws of physics are same in all inertial frames of reference. But the fundamental equations of electromagnetism assume very different forms when these equations are used to convert quantities measured in one frame with the equivalent quantities in the other. Again, if we measure the speed of light along the x-direction in the frame S to be c, then in the moving frame S′ it will be c ′ = c − v. This violates the second postulate of special relativity. (The speed of light in free space is same in all inertial frame of reference). Hence, a different transformation is required if the postulates of special relativity are to be satisfied. Example 1: The position of a point in the frame S′ moving relative to S, with a constant velocity of 10 cm/sec along the X-axis is given by (11, 9, 8). Calculate its position with respect to the frame S, if the two frames were in coincidence only 0.5 second before. →

Solution: If the co–ordinates of a point in frame S′ moving with velocity v along X-axis relative to a frame S at rest are ( x ′ , y ′ , z ′ ) and ( x, y, z) respectively, then the position of the point in stationary frame S are

8

x = x ′ + vt = 11 + 10 × 0 . 5 = 11 + 5 = 16 y = y ′ = 9 and z = z ′ = 8 Hence in stationary frame S the position of the particle is given by (16, 9, 8). Example 2: Use Galilean transformation to prove that the distance between two points ( x1, y1, z1 ) and ( x2 , y2 , z2 ) is invariant in two inertial frames.

or

Show that the distance between any two points in two inertial frames is invariant under Galilean transformation. [UPTU, B.Tech. II Sem. 2008] Solution: Suppose a frame of reference S′ is moving with velocity v relative to a frame S at rest, such that →

v = iv x + jv y + kvz . Let the co–ordinates of two points in frame S be ( x1, y1, z1) and ( x2 , y2 , z2 ); while

those in frame S′ be ( x1′, y1′, z1′ ) and ( x2′, y2′, z2′ ) . From Galilean transformations, we have x1′ = x1 − v x t, and

y1′ = y1 − v y t,

z1′ = z1 − vz t

...(1)

x2′ = x2 − v x t, y2′ = y2 − v y t, z2′ = z2 − vz t

...(2)

The distance between two points in moving frame S′ d = √ [( x2′ − x1′ )2 + ( y2′ − y1′ )2 + (z2′ − z1′)2 ]

...(3)

Substituting the values of x1′, y1′, z1′, x2′ , y2′ and z2′ from equations (1) and (2) in equation (3), we get d = {[( x2 − v x t) − ( x1 − v x t)]2 + [( y2 − v y t) − ( y1 − v y t)]2 + [(z2 − vz t) − (z1 − vz t)]2 }1 /2 d = √ [( x2 − x1)2 + ( y2 − y1)2 + (z2 − z1)2 ] = The distance between the two points in stationary frame S Hence, the distance between any two points is invariant under Galilean transformation. Example 3: A ball has velocity (4 i − 5 j + 10 k ) m/sec relative to a train moving with velocity (3 i + 4 j) m/sec relative to an observer on the ground. Calculate the velocity of the ball relative to the ground. Solution: Suppose the train is equivalent to a moving frame of reference S′ and the ground as stationary frame of reference S. According to the given problem, →

→

V = (3 i + 4 j) m/sec and u′ = (4 i − 5 j + 10 k) m/sec Therefore the velocity of the ball relative to the ground is given as →

→

u = u′ + V

= (4 i − 5 j + 10 k) + (3 i + 4 j) or

→

u = 7i − j + 10 k ms −1

9

Example 4: Show that the motion of one projectile as seen from another projectile will always be a straight line motion. Solution: Let the two projectiles be projected from the origin O with initial velocities u1 and u2 at angles of projections α1 and α 2 respectively as shown in Fig. 3. After t sec the position co–ordinates of points P1 ( x1, y1) and P2 ( x2 , y2 ) along X- and Y –axis are given as x1 = (u1 cos α1) t, y1 = (u1 sin α1) t − 12 gt2 x2 = (u2 cos α 2 ) t, y2 = (u2 sin α 2 ) t − 12 gt2 ∴

x2 − x1 = (u2 cos α 2 − u1 cos α1) t

and

y2 − y1 = (u2 sin α 2 − u1 sin α1) t

or

y2 – y1 u sin α 2 – u1 sin α1 = 2 =m x2 – x1 u2 cos α 2 – u1 cos α1

where m is constant as u1, u2 , α1 and α 2 are constants ∴ Suppose ∴

( y2 − y1) = m ( x2 − x1) y2 − y1 = Y

and x2 − x1 = X

Y = mX

This is the equation of a straight line. Therefore, the motion of one projectile as seen from another projectile will always be a straight line motion.

Michelson-Morley Experiment and Its Outcome Background (The ether hypothesis) Previous experience regarding the necessity of a medium for the propagation of mechanical wave forced the nineteenth century physicists to think that the existence of a medium that fills all space and penetrates all matter is essential for the propagation of light and other electromagnetic wave in free space. Therefore, they assumed that the entire space of the universe including vacuum is filled by a hypothetical light transmitting medium, called ether which is rigid, invisible, massless, perfectly transparent, perfectly non-resistive, continuous and stationary solid like steel having a very high elasticity and negligible density. All bodies (light or heavy) including earth move freely through this hypothetical medium (ether) without disturbing it. Thus, ether provides a fixed frame of reference which was called ether frame or rest frame or absolute frame of reference. Upto the end of the nineteenth century the ether hypothesis, was considered as the most promising and even necessary hypothesis, as it was very successful in the explanation of the phenomena of interference and diffraction. At that time, no one seemed to object to the existence of a medium.

10

On the necessity of the medium scientists were of the view that if the ether hypothesis is correct then it should be possible to determine the absolute velocity of the earth with respect to stationary ether frame. Many experiments with sophisticated instruments were performed in this direction. The most famous among them was performed by Michelson and Morley using Michelson Interferometer which is as follows:

Experiment The main objective of conducting the Michelson – Morley experiment was to confirm the existence of a stationary ether (frame). Michelson and Morley in 1887 performed an extremely sensitive experiment, for measuring the absolute velocity of the earth with respect to stationary ether. This experiment has long been regarded as one of the greatest experiment in physics and one of the main experimental pillars of special theory of relativity. The essential features of this apparatus, universally known as Michelson interferometer, as shown in Fig. (4). The two plane mirrors M1 and M2 used in the apparatus are highly silvered on their front surfaces to avoid multiple internal reflections. A beam of light from an extended source S is incident on a semi-silvered glass plate, P inclined at an angle 45° to the beam. This plate splits the light into two parts. One part of the beam travels through the plate P and falls normally on the mirror M2 which reflects it back to the point P . The other part of the beam after reflection from the plate P falls normally on the mirror M1 which also reflects it back to P. The two parts of the beam returned to P are directed towards telescope T. During their journey toward the telescope the two beams interfere and form interferance fringes that can Fig. 4 be observed by the telescope T. Let the two mirrors M1 and M2 be at the equal distance l from the plate P. If the apparatus is at rest in ether then the two rays (reflected and transmitted) would take equal time to return to P. But, in fact the earth, and hence, the whole apparatus is moving with earth through the ether with a velocity, say, v. Suppose the direction of motion of the earth is along the direction of incident light, that is, from P to M2 . If the incident beam strikes the glass plate P in the position shown in Fig. 1, then the paths of the two rays and the positions of their reflections from the mirrors M1 and M2 will be shown by the dotted lines in the figure. Due to the motion of apparatus with earth, the time taken by two rays in their journeys would not be the same. This time difference may be measured as follows: Let c be the velocity of light through ether. According to Galilean transformations, the velocity of light with respect to the apparatus along the path PM2 is (c − v) in the forward trip and is (c + v) in the backward trip. If t1 be the time taken by this ray to travel from P to M2 and back, that is PM2 P, then

11

 2 lc 2l  l l 1   + = = c − v c + v c2 − v2 c 1 − v2 / c2 

t1 =

...(1)

The part of the beam moving towards the mirror M1 with respect to the apparatus retains its velocity c. If t′ be the time taken by the beam in going from the point P to M1, then the distance travelled by it is ct′. In the same time t′, the mirror M1 shifted to M1′ after covering a horizontal distance vt′. Therefore, in the right angled triangle PM1 M1′ (PM1′ )2 = (PM1)2 + ( M1 M1′ )2 Here ∴ or

PM1 = l, M1 M1′ = vt′ and PM1′ = ct′ (ct′ )2 = l2 + (vt′ )2 l

t′ =

2

2 1 /2

(c − v )

=

l c 1 − (v / c)2

Therefore, the total time taken by the beam in travelling from P to M1′ and then from M1′ to P′ would be t2 = 2 t′ =

2l 1 c 1 − (v / c)2

...(2)

Hence, the time difference, ∆ t between the times of travel of the two beams is, given by ∆ t = t1 − t2 =

2l 2

2

c (1 − v / c )

−

2l c 1 − v2 / c2

2l [(1 − v2 / c2 )−1 − (1 − v2 / c2 )−1 /2 ] c

=

Using binomial expansion [(1 + x)n = 1 + nx + ...] and neglecting higher terms , we get ∆t =

=

    v2 1 v2 + ...  1 + 2 + ... − 1 + 2 2 c c     

2l c

2l c

 1 v2  l v2  2= 3 2 c  c

∴ The corresponding path difference, ∆ = c . ∆ t = c or

∆=

lv

...(3) lv2 c3

2

c2

...(4)

We know that if the path difference between the two interfering rays changes by λ, the shifting of one fringe in the field of view of the telescope is observed. Therefore, if n be the number of fringes that shift when interferometer is suddenly brought to rest (that is, v is made zero), then, from eqn. (4), we have n=

2 ∆ lv = 2 λ c λ

In the actual experiment, the whole apparatus, which was placed on a block of stone floated on mercury, was turned through 90° so that the path PM1 became longer than the path PM2 by an amount (lv2 / c2 ). Thus, the

12

rotation of apparatus through 90° introduces a path difference of the same amount in opposite direction so that the total path difference between the two rays became 2 lv2 / c2 . Hence, a shift of

2 l v2 c2 λ

was expected.

To get accurate results, the distance l was effectively increased to a value upto 11 meters by Michelson and Morley by the method of multiple reflections by using a system of mirrors. Taking earth’s velocity through ether equal to its orbital velocity, that is, v = 3 × 104 m/sec, the expected fringe shift for visible light (λ = 5 . 5 × 10 −7 m) is ∆n=

2 lv2 1 2 × 11 × (3 × 104 )2 1 . = × = 0 .4 2 8 2 λ c (3 × 10 ) 5 . 5 × 10 −7

or a shift of four-tenths a fringe. A shift of this magnitude can be easily measured with the help of Michelson-Morley set up. Michelson and Morley were extremely surprised to see that no shift in the fringe was observed when the interferometer was rotated through 90°. Michelson and Morley repeated the experiment at different places, different times of the year, and at different heights, but they always found no shift. That is, they could not detect the relative velocity of the earth with respect to stationary ether. Trouton and Noble, in the year 1902, performed an electromagnetic experiment for the same purpose but failed to achieve positive result. It means that the relative velocity between the earth and the ether is zero. Thus, the motion of the earth through the ether could not be detected experimentally. Hence, the hypothesis of the existence of stationary medium was disapproved.

Explanation and Interpretation of the Negative Result A number of explanations were offered to interpret the negative results of Michelson-Morley experiment and to preserve the concept of stationary ether. Nevertheless all of them failed. Here, we are giving a summary of three main explanations. 1.

Ether-Drag Hypothesis: This hypothesis assumed that there is an ether medium which is centered on the earth and moves with the earth in its motion through space. Therefore, there should be no relative motion between the earth and ether and hence, the question of shift does not arise. But this explanation was discarded due to following two arguments: (i) Ether-drag hypothesis goes against the observed aberration of light from stars, that is, it is against the phenomenon of stellar aberration. (ii) Fizeau’s experimental conclusion revealed that a moving body could drag the light waves only partially. This partial dragging of light waves was explained on the basis of electromagnetic theory, without using the ether-drag hypothesis.

2.

Fitzgerald-Lorentz Contraction Hypothesis: Fitzgerald proposed a hypothesis to explain the negative results of Michelson-Morley’s experiment and to retain the concept of preferred ether frame. According to this hypothesis, all material bodies are contracted in the direction of motion relative to stationary ether by a factor √ (1 − v2 / c2 ). It can be easily observed that such a contraction in the interferometer arm would equalize the two times t1 and t2 taken by the ray in travelling towards the mirrors M1 and M2 , and thus no fringe-shift would be expected. This explanation also

13

could not gain acceptance because contraction hypothesis was purely mathematical without any logic behind it and without any experimental confirmation. 3.

Constancy of Speed of Light Hypothesis: It was proposed that light travels with a constant velocity not with respect to the stationary ether but with respect to the source. Thus, the light from a moving source has a velocity which is the vector sum of its natural velocity and the velocity of source. This explains the negative results, but it was in conflict with the wave theory of light and there is an astronomical evidence concerning double stars which goes against this hypothesis. Hence, this explanation was also rejected.

Einstein Novel and Revolutionary Idea True explanation of negative results of failure of Michelson-Morley and other like experiments was offered by Einstein. He proposed, in the year 1905, a radically new profound idea that represented a vast revolution in physical thought. Einstein put forward that the motion through stationary ether is a meaningless concept; only motion relative to material bodies has physical significance. He announced to the world his fascinating special theory of relativity.

Significance of Negative Results Following important conclusions can be drawn from the negative results of Michelson-Morley experiment. 1.

The velocity of light is constant in all directions.

2.

The effects of presence of ether in the entire space of the universe are undetectable. Therefore, all efforts to make ether a universal frame of reference are meaningless.

3.

A new theory with different concepts of space, time and mass is needed. Thus, we must think of different set of transformations in contrast to Galilean transformation which failed to give correct results.

Example 5: What will be the fringe-shift according to the ether theory in the Michelson-Morley experiment, if the effective path length of each path is 7 meters and light has 7000 Å wavelength ? The velocity of earth is 3 × 10 4 m/sec. Solution: The expected fringe shift ∆ n, according to the ether theory in Michelson-Morley experiment, is given by ∆n=

2 l v2 1 . λ c2

Here l = 7 m, v = 3 × 104 m/sec, c = 3 × 108 m /sec and λ = 7000 Å = 7 × 10 −7 m ∴

∆n=

2 × 7 × (3 × 104 )2 (3 × 108 )2 × (7 × 10 −7)

or ∆ n = 0.2

14

Example 6: In Michelson-Morley experiment the length of the paths of the two beam is 11 meter each. The wavelength of the light used is 6000 Å. If the expected fringe-shift is 0.4 fringe. Calculate the velocity of earth relative to ether. Solution: The expected fringe shift is given by : ∆ n =

2 l v2 c2 λ

or v = c

λ∆n 2l

Here

l = 11 m, λ = 6000 Å = 6 × 10 −7 m and ∆ n = 0 .4

∴

v = (3 .0 × 108 )

or

v = 3 .13 × 10 4 m/sec

6 × 10 −7 × 0 .4 2 × 11

Example 7: Calculate the expected fringe shift in a Michelson-Morley experiment if the distance of each path is 11 metre and the wavelength of light is 5 . 6 × 10 −7 m. The experimental set up was not rotated through 90°. The linear velocity of earth may be taken as 30 kms −1. Solution: When the set up was not rotated through 90°. Then expected fringe–shift is given as, ∆ n= Here ∴

l v2 c2 λ

l = 11m, v = 30 × 103 m/s, c = 3 .0 × 108 m/s and λ = 5 .6 × 10 −7m ∆n=

11 × (30 × 103 )2 5 .6 × 10 −7 × (3 .0 × 108 )2

= 0.196

Einstein’s Postulates of Special Theory of Relativity The consequences of the absence of stationary ether (or universal frame of reference) led to the development of the special theory of relativity by Albert Einstein in 1905. The formulation of the special theory of relativity is based upon two basic postulates which are as follows :

Postulate I: The Principle of Equivalence (or Relativity) The principle of equivalence states that the laws of physics are same in all inertial frames of reference moving with a constant velocity (without any acceleration) with respect to one another.

Postulate II: The Principle of Constancy of the Speed of Light The second postulate states that the speed of light in free space (vacuum) is always same in all inertial frames of reference and is equal to c, that is, it is independent of the relative motion of the inertial frames, the source and the observer. The first postulate is the extension of the consequence drawn from Newtonian mechanics; since velocity is not absolute but relative. It is a direct consequence of the absence of an absolute frame of reference. The second postulate states an experimental fact and is responsible for differentiating the classical and Einstein's theories of relativity. The theory based on these two postulates is called special theory of relativity.

15

Lorentz Transformation Equations The Galilean transformation equations are not suitable under the new concept of special theory of relativity, where the speed of the object or observer is comparable with the velocity of light, therefore, Galilean transformation equations must be replaced by new ones consistent with experiment. We shall now derive new transformation equations, based on two fundamental postulates of Einstein special theory of relativity. The equations relating the co–ordinates of a particle in two inertial frames are needed when one goes from one inertial frame to another in uniform motion. New transformation equations were discovered by Lorentz and are known as Lorentz transformation equations of space and time. For speeds much smaller than, the velocity of light c, the Lorentz transformation equations reduce to Galilean transformation equations. Therefore, the Einstein’s theory of relativity does not overthrow, the classical theory, but rather extends and modifies it. To derive Lorentz transformation equations for space and time, let us suppose a system of two inertial frames of reference S and S′. Let S′ is moving with uniform velocity v relative to S and S is at rest as shown in Fig. 5. Let two observers situated at O and O′ (origins of frames of reference S and S′ respectively) are observing any event P. For simplicity, let us assume that the X-axis of two systems coincide permanently and the velocity v is parallel to X–axis. The event P is determined by the co–ordinates x, y, z, t for an observer O on the stationary frame S, while the same event is determined by the co–ordinates x ′ , y′ , z ′, t ′ for an observer O′ on the moving frame S′. Let the time be counted from the instant when the origins O and O′ momentarily coincide. In our new transformation, the measurement in the X-direction made in frame S must be linearly proportional to that made in S′. That is, ...(1)

x ′ = γ ( x − vt) where γ is the proportionality constant.

According to the first postulate of special theory of relativity, the equations relating the physical quantities (or laws of physics) have the same form in both primed (S′ ) and unprimed (S) frames of reference. Therefore, the equation corresponding to equation (1) for x in terms of x′ and t′ will be same as equation (1) except that v will be replaced by − v. That is, ...(2) x = γ ( x ′ + vt ′ ) where t ≠ t′ Substituting the value of x ′ from equation (1) in equation (2), we get x x = γ [γ ( x − vt) + vt ′ ] or = γ x − γ vt + vt′ γ ∴

Similarly, we can achieve t = γ t ′ +

t′ =

γx x − + γt γv v

γ x′  1 1 − 2  v  γ 

or t ′ = γ t −

γx  1 1 − 2  v  γ 

...(3)

...(4)

16

The value of γ can be evaluated with the help of second postulate. Let a flash of light is emitted from the common origin of S and S′ at time t = t ′ = 0. The flash travels with the velocity of light c which is same in both frames (2 nd Postulate). After some time the position of flash as seen from the observers in frames S and S′ is given by x = c t and x ′ = c t′ Substituting these values of x and x ′ in equations (1) and (2), we get c t ′ = γ t (c − v) and c t = γ t ′ (c + v) Multiplying both these equations with each other, we get c2 = γ2 (c2 − v2 )

c2 − v2

1

γ=

∴

c2

γ2 =

or

...(5)

1 − (v2 / c2 )

Substituting the value of γ from equation (5) in equation (1), we get x − vt x′ = (1 − v2 / c2 ) Squaring equation (5) γ2 = Substituting this value of 1 −

1 γ2

1 2

2

1 − v /c

or

1 γ

2

=1 −

v2 2

c

...(6)

or

v2 2

c

=1 −

1 γ2

in equations (3) and (4), we get t′ = γ t −

or

t′ =

γ x  v2  x v   2  = γ t − 2   v c  c 

t − xv / c2 (1 − v2 / c2 )

...(7)

In cases where flash of light is not restricted to travel along the X-axis, we have y′ = y

and

z′ = z

...(8)

The equations (6), (7) and (8) are known as Lorentz transformation equations for space and time. These transformation equations are frequently used in the transformation of the space and time co–ordinates of an event from one frame of reference into other frame of reference.

Inverse Lorentz Transformation Equations The space and time co–ordinates (x, y, z, t) of an event in the stationary system S can be obtained from the co–ordinates ( x ′, y ′ , z ′ , t ′ ) in the moving system S′ by replacing v by − v and by interchanging the primed and unprimed co–ordinates in equations (6), (7) and (8). The resulting transformation equations are : x=

x ′ + v t′ (1 − v2 / c2 )

, t=

t ′ + x ′ v / c2 (1 − v2 / c2 )

, y = y ′ and z = z ′

...(9)

The transformation equations represented by equation (9) are identical with Lorentz transformation equations represented by equations (6), (7) and (8) and are called Inverse Lorentz transformation equations. (For their derivations see example 8).

17

The most important aspect of the Lorentz transformation equations is that the measurements in space and time are no longer absolute, but relative and depend upon the frame of reference of the observer. The two events which are simultaneous when viewed by an observer in one frame of reference not simultaneous when viewed by an another observer sitting in second frame of reference which is moving with constant velocity relative to first. The most significant conclusion of Lorentz transformations is that it limits the maximum velocity of the material bodies. According to the conclusion, nothing can move with a velocity greater than the velocity of light, c or v should always be less than c. The Lorentz transformation equations reduce to the classical Galilean transformation equations xv when v av = < E y Hz > av = (1/2) E0 H0 = (1/2) × 0 .05 × 1.33 × 10 −4 = 3.325 × 10 −6 Wb m −2

139

Example 30: A plane electromagnetic wave propagating in the X-direction has a wavelength 7.0 mm. The electric field is in the Y -direction and its maximum magnitude is 42 V/m. Write suitable equations for the electric and magnetic fields as a function of x and t. Solution: The equations of electric and magnetic fields of a plane electromagnetic wave propagating in X-direction is given by E = E0 sin ω (t − x /c) and B = B0 sin ω (t − x /c) 2π We know that, ω = 2 πυ = c λ 2 π  2 π  ∴ E = E0 sin  (ct − x) and B = B0 sin  (ct − x) λ  λ  Here, E0 = 42 (V/m) and λ = 7.0 mm 2 π  ∴ E = 42 sin  (ct − x ) 7  The maximum magnetic field B0 is E 42 B0 = 0 = = 14 . × 10 −7 Weber/m2 or 1.4 × 10 −7 tesla c 3.0 × 108 2 π  B = 14 . × 10 −7 sin  (ct − x ) 7  The magnetic field is along Z-axis. Hence,

Example 31: The relative permittivity of distilled water is 81. Calculate refractive index and velocity of light in it. [UPTU, B. Tech. I, Q. Bank 2000] Solution: We know that µ = (µε / µ0 ε0 ) Here ∴

and

v = c /µ

ε = 81 ε0 and for distilled water µ ≈ µ0 3 × 108 µ = (81) = 9 and v = = 3.33 × 10 7 m s−1 9

Example 32: Calculate the magnitude of Poynting vector at the surface of the sun. Given that power radiated by sun = 5.4 × 10 28 watts and radius of sun is 7 × 10 8 m. Solution: Power radiated by the surface area (4 πR 2 ) of the sun is P = S ×4 π R 2 where S is the poynting vector and R is the radius of the sun P

5 .4 × 1028

∴

S=

or

S = 8.77 × 10 9 watt /m2

4 πR 2

=

4 × 3 .14 × (7 × 108 )2

Example 33: If the average distance between the sun and earth is 1.5 × 1011 meter and the power radiated by the sun is 5.4 × 10 28 watt, find the average solar energy incident on the earth.

140

Solution: If r is the distance between the centres of the sun and earth. If SE is the poynting vector at the surface of the earth, then SE .4 πr 2 = P SE =

∴

or

SE =

P 4 πr 2

5.4 × 1028 4 × 3 .14 × (1.5 × 1011)2

watt /m 2

= 1.91 × 105 watt /m 2 or

SE =

191 . × 105 4 .2 × 104

cal /cm 2 − min = 4.54 cal /cm2 – min

Example 34: The electric field intensity of a plane electromagnetic wave in free space is expressed as → z ^ E = E0 cos ω t −  j  v →

Find the expression for the components of magnetic field intensity H with the help of Maxwell’s equations for free space. Solution : The intensity of electric field associated with an electromagnetic wave is given as z E = E0 cos ω  t −  ^j  v

→

…(1) →

It is clear from equation (1), that the wave is propagating along Z-direction with E vector existing in Y-direction. Therefore, …(2) E x = 0, E y = E0 cos ω [t − (z /v)] and Ez = 0 The Maxwell’s equation for free space is →

→

∇ × E = − µ0

→

∂H ∂t

…(3)

L.H.S. of equation (3) may be written as ^  i → →  ∂ ∇× E=  ∂x  Ex 

^

j ∂ ∂y Ey

^

k ∂ ∂z  Ez  

…(4)

Substituting the values of E x , E y and Ez from equation (2), we get ^  i → →  ∂ ∇× E=  ∂x 0  or

^

j ∂ ∂y E0 cos ω (t − z /v)

^

k ∂ ∂z  0 

∂  → ∂ ∂ ∂  ∇ × E = ^i  (0) − [ E0 cos ω (t − z /v)] + ^j  (0) − (0) + ∂z ∂x  ∂z   ∂y 

→

^

 ∂ ∂ k  [ E0 cos ω (t − z /v)] − (0) ∂y  ∂x 

141

∂ ^ ∂ [ E cos ω (t − z /v)] + k [ E cos ω (t − z /v)] ∂z 0 ∂x 0 ω = − ^i E0 sin ω (t − z /v) + 0 v → → ω ∇ × E = − E0 sin ω (t − z /v)^i v = − ^i

or →

…(5)

→

Substituting this value of ∇ × E in equation (3), we obtain −

ω ∂ ^ E sin ω (t − z /v) ^i = − µ0 ( H x ^i + H y ^j + Hz k) v 0 ∂t

Comparing various components on both sides of equation (6), we get ∂H x ω − µ0 = − E0 sin ω (t − z /v) ∂t v ∂H y ∂Hz − µ0 = 0 and − µ0 =0 ∂t ∂t

…(6)

…(7) …(8)

From equation (7), we have ∂H x ωE0 = sin ω (t − z /v) ∂t µ0 v Integrating with respect to t, we get ωE0 ∂H x ∫ ∂t dt = µ0 v ∫ sin ω (t − z /v) dt ωE 1 Hx = − 0 cos ω (t − z /v) µ0 v ω − E0 or Hx = cos ω (t − z /v) µ0 v From equation (8), we get H y = 0 , Hz = 0 → − E0 ^ ^ ∴ H = Hx i = cos ω (t − z /v ) i µ0 v The velocity of propagation in free space is given by 1 v= µ0 ε0 Therefore, equation (10) may be rewritten as, → E ε H = − 0 µ0 ε0 cos ω (t − z /v)^i = − E0 0 cos ω (t − z /v)^i µ0 µ0 µ0 But = η0 , the intrinsic impedance of wave in free space. ε0 → E ^ ∴ H = − 0 cos ω (t − z /v ) i η0

…(9)

…(10)

Example 35: The electric field intensity of a uniform plane electromagnetic wave in air is 7.5 kV/m in the y-direction. The wave is propagating in the x-direction at a frequency of 2 × 10 9 rad/sec. Determine; (i) wavelength of electromagnetic wave, (ii) frequency, (iii) time period and (iv) the amplitude of magnetic field intensity. (µ0 = 4 π × 10 −7 N /A 2 and ε0 = 8.854 × 10 −12 C2 /N- m2 )

142

Solution : Intensity of electric field E of the wave along y-direction in air is given by E = E y = E0 cos (ωt − βx) = 7.5 × 103 cos (2 × 109 t − βx) (i) Wavelength of the electromagnetic wave 8 8 c 3 × 10 m/sec 3 × 10 × 2 × 3 .14 = = = 0 .942 m f (ω /2 π ) 2 × 109

λ=

(ii) Frequency,

f =

(iii) Time period

T =

2 × 109 ω = = 318 .5 × 106 Hz = 318.5 MHz 2 π 2 × 3 .14 1 1 = = 3.14 × 10 −9 sec f 318 .5 × 106

(iv) We know that,

4 π × 10 −7 E0 µ0 = = = 376 .72 Ω H0 ε0 8 .85 × 10 −12

or

H0 =

∴

Hz = 19.91 cos (2 × 10 9 t − βx )

75 . × 103 E0 = = 19.91 A - m −1 376 .72 Ω 376 .72 Ω

Example 36: The electric field of a uniform plane electromagnetic wave in free space is given by →

^

^

E = 45 sin (6 π × 10 8 t − 2 πx ) j + 15 cos (6 π × 10 8 t − 2 πx ) k Volt/m

Determine : (i) phase constant β, (ii) angular frequency ω, (iii) frequency f , (iv) intrinsic impedance η0 →

and (v) the corresponding magnetic field H. Solution : According to the given problem, the electric field E associated with a plane electromagnetic wave in free space is given as →

^

E = 45 sin (6 π × 108 t − 2 πx)^j + 15 cos (6 π × 108 t − 2 πx) k

…(1)

The above equation is of the form →

^

E = E y sin (ωt − βx)^j + Ez cos (ωt − βx) k

Comparison of equations (1) and (2), gives (i)

phase constant, β = 2 π rad s−1 and

(ii) angular frequency ω = 6 π × 10 8 rad s−1 (iii) frequency f =

6 π × 108 ω = = 3 × 10 8 Hz 2π 2π

(iv) Intrinsic impedance for free space η0 =

4 π × 10 −7 ~ µ0 = − 377 ohm ε0 8854 . × 10 −12

…(2)

143

→

(v) Since the electromagnetic wave is propagated in x-direction, E y component of E produces Hz component of H, and Ez component produces H y component. Ey 45 ∴ Hz = = sin (6 π × 108 t − 2 πx) = 0119 . sin (6 π × 108 t − 2 πx) η0 377

Hy =

and

Ez 15 = cos (6 π × 10 8 t − 2 πx) = 0 .0398 cos (6 π × 10 8 t − 2 πx) η0 377

→

^

^

H = 0 .0398 cos (6 π × 10 8 t − 2 πx ) j + 0 .119 sin (6 π × 10 8 t − 2 πx ) k

Thus,

Example 37: If in a time invarient field, the magnetic field associated with a plane electromagnetic wave has only z-component which is expressed as Hz = 6 x cos β + 12 y sin γ →

then obtain an expression for current density J . →

→

→

∂D ∂t

Solution: Maxwell’s equation relating H and J is expressed as →

→

∇×H= J +

→

…(1)

Since the field is time invarient, therefore →

∂D =0 ∂t Equation (1) is reduces to →

→

→

∇×H= J

^  i → →  ∂ ∇×H=  ∂x  Hx  Here,

∴

^

j ∂ ∂y Hy

 k  ∂  ∂z  Hz   ^

H x = 0, H y = 0 and Hz = 6 x cos β + 12 y sin γ ^  i → →  ∂ ∇×H=  ∂x 0 

 ^ j k  ∂ ∂   ∂y ∂z 0 (6 x cos β + 12 y sin γ)  ^

∂  ∂ ∂ ∂  = ^i  (6 x cos β + 12 y sin γ) − (0) + ^j  (0) − (6 x cos β + 12 y sin γ) ∂ y ∂ z ∂ z ∂ x      ∂ ^ ∂ + k  (0) − (0) ∂ x ∂ y   →

→

→

∇ × H = J = (12 sin γ)^i − 6 cos β ^j

∴

→

^

^

Current density, J = (12 sin γ ) i − (6 cos β ) j

144

Momentum of Electromagnetic Waves Maxwell had also predicted that electromagnetic waves transport linear momentum in the direction of propagation. The momentum of a particle of mass m moving with velocity v is given by …(1)

p = mv According to Einstein’s mass-energy relation U = mc2

Energy,

→

p=

∴

U

or m =

U c2

→

…(2)

v

c2

The energy density in plane electromagnetic wave in free space is given by u = ε0 E2

…(3)

where E is the magnitude of electric field. Thus, the momentum density or momentum per unit volume associated with an electromagnetic wave is → u→ …(4) p= 2 v c If the electromagnetic waves are propagating along X-axis, then →

v = c ^i

→

p=

∴ The poynting vector

…(5)

→

1 → → ( E × B) µ0

→

E2 ^ i µ0 c

S= S=

∴

u^ i c →

→

But E × B =

E2 ^ i c …(6)

Substituting the value of E2 from equation (3) in equation (6), we get →

S=

u ^ i = uc ^i ε0 cµ0

 1  ∵ c =  µ0 ε0  

→

u^i =

or

S c

…(7)

Putting the value of u^i from equation (7) in equation (5), we get →

p=

or

→

S

2

c

→

=

1

→

→

2

( E × B)

→

→

µ0 c

p = ε0 ( E × B)

  1 ∵   µ c2 = ε0  …(8)   0

Equation (8) represents momentum per unit volume for an electromagnetic wave. The value of this momentum is, u or u = pc p= c that is, Energy density = wave momentum × wave velocity.

145

Radiation Pressure and Energy Density When electromagnetic wave strikes a surface, its momentum changes. The rate of change of momentum is equal to the force. This force acting on the unit area of the surface exert a pressure, called radiation pressure. Let a plane electromagnetic wave incident normally on a perfectly absorbing surface of area A for a time t. If energy U is absorbed during this time, the momentum p delivered to the surface is given, according to Maxwell’s prediction, by U p= c If S is the energy passing per unit area per unit time, then U = SA t S At ∴ p= c where S is the magnitude of Poynting vector. S But = u (energy density) c ∴ p = uAt From Newton’s law average force F acting on the surface is equal to the average rate at which momentum is delivered to the surface. Therefore, p F = = uA t The radiation pressure Prad exerted on the surface. F Prad = =u A Hence, the radiation pressure exerted by a normally incident plane electromagnetic wave on a perfect absorber is equal to the energy density in the wave. For a perfect reflector or for a perfectly reflecting surface, the radiation after reflection has a momentum equal in magnitude but opposite in direction to the incident radiation. The momentum imparted to the surface will therefore be twice as on perfect absorber. That is, Prad = 2 u Example 38: Let you are sitting in sun for 3 hours. The area of your body exposed normally to sun rays is 1.3 m2 . The intensity of sun rays is 1.1 kW/m2 . If your body completely absorbs the sun rays, what will be the momentum transferred to your body ? Solution: Momentum transferred to the absorbing body of area A, when exposed to the time t or radiation pressure is given by p Pr = = u (energy density), for a perfectly absorbing surface, At and intensity or energy flux, S = cu AtS S  ∴ p = Atu or p = ∵ u =   c c Here A = 1. 3 m 2 , t = 3 hrs = 3 × 3600 sec, S = 1.1 kW/m 2 = 1100 W/m 2 and c = 3 × 108 m/sec ∴

p=

1.3 × 3 × 3600 × 1100 3 × 108

= 514 . 8 × 10 −4 kgms −1

146

Example 39 : The energy flux of 10 watt/m2 of a laser beam is incident on an ideal plane mirror for one hour. Determine the momentum imparted in the mirror during this time and the force. Solution : The momentum imparted per second per unit area by electromagnetic wave is given by p Pr = = 2 u (energy density), for a perfectly reflecting surface At Intensity of radiation or energy flux, S = cu 2 SAt ∴ p = 2 uAt or p = c −2 2 Here S = 10 watt m , A = 1 m , t = 1 hr = 3600 sec and c = 3 × 108 ms−1 Therefore, the momentum imparted to the mirror, that is, 2 × 10 × 1 × 3600 p= = 2.4 × 10 −4 kgms−1 3 × 108 dp p 2 SA 2 × 10 × 1 Force acting on the mirror, F= = = = = 6.66 × 10 −8 N (newton) dt t c 3 × 108

Exercise Question Bank 1. Write the Maxwell’s equation in integral and differential forms. Explain the physical significance of each equation. [UPTU, B. Tech. II Sem. 2005, 2003] 2. Deduce four Maxwell’s equations in free space. Explain the concept of Maxwell’s displacement current and show how it lead to the modification of Ampere’s law. [UPTU, B. Tech. I Sem. 2008] 3. Derive Maxwell's equations. Explain the physical significance of each equations. [GBTU, B.Tech. II Sem. 2010] →

→

4. Show that if B is a magnetic field and its divergence B = 0 everywhere, it means that an isolated magnetic pole is not possible.

[UPTU, B. Tech. I, Q. Bank, 2001] →

→

5. Show that the Faraday’s law of electromagnetic induction can be expressed as, ∇ × E = − down its integral form.

→

∂B . Write ∂t

[UPTU, B. Tech. II Sem. 2001]

6. State and explain Poynting theorem for the flow of energy in electromagnetic waves. [UPTU, B. Tech. Special (C.O.)Exam. Aug. 2008]

7. Write the Maxwell’s equations in integral as well as in differential forms and explain their physical significance. Show that the velocity of plane electromagnetic waves in the free space is given by, c = 1/ (µ0 ε0 ). [UPTU, B.Tech. Special C.O. Exam. Aug. 2008, I Sem. (C.O.) 2007, I Sem. 2001] 8. Prove that electromagnetic waves are transverse in nature.

[UPTU, B.Tech. II Sem. 2008]

147

9. Starting from Maxwell’s equations and constitutive equations obtain electromagnetic wave equation in free space. [UPTU, B. Tech. II Sem. 2006] 10. Write down Maxwell's equation in free space and using these equation derive wave equations for both electric d magnetic field. [UPTU, B.Tech. II Sem. 2009] 11. Write down Maxwell’s equation in free space and show that E, H and direction of propagation form a set of orthogonal vectors. [UPTU, B.Tech. I Sem. (C.O.) 2003] 12. Deduce the equation for the propagation of plane electromagnetic wave in the free space. Show that the electric and magnetic vectors are normal to each other and to the direction of the propagation of the wave. [U.K. B.Tech. I Sem. 2010, UPTU, B.Tech. I Sem. 2007, II Sem. 2004] 13. Deduce the Maxwell’s equation for free space and prove that the electromagnetic waves are transverse. [UPTU, B. Tech. I, Q. Bank, 2001] 14. Write down the maxwell's equation in free space and Prove that the velocity of plane electromagnetic wave in the vacuum is given by c = 1/ (µ0 ε0 ). [GBTU, B.Tech. II Sem 2010, II Sem. (C.O.) 2010, UPTU, B.Tech. II Sem. (C.O.) 2006, I Sem. 2005, II Sem. 2002]

15. Discuss the physical significance of poynting theorem.

[UPTU B. Tech. I Sem. 2008]

16. What is poynting vector ? Discuss the work-energy theorem for the flow of energy in an electromagnetic field. [UPTU B. Tech. II Sem. 2008] 17. State and Deduce poynting theorem for the flow of energy in an electromagnetic field. [GBTU, B.Tech. II Sem. (C.O.) 2010, UPTU, B.Tech. I Sem. 2007, II Sem. 2007]

18. Define Poynting Vector. Derive an expression for it and explain its physical significance for electromagnetic wave in free space. [UPTU, B. Tech. I Sem 2004] 19. What is poynting vector? Write down poynting theorem and explain its physical significance. [MTU, B.Tech. II Sem 2012]

20. Derive and explain poynting theorem. 21. Discuss the propagation of electromagnetic waves in free space.

[UPTU, B.Tech. II Sem 2009] [UPTU, B. Tech. I, Q. Bank, 2001]

 2 → → ∂ E 22. Show that the wave equation for electric field E is given by ∇2 E = µ0 ε0  2  . ∂ t   [UPTU, B.Tech., II Sem. 2007, I Sem. 2006, II Sem. 2003, I Sem. 2003 II Sem. 2001]

Short Answer Type Questions 1. What is displacement current ? 2. What is equation of continuity ? 3. Write Maxwell’s equation in differential form. 4. Write Maxwell’s equation in integral form. 5. What is poynting theorem and poynting vector ?

148

6. What are electromagnetic waves ? 7. Discuss the nature of electromagnetic waves. 8. Show that the radiation pressure exerted by an electromagnetic wave is equal to the energy density. 9. What is the physical significance of poynting theorem ? 10. What happens to an electromagnetic wave when it enters a conducting medium. [GBTU, B.Tech. II Sem. 2012]

Unsolved Numerical Problems 1. A light beam travelling in the X-direction is described by the electric field, E y = (300 V/m) sin ω (t − x c). An electron is constrained to move along the Y -direction with a speed of 2 . 0 × 107 m/sec. Find the maximum electric force and the maximum magnetic force on the electron. 2. A mercury lamp is radiating monochromatic light of 10 watts radiate power. Calculate the electric field strength at a distance 5 m from the lamp. 3. If the electric amplitude of the wave is 5 V/m, what is the magnetic amplitude of this wave ? 4. The maximum electric field in a plane electromagnetic wave is 600 N/coul. The wave is going in the X-direction and the electric field is in the Y -direction. Find the maximum magnetic field in the wave and its direction. 5. A plane monochromatic linearly polarized light wave is travelling eastward. The wave is polarized with E directed vertically up and down. Write expressions for E, H and B provided that E0 = 01 . V/m and frequency is 20 MHz. 6. The electric field in an electromagnetic wave is given by E = (50 N/C) sin ω (t − x /c). Find the energy contained in a cylinder of cross-section 10 cm2 and length 50 cm along the X-axis. 7. The magnetic field in a plane electromagnetic wave is given by B = (200 µT ) sin [(4 .0 × 1015 sec −1) (t − x /c). Find the maximum electric field and the average energy density corresponding to the electric field.

Answers Unsolved Numerical Problems 1. 4.8 × 10 −17 N, (ii) 3.2 × 10 −18 N

2. 4.9 V/m

3. 167 . × 10 −8 Weber/m2

4. 2 × 10 −6 Weber/m2 , along Z-axis

5. Ez = 0 .1 sin (4 π × 107 t − 0 .419 x), B y = − Ez /c, H y = B y /µ0

6. 5.5 × 10 −12 J

7. 6 × 104 N/C, 0.016 J/m3 ❍❍❍

Unit

III

I nterference & D iffraction

Section-A: Interference ...................................149–198 S patial and T emporal C oherence

In terference in thin F ilms of U niform T hickness W edge-shaped F ilm (Qualitative) N ewton's R ings A nti R eflection and H igh R eflection C oatings (Qualitative) In terference F ilters (Qualitative)

Section-B: Diffraction......................................199–256 S ingle and n-slit D iffraction

G rating S pectra R ayleigh C riterion of R esolution R esolving P ower of G rating

Exercise .......................................................257–266 Q uestion B ank S hort A nswer T ype Q uestions U nsolved N umerical P roblems A nswers to Unsolved Numerical Problems

151

U nit-III

S ection

A

I nterference

Introduction

T

h e wave theory of light was first put forward by Huygens in 1678. On the basis of his wave theory, Huygens explained satisfactorily the phenomena of reflection, refraction and total internal reflection. According to his theory a luminous body is a source of disturbance in a hypothetical medium called ether. This medium pervades all space. The disturbance from the source is propagated in the form of waves through space and the energy is distributed equally in all directions. When these waves carrying energy are incident on the eye, the optic nerves are excited and the sensation of vision is produced. Huygen's theory predicted that the velocity of light in medium shall be less than the velocity of light in free space, which is just converse of the prediction made from Newton's corpuscular theory. The experimental evidence for the wave theory in Huygen's time was very small. In 1801, however, Thomas Young obtained evidence that light could produce wave effects. Very shortly, diffraction was explained by Fresnel and Fraunhofer, while the transverse nature of light was explained by polarization experiments. The subject of interference, diffraction, and polarization is called physical optics or wave optics and should be explained by using the wave theory of light.

Interference of Light The phenomenon of interference of light has proved the validity of the wave theory of light. According to it, when the two light waves of the same frequency and having a constant phase difference traverse simultaneously in the same region of a medium and cross each other, then there is a modification in the intensity of light in the region of superposition, which is in general, different from the sum of intensities due to individual waves at that point. This modification in the intensity of light resulting from the

152

superposition of two (or more) waves of light is called interference. At certain points the waves superimpose in such a way that the resultant intensity is greater than the sum of the intensities due to individual waves. The interference produced at these points is called constructive interference or reinforcement, while at certain other points the resultant intensity is less than the sum of the intensities due to individual waves. The interference produced at these points is called destructive interference. Beyond the region of superposition the waves come through completely uninfluenced by each other.

Coherent Sources Two sources of light are said to be coherent if they emit light which have always a constant phase difference between them. It means that the two sources must emit radiations of the same wavelength. The two independent sources cannot be coherent because of the fact that independent sources cannot maintain a constant phase difference between them. For experimental purposes, two virtual sources obtained from a single parent source can act as coherent. In such case all the random phase-changes occurring in the parent source are repeated in the virtual sources also, thus maintaining a constant phase difference between them. Since the wavelength of light waves is extremely small, the two sources must be narrow and must also be close to each other.

No Interference by Two Independent Light Sources Sustained interference can never be obtained with two independent sources of light, such as two bulbs or two candles. It is due to the fact that any two independent beams of light are always incoherent. Actually a beam of light is built up of waves radiated from millions of excited atoms or molecules, whose vibrations are completely independent of each other, so that the initial phase of beam is absolutely governed by chance. After a time interval of 10 −8 sec, this phase will be randomly changed because of the fact that the excited atoms which are responsible for vibrations are replaced by other excited atoms. Thus every source undergoes haphazard changes of phase in every billionth of a second. Hence the resultant intensity on the screen vary rapidly with time and two incoherent sources cannot produce any stationary interference pattern.

Incoherent Sources Two sources of light are said to be incoherent if they emit light waves whose phases change with time, the phase difference at a point at which they arrive also vary with time in a random way. Hence, the intensity at a point will change with time and there will be no modification in intensity due to superposition of waves. In other words, the incoherent waves would not produce any observable interference pattern.

Spatial and Temporal Coherence During the studies of interference and diffraction of light we had assumed that the displacement associated with a wave train was sinusoidal for all values of time, but it was not true. If any wave appears as a pure sine wave for infinitely long time and for infinite space, then it is called perfectly coherent wave. Practically no wave is perfectly coherent because no source of light including laser, is capable of producing absolute or perfectly monochromatic beam of light. All waves are only partially coherent and appears as a

153

pure sine wave only for limited time or space. In fact, the light that emerges from an ordinary light source is a jumble of tiny, separate waves that reinforce or cancel each other in random manner; the wavefront thus produced varies from point to point and changes from instant to instant. There are, therefore, two independent concepts of coherence namely, temporal coherence and spatial coherence. We will discuss both of them one by one as follows:

Temporal coherence The temporal coherence correlates two oscillating electric fields at a given point at two different times, that is, it relates E ( x, y, z, t1) and E ( x, y, z, t2 ). In ideal conditions the oscillating electric field of a perfectly coherent light would have a constant amplitude of vibrations at all the times while its phase would vary with time or it will appear as ideal sinusoidal function of time [Fig. 1(a)]. However, none of the actual source of light is able to produce an ideal coherent wave. This is because every excited atom during their return journey to its

t

E (a)

t

E lc =cτ (b)

Fig. 1

initial state emits a light pulse of short duration (of the order of 10 −10 sec). Thus, the field remains sinusoidal for time intervals of the order of 10 −10 sec. After this duration, the phase of next emitted pulse changes abruptly as shown in Fig. 1(b). Over an interval of time shorter than the duration of one pulse or wave packet, the wave will appear to be pure sine wave. The average time during which the field remain sinusoidal (or definite phase relation exists) or, that is, an ideal sinusoidal emission exist is called the coherence time τ c or temporal coherence of the given light wave. The average length for which the filed remains sinusoidal is called the coherence length lc of the wave, and is given by l c = cτ c where c is the velocity of light. The coherence time and the coherence length can be

M1

d

measured with the help of Michelson interferometer (Fig. 2).

M2

Light from an extended source, S falls on a semi–silvered glass plate P, which is inclined at an angle 45° to the beam,

1

at which it is partly reflected and partly transmitted.The

O

reflected beam ‘1’ moves towards movable mirror M1 and falls normally on it and hence it is reflected back to P. The

S

P

2 1

M2

transmitted beam ‘2’ moves towards fixed mirror M2 and falls on it normally, and reflected back towards P. The

2

beams 1 and 2 finally enter the telescope T in which T

interference fringes are observed. We know that the two beams can produce a stationary interference pattern only if there is a definite phase relationship between them.

Fig. 2

154

Let M2′ be the image of M2 formed by reflection at plate P so that OM2 = OM2′ . The arrangement is then equivalent to an air film enclosed between the mirror M1 and virtual mirror M2′ . If d is the distance between M1 and M2′ , then 2d will be the path difference between the interfering beams. Now if, 2d > l c

then, in general, there is no definite phase relationship between the two beams and no interference pattern is observed.Therefore, starting with equal path lengths (or d = 0 ), as the distance d increases (by moving mirror M1), the contrast of the fringes become gradually poorer and eventually the fringe system disappears. The path difference at disappearance of interference pattern gives an estimation of coherence length and hence the coherence time. For the sodium yellow line (5896 Å) the disappearance occurs when the path difference is about 3 cm giving coherence time τ c = l c /c = 3 /(3 × 1010 ) = 10 −10 sec. On the other hand, for the red cadmium line (λ = 6438 Å) the coherence length is of the order of 30 cm giving τ c ~ 10 −9 sec. For helium-neon laser, coherence time, τ c ~50 n sec implying coherence length of about 15 m.

Purity of a spectral line For a perfectly monochromatic light the frequency spread (∆υ) is zero. But frequency spread of waves emitted by an actual light source (even laser) has a finite value. Every spectral line has a finite width which means that it corresponds to a continuous distribution of wavelength in some narrow interval between λ and λ + dλ . Thus, the decrease in the contrast of the fringes in Michelson interferometer experiment can also be interpreted as being due to the fact that the source is not emitting at a single frequency but over a narrow band of frequencies. Therefore, the poor fringe visibility for a large optical path difference is due to the non–monochromaticity of the source. We also know that the pattern disappears when the path difference exceeds the coherence length. Thus, the concept of coherence length is directly related to the width (or purity) of the spectral line. In the measurement of two closely spaced wavelengths using Michelson interferometer, we use sodium lamp which emits predominantly two closely spaced wavelengths λ1(= 5890 ° Å) and λ 2 = (5896 Å). The interferometer is first set corresponding to the zero path difference, that is, nearly d = 0, both the fringe pattern will overlap. Let us adopt the criterion that when the path difference (which changes by moving movable mirror M1 away or towards the plate P) between the interfering beams becomes equal to the coherence length l c , the bright fringes due to λ1 coincides with dark fringes due to λ 2 . If nth bright fringe due to λ1 coincides with (n + 1)th dark fringe due to λ 2 , then 1 l  l c = nλ1 =  n +  λ 2 or n = c  2 λ1 Eliminating n, we get

l 1 l l 1 l c =  c +  λ 2 or c − c = λ 2 λ1 2  λ1 2 

or lc =

λ1λ 2 2 (λ1 − λ 2 )

155

If instead of two discrete wavelengths λ1 and λ 2 , the beam consists of all wavelength lying between λ1 and λ 2 , then the pattern would disappear if lc =

λ1λ 2 λ2 λ2 or l c ~ or ∆ λ ~ λ1 − λ 2 ∆λ lc

Thus, if the fringes become indistinct when the path difference exceeds l c, we infer that the spectral line of mean wavelength λ has a wavelength spread or the spectral width of the source ∆λ given by ∆λ ~

λ2 λ2 = lc c τc

Thus, the temporal coherence τ c of the beam is directly related to the spectral width. For example, for cadmium red line, λ ~ 6438 Å, lc ~ 30 cm (τ c ~10 −9 sec) giving ∆λ ~

(6 .438 × 10 −7)2 λ2 = ~ 0 .01 Å c τ c 3 × 108 × 10 −9

Similarly, for the sodium line λ ~ 5890 Å, l c ~ 3 cm (τ c = 10 −10 sec) giving ∆λ ~ 0.1 Å. c , the frequency spread ∆ν of a spectral line would be, λ c ∆ υ = − 2 ∆λ λ c The magnitude of ∆ν, that is, | ∆ υ| = 2 ∆λ λ Further, since υ =

But We know that,

λ2 = lc ∆λ τc =

∴ ∆ υ~

c lc

lc , ∴ ∆υ ~1 /τ c c

Therefore, the frequency spread (∆υ) of a spectral line is of the order of inverse of the coherence time (τ c ) . Hence for a perfect monochromatic spectral line (∆υ = 0 ), the coherence time is infinite. Hence, the concept of temporal coherence is intimately connected with monochromaticity. For example frequency spread of He–Ne laser line is ~ 500 Hz corresponding to l c ~600 km and τ c ~2 × 10 −3 sec. The quantity ∆υ/υ or λ /∆λ represents the monochromaticity or the spectral purity of the source.Thus, λ ∆υ spectral purity, Q= = ∆λ υ For a commercially available laser beam τ c = 50 n sec implying ∆ υ / υ ~4 × 10 −8 . In terms of purity factor Q the coherence length l c may be expressed as, l c = c τc =

λ2 ∴ l c = Qλ = purity factor (Q) × wavelength (λ ) ∆λ

Spatial coherence

C

The spatial coherence is a measure of phase relationship between the waves reaching at two different points in space. Let us consider that S light waves are originating from a point source S as shown in Fig. 3.

A Fig. 3

B

156

Let A and B be two points lying on a line joining them with source S. The phase relationship between points A and B depends on the distance AB as well as on the temporal coherence of the beam. If AB < < lc

(coherence length)

there will be a definite phase relationship between A and B. On the other hand, if

AB >> lc

there will be no coherence between A and B. Let us now consider two non–linear points A and C which are equidistant from light sources S. If S is a true point source, then the waves shall reach at A and C in exactly the same phase, that is, the two points will have perfect spatial coherence. If however, the source, S is an extended source, the points A and C will no longer remain in coherence. This indicates that spatial coherence is associated with finite dimension of the source. This may be demonstrated by Young's double hole or double slit experiment as shown in Fig. 4. Practically no source of light is a perfect point source but it is an extended source. The size of the source can be reduced by placing an adjustable slit in front of it.

Y S1

l

S y

Light emitting from a narrow slit, S falls on two slits S1 and S2 placed symmetrically with respect to S. If l is the width of

a

slit S and Y be the separation between S1 and S2 , then the maximum value of Y for which the sources S1 and S2 remain

O

S2

Fig. 4

coherent or a good contrast of stationary interference pattern is observed on the screen, is defined as spatial coherence length (lω ) , l ω = ymax ∴ The angular spread of the light waves emitted by each excited atom in the slit of width l is, λ angular spread = l λ Therefore, if < S1SS2 < l then S1 and S2 will be coherent and good contrast interference pattern is observed on the screen. y If a be the coherent distance of source, S from the plane S1S2 , then < S1SS2 ~ a y λ λa So or y ≤ ≤ a l l aλ So that the spatial coherence length, l ω = ymax = l But l /a be the angle θ subtended by the slit S at the point of observation in the plane S1S2 , that is, l θ= a λ ∴ Spatial coherence length, lω = θ 1.22 λ For circular aperture, lω = θ

Y'

157

Non-realization of strictly monochromatic light in practice An strictly monochromatic light corresponds to a perfect sinusoidal wave for infinitely long time or over infinitely long distance with frequency spread (∆υ) zero. Since no light source emits pulse for infinite duration, the strictly monochromatic light is not realizable in practice. However, a laser light is nearly monochromatic.

Interference by two independent laser beams In laser sources, a large number of atoms during de–exitation process emit radiation simultaneously which has constant phase difference for a large interval of time as compared to ordinary light sources. Therefore, laser beam is essentially monochromatic and spatially coherent. Hence, two independent laser beams can produced sustained interference pattern. Example 1: In Michelson interferometer experiment with He–Ne laser, fringes are visible up to path difference spread of 8 m. Determine the lower limit of the following: (i) coherence length (ii) coherence time, (iii) spectral line width, and (iv) purity factor, given (λ = 11. 5 × 10 −7 m). Solution: (i) Path difference spread = Coherent length lc = 8 m l 8 (ii) Coherence time, τc = c = = 2.67 × 10 −8 sec c 3 × 108 1 1 (iii) Spectral line width, ∆υ = = = 3.75 ×10 7 Hz τ c 2 .67 × 10 −8 (iv) Purity factor,

Q=

3 × 108 υ c = = = 6 . 96 × 10 6 ∆ υ λ ∆ υ 11. 5 × 10 −7 × 3 . 75 × 107

Example 2: Michelson interferometer is illuminated by a source of light whose wavelength is (5890 ± 0 .1) Å. Initially interferometer is set for zero optical path difference. Through what displacement the mirror of interferometer should be moved so that the interference fringes may disappear. Solution: If the optical path difference is larger than coherence length, the interference fringes disappear. In Michelson interferometer optical path difference is twice the distance (d) moved by the movable mirror. Therefore, l c = 2d =

λ2 or ∆λ

d=

λ2 2∆λ

Here

λ = 5890 Å = 5 .89 × 10 −7m and ∆λ = 0.1 Å = 0 .1 × 10 −10 m

∴

d=

5 .89 × 10 −7 × 5 .89 × 10 −7 2 × 0 .1 × 10 −10

=

34 . 7 × 10 −14 0 . 2 × 10 −10

= 0 .0173 m

Example 3: The figure shows a Young's double slit experimental arrangement. What should

be the

maximum distance between S1 and S2 , so that distinct interference fringes may be obtained ? The wavelength of light used is 6438 Å.

158

Solution: For distinct interference fringes the maximum distance between S1and S2 or spatial coherence length is, S1S2 = l ω =

λ l 6 .438 × 10 −7 × 1 = a 1 × 10 −3

= 6 .438 ×10 −4 m

Fig. 5

Example 4: The frequency range of neon spectral line (6328 Å ) is 1010 Hz. Determine coherence time and coherence length. Solution: We know that, Coherence time,

1 1 = 10 = 10 −10 sec ∆υ 10 l c = cτ c = 3 × 108 × 10 −10 = 3 × 10 −2 m = 0 . 03 m

τc =

and coherence length,

Example 5: Determine the coherence 14

υviolet = 7 .7 × 10

length and coherence time for white 14

Hz and υred = 3 . 85 × 10

light (Giving that

Hz.

Solution: Frequency spread, ∆υ = υ violet − υred = (7 .7 − 3 .85) × 1014 = 3 .85 × 1014 Hz 1 1 = = 2 . 6 × 10 −15 sec ∆ υ 3 .85 × 1014

∴ Coherence time,

τc =

and coherence length,

l c = cτ c = 3 × 108 × 2 .6 × 10 −15 = 7 . 8 × 10 −7 m

Example 6: The coherence length of helium–neon laser is 15000 km and wavelength of light used is 632 .8 nm Å. Calculate the band width and frequency spread. Solution: If l c is the coherence length and λ the wavelength, then Band width,

∆λ =

−7 2 λ2 (6 .328 × 10 ) = = 2.67 ×10 −20 m lc 15000 × 103

and frequency spread,

∆υ=

3 × 108 1 c = = = 20 Hz τ c l c 15000 × 103

Types of Interference (Formation of Coherent Sources in Practice) The phenomenon of interference may be grouped into two categories depending upto the formation of two coherent sources in practice. (i)

Division of wavefront: Under this category, the coherent sources are obtained by dividing the wavefront, originating from a common source, by employing mirrors, biprisms or lenses. This class

159

of interference requires essentially a point source or a narrow slit source. The instruments used to obtain coherent sources and hence interference by division of wavefront are Fresnel biprism, Fresnel mirrors, Lloyd's mirror, Laser etc. (ii) Division of amplitude: In this method, the amplitude of the incident beam is divided into two or more parts either by partial reflection or refraction. Thus we have coherent beams produced by division of amplitude. These beams travel different paths and are finally brought together to produce interference. The effects resulting from the superposition of two beams are referred to as two beam interference and those resulting from superposition of more than two beams are referred to as multiple beam interference. The interference in thin films, Newton's rings and Michelson's interferometer are the examples of two beam interference and Fabry-Perot interferometer is the example of multiple beam interference.

Methods of Obtaining Coherent Sources In actual practice it is not possible to have two independent coherent sources of light, but for experimental purposes two virtual coherent sources are derived from a single source by some devices. Any phase change in one is simultaneously accompanied by the same phase change in the other. So that the phase difference between the two sources remains constant. In some experiments a real source and its virtual image act as coherent sources and in some others two virtual images of a single source serve the purpose. In some famous experimental set up the coherent sources are obtained in the following manner : 1.

Young's Double-Slit Experiment: In this method monochromatic light is passed from a small source through a slit S and the light emerging from this slit is used to illuminate two other narrow slits S1 and S2 which are very closed together and parallel to S. S1 and S2 act as coherent sources, both being derived from S. Alternate bright and dark equally spaced bands

S1 S S2

(interference fringes) are observed on a screen placed at some distance from the coherent sources S1 and S2 . 2.

Fresnel's Biprism Experiment: In Fresnel's biprism Fig. 6 experiment two coherent sources are obtained with the help of biprism in the following way : The biprism is B constructed as a single prism of obtuse angle of 30° S1 about 179º and remaining two acute angles 30′ A each. A narrow vertical slit S is illuminated by S monochromatic light. The light from S is allowed to fall symmetrically on the biprism S2 ABC with its refracting edge vertical (Fig. 7). 30° When light falls from slit S on lower half C (surface AC) of the biprism it is bent upwards Fig. 7 and appears to come from S2 . Similarly, the light from S, which falls on upper half (surface AB) of the prism, is bent downwards and appears to come from S1. The virtual images S1 and S2 of S act as two coherent sources.

160

3.

Lloyd's Mirror: It is an arrangement to obtain two coherent sources of light to produce interference. A plane glass plate ( AB, acting as mirror) is illuminated at almost grazing incidence by a monochromatic light from a narrow slit S1. A virtual image S2 of S1 is formed closed to S1 by reflection (Fig. 8) from the mirror. The real source S1 and virtual source S2 act as coherent sources for the study of interference.

4.

Thin Film (Reflected System): In this case two coherent beams are obtained by division of the amplitude of a incoming beam by partial reflection and refraction. A single wave train SA of monochromatic light be incident on the upper surface of a thin film of thickness t and refractive index µ (> 1) at an acute angle (< 90°). This ray is partly reflected along AB and partly refracted along AC. At point C the ray AC is again partly reflected from the second surface along CD, then part of it is transmitted along DE (Fig. 9). As the rays AB and DE are derived from the same incident wave SA, they are coherent and act as they are originating from two coherent sources.

Change of Phase on Reflection It is an important conclusion of Stoke that when a wave of light is reflected at the surface of denser medium, it always suffers a phase change of π or a path change of λ /2. However, no such change in phase or path is observed when the wave is reflected at the surface of rarer medium.

Interference in Thin Films A film is said to be thin when its thickness is about the order of one wavelength of visible light which is taken to be 5500 Å. A thin film may be an air film enclosed between two transparent sheets or a soap bubble. A thin sheet of transparent material such as glass, mica etc are also treated as thin film. There are many interesting examples of interference produced in thin films. When a thin film of oil spreads on the surface of water and is exposed to white light beautiful colours are seen. Similar colour phenomenon is observed when a soap film is illuminated with white light. These phenomena can be explained on the basis of interference between light reflected from the upper and lower surfaces of a thin film.

Interference Due to Reflected Light To understand the phenomenon of interference in thin film due to reflected light, consider a ray SA of

161

monochromatic light of wavelength λ be incident on the upper surface of a thin transparent film of uniform thickness t and index of refraction µ, at an angle i as shown in Fig. 10. The ray SA is partly reflected along AB and partly refracted along AC at an angle r. The refracted part AC is reflected from the point C on the lower surface of the film along CD and finally emerges out along DE. As the rays AB and DE are derived from the same incident ray, they are coherent. If the film is thin, the rays AB and DE will be sufficiently close to each other and in a position to interfere. To evaluate the path difference between AB and DE, the perpendiculars DL and CN are drawn on AB and AD respectively. As the paths of the rays AB and DE beyond DL are equal, the optical path difference between these two rays is given by

Source of light

B

S

E i

A

t

L

A

r

N

i

Air

r r

D

µ

C

Air

Fig. 10

∆ = path ( AC + CD) in film − path AL in air ∆ = µ( AC + CD) − AL ( ∵ path in air = µ times path in the film) In right angled triangle ACN, we have CN CN t = cos r or AC = = AC cos r cos r

...(1)

...(2)

where ∠ ACN = r and CN = t, the thickness of the film. Similarly, in other right angled triangle CND, we have CN CN t = cos r or CD = = CD cos r cos r

...(3)

where ∠ NCD = r In right angled triangle ADL sin i =

AL or AL = AD sin i = ( AN + ND) sin i AD

...(4)

where ∠ ADL = i Again in triangles ACN and CND AN ND = tan r and = tan r NC NC or

AN = NC tan r = t tan r and ND = NC tan r = t tan r

Putting these values of AN and ND in eqn. (4), we get AL = (t tan r + t tan r ) sin i = 2 t tan r sin i = 2 t tan r (µ sin r )

(From Snell's law)

2

= 2 µt

sin r cos r

...(5)

Substituting values of AC, CD and AL from eqns. (2), (3) and (5) in eqn. (1), we get  t t  sin2 r  − 2 µt ∆ =µ + cos r  cos r cos r  2µt 2µt [1 − sin2 r ] = . cos2 r = 2 µt cos r cos r cos r

or

∆=

∴

∆ = 2µt cos r

...(6)

162

As the ray AB is the reflected ray from a denser medium, therefore, there occurs an addition path difference of λ /2 or phase change of π (according to Stoke's law) Thus, the total path difference = ∆ − λ /2 = 2 µt cos r − λ / 2

...(7)

Condition of Maximum and Minimum Intensities (i)

For constructive interference, path difference should be an even multiple of λ /2, that is, for maxima 2 µt cos r − λ /2 = 2 nλ /2 or

2µt cos r = (2 n + 1) λ /2, where n = 0, 1, 2, 3 . . .

or

2µt cos r = (2 n − 1)λ /2, where n = 1, 2, 3, . . .

...(8)

When condition (8) is satisfied, the film appears bright. (ii)

For destructive interference, path difference should be an odd multiple of λ /2, that is, for minima 2µt cos r − λ /2 = (2 n − 1) λ /2 or 2 µt cos r = 2 nλ /2 = nλ Therefore, the film will appear dark if 2µ tcos r = n λ , where n = 0, 1, 2, 3 ...

...(9)

From equations (8) and (9) it is clear that the maxima and minima in the interference pattern depend upon two factors (1) the thickness of the film t and (2) the cosine of the angle r. When t = 0, the path difference is λ /2 and the condition of minimum intensity is satisfied. Hence, when t = 0, the film will appear dark and as the thickness is increased gradually, maxima and minima occur alternately. When t is kept constant and r is gradually varied, we again get an interference pattern with bright and dark bands.

Interference Due to Transmitted Light Let a ray of light of wavelength λ be incident on the upper surface of a thin transparent film of uniform thickness t and index of refraction µ at an angle i as shown in Fig. 11. The ray SA is refracted along AB at an angle r. The refracted part AB is partly reflected along BC and partly refracted along BP at an angle i. The reflected part BC is again reflected from the point C on the upper surface of the film, along CD and finally emerges out along DQ. If the film is thin then the rays BP and DQ, which are derived from same incident ray and hence are coherent, will be very close to each other and in a position to interfere. To evaluate the path difference between BP and DQ,the perpendiculars DN and CM are drawn on BP and

Source of light S i Air

C A t

r r r

r

µ

r D

B

M i i

Air

N Q

Fig. 11

P

163

BD respectively. As the paths of the rays BP and DQ beyond DN are equal, the optical path difference between these two rays is given by ∆ = path (BC + CD) in film − path BN in air = µ(BC + CD) − BN (∵ path in air = µ times path in the film) ...(1) In right angled triangle BCM, we have MC MC t = cos r or BC = = BC cos r cos r

...(2)

where ∠BCM = r, MC = t, the thickness of the film. In other right angled triangle MDC, we have MC MC t = cos r or CD = = CD cos r cos r where

...(3)

∠ MCD = r

In right angled triangle BND sin i = where

BN or BN = BD sin i = (BM + MD) sin i BD

...(4)

∠ BDN = i

Again in right angled triangles BMC and CDM BM MD = tan r and = tan r CM CM or

BM = CM tan r = t tan r and MD = CM tan r = t tan r

Putting these values of BM and MD in eqn. (4), we have BN = (t tan r + t tan r ) sin i = 2 t tan r sin i But from Snell's law (sin i)/(sin r ) = µ ∴

sin2 r BN = 2 t tan r µ sin r = 2µt cos r

...(5)

Substituting values of BC, CD and BN in eqn. (1) from eqns. (2), (3) and (4), we get  t t  sin2 r 2µt 2µt  − 2µt ∆ =µ + = (1 − sin2 r ) = cos2 r cos r cos r cos r cos r cos r   or

∆ = 2µt cos r

...(6)

Condition for Maximum and Minimum Intensities (i) For constructive interference or Maxima ∆ = 2 nλ /2 or 2 µt cos r = nλ , where n = 0, 1, 2, 3, ...

...(7)

(ii) For destructive interference or Minima ∆ = (2 n − 1) λ /2 or

2 µt cos r = (2n −1) λ / 2, where n = 1, 2, 3, ...

or

2 µt cos r = (2n + 1) λ / 2, where n = 0, 1, 2, 3, ...

...(8)

Thus the conditions of maxima and minima in the transmitted light are just opposite to those for reflected light. Hence, the point of the film which appears bright in reflected light, appears dark in transmitted light. Hence, the interference patterns of reflected and transmitted monochromatic light are complementary.

164

Colours in Thin Films When an oil film on water or a soap film or wedge shaped air film between the two glass plate is seen in reflected light it shows brilliant colours. The explanation of the origin of this coloured phenomenon was offered by Young on the basis of interference of light waves. When a beam of white light from an extended source is incident normally on a thin film of transparent material and seen in reflected light then coloured fringes will be observed. These colours arise due to the interference of the light waves reflected from the upper and lower surfaces of the film. The path difference between these interfering rays depends upon the thickness of the film t and upon the inclination of the incident or reflected ray r. At a particular point of the film (t) and for a particular position of the eye (r ) the waves of only certain wavelengths satisfy the condition of maxima {i. e. 2µt cos r = (2 n ± 1) λ /2}. Hence only those colours (wavelengths) which satisfy the condition of maxima will be present with maximum intensity. Other neighbouring colours (wavelengths) will be present only with diminished intensity, while other wavelengths, which satisfy the condition of minima (2 µt cos r = n λ ), will be absent from the reflected system. Hence the point of the film will appear coloured. Similarly, if the same point of the film is observed with an eye in different positions or at different points by keeping the eye at a fix position, a different set of colours is observed at each time. If white light falls as a parallel beam on a uniformly thick film (everywhere), then the path difference at each point of the film will be same and film will appear uniformly coloured. As the condition of maxima and minima for transmitted system are just reverse to those for reflected system, hence the colours visible in the reflected system are absent in the transmitted system or the colours visible in reflected system will be complementary to the colours visible in the transmitted system.

Colours in Thick Film and Blackness of an Excessively Thin Film in Reflected White Light A film of thickness about 10 µm to 50 µ is considered a thick film. A thick film shows no colour in reflected system when illuminated with an extended source of white light. (A thick film means a film having thickness more than a few wavelength of light). Suppose white light is incident on a plane parallel thick film, which is viewed in reflected system. The bright and dark appearance of the reflected light depends upon the values of µ , t and r. In case of white light even if t and r made constant µ varies with wavelength. At any given point or for any value of r, due to large thickness (t) a large number of wavelengths (different values of wavelength has different values of n) can be found to satisfy the condition of constructive interference [2 µt cos r = (2 n − 1) λ /2] for every colour in the spectrum and on the other hand, at the same point the condition of destructive interference [2 µt cos r = nλ ] is also satisfied for another set of large number of wavelengths (or colours). Moreover, at a given point the number of wavelengths (or colours) sending maximum intensity are almost equal to the number of wavelengths sending minimum intensity. In spite of this, fringes due to colours satisfying the condition of maximum intensity will superimpose and produce uniform illumination.Thus, the accumulative effect at any point will be a general white illumination. Hence, a thick film shows no colour but appears white in reflected system. A converse effect appears in the transmitted system.

165

An excessively thin film appears black in reflected system when illuminated with an extended source of white light (An excessively thin film means a film having thickness very small than the wavelength of light or whose thickness approaches to zero, that is, t → 0 ). For an excessively thin film the effective path difference in reflected system (2 µt cos r + λ /2) becomes λ /2, as 2 µt cos r is negligible in comparison to λ /2. This is a condition of minimum intensity for all wavelengths. Thus, when such an excessively thin film is illuminated with white light every wavelength will be absent in the reflected system, the film therefore, reflects no light whatsoever. All light is transmitted. Hence, the film will appear black in reflected system even when illuminated with white light. A converse effect appears in transmitted system.

Interference Colour Pattern on the Surface of a Soap Bubble When a white light is incident on a soap bubble and the reflected light is observed, the bubble appears coloured, and the colouration of the bubble varies with the thickness of the surface of the bubble. As, in bubble, the soap solution drains to the bottom, the soap film gets thinner at the top and the colours become more and more brilliant. When the thickness becomes less than the order of wavelength, a black band is formed at the top. Therefore, the interference pattern on the surface of a soap bubble changes continuously. Example 7: A parallel beam of sodium light of wavelength 5880 Å is incident on a thin glass plate of refractive index 1.5 such that the angle of refraction in the plate is 60°. Calculate the smallest thickness of the plate which will make it appear dark by reflection. Solution: We know that the condition for destructive interference of light in the reflected system is given by 2µt cos r = 2 nλ /2 Here,

µ = 1. 5, r = 60°, λ = 5880 Å = 5880 × 10 −8 cm cos 60° = 12 / , for the smallest thickness n = 1

then

t=

1 × 5880 × 10 −8 nλ = = 3920 × 10 −8 cm = 3920 Å 2µ cos r 2 × 1.5 × 0 .5

Example 8: Calculate the thickness of the thinnest film (µ = 1.4 ) in which interference of violet component (λ = 4000 Å) of incident light can take place by reflection. Solution: We know that the condition of constructive interference in the reflected system is 2 µt cos r = (2 n + 1) λ /2, n = 0, 1, 2, 3, ... For thinnest film n = 0 and for normal incidence r = 0

or cos r = 1,

Here,

λ = 4000 Å = 4000 × 10 −8 cm and µ = 1.4

Thus,

t=

4000 × 10 −8 λ = = 714 .3 Å 4 µ cos r 4 × 1.4 × 1

166

Example 9: A man whose eyes are 150 cm above the oil film on water surface observes greenish colour at a distance of 100 cm from his feet. Calculate the probable thickness of the film. (λ green = 5000 Å, µ oil = 1.4, µ water = 1. 33)

[UPTU, B.Tech. I Sem. 2008] Eyes

Solution: If t is the thickness and r the angle of refraction, then the condition of maxima in the reflected light is given by λ 2 µt cos r = (2 n − 1) , where n = 1, 2, 3, ... 2 (2 n − 1)λ or ...(1) t= 4 µ cos r

i

i

Water

From the adjacent Fig. 12; 100 2 2 tan i = = ∴ sin i = 150 3 13

100 cm

Fig. 12

sin i 2 / 13 sin i µ= , ∴ sin r = = sin r µ 1.4

But

150 cm

Oil

λ = 5000 Å = 5 × 10 −5 cm, µ oil = 1.4

Here

i

Air

2 2 = = 0 . 3962 1.4 × 3 .6055 5 .0477

or

sin r =

∴

cos r = 1 − sin2 r = 1 − (0 . 3962)2

= 1 − 0 .1569 or cos r = 0 . 8431 = 0 . 9182 Substituting this value of cos r in equation (1), we get (2 n − 1) 5 × 10 −5 t= = 9 .725 × 10 −6 (2 n − 1) cm, n =1, 2, 3, ... 4 × 1.4 × 0 .9182 Example 10: White light falls normally on a film of a soapy water whose thickness is 1.5×10 –5 cm and refractive index 1.33. Which wavelength in the visible region will be reflected strongly? [GBTU, B.Tech I Sem 2012, I Sem (C.O.) 2012]

Solution: When light falls normally (r = 0 ° ) on a film the condition of maxima is, 2µt = (2 n + 1) λ / 2, where n = 0,1, 2,... Here

t = 1.5 × 10 –5 cm and µ = 1.33

∴

λ=

For

4 × 1.33 × 1.5 × 10 4µt = (2 n + 1) (2 n + 1)

–5

n=0

λ=

798 . × 10 –5 1

or

λ = 798 . × 10 –5 cm

n =1

λ1 =

798 . × 10 –5 3

or

λ1 = 2.66 × 10 –5 cm

=

7. 98 × 10 –5 (2 n + 1)

167

n=2

λ2 =

798 . × 10 –5 5

or

λ 2 = 1596 . × 10 –5 cm

n=3

λ3 =

798 . × 10 –5 7

or

λ 3 = 114 . × 10 –5 cm

Out of above wavelengths only 7.98×10 –5 cm or 7980×10 –8 cm lies near the upper limit of visible region (4.0 × 10 –5 cm to 7.5 × 10 –5 cm). Hence, wavelength 7980Å is most strongly reflected. Example 11: White light is incident normally on a thin film of thickness 0 .50 × 10 –6 m and index of refraction 1.50 . Find the wavelengths in visible region (400 nm – 700 nm) reflected most strongly. [UPTU, B.Tech. I Sem (C.O.) 2009]

Solution: For normal incidence (r = 0 ° ), the condition of maxima is given by, 2 µt = (2 n + 1)λ / 2, where Here µ = 1. 5 and t = 0 . 5 × 10 –6 m = 5.0 × 10 –5 m ∴ For

λ= 30 × 10 –5 1

n = 0, 1, 2, 3,....

4 × 1. 5 × 5 . 0 × 10 4µt = (2 n + 1) (2 n + 1)

–5

=

30 × 10 –5 (2 n + 1)

or

λ = 30 × 10 –5 m =3000nm

30 × 10 –5 3

or

λ 1 = 10 × 10 –5 m =1000nm

λ2 =

30 × 10 –5 5

or

λ 2 = 6 × 10 –5 m =600nm

n=3

λ3 =

30 × 10 –5 7

or

λ 3 = 4 . 286 × 10 –5 m =428nm

n=4

λ4 =

30 × 10 –5 9

or

λ 4 = 3 . 333 × 10 –5 m =333nm

n=0

λ=

n =1

λ1 =

n=2

Thus, within the visible region (400nm – 700nm), the light of wavelengths 428.6 nm and 600 nm are reflected most strongly. Example 12: A soap film of refractive index 1.43 is illuminated by white light incident at an angle of 30°. The refracted light is examined by a spectroscope in which dark band corresponding to the wavelengths 6 × 10 −7 m is observed. Calculate the thickness of the film.

[UPTU, B.Tech. II Sem. 2002]

Solution: The condition of destructive interference or dark bands in reflected light is, nλ 2 µt cos r = 2 n λ /2 or t = 2 µ cos r Here, From Snell's law,

i = 30°, µ = 1.43, n = 1, and λ = 6 × 10 −7 m sin i sin i µ= ∴ sin r = sin r µ

168 sin 30 o (12 / ) = = 0 . 38 1.43 1.43

or

sin r =

∴

cos r = √ (1 − sin2 r ) = √ [1 − (0 . 38)2 )] = 0 .92

Thus,

t=

1 × 6 × 10 −7 = 2 . 28 × 10 −7 m or 2 . 28 × 10 −5 cm 2 × 1.43 × 0 .92

Example 13: A parallel beam of sodium light (λ = 5890 Å) strikes a film of oil floating on water. When viewed at an angle of 30 o from the normal, 8 th dark band is seen. Determine the thickness of the film. (refractive index of oil = 1. 5). Solution: We know that the condition of destructive interference or dark band in reflected light is 2µt cos r = 2 nλ /2 or In this problem, From Snell's law,

nλ 2 µ cos r i = 30 o, µ = 1. 5, n = 8

...(1)

t=

µ=

sin i sin r

∴ sin r =

sin i µ

  sin i 2  1 −      µ  

so

cos r = 1 − sin2 r =

∴

  (0 . 5)2  sin2 30 o  1 −  = 0 . 943 cos r = 1 − =  (1. 5)2  (1. 5)2    λ = 5890 Å = 5890 × 10 −8 cm

Therefore, from eqn. (1) t=

8 × 5890 × 10 −8 = 1.665 × 10 −4 cm 2 × 1. 5 × 0 . 943

Example 14: Calculate the thickness of a soap bubble film (refractive index = 1.46) that will result in constructive interference in the reflected light, if the film is illuminated with light whose wavelength in free space is 6000 Å.

[UPTU, B.Tech I Sem. 2001]

Solution: The condition for the constructive interference in the reflected light or for the brightness of soap bubble film is 2 µt cos r = (2 n − 1)λ /2, where n = 1, 2, 3, ... If the light is incident normally on the film, then r = 0 and cos r = 1 ∴

2 µt = (2 n − 1)λ /2 or t =

For the least thickness of the film, n = 1 λ 6000 = = 1027 .4 Å ∴ least thickness of the film, t = 4 µ 4 × 1.46

(2 n − 1) λ 4µ

169

Example 15: Light of wavelength 5893 Å is reflected at nearly normal incidence from a soap film of refractive index (µ ) = 1.42. What is the least thickness of the film that will appear (i) dark, (ii) bright ? or [UPTU, B.Tech. I Sem. 2005, I Sem. (C.O.) 2003] Light of wavelength 5893 Å is reflected at nearly normal incidence from a soap film of refractive index 1.42. What is the least thickness of the film that will appear black ? [UPTU, B.Tech. II Sem. (C.O.) 2004] Solution: (i) The condition for the darkness of the film in reflected system is given by, 2 µt cos r = nλ For normal incidence ,

r = 0 and cos r = 1 2 µt = nλ or t = n λ /2 µ

∴

For the least thickness of the film, n = 1 ∴

the least thickness of the film, t =

5893 × 10 −8 λ = = 2 .075 × 10 −5 cm = 2075 Å 2µ 2 × 1.42

(ii) The condition for the brightness of the film in reflected system is given by, 2 µt cos r = (2 n − 1) λ /2 For normal incidence, r = 0 and cos r = 1 (2 n − 1) λ ∴ 2 µt = (2 n − 1) λ /2 or t = 4µ For least thickness, n = 1 ∴ the least thickness of the film, t=

5893 × 10 −8 λ = = 1037 . 5 × 10 −5 cm = 1037 .5 Å 4µ 4 × 1.42

Example 16: White light is incident on a soap film at an angle sin−1

4 and the reflected light is observed 5

with a spectroscope. It is found that two consecutive dark bands correspond to wavelengths 6 .1 × 10 −5 and 6 .0 × 10 −5 cm. If the refractive index of the film be 4 /3, calculate the thickness. [UPTU, B.Tech. II Sem. 2007]

Solution: We know that the condition for dark band or fringe in the reflected light is 2 µt cos r = nλ If n and (n + 1) are the orders of consecutive dark bands for wavelengths λ1 and λ 2 respectively, then 2 µt cos r = nλ1 and 2 µt cos r = (n + 1) λ 2 or

2µt cos r = nλ1 = (n + 1) λ 2

∴

nλ1 = (n + 1)λ 2 or nλ1 = nλ 2 + λ 2 λ2 n(λ1 − λ 2 ) = λ 2 or n = (λ1 − λ 2 )

or

...(1)

Substituting this value of n in eqn. (1), we get 2 µt cos r =

But

λ1λ 2 λ1λ 2 1 or t = (λ1 − λ 2 ) (λ1 − λ 2 ) 2 µ cos r

  sin i 2  cos r = (1 − sin2 r ) = 1 −      µ  

...(2)

 sin i  ∵ From Snell' s law µ =  sin r  

170

Given µ = 4 /3 and sin i = 4 /5 ∴

cos r = 1 −

(4 /5)2 2

(4 /3)

 3 = 1−    5

2

=

16 4 = 25 5

Substituting this value of cos r in eqn. (2) and also λ1 = 6 .1 × 10 −5 cm, λ 2 = 6 .0 × 10 −5 cm, µ = 4 /3, we have t=

∴

t=

6 .1 × 10 −5 × 6 .0 × 10 −5 (6 .1 × 10 −5 − 6 .0 × 10 −5 ) × 2 × (4 /3) × (4 /5) 6 .1 × 10 −5 × 6 .0 × 10 −5 × 3 × 5 0 .1 × 10 −5 × 2 × 4 × 4

= 0 .0017cm

Example 17: A film of refractive index µ is illuminated by white light at an angle of incidence i. In reflected light two consecutive bright fringes of wavelengths λ1 and λ 2 are found overlapping. Obtained an expression for the thickness of the film.

[UPTU, B.Tech. II. Sem. 2006]

Solution: The condition for bright fringe in the reflected light is given by 2 µt cos r = (2 n − 1)λ /2, where n = 1, 2, 3, ...

...(1)

where t is the thickness of the film If n and (n + 1) are the orders of consecutive bright fringes corresponding to the wavelength λ1 and λ 2 respectively which are overlapped, then λ λ ...(2) 2 µt cos r = (2 n − 1) 1 and 2 µt cos r = [2 (n + 1) − 1] 2 2 2 λ λ (2 n − 1) 1 = (2 n + 1) 2 or (2 n − 1)λ1 = (2 n + 1)λ 2 ∴ 2 2 λ1 + λ 2 or ...(3) 2 n(λ1 − λ 2 ) = λ1 + λ 2 or n = 2 (λ1 − λ 2 ) White light is incident at an angle i, therefore cos r = 1 − sin2 r and

∴

 sin i cos r = 1 −    µ 

sin i =µ sin r

2

or cos r =

µ2 − sin2 i µ

...(4)

Substituting the value of n from equation (3) and value of cos r from equation (4) in equation (2), we get t=

∴

t=

 (2 n − 1) λ1 /2 2 λ2  2 λ 2 λ1µ = ∵ 2 n − 1 =  2 2 2 µ cos r (λ1 − λ 2 )  (λ1 − λ 2 ) 2 µ µ − sin i  λ 1λ 2 (λ1 − λ 2 ) µ2 − sin2 i

171

Example 18: White light is incident on two parallel glass plates separated by an air film of 0 .001 cm thickness and the reflected light is examined by the spectroscope. Find the number of dark bands seen in the spectrum between the wavelength 4 × 10 −5 cm and 7 × 10 −5 cm when light is incident at an angle of 30 o to the normal to surfaces. Solution: We know that the condition of dark bands in the reflected light is ...(1) 2 µt cos r = nλ o −5 In the given problem, µ = 1, t = 0 .001 cm, i = 30 and the range of visible spectrum is from 4 × 10 to 7 × 10 −5 cm. Let the number of dark bands seem at λ = 4 × 10 −5 cm be n1 and at λ = 7 × 10 −5 cm be n2 . sin i 3 = 1, ∴ sin i = sin r or cos r = cos 30 o = sin r 2 For wavelength λ = 4 × 10 −5 cm, from eqn. (1), we have From Snell's law, µ =

 3  = 4 × 10 −5 n1 ∴ n1 = 43 2 × 1 × 0 .001 ×   2  Similarly for wavelength λ = 7 × 10 −5 cm  3  = 7 × 10 −5 n2 2 × 1 × 0 .001 ×   2 

∴ n2 = 24

Hence the number of dark bands observed within the given spectral range = (n1 − n2 ) = 43 − 24 = 19.

Wedge Shaped Film A wedge shaped thin film is one whose plane surfaces (OA and OB) are slightly inclined to each other at a small angle (θ) and encloses a film of transparent material of refractive index µ as shown in fig. 13. The thickness of the film increases gradually from O to A. At the point of contact thickness is zero. When the upper surface (OB) of the film is illuminated by a parallel beam of monochromatic light and the surface is viewed by reflected light, then the interference between the two rays; one (PQ) reflected from the upper surface of the film (glass to film boundary) and the other ( FG) obtained by internal reflection (Film to glass boundary) at the back surface and consequent transmission at the film surface ( AB). since both the rays PQ and FG(or PEFG) are derived from the same incident ray SP (by the division of amplitude) they are coherent and on overlapping produce a system of equidistant bright and dark fringes. The fringes are straight and parallel to the contact edge of the wedge. With white light coloured fringes are observed.

172

When a beam of monochromatic light is incident normally at point P on the upper surface of the film, the path difference between the rays reflected from the upper and lower surfaces of the film is 2µ t, where t is the thickness of the film at P. At point P, reflection occurs from the interface between the optically denser medium and optically rarer medium, therefore, there occurs an additional path difference of λ / 2 or phase change of π. Thus an additional path difference of λ / 2 is introduced in the ray reflected from the upper surface. Hence the effective path difference between the two rays = 2µt +

λ 2

The condition for bright fringe or maximum intensity. 2µt + λ / 2 = 2 n λ / 2 or

2µt = (2 n – 1)λ / 2, where n = 1, 2, 3, ....

...(1)

Similarly, the condition for dark fringe or minimum intensity is λ 2µt + = (2 n + 1) λ / 2 2 or

2µt = nλ , where n = 0 ,1, 2, . . . .

...(2)

Fringe Width Fringe width ω or the separation between the two successive bright fringes or between two successive dark fringes may be obtained as follows: Let x n be the distance of nth dark fringe from the edge O of the film (Fig. 14), then t tan θ = 1 xn

or

B

t1 = x n tan θ t1

Putting this value of t1 in eqn. (2), we get ...(3)

2µx n tan θ = nλ Similarly, if x n+1 is the distance of (n + 1)

th

O

dark fringe,

then

t2

θ xn

A xn+1 Fig. 14

2µ x n+1 tan θ = ( n + 1) λ

...(4)

Subtracting eqn. (3) from eqn. (4), we get 2µ( x n+1 – x n) tan θ = λ or

x n+1 – x n =

λ 2µ tan θ

...(5)

For very small value of θ, tan θ ≈ θ ∴

Fringe width, ω = xn +1 – xn =

λ 2µ θ

...(6)

where θ is measured in radian. Similarly, we can obtain same formula for the fringe width of bright fringes, that is the fringe width of bright fringe is expressed as,

173

ω=

λ 2µθ

....(7)

It is clear from eqns (6) and (7) that for a given wedge angle θ, the fringe width of dark or bright fringes is constant (as λ and µ is constant). It means that the interference fringes are equidistant from one another. Eqn (6) or (7) reveals that, an increase in wedge angle θ fringes move closer. At an angle θ ≈ 1, the interference pattern vanishes. On the other hand is wedge angle θ is gradually decrease, the fringe width ( ω) increases and finally fringes disappear when θ ≈ 0 or when the faces of the film become parallel. At the edge or at the apex of the wedge the thickness of the wedge is almost zero, therefore at the apex the optical path difference is λ ∆= 2 Thus at the apex interfering rays will always be 180° out of phase and destructive interference occurs. Therefore, at the apex or edge of the film always dark fringe is observed. hence, the wedge fringe pattern will always begin with a dark fringe. As each bright and dark fringe is a locus of constant film thickness the fringes obtained on the wedge shaped thin film are called fringes of equal thickness.

Determination of Wedge Angle θ The wedge angle θ of the wedge shaped thin film can be experimentally determined by a travelling microscope. The positions of the dark fringes at two distant points P and Q situated at distances x n1 and x n respectively from (Fig. 15). the edge O of the film is measured. 2

If t1 is the thickness of the wedge at P, then for a dark fringe, for normal incidence ....(1)

2µt1 = mλ In terms of wedge angle θ, the thickness at P is t1 = x n tan θ ≈ x n θ as θ is very small (tan θ ≈ θ) 1

∴

2µx n θ = mλ 1

1

...(2)

174

Similarly if t2 be the thickness of the wedge at Q, then again for a dark fringe ...(3)

2µt2 = (m + n)λ where n is the number of dark fringes between points P and Q. In terms of wedge angle t2 = x n θ 2

...(4)

2µx n θ = (m + n) λ

∴

2

Subtracting eqn. (2) from eqn. (4), we get 2µ( x n – x n )θ = nλ 2

or

1

θ=

nλ 2µ( xn – xn ) 2

For air film µ = 1

∴

θ=

nλ 2( xn – xn ) 2

...(5)

1

...(6)

1

Thus by measuring x n and x n from travelling microscope and counting number of dark fringes between P 2

1

and Q, angle of wedge is determined.

Necessity of an Extended Source To see the interference effects over the entire film simultaneously, a broad source of light is necessary. This fact may be understood as follows : When a thin film is illuminated with monochromatic light from a point source and is viewed with a lens of small aperture, the light reflected from all corresponding points on the film does not reach the eye simultaneously as shown in Fig. 16(a). Thus only a small portion of the film will be visible. To see the whole film, the eye will have to be moved from one position to the other. Hence, with a point source the entire film cannot be viewed at a glance. If we employ an extended source, the light reflected by every point of the film reaches the eyes (as shown in Fig. 16 (b)). Hence the entire film can be viewed simultaneously by keeping the eye at one place only. Hence an extended source of light is necessary to view a film simultaneously.

Newton’s Rings If we place a plano-convex lens of large focal length on a plane glass plate, a thin film of air is developed between the curved surface of the lens and the plane glass plate. The thickness of the air film is zero at the point of contact O and gradually increases as we move away from the point of contact on either side. When

175

the monochromatic light falls normally on the surface of the lens, then the light reflected from the upper surface of the film AOB interferes with the light reflected from the lower surface of the film POQ (Fig. 17). As a result, an alternate dark and bright circular rings concentric around the point of contact are seen. The interference rings of equal thickness so formed were

A

B

P

O

first investigated by Newton and hence called Newton's ring. If the incident light is white, a series of concentric coloured rings is seen around the point of contact in both the reflected and transmitted systems.

Q

Fig. 17

Experimental Arrangement Newton's rings can easily be observed in the laboratory by using an

M

apparatus as shown in Fig. 18. Light from an extended

L1

G

monochromatic source S rendered parallel by a lens L1 and then falls on a 45° inclined thin transparent glass plate G, which partially reflects the light in the downward direction. These reflected beams

S

fall normally on an air film formed between the convex surface of the plano-convex lens and glass plate. The light transmitted through the plano-convex lens on reflection from the surface of the air film in contact with the glass plate PQ interferes with the light reflected from the surface of air film in contact with the lower surface of the

A

P

B

O

Q

Fig. 18

plano-convex lens. These reflected beams proceed upwards and enter the observer's eye through a low power travelling microscope M. On focussing a large number of Newton's rings, alternately bright and dark can be seen in the field of view of the microscope.

Theory (or Explanation of the Formation of Circular Newton's Ring) The formation of Newton's rings was satisfactorily explained by Young. According to him, these circular rings are formed due to the interference of light rays reflected from the upper and lower surfaces of the air film formed between the convex surface of a plano–convex lens and the glass plate. In Fig. 19, the incident ray SF is divided into two coherent rays 1 and 2 by reflection from the upper and lower surfaces of the wedge shaped air film. The reflected rays 1 and 2 interfere and produce bright and dark circular rings around the point of contact. The effective path difference between the interfering rays in reflected light is 2 µt cos r + λ /2 or 2 µt cos r − λ /2, where µ is the refractive index of the film, t the thickness of the film at the point of incidence and r the angle of refraction. For normal incidence r = 0, cos r = 1

176

Therefore, the effective path difference = 2µt + λ /2 At the point of contact, t = 0, the effective path difference = λ /2. This is the condition of minimum intensity. Hence the central spot of the rings system appears dark. For constructive interference or for a bright ring the effective path difference = 2 nλ /2 or 2 µt = (n − 12 / )λ

∴

2 µt + λ /2 = nλ

or

2 µt = (2 n − 1) λ / 2, where n = 1, 2, 3, 4, . . .

or

2 µt = (2 n + 1) λ / 2, where n = 0,1, 2, 3, 4, . . .

...(1)

For destructive interference or for a dark ring. 2 µt + λ /2 = (2 n + 1) λ /2 or 2µt = nλ , where n = 0, 1, 2, 3, 4 ...

...(2)

Thus, eqns. (1) and (2) indicate that for a particular bright or dark ring, t should be constant. Every fringe is the locus of points having the constant thickness air film enclosed between the convex surface of the plano-convex lens and glass plate. These loci are concentric circles with the centre at the point of contact of the two surfaces. Therefore, the fringes are circular with its centre at the point of contact. Hence the fringes are due to equal thickness in the shape of concentric rings, with point of contact of the film at a centre of the rings.

Diameters of Bright and Dark Rings To evaluate the diameters of bright and dark rings, consider a plano-convex lens AOB placed on a glass plate POQ (Fig. 20). Let R be the radius of curved surface AOB of the lens and t the thickness of the film at any point F From the property of a circle, EF × DE = OE × EG

....(3)

But EF = ED = r (radius of the ring passes through the point F) Therefore from eqn. (3), we have r . r = t.(2 R − t) = 2 Rt − t2 t R1 ). Consider the case in which the concave surface of a plano-concave lens and the convex surface of a plano-convex lens are in contact at point O, as shown in Fig. 25. In this case a curved air film is formed between the two curved surfaces and hence the formula for the diameters of bright and dark rings so formed are to be modified as under. Let t be the thickness of air film at point A and nth dark ring of radius rn passes through A. From Fig. 25, we have t = AB = AC − BC From the property of the circle, AC = ∴

t=

rn2 r2 and BC = n 2 R1 2 R2

rn2 r2 r2  1 1  − n = n  − 2 R1 2 R2 2  R1 R2 

...(1)

where R1 is the radius of curvature of plano-convex less and R2 that of plano-concave lens. We know that the effective path difference between the two interfering rays in reflected light, for normal incidence and for bright ring is ∆ = 2 µt− λ /2 For dark rings,

...(2)

2 µt = nλ

Substituting the value of t from eqn. (1) in eqn. (2), we get 2

2 µ. For bright rings

rn  1 1  2 − = nλ or µrn 2  R1 R2 

2 µt = (2 n − 1) λ /2

1 1   R − R  = nλ  1 2

...(3)

182

Therefore,

1 1  µrn2  −  = (2 n − 1)λ /2, where n = 1, 2, 3, ... R R  1 2

...(4)

For air film, µ = 1, and if Dn represents the diameter of nth ring then rn = Dn /2 For dark ring,

D 2n =

4 nλ 1 1 4nλ or − = 2 [1 / R1 − 1 / R 2 ] R1 R 2 Dn

For bright ring

D 2n =

2 (2n −1) λ 1 1 2 (2n −1)λ or − = [1 / R1 − 1 / R 2 ] R1 R 2 D 2n

(ii) When both the lenses are plano-convex: When both the convex surfaces of plano-convex lenses are placed in contact, the air film so formed between the two surfaces is shown in Fig. 26. Let the radii of curvature of the convex faces of the above and lower lenses be R1 and R2 , and the nth ring of radius rn passes

..(5)

R1

2

rn

R2

through the point A, then the thickness t of the film will be

t = AB = AC + CB =

...(6)

A C B

Fig. 26

2

rn r + n 2 R1 2 R2

(From the property of a circle) ...(7)

We know that the effective path difference between the two interfering rays in reflected light for normal incidence and for bright rings is ∆ = 2µt − λ /2 For dark rings 2µt = nλ ...(8)   1 1 Substituting the value of t in eqn. (8) from eqn. (7), we get, µrn2  +  = nλ  R1 R2  For air film µ = 1, the diameter of the nth dark ring Dnis 4nλ 1 1 4nλ or D 2n = + = 2 [1 / R1 + 1 / R 2 ] R1 R 2 Dn 2 (2n − 1) λ Similarly for bright ring D 2n = [1 / R1 + 1 / R 2 ] or

2 (2n − 1) λ 1 1 + = , where n = 1, 2, 3,... etc. R1 R 2 D 2n

...(9)

...(10)

Effect of Introduction of Liquid Between the Plate and Lens on Newton's Rings When the liquid of refractive index µ is introduced between the plano-convex lens and plane glass plate, a liquid film is developed in place of air. With liquid film the diameter of nth dark ring is (D2 n)liquid =

4 nλR µ

...(1)

where R is the radius of curvature of lens and λ the wavelength of incident monochromatic light. With air film the diameter of nth dark ring is (D2 n)air = 4 nλR Dividing equation (1) by (2), we get

...(2)

183

(D2n )liquid (D2n )air As µ > 1,

=

1 (Dn)air or (Dn)liquid = µ µ

(Dn)liquid < (Dn)air

Thus, when the liquid of refractive index µ is introduced between the glass plate and the plano-convex lens, the diameter of the ring decreases, that is, the rings contract.

Effect of increasing the Distance Between Lens and Plate or Lifting up the Lens from the Flat Surface As the distance between the lens and the plate is increased or the lens is lifted up slowly from the flat surface, the order of the ring at a given point increases.The rings, therefore, come closer and closer until they can no longer be separately observed.

Effect of Placing the Lens on Silver Glass Plate or Mirror If the top surface of the glass plate on which lens is kept is highly silvered, the ring on the reflected system would disappear and a uniform illumination is observed. It is due to the fact that, there would be no transmission of the rays, but transmitted rays will also be reflected at the silvered surface and the two complementary systems of rings superimpose on each other and gives a uniform illumination.

Newton's Rings are Circular but Air-wedge Fringes and Straight In both Newton's rings and Air wedge fringe arrangements, each fringe is the locus of points of equal thickness of the film. In Newton's rings arrangement, the locus of points of equal thickness of air film lie on a circle with the point of contact of the plano-convex lens and the glass plate as centre. Hence the fringes are circular and concentric. In case of wedge-shaped air film, the loci of points of equal thickness are straight lines parallel to the edge of the wedge. Hence, the fringes are straight and parallel.

The Effect of Placing the Concave Surface of the Plano-Concave Lens Towards the Plane Glass Plate Plano concave In this situation the fringes are still circular. In this case the lens lens and the plate will be in contact along the circumference of a circle where the thickness of the air film is zero. The Plane glass plate thickness of air film increases as we move from either side Fig. 27 towards the centre as shown in adjacent fig. 27. Therefore, in this case the order of ring is reverse, that is, order of the ring is maximum at the centre and zero at the periphery of the lens. Hence, the spacing between two consecutive rings go on decreasing as we move towards the centre of lens.

Effects of Using a Lens of Small Radius of Curvature The expressions for diameters of bright and dark rings [(Dn)Dark ∝ √ (2λR ) and (Dn)Bright ∝ √ (4λR ) clearly indicate that the diameter of a dark or bright ring is directly proportional to the square root of the radius of curvature, R of the lens. Hence, if we use a lens of small radius of curvature, the diameter of the rings will be small. This may cause large percentage error in the measurements.

184

Effect of not Using Monochromatic Light The two sources should give monochromatic or very nearly monochromatic light. If the sources give polychromatic light (light having a mixture of wavelengths) consisting of a number of wavelengths, then the light of each wavelength gives its own set of fringes, the fringe width (Dλ /2 d) being different for each wavelength. Hence, only zero order or central fringes of all wavelengths lie in the same position, the others being at different positions for each wavelength. The fringes of different wavelengths are intermixed. When the path difference is large, there is very much intermixing at any point so a uniform illumination is obtained. Hence, to obtain interference pattern the light should be monochromatic.

Newton's Rings with Bright Centre due to Reflected Light Ordinarily the centre of Newton's ring is dark. At the centre, the geometrical path difference between the interfering beams is zero but the effective path difference between the two interfering rays at the point of contact is λ /2 which is the condition of minimum intensity. This path difference (λ /2) is due to a phase change π which occurs in one of the interfering rays reflected from the lower denser surface of the air film (that is, from glass plate) developed between the plano-convex lens and glass plate. Hence the centre of Newton's ring is dark. To obtain the bright centre, a liquid of refractive index µ is introduced between a lens of refractive index µ1 and a plate of refractive index µ2 such that µ2 > µ > µ1. Under this condition the reflection of both the interfering rays will be from denser to rarer medium. Hence the effective path difference between both the interfering rays at the point of contact becomes zero which is the condition of maximum intensity. Hence the centre of Newton's rings appears bright. This situation was achieved by Young. In this experiment, he poured few drops of sassafras oil (µ = 1. 57 ) between a crown glass (µ = 1. 50 ) lens and a flint glass (µ = 1.65 ) plate and obtained a bright centre of the rings system.

Newton's Ring with White Light When an air film between the plano-convex lens and glass plate is illuminated with an extended source of white light, a few mixed coloured rings around a black centre are observed and beyond it a uniform illumination is obtained. This is because the diameters of the ring is a function of wavelength and white light is composed of a number of colours (wavelengths). Thus the diameters of rings of different colours will be different. As we know that λ r > λ v , therefore the diameter of violet ring of the same order will be smallest and those for red ring will be the largest, and the diameters of other coloured rings shall occupy the intermediate positions. Due to overlapping of the rings of different colours over each other, only first few rings will be clearly seen while other rings cannot be observed. Hence the ring system with reflected white light will have concentric mixed coloured rings from violet to red around a black centre and beyond it a uniform illumination due to overlapping of colours is observed. Example 19: Two glass plates enclose a wedge-shaped air film, touching at one edge and are separated by a wire of 0 .05 mm diameter at a distance of 15 cm from the edge. Calculate the fringe width. Monochromatic light of λ = 6,000 Å from a broad source falls normally on the film.

185

Solution: In Fig. 28, let θ be the angle of the wedge, then the fringe-width ω in air-wedge for normal B λ AB Arc incidence is given by ω = , From Fig. 28. θ = = 2θ OA Radius 0 .005 Air film ∴ θ= θ 15 O A Here λ = 6000 Å = 6000 × 10 −8 cm X =15 cm ∴

ω=

6000 × 10 −8 × 15 = 0 .09 cm 2 × 0 .005

Fig. 28

Example 20: Light of wavelength 6000 Å falls normally on a thin wedge-shaped film of refractive index 1.4 forming fringes that are 2.0 mm apart. Find the angle of wedge in seconds. [UPTU, B.Tech. I Sem (Old) 2010, I Sem. (C.O.), 2005]

Solution: If θ is the angle of the wedge formed by a medium of refractive index µ, then for normal incidence the fringe width for wavelength λ is given by λ λ ω= ∴ θ= 2µθ 2µω In the given problem, λ = 6000 Å = 6000 × 10 −8 cm, µ = 1. 4 and ω = 2 . 0 mm = 0 . 20 cm 6000 × 10 −8 180 o = 10 .71 × 10 −5 rad = 10 .71 × 10 −5 × 2 × 1.4 × 0 . 20 π

∴

θ=

or

θ = 0 .0061o

or

θ = 0 .0061 × 60 × 60 second = 21. 96 sec

 180 o  ∵ rad =  π  

Example 21: Two plane glass surfaces in contact along one edge are separated at the opposite edge by a thin wire. If 20 fringes are observed between these edges in sodium light for normal incidence, what is the thickness of the wire ? or Two plane glass surfaces in contact along one edge are separated at the opposite edge by a thin wire. If 20 interference fringes are observed between these edges in sodium light of λ = 5890 Å of normal incidence, find the diameter of the wire. [UPTU, B.Tech. I Sem., 2011, I Sem 2004] Solution: The fringe width in air-wedge for normal incidence is given by λ ...(1) ω= 2θ Let t be the thickness of the wire and x be the length of the glass surface from the point of contact as shown in Fig. 29, then the wedge angle θ = t / x, therefore eqn. (1) λx , If n fringes are seen in the entire film, then x = nω ω= 2t λnω nλ Therefore or t = ω= 2t 2 In the given problem,

t

θ

x Fig. 29

186 n = 20 and λ = 5890 Å (for sodium light) = 5890 × 10 −8 cm ∴

t=

20 × 5890 × 10 −8 = 5 .89 × 10 −4 cm 2

Example 22: Two plane rectangular piece of glass are in contact at one edge and separated by a hair at opposite edge so that a wedge is formed. When light of wavelength 6000 Å falls normally on the wedge, nine interference fringes are observed. What is the thickness of hair ? [UPTU, B.Tech. I Sem., Q. Bank, 2000] Solution: The fringe width in air-wedge for normal incidence is given by ...(1)

ω = λ /2θ

If t be the thickness of the hair and x be the length of the glass surface from the point of contact, then the wedge angle, θ = t / x. If n fringes are seen in the entire film, then x = nω. Therefore, θ = t /nω Substituting this value of θ in equation (1), we get ω=

nλ λnω or t = 2 2t

Here n = 9 and λ = 6000 Å ∴

t=

9 × 6000 = 27000 Å 2

Example 23: A square piece of cellophane film with index of refraction 1.5 has a wedge shaped section so that its thickness at two opposite sides is t1 and t2 . If the number of fringes appearing with wavelength λ = 6000 Å is 10, calculate the difference (t1 − t2 ).

[UPTU, B.Tech. II Sem. 2008]

Solution: If the order of the fringe appearing at one end of the film be n, then the order of the fringe appearing at other end will be (n + 10). For nth and (n + 10)th dark fringes. 2 µt1 cos r = nλ

...(1)

2 µt2 cos r = (n + 10)λ

...(2)

Subtracting equation (1) from equation (2), we get ...(3)

2µ (t2 − t1) cos r = 10 λ If the fringe is seen normally and the angle of wedge is very small, then r = 0 , so that

...(4)

cos r = 1 Substituting the value of cos r from eqn. (4) to eqn. (3), we get 10 λ 5 λ 2µ (t2 − t1) = 10 λ or t2 − t1 = = 2µ µ Here µ = 1. 5 and λ = 6000 Å = 6000 × 10 −3 cm ∴

t2 − t1 =

5 × 6000 × 10 −3 1. 5

= 2 × 10 −4 cm

187

Example 24: Between two optically plane glass plates, at one edge a foil is introduced to get

a

wedge-shape air film. Viewed in mercury green light (λ = 5.46 × 10 −5 cm) along the normal, we see 12 fringes in 0.40 cm width. Deduce (i) the angle of the wedge (ii) the thickness of the foil if plate length is 3.0 cm in all (iii) fringe width if water is introduced in the wedge space. Solution: In the given problem, the spacing between 12 fringes is 0.40 cm, therefore 12 ω = 0.40 cm or (i)

ω=

0 . 40 = 0 . 33 × 10 −2 cm 12

The fringe width in air wedge for normal incidence is given by ω = λ / 2 θ or θ = λ / 2ω Here, λ = 5 .46 × 10 −5 cm and ω = 0 .33 × 10 −2 cm

∴ (ii)

θ=

5 .46 × 10 −5 2 × 0 .33 × 10 −2

= 8 .2 × 10 −4 rad

Let t be the thickness of the foil and l be the length of plate, therefore θ = Given l = 3 .0 cm

∴

t or t = θl l

t = 8 .2 × 10 −4 × 3 .0 = 2 .46 × 10 −3 cm

(iii) We know that if θ is the angle of the wedge formed by a medium of refractive index µ, then for normal incidence λ , Here µ = 1.33 for water ω′ = 2 µθ ∴

ω′ =

5 .46 × 10 −5 2 × 1.33 × 8 .2 × 10 −4

= 0 .025 cm

Example 25: Newton's ring are observed with sodium light, between a plane surface and convex lens whose separation from the surface can be varied. Calculate the minimum separation beyond which the rings would be invisible and explain it. The difference in wavelengths of D1 and D2 lines of sodium is 6 Å and their average wavelength is 6000 Å. Solution: Suppose λ1 and λ 2 are the wavelengths of D1 and D2 lines respectively. In general, each wavelength produces its own system of rings. At the point of contact of the lens and glass plate the path difference between the interfering waves due to both wavelengths is very small and the ring systems due to λ1 and λ 2 almost concide, since λ1 and λ 2 are nearly equal. On increasing the separation between the lens and the plate, path difference increases and ring systems due to both wavelengths appear separated. When bright ring due to λ1 concides with dark ring due to λ 2 or vice-versa, the ring system disappear and a uniform illumination appears. Suppose this happens when the separation between lens and plate at the centre is t. Then the path difference between the interfering rays at the centre is (2 t + λ1 /2) for λ1 and (2 t + λ 2 /2) for λ 2 . Addition path λ1 /2 (or λ 2 /2) is due to reflection at the plate (denser medium). For the disappearance of rings, we have

188

and

2 t + λ1 /2 = nλ1 for maxima due to λ1

...(1)

2 t + λ 2 /2 = (2 n + 1) λ 2 /2 for minima due to λ 2

...(2)

From eqn. (2),

n=

2t λ2

Substituting this value of n in eqn. (1), we get 2 t + λ1 /2 = or

λ  2 t λ1 or 2 t  1 − 1 = λ1 /2 λ2  λ2 

 λ − λ 2  λ1 λ + λ2 λ1λ 2 λ2  = or t = , where λ = 1 2t  1 = 2 4 (λ1 − λ 2 ) 4 ∆λ 2  λ2 

Here,

λ = 6000 Å and ∆ λ = λ1 − λ 2 = 6 Å

∴

t=

(6000)2 = 1. 5 × 106 Å = 0 ⋅ 015 cm 4 ×6

Example 26: If the angle of wedge is 0 .25 o of arc and the wavelength of sodium D lines are 5890 × 10 −8 and 5896 × 10 −8 cm, find the distance from the apex of the wedge at which the maximum due to each wavelength first coincide. Solution: Let t be the thickness of the wedge at a point where the maximum due to λ1 = 5896 × 10 −8 cm and λ 2 = 5890 × 10 −8 cm coincide when viewed by the reflected light. Thus, for normal incidence and air film. This condition can be given by or

...(1)

2 t = (2 n + 1) λ1 /2 = (2 n + 3) λ 2 /2 (3 λ 2 − λ1) (2 n + 1) λ1 = (2 n + 3) λ 2 or n = 2 (λ1 − λ 2 )

Substituting the value of n in eqn. (1), we get 2t =

λ1λ 2 (λ1 − λ 2 )

...(2)

If θ is the angle of wedge and x is the distance from the apex of the wedge at which the maximum due to λ1 and λ 2 coincide, then t or t = x tan θ or t = x θ ...(3) tan θ = x 0 .25 π Given θ = 0 .25 o = rad. Therefore, from eqns. (2) and (3), we get 180 λ1λ 2 λ1λ 2 or x = ...(4) 2xθ = (λ1 − λ 2 ) (λ1 − λ 2 ) 2 θ Given λ1 = 5896 × 10 −8 cm, λ 2 = 5890 × 10 −8 cm, therefore (λ1 − λ 2 ) = 6 × 10 −8 cm Substituting these values in eqn. (4), we get x=

5896 × 10 −8 × 5890 × 10 −8 × 180 6 × 10 −8 × 2 × 0 .25 × 3 .14

= 6 .63 cm

189

Example 27: Light containing two wavelengths λ1 and λ 2 fall normally on a plano-convex lens of radius of curvature R resting on a glass plate. If the nth dark ring due to λ1 coincides with the (n + 1)th dark ring due to λ 2 , prove that the radius of the nth dark ring of λ1 is λ1λ 2 R (λ1 − λ 2 ) Solution: Radius of the nth dark ring in the Newton's ring pattern is given by ...(1)

rn = n λ R th

n dark ring due to λ1 coincides with the (n + 1)th dark ring due to λ 2 . That is, D2n = 4 nλ1R = 4 (n + 1)λ 2 R or

nλ1 = (n + 1)λ 2 or

n=

λ2 λ1 − λ 2

Substituting this value of n in equation (1), we get rn = nλ1R =

λ 1λ 2 R ( λ1 − λ 2 )

Example 28: Newton's rings are observed normally in reflected light of wavelength 6000 Å. The diameter of the 10 th dark ring is 0 . 50 cm. Find the radius of curvature of the lens and the thickness of the film. [UPTU, B.Tech I Sem., 2002, I Sem. 2005]

Solution: The diameter of nth dark ring is given by D2n 4 nλ In the given problem, Dn = 0 . 50 cm, λ = 6000 Å = 6 .0 × 10 −5 cm and n = 10. 0 . 50 × 0 . 50 ∴ R= = 106 cm 4 × 10 × 6.0 × 10 −5 D2n = 4 nλR or R =

If t is the thickness of the film corresponding to a ring of D diameter, then D2 D2 0 . 50 × 0 . 50 or t = 2t = = = 3 × 10 −4 cm 4R 8R 8 × 106 Example 29: The lower surface of a lens resting on a plane glass plate has a radius of curvature of 400 cm. When illuminated by monochromatic light, the arrangement produces Newton's rings and the 15 th bright ring has a diameter of 1.16 cm. Calculate the wavelength of the monochromatic light. Solution: The diameter of nth bright ring is given by Dn2 =

2 (2 n − 1) λR where n = 1, 2, 3, ... µ

In the given problem, n = 15, D15 = 116 . cm, R = 400 cm and µ = 1

...(1)

190

∴

λ=

116 . × 116 . Dn2 = = 5429 Å = 5429 × 10 −8 cm 2 (2 n − 1)R 2 × 29 × 400

Example 30: Newton's rings are observed by keeping a spherical surface of 100 cm radius on a plane glass plate. If the diameter of the 15 th bright ring is 0 . 590 cm and the diameter of the 5 th ring is 0.336 cm, what is the wavelength of light used ?

[UPTU, B.Tech. I Sem. (C.O.) 2006]

Solution: If Dn + p and Dn be the diameters of (n + p) λ=

th

th

and n

D2 n + p − Dn2 4 pR

bright ring, then

, Here D15 = 0 .590 cm, D5 = 0 .336 cm,

p = 10 and R = 100 cm (0 .590)2 − (0 .336)2 (0 .590 + 0 .336) (0 .590 − 0 .336) λ= = 4 × 10 × 100 4 × 10 × 100

∴

= 5 .88 × 10 −5 cm = 5880 Å Example 31: In Newton's ring experiment the diameter of 4 th and 12 th dark rings are 0 .400 cm and 0 .700 cm respectively. Deduce the diameter of 20 th dark ring. [UPTU, B.Tech. II Sem (Old) 2009, I Sem. (C.O.) 2005]

Solution: If Dn + p and Dn be the diameters of (n + p)th and nth dark rings respectively, then D2n + p − Dn2 = 4 pλR

...(1)

In the given problem, n = 4, n + p = 12, D4 = 0 .400 cm and D12 = 0 .700 cm D122 − D42 = 4 × 8 × λ × R

∴ Suppose the diameter of 20

th

...(2)

dark ring is D20 , then D202 − D42 = 4 × 16 × λ × R

...(3)

Dividing eqns. (2) by (3), we get 2 D12 − D42 2 D20

∴

−

D42

=

4 ×8 1 2 or 2 (D12 − D42 ) = (D202 − D42 ) or D202 = 2 D122 − D42 = 4 × 16 2

D202 = 2(0 .700) × (0 .700) − (0 .400) × (0 .400) ∴ D202 = 0 .98 − 0 .16 = 0 .82

so the diameter of 20 th Dark ring = √ 0 .82 = 0 .0906 cm Example 32: Newton's rings are formed in reflected light of wavelength 6000 Å with a liquid between the plane and curved surfaces. If the diameter of the 6 th bright ring is 3.1 mm and the radius of curvature of the curved surface is 100 cm, calculate the refractive index of the liquid. [UPTU, B. Tech. I Sem. 2007]

191

Solution: The diameter of nth bright ring is given by 2 (2 n − 1) λR 2 (2 n − 1)λR 2 or µ = Dn = µ D2n In the given problem, n = 6, λ = 6000 Å = 6000 × 10 −8 cm, R = 100 cm, and D5 = 3 .1 mm = 0 .31 cm ∴

µ=

or

µ=

2 (2 n − 1) λR D2n

=

2 (12 − 1) × 6000 × 10 −8 × 100 (0 .31) × (0 .31)

2 × 11 × 6000 × 10 −8 × 100 = 1. 373 0 .31 × 0 .31

Example 33: Newton's rings are made with light of λ = 6400 Å and a thin layer of oil (µ = 1.60 ) formed between the curved surface of a plano-convex lens (radius of curvature 80 cm, µ = 1.65) and a plane glass plate (µ = 1. 55). Calculate the radius of the smallest dark ring. Solution: In this case both the interfering waves are reflected from denser medium, therefore, there will be no addition path on reflection. Hence, the path difference between interfering rays in simply 2µ oil t for (normal incidence).The condition for smallest dark ring is ...(1) 2 µ oil t = λ /2 If ρ is the radius of this ring, then from the property of the circle ...(2) 2 t = ρ2 /R Eqns (1) and (2) gives

µ oil

ρ2 λ = R 2

or ρ =

Rλ 2 µ oil

Here, R = 80 cm, λ = 6400 Å = 6400 × 10 −8 cm and µ oil = 1.60 ∴

ρ=

80 × 6400 × 10 −8 = 0 .04 cm 2 × 1.60

Example 34: In an arrangement for observing Newton's rings with two different media between the glass surfaces, the nth rings have diameters as 10 : 7. Find the ratio of the refractive indices of the two media. Solution: The square of diameters of rings are inversely proportional to the refractive index of the film. In the diameter of nth dark ring in one medium is Dn and in other medium is D′ n, then µ′  Dn  = = Dn′2 µ  Dn′  Dn2

2

In the given problem, Dn : Dn′ = 7 : 10 Therefore

 7   10 

2

=

µ′ 49 = µ 100

Thus, the ratio of the refractive indices of two media is 49 : 100.

192

Example 35: A Newton's ring arrangement is used with a source emitting two wavelengths λ1 = 6 × 10 −5 cm and λ 2 = 4 . 5 × 10 −5 cm and it is found that the nth dark ring due to λ1 coincides with (n + 1)th dark ring due to λ 2 . If the radius of curvature of the curved surface is 90 cm, find the diameter of the nth dark ring for λ1. Solution: Let Dn be the diameter of nth dark ring and Dn +1 is that of (n + 1)th dark ring, then 2

Dn =

4 (n + 1) λ 2 R 4 nλ1R and D2 n +1 = µ µ

According to the problem, nth dark ring due to λ1 coincides with (n + 1)th dark ring due to λ 2 . Thus 2

Dn = D2 n + 1 or

4 nλ1R 4 (n + 1) λ 2 R = µ µ or n(λ1 − λ 2 ) = λ 2

nλ1 = (n + 1) λ 2 λ2 . Here λ 2 = 4 .5 × 10 −5 cm and λ1 = 6 × 10 −5 cm n= (λ1 − λ 2 )

∴ or

n=

∴ The diameter of nth dark ring, Dn = 2

4 .5 × 10 −5 (6 − 4 .5) × 10 −5

=

4 .5 × 10 −5 1.5 × 10 −5

= 3 Here R = 90 cm

4 × 3 × 6 × 10 −5 × 90 µ

for air film µ = 1, Dn = √ (4 × 3 × 6 × 10 −5 × 90) = 0 . 2538 cm Example 36: In a Newton's ring arrangement with a film observed with light of wavelength 6 × 10 −5 cm, the difference of square of diameters of successive rings are 0.125 cm2 . What will happen to this quantity if : (i) Wavelength of light changed to 4 . 5 × 10 −5 cm. (ii) A liquid of refractive index 1.33 introduced between the lens and the plate. (iii) The radius of curvature of convex surface of the plane convex lens is doubled. [UPTU, B.Tech. I Sem. Q. Bank, 2000]

Solution: In Newton's ring, if Dn and Dn + p are the diameters of nth and (n + p)th rings, then the difference of squares of diameters of rings is given by 4 p λR D2n + p − D2 n = µ where R is the radius of curvature of convex lens and µ the refractive index of the film. For successive rings p = 1, therefore 4 λR D2n +1 − D2n = µ (i) When wavelength of light changes from λ to λ′, we have 4λ ′ R D′2 n + 1 + D′2 n = µ Dividing equation (2) by equation (1), we get D′2n +1 − D′2n λ ′ λ′ or D′2n + 1 − D′2n = = (D2n + 1 − D2n ) λ λ D2n + 1 − D2n Here λ′ = 4 .5 × 10 −5 cm, λ = 6 × 10 −5 cm and D2 n + 1 − D2n = 0 .125 cm2

...(1)

...(2)

193

D′2n + 1 − D′2n =

∴ (ii)

4 .5 × 10 −5 6 .0 × 10 −5

× 0 .125 = 0 .0937 cm2

When a liquid of refractive index µ′ is introduced between the lens and plate, then we have 4 λR D′2n +1 − D′2n = µ′

...(3)

Dividing equation (3) by equation (1), we get D′2n +1 − D′2n D2n + 1

−

=

D2n

µ µ′

D′2n + 1 − D′2n =

or

µ (D2n + 1 − D2n ) µ′

Here µ = 1, µ′ = 1.33 and D2n + 1 − D2n = 0 .125 cm 2 D′2n +1 − D′2n = (11 / .33) × 0 .125 = 0 .094 cm2

∴

(iii) When the radius of curvature of the convex surface of the plano-convex lens is changed to R′ , then we have 4 λR ′ ...(4) ∴ D′2n +1 − D′2n = µ Dividing equation (4) by equation (1), we get D′2n +1 − D′2n D2n + 1 − D2n Here R ′ = 2 R

=

R′ R′ or D′2n + 1 − D′2n = (D2n + 1 − D2n ) R R

and D2n + 1 − D2n = 0 .125 cm 2

∴

D′2n +1 − D′2n = 2 × 0 .125 = 0 .250 cm2

Example 37: Two plano-convex lenses, each of radius of curvature 100 cm, are placed with their curved surfaces in contact with each other. Newton's rings are formed by using a light of wavelength 6 × 10 −5 cm. Find the distance between 10 th and 20 th rings. Solution: In Fig, 30, let t be the thickness of air film and diameter of nth ring is Dn, then 2  D2 D  t = t1 + t2 =  n + n   8 R1 8 R2 

R1

For normal incidence the condition of nth dark ring is  Dn2

 2 t = nλ ∴ 2  +  = nλ  8R1 8 R2  2

or or or radius of nth dark ring,

Dn  1 1  + = nλ 4  R1 R2  4 nλR1R2 2 Dn = R1 + R2 rn =

Dn t1

2 Dn

nλR1R2 R1 + R2

If r20 is the radius of 20 th and r10 is the radius of 10 th dark ring, then

t2 R2

Fig. 30

t

194

r20 − r10 =

λR1R2 [√ (20) – √ (10)] (R1 + R2 )

Here λ = 6 × 10 −5 cm, R1 = 100, R2 = 100 cm Since R1 = R2 ∴

r20 − r10 =

λR 2 [√ (20) − √ (10)] or r20 − r10 = 2R

r20 − r10 =

6 × 10 −5 × 100 .[√ (20) − √ (10) ] = 0 .717 cm 2

λR . [√ (20) − √ (10)] 2

Example 38: The convex surface of radius 300cm of a plano-convex lens rests on the concave spherical surface of radius 400 cm. If Newton's rings are viewed with reflected light of wavelength 6 × 10 −5 cm, calculate the radius of 12 th dark and 13 th bright ring. Solution: When a plano-convex lens of radius of curvature R1 is placed on the plano-concave lens of radius of curvature R2 , then the radius of nth dark ring and nth bright rings are rn =

nλR1R2 (R2 − R1)

rn =

(2 n − 1) λR1R2 2 (R2 − R1)

(for dark ring) (for bright ring)

Here, R1 = 300 cm, R2 = 400 cm, n = 12 and λ = 6 × 10 −5 cm Therefore, the radius of 12 th dark ring

r12 = = The radius of 13 th bright ring, n = 13 r13 = = ∴

12 × 6 × 10 −5 × 300 × 400 (400 − 300) 12 × 6 × 10 −5 × 300 × 400 ∴ r12 = 0 .928 cm 100 (2 × 13 − 1) × 6 × 10 −5 × 300 × 400 2 (400 − 300) 25 × 6 × 10 −5 × 300 × 400 2 × 100

r13 = 0 .948 cm

Anti-reflection Coatings or Non-reflecting Films A simple and very interesting application of thin film interference phenomenon is the deposition of antireflection coatings of suitable material and of suitable thickness on the surface of lens, prism, solar cells etc. for reducing reflectivity and increasing the transmission. In fact anti-reflection coating is a type of optical coating which improves the efficiency of the system since less light is lost. In complex system such as telescope, the reduction in reflection also improves the contrast of the image by the elimination of stray light. Most of the optical instruments such as microscope, telescopes and cameras use glass lenses as

195

multi components. When light passes through a lens or lens system, a part of it is lost due to reflection at each surface, thus the intensity of transmitted light is reduced. when a beam of light incident normally on a glass lens or plane glass plate nearly 4% of the incident light is reflected and lost. In many optical instruments there are many interfaces and the loss of intensity due to reflection can be severe. As an example, an achromatic lens has two reflecting surfaces, and nearly 8% of the incident intensity will be lost by reflection. If there is an assembly of such lenses, the loss will be many fold. To minimize such losses a anti-reflection coating is applied to lens surfaces. Now-a-days, all optical parts of high quality are coated with non-reflecting films. Thus, non-reflecting coatings on the lens surfaces appreciably reduces the loss of light by reflection and stray light reaching the image. A thin transparent film coated on a surface of lens or lens system to suppress the reflection and to improve the transmission is called anti-reflection or non-reflecting film. A thin film can act as an anti-reflection film if the rays reflected from its upper and lower surfaces are exactly 180° out of phase and have equal amplitude so that a complete destructive interference occurs between them. For the reduction of reflectivity of the surface, the selection of the thickness of the film and the refractive index of the coating material are very important factors. For proper anti-reflection effects the lens surfaces are often coated with λ /4µ f thick film, for a particular wavelength λ. The refractive index of the coating should be roughly equal to the square-root of the refractive index of glass (µ f ≈ µ g ). clearly, the refractive index of the material (µ f ) coated on the glass should be intermediate between that of air and glass. µ g > µ f > µ a . If a film of thickness λ /4µ f with µ g > µ f > µ a is coated on a lens surface then rays reflected from the top surface of the film interfere destructively with the rays reflected from the bottom surface of the film (Fig. 31). Moreover the two reflected beams must be of equal intensity to ensure that the two reflected components nullify the effect of each other. Hence, a coating of material of refractive index µ f (= µ g ) and thickness λ /4 µ f , for particular wavelength λ, does not allow any light to be reflected. Of course, the coating material should not only be transparent but also should be insoluble in ordinary solvent and scratch resistance.

Destructive Interference

S Rarer to denser interface

1 2

Air (µa=1) Antireflection Coating (µf µg) is deposited on the glass surface. When a beam of light is incident normally on such a film (Fig. 32), the beams reflected from the air-film interface and the film-glass interface interfere constructively because an abrupt phase change of π or path difference λ / 2 occurs only at air-film interface Thus a λ / 4µ f thick film of a dielectric material with µ f > µ g would increase the reflectivity of the surface. Larger the difference between the refractive indices, greater will be reflectivity. A single layer of zinc sulphide (µ = 2.37) film reflects 30% of incident light. With multiple coatings, it is possible to achieve almost 100% reflectivity. To increase the reflectivity of a surface by a large amount, multilayer structure of alternate higher and lower refractive index materials are used. These multilayer periodic medium finds wide applications in volume holography etc. High reflectivity window films help in blocking summer heat gain. They are best used in climates with long cooling seasons because they also block the sun's heat in the winter

Interference Filters An interference filter is an optical system that transmits nearly a monochromatic beam of light in a very narrow range of wavelength and is based on the principle of Fabry-Perot interferometer in which the interference pattern is obtained in the transmitted system by means of multiple reflection between two parallel plates silvered from their inner side and separated from each other. When a parallel beam of white light falls normally on such a system of parallel glass plates, interference occurs between all the component wavelengths of incident light. Transmitted system shows a series of bright fringes in the spectrum only for those wavelengths which satisfy the relation. 2µt = nλ where t is the thickness of air film between the plates, n is the order of maxima and µ is the refractive index of the film between the plates.

198

If t is large, a large number of maxima will be observed in the visible region, if we go on reducing t, we reach a situation in which only one or two maxima are observed in the visible region. For µ = 1. 5

and

t = 6 × 10 –3 cm, there are only two maxima in the visible region corresponding to

λ = 6000 Å(n =3) and λ = 4500 Å (n = 4). These maxima are widely separated from each other and any one of them can be covered so as to transmit only one wavelength. Thus, with this structure it is possible to filter a particular wavelength out of white light beam. Hence, a structure which is capable of filtering one particular wavelength out of white light beam is called an interference filter. Interference filters based on this principle can be Cover plate fabricated by modern vaccum deposition Metallic techniques. In this process first of all a thin film of Dielectric Film Coatings aluminium or silver metal is deposited on a glass Substrate substrate by vacuum deposition techniques. On Interference Filter this metallic film, a thin layer of dielectric cryolite (3NaF. AlF3 )is deposited. This structure is further Fig. 33 covered by another similar metallic film and to protect it by any damage another glass plate is placed over it (Fig 33.). This film structure is serve as an interference filter. By varying the thickness of the dielectric between the metallic films any particular wavelength can be removed from the white light beam. However, the transmitted system has a narrow spectrum sharply peaked about one wavelength. The sharpness of the transmitted spectrum depends on the reflectivity of the surfaces. The larger is the reflectivity the narrower is the transmitted system and lower is the intensity of transmitted light. Therefore, it is not possible to increase the thickness of the metallic film beyond a limit, since this would reduce the intensity of the transmitted light to a untolerable value. To overcome this difficulty, the metallic film are replaced by all dielectric structures. In an all-dielectric structure layers of different dielectric materials of appropriate refractive indices are coated. To obtain such type of filters, first of all, a λ / 4 thick film of titanium oxide (µ = 2.8) or zinc sulphide (µ = 2.3) is grown. On a glass substrate. Then a layer of dielectric with low refractive index, such as cryolite or magnesium fluoride, is deposited. On this again a λ / 4 thick film of a material of higher refractive index is developed as shown in fig. 34. To enhance the reflectivity, multilayer structures of alternate higher and lower refractive index are used. With this multilayer arrangement it is possible to achieve a reflectivity of more than 90% over any particular wavelength. Such interference filters are capable of transmitting over a bandwidth as small as 11Å or even less. Interference filters are preferred over coloured glass filters because in the case of interference filters no light is absorbed and hence there is no overheating. Interference filters are frequently used in the study of spectra in a narrow range of wavelengths. Interference filters are widely used in instrumentation for clinical chemistry, environmental testing, colorimetry, flame photometry, fluorescence spectroscopy. Elemental and laser line separation optical interference filters are effectively used in LCD projector, microscopy, semiconductor microlithography, opto-electronics, sensors, telecommunication, safty and monitoring systems etc. In medical application interference filters are used in blood and urine diagnostics, immunodiagnostics, opthalmology, dentistry etc. ❍❍❍

199

U nit-III

S ection

B D iffraction

Introduction (Diffraction of Light)

I

n the preceding chapter we have read about the interference phenomenon which arises due to two or more coherent beams of light obtained either by wavefront division or by amplitude division or both. In this chapter, we shall deal with the interference phenomenon occurring between light beams from various parts of one aperture. Ordinarily, when a plane wave of light is incident on a long narrow slit placed in front of a screen, a sharp patch of light and a shadow is obtained on the screen (as shown in Fig. 1). This shows that the light travels approximately in straight lines. But a very close examination of the light distribution on the screen reveals dark and bright fringes near the edges. It is observed that, if the aperture is reduced in width, the patch of light on the screen becomes distinctly wider than given by laws of geometrical optics and the fringes become more and more distinct till they become broad and practically cover the entire patch. Similarly when an opaque obstacle is placed in the path of light, instead of a sharp shadow, bright and dark fringes are obtained on the screen. This observation can be explained by assuming that light waves bend around an obstacle. The amount of bending, however, depends upon the size of the obstacle and the wavelength of the wave. Thus, we see that when light passes through a small opening (or aperture) or by the sides of a small obstacle, it bends, to some extent, into the region of geometrical shadow and its intensity falls off rapidly.

200

This deviation is extremely small when the wavelength is small in comparison to the dimensions of the obstacle. But the deviation becomes much pronounced when the dimensions of the aperture (or opaque disc) are comparable with the wavelength of light. Thus the phenomenon of bending of light round the corners of an obstacle and their spreading into the geometrical shadow (of an object) is called diffraction and the distribution of light intensity resulting in dark and bright fringes (that is, with alternate maxima and minima) is called a Diffraction Pattern. This phenomenon was first discovered in 1665 by an Italian scientist named Grimaldi and was studied by Newton. In terms of the wave theory, the first attempt to explain the diffraction phenomenon was made by Young. He considered the diffraction as interference between the direct light waves which passes through the edge of the obstacle and the wave of light reflected from the edge of the obstacle. This idea was not accepted on the ground that diffraction pattern does not depend upon the sharpness of the edge, the material of obstacle and its degree of polish. Later in 1815 A.J. Fresnel was able to explain successfully the phenomenon of diffraction by considering that the diffraction phenomenon is caused by the interference of innumerable secondary wavelets produced by the unobstructed portions of the same wave front. With this explanation he not only explained the phenomenon of diffraction but also rectilinear propagation of light.

Distinction between Fresnel and Fraunhofer Diffractions The diffraction phenomenon is usually divided into two categories: (i) Fresnel diffraction and (ii) Fraunhofer diffraction. In Fresnel class of diffraction the source of light or the screen or both are, in general at a finite distance from the diffracting element but no lenses are needed for rendering the rays parallel or convergent. Therefore the incident wavefront is spherical or cylindrical instead of being plane. In Fraunhofer class of diffraction, the source and the screen are effectively at infinite distance from the diffracting element. It is observed by employing two convergent lenses: one to render the incoming light parallel and the other to focus the parallel diffracted rays on the screen. Therefore the incident wavefront is plane. In Fresnel diffraction lateral distances are important, whereas in Fraunhofer diffraction the angular inclinations are important. In Fresnel diffraction the observed pattern is a projection of a diffracting element modified by diffracting effect and the geometry of the source, whereas in Fraunhofer diffraction the observed diffraction pattern is an image of the source modified by the diffraction at the diffracting element. In Fresnel diffraction, the centre of diffraction pattern may be bright or dark depending upon the number of Fresnel zones, whereas in Fraunhofer diffraction, the centre of the diffraction pattern is always bright for all paths parallel to the axis of the lens.

201

Resultant of n Simple Harmonic Motions of Equal Amplitude and Period, and Phases Increasing in Arithmetic Progression P

Let a particle be affected simultaneously by n simple harmonic

(n – 1) δ

a

vibrations of equal amplitude a and having common phase difference δ between successive vibrations. To find the resultant amplitude R and phase φ of the particle, we construct a polygon of

R

vector amplitudes having each side equal to a as shown in Fig. 2

a

(on the next page). The closing side OP of the polygon gives the resultant amplitude R and the angle which this closing side

2δ

subtends with the first vibration gives the resultant phase φ .

a

φ

Resolving the amplitudes parallel and perpendicular to OA , we get

3δ

a

O

a

δ

A

Fig. 2

R cos φ = a + a cos δ + a cos 2 δ + … + a cos (n − 1) δ and

...(1)

R sin φ = 0 + a sin δ + a sin 2 δ + … + a sin (n − 1) δ

...(2)

Multiplying eqn. (1) by 2 sin δ / 2 , we get 2 R cos φ sin (δ / 2) = a [2 sin (δ / 2) + 2 sin (δ / 2) cos δ + 2 sin (δ / 2) cos 2 δ + … + 2 sin (δ / 2) cos (n − 1) δ] Applying trigonometrical formula, 2 cos A sin B = sin ( A + B) − sin ( A − B), we get 2 R cos φ sin

3δ δ δ  δ   = a 2 sin + sin − sin   +   2 2 2 2 

5δ 3 δ  − sin sin  + ... +  2 2 

δ 1   = a sin + sin  n −    2 2 

or

 sin 

1   n −  δ − sin  2

3  n −   2

 δ  

nδ (n − 1) δ  δ  = 2 a sin cos 2 2 

 nδ  a sin    2 (n − 1) δ  R cos φ = cos   δ  2 sin 2

...(3)

Similarly the second series [eqn. (2)] after multiplication by 2 sin (δ / 2) and on simplification gives  nδ  a sin    2 (n − 1) δ  R sin φ = sin   δ  2 sin 2 Squaring eqns. (3) and (4), and adding we get R 2 =

a2 sin2 (nδ / 2) sin2 (δ / 2)

...(4)

...(5)

202

or

R=

a sin (nδ / 2 ) sin (δ / 2 )

...(6)

Dividing eqn. (4) by eqn. (3), we obtain tan φ = tan

(n − 1) δ (n − 1)δ or φ = 2 2

...(7)

Let us suppose that the number of vibrations n are infinitely large but the amplitude a and the common phase δ are infinitely small, so that na and nδ is finite. Also suppose n δ = 2 α ; then eqn. (6) gives R=

a sin α a sin α = sin (α / n) (α / n)

R = na

(because α / n is very small)

sin α sin α , Now putting na = A, we obtain R = A α α

...(8)

From eqn. (7), we have φ=

(n − 1) δ n δ = =α 2 2

 ∵ 

For large n,

n δ (n − 1) δ  ≈  2 2 

...(9)

Equations (8) and (9) give the resultant amplitude and phase of n simple harmonic vibrations of equal amplitude and period, and phases increasing in arithmetic progression.

Fraunhofer Diffraction at a Single Slit (Single Slit Diffraction) Let a collimated beam of monochromatic light of wavelength λ, produced by a point source S placed at the focus of a spherical lens L1, be incident normally on a narrow slit AB of width a (Fig. 3). Let the diffracted light be focused by another convex lens L2 on the screen XY placed in the focal plane of lens L2 and perpendicular to the plane of the paper. According to the laws of geometrical optics, we should get a sharp image of the slit ( AB) on the screen. But this is not the case, actually the transmitted beam spreads out perpendicularly to the length of the slit. Therefore when it is focused on the screen a diffraction pattern having alternate dark and bright fringes of decreasing intensity is obtained on both the sides of the central point C. The occurrence of diffraction pattern can be explained as follows:

203

Explanation Let a plane wavefront of monochromatic light of wavelength λ be propagating normally to the slit AB . According to Huygen’s theory, every point within the slit becomes the source of secondary wavelets, which spread out in all directions. These wavelets then interfere to produce the diffraction pattern. All the secondary wavelets diffracted along the direction of incident waves are focused at C and those which are diffracted at an angle θ with the normal, are brought to focus at P. The intensity of the diffracted beams will be different in different directions and there will be some directions in which there will be no light. The point C is equidistant from all points on either side of the centre of slit O. Therefore at the point C intensity is maximum. In order to find the resultant intensity at P, draw a perpendicular AG on BE . The path difference between the rays emanating from extreme points A and B of the slit AB is given by ∆ = BG = AB sin θ = a sin θ where a is the width of slit AB and corresponding phase difference is

2π (a sin θ) . λ

Let the width AB of the slit may be imagined to be divided into a large number of n equal parts, each part being the source of secondary wavelets. The amplitude of the wave due to each part is equal to a but their 2π phases will vary gradually from zero to (a sin θ), as the path difference between the rays originating λ from points in AB vary from O to a sin θ . Thus, the phase difference between the waves from any two successive parts of the slit AB would be, 1  2π  a sin θ = δ   n λ According to the theory of composition of n simple harmonic motions of equal amplitude (a) and common phase difference between successive vibrations, the resultant amplitude at P is given by,

or

R =a

sin (nδ / 2) sin {(πa sin θ) / λ } =a sin (δ / 2) sin {(πa sin θ) / nλ}

R =a

sin α a sin α πa sin θ = , where α = sin (α / n) (α / n) λ

and for large value of n, sin (α / n) ≈ α / n ∴

R = na

sin α A sin α = α α

...(1)

where A = na The resultant intensity at P, which is proportional to the square of the resultant amplitude R, is given by  sin α  I = R2 = A 2    α 

2

...(2)

For simplicity, the constant of proportionality being taken as unity. This is the required expression for the intensity distribution due to Fraunhofer diffraction at a single slit. The eqn. (2) gives the variation of intensity with θ (or α) .

204

Positions of Maxima and Minima Principal Maximum: The resultant amplitude given by eqn. (1) can be expanded as R=

  α2 α4 α6  A A α3 α5 α7 sin α = + − + .. = A 1 − + − + … α − α α  3 ! 5 ! 7! 3 ! 5 ! 7!   

R will be maximum if the negative terms vanish. This is possible only when α =

π (a sin θ) = 0 or θ = 0. λ

Thus the maximum value of resultant amplitude R at C is A and the corresponding maximum intensity I max is proportion to A2 . Positions of Minima: From eqn. (2) it is clear that the intensity is minimum when sin α = 0 and α ≠ 0 i.e.

α = ± π , ± 2 π , ± 3 π , … ± nπ or α = ± nπ ,

∴

α=

or

a sin θ = ± n λ

where n = 1, 2, 3, …

πa sin θ = ± nπ λ ...(3)

where n = 1, 2, 3 etc. gives the directions of first, second, third, ... minima. Here n ≠ 0, because n = 0 or θ = 0 corresponds to principal maximum.

Secondary Maxima In addition to principal maximum, there are less intense secondary maxima between equally spaced minima. In order to locate the positions of secondary maxima we apply the mathematical method for finding maxima and minima. Let us differentiate eqn. (2) with respect to α and equate it to zero, that is, dI d = dα dα

 2 sin2 α  2 sin α A  = A2  2  α α  

that is, either

sin α α cos α − sin α = 0 or =0 α α2

∴

sin α = 0 or α cos α − sin α = 0

or

α cos α − sin α  = 0   α2

α = tan α

The equation, sin α = 0 gives the positions of minima except when α = 0 . Thus the positions of secondary maxima is given by the equation α = tan α

...(4)

The eqn. (4) can be solved graphically by plotting the curves according to the simultaneous equations y = α and y = tan α The eqn. y = α, is a equation of straight line passing through an origin and making an angle 45° with the axis as shown in Fig. 4. The equation y = tan α represents a discontinuous curve having a number of branches with asymptotes at an interval α = π (as shown in Fig. 4). The point of intersection of these two curves gives the value of α satisfying the equation α = tan α .

205

The first value of a = 0 gives principal maximum. The remaining values of α which give the secondary maxima are 3 π 5 π 7π 9 π ...(5) α= , , , ,… 2 2 2 2 The more exact values of α are α = 1.430 π , 2 .46 π , 3 .47 π , 4 .471 π , … The directions of secondary maxima are approximately given by the relation (2 n + 1) π , where n = 1, 2, 3, … 2 (2n + 1) λ π because,α = a sin θ a sin θ = ± 2 λ α=±

∴

...(6)

Substituting the approximate values of α from eqn. (5) in eqn. (2), we get intensities of various secondary maxima. (i)

For Principal maximum (central maximum), α = 0 I = I0

∴

(ii) The intensity of first secondary maximum, I1 = A2

3π   sin   2   3π     2

2

=

2

4 9 π2

A2 =

I A2 = 0 22 22

where I0 is the intensity of primary maximum. Thus, I1 is 4 . 5% of I0 or the intensity of first secondary maximum is about (1 / 22 )th of intensity of the central maximum. The ratio of the intensity of the secondary maximum that is adjacent to the central maximum relative to the central maximum is [UPTU, B.Tech I Sem. 2001] I1 4 1 = = I0 9 π 2 22 (iii) The intensity of second secondary maximum, I2 = A2 I2 4 1 = = I0 25 π 2 62

or

5π   sin   2   5π     2 

2

2

=

4 25 π

2

A2 =

I A2 = 0 62 62

Thus, I2 is 1.61% of I0 or the intensity of second secondary maxima is about (1 /62) th of the intensity of central maximum. Similarly, the intensity of third secondary maximum, I3 =

4 49 π 2

I0 =

I0 and so on. 120

If we assume I0 = 0, then the relative intensities of the principal, the first, the second . . . . . .maximum are, 1,

4 2

9x

;

4 25 π

2

;

4 49 π 2

, . . . , etc.

206

From the above various expressions of intensities of secondary maxima, it is clear that most of the light is concentrated in the principal maximum, intensity of the first subsidiary maximum on either side of the principal maximum have about 4 . 5% of the light that of principal maximum. Thus, the intensity of secondary maxima decreases very rapidly. The minima are equally spaced and their intensity is zero. The intensity distribution curve in case of single slit diffraction as depicted in Fig. 5 indicates that the diffraction pattern consists of a bright central maximum surrounded alternately by minima of zero intensity and less intense secondary maxima of rapidly decreasing intensity.

Spread of Central Diffraction Maximum The direction of first minima is given by a sin θ = ± λ or

(∵ n = 1)

sin θ = ± λ / a or θ = sin −1 (± λ / a)

Therefore, central maximum extends between θ = sin

−1

(λ / a) and θ = sin

...(7) −1

(− λ / a) . It means θ is the

angular half width of the central maximum. If the lens towards the screen, that is, L2 is very near to the slit AB or the screen is far away the lens L2 (Fig. 3), then, we have, r ...(8) sin θ = ± f where r is the linear half width of the central maximum and f the focal length of lens L2 . Comparison of eqns. (7) and (8) gives λ f r λ or r = = f a a Hence, the width of central maximum = 2r =

2f λ a

...(9)

It is obvious from eqn. (9) that the width of central maximum is directly proportion to λ . As the wavelength (λ ) of red light is longer than that of violet light, the spread of central maximum with red light is more than that with violet light.

Effect of Making Slit Narrower The first minima on either side of the central maximum occurs in the direction θ given by a sin θ = ± λ

207

When the slit is narrowed by reducing a, the angle of diffraction θ increases which means that the central maximum becomes wider. When the width of slit is made equal to the wavelength of light, that is, a = λ, the first minimum occurs at θ (sin θ = λ / λ ) = 90 ° , it means that central maximum occupies the whole space.

Distinction Between Single-slit Diffraction Pattern and Duble Slit Interference Pattern The diffraction pattern due to single slit of width a differs is many respect with the interference pattern due to double slit of separation a. In the diffraction pattern, the principal maximum has highest intensity and brightest and has non–symmetrical weak secondary maxima on either side. In the interference pattern, the maxima and minima are equidistant and equally wide and all the maxima have the same intensity. Example 1: Intensity distribution due to diffraction at a slit is given by I θ = I0

sin2 α α2

, where α =

π a sin θ λ

Plot typical graph showing I θ / I0 against sin θ for the case a = 2 . 5 λ . Show that the intensity of the first maximum on either side is approximately 4 / 9 π 2 times the intensity of the central maximum. Solution: According to the theory of Fraunhofer diffraction due to a single slit, the position of central maximum is given by the expression as α=

π a sin θ = 0 , that is, sin θ = 0 λ

Iθ I0

where a is the width of the slit and θ the angle of diffraction.

1.0

Therefore, the intensity at central maximum is, I θ = I0

0.5

The positions of secondary maxima are given as α= where or

π a sin θ (2 m + 1) π =± , λ 2 m = 1, 2, 3, …

–3λ/a –2λ/a –λ/a O λ/a sin θ

(2 m + 1) λ sin θ = ± . 2 a

The position of first secondary maximum (m = 1) is sin θ = ±

3λ λ = ± 1.5 2a a

Similarly, the position of second secondary maximum (m = 2) is sin θ = ±

5λ λ = ± 2 .5 2a a

Fig. 6

2λ/a 3λ/a

208

The positions of minima are given by α=

π a sin θ = ± mπ , m = 1, 2, 3, … λ

The position of first minimum is given as sin θ = ± λ / a Similarly, the position of second minimum is given as sin θ = ± 2 λ / a As the width of the slit, a is 2 .5 λ and maximum value of θ is 90°, that is, sin θ = 1. The typical graph between I θ / I0 and sin θ is limited to ± 2 .5 λ / a as shown in Fig. 6. For first secondary maximum, α =

π a sin θ π a 3 λ 3 = . = π (approx.) λ λ 2a 2

Therefore, the intensity of first secondary maximum is given by I θ = I0

sin2 (3 π / 2) 4I 1 = I0 . = 02 2 (3 π / 2) (9 π / 4) 9 π

Hence, the intensity of first maximum or either side is approximately 4 /9 π 2 times the intensity of central maximum. Example 2: Light of wavelength 5500 Å falls normally on a slit of width 22 . 0 × 10 − 5 cm. Calculate the angular position of the first two minima on either side of the central maximum. [U.P.T.U. B.Tech. I Sem. (C.O.) 2003]

Solution: The angular positions of minima in a single slit diffraction pattern are given by a sin θ = ± nλ or sin θ = nλ / a, where n = 1, 2 , 3, … Here,

a = 22 .0 × 10 − 5 cm and λ = 5500 × 10 − 8 cm

For first order minimum, n = 1 ∴

sin θ1 =

−8 λ 5500 × 10 = = 0 .25 or θ1 = sin −1 (0 .25) = 14 °29 ′ a 22 .0 × 10 − 5

For second order minimum, n = 2 ∴ sin θ2 = 2 λ / a = 2 × 0 .25 = 0 .5 ∴ θ2 = sin −1 (0 .5) = 30 ° Therefore, the first two minima will occur at an angle 14 °29′ and 30 ° on either side of central maximum. Example 3: In Fraunhofer diffraction due to a narrow slit, a screen is placed 2 m away from the lens to obtain the pattern. If the slit width is 0 . 2 mm and first minima lie 5 mm on either side of the central maximum, find the wavelength of light.

209

Solution: In the Fraunhofer diffraction pattern due to a single slit of width a, the directions of minima are given by

Ist minimum A

θ

Central maximum

P

a sin θ = ± nλ where n = 1, 2, 3, … or

sin θ = nλ / a

a θ

If θ is measured in radian, than sin θ = θ ∴

x=0.5 cm

θ

O

θ = nλ / a

C G

The angular separation θ between the first minimum on either side of the central maximum is

P'

B

θ = λ / a, Here n = 1

D = 200 cm

Ist minimum

In the given problem, a = 0 .2 mm = 0 .02 cm ∴

...(1)

θ = (λ / 0 .02) rad

Fig. 7

Linear separation between the first minimum and central maximum is as CP = 5 mm = 0 .5 cm According to Fig. 7 θ=

CP 0 .5 = rad OC 200

...(2)

Comparing eqns. (1) and (2), we get 0 .5 λ = 0 .02 200 or

λ=

0 .5 × 0 .02 = 5 × 10 − 5 = 5000 Å 200

Example 4: Light of wavelength 5000 Å is incident normally on a single slit. The central maximum falls out at 30 ° on both sides of the direction of the incident light. Calculate the slit width. For what width of the slit the central maximum would spread out to 90 ° from the direction of the incident light ? Solution: In the Fraunhofer diffraction due to single slit, the direction of minima are given by a sin θ = ± nλ , where n = 1, 2 , 3, … Therefore, the angular spread of the central maximum on either side of incident light is, λ λ sin θ = or a = a sin θ Here θ = 30 ° , so sin θ = sin 30 ° = 0 .5 and λ = 5000 Å = 5000 × 10 − 8 cm ∴

a=

5000 × 10 − 8 = 10 − 4 cm 0 .5

For θ = 90 ° , we have

a=

λ = λ = 5000 Å sin 90 °

210

Example 5: Light of wavelength 5000 Å is incident normally on a slit. The first minimum of diffraction pattern is observed to lie at a distance of 5 mm from the central maximum on a screen placed at a distance of 2 m from the slit. Calculate the width of the slit.

[U.P.T.U., B.Tech. I.Q. Bank, 2000]

Solution: In the Fraunhofer diffraction due to a single slit of width a, the directions of minima are given by a sin θ = ± nλ , where n = 1, 2 , 3, … The angular separation θ between the first minimum (n = 1) on either side of central maximum is given by or a sin θ = λ sin θ = a / λ If θ is small, then sin θ ≈ θ

∴

...(1)

θ = λ /a

Linear separation between the first minimum and central maximum (given) = 5 mm = 0 .5 cm The distance of the screen from the slit = 2 m = 200 cm Then, from Fig. 7,

θ=

CP 0 .5 = rad OC 200

...(2)

Comparing eqn. (1) and (2), we get 0 .5 λ = 200 a

or

 200   a=λ   0 .5 

Here

λ = 500 Å = 5000 × 10 − 8 cm

∴

a=

5000 × 10 − 8 × 200 = 0.02 cm 0 .5

Example 6: A light of wavelength 6000 Å falls normally on a strength slit of width 0.10mm. Calculate the total angular width of the central maximum and also the linear width as observed on a screen placed 1 metre away.

[GBTU, B.Tech I Sem. 2010, UPTU, B.Tech. I Sem. 2009]

Solution: In a single slit diffraction pattern, the directions of minima are given by a sin θ = ± nλ , where n = 1, 2, 3, ..... For the first order minimum, If θ is small,

n =1

∴

a sin θ = λ

sin θ ≈ θ

∴

sin θ ≈ θ ≈

Here θ is the angular half width of the central maximum Here a = 0 .10 mm = 0 . 01 cm and λ = 6000 Å = 6000 × 10 −8 cm

λ a

211

∴ angular half width,

θ=

6000 × 10 −8 = 6 × 10 −3 radian. 0 . 01

Hence, the total angular half width 2θ = 2 × 6 × 10 −3 = 1.2 × 10 –2 rad According to Fig. 7 (shown in example 3) the linear half width CP = x = θ . D Total linear width of central maximum, PP' = 2 x = 2θD linear width = 2 × 1. 2 × 10 −2 × 100 = 2.4cm

∴

Example 7: Plane wave of λ = 6 . 0 × 10 − 5 cm fall normally on a slit of width 0 . 20 mm. Calculate (i) the total angular width of the central maximum (ii) the linear width of the central maximum on a screen placed 2 m away. Solution: (i) In case of a single slit diffraction pattern, the direction of minima are given by a sin θ = ± nλ For the first order minimum, n = 1 ∴

a sin θ = λ

If θ is small, then sin θ = θ

sin θ = θ ≈ λ / a

∴

Here, a = 0 .2 mm = 0 .02 cm and λ = 6 × 10 − 5 cm

or

∴

θ=

sin θ = λ / a

6 × 10 − 5 = 3 × 10 − 3 radian 0 .02

This is the angular half width of the central maximum. Therefore, the total angular width of central maximum = 2 θ = 2 × 3 × 10 − 3 = 6 × 10 –3 rad (ii) According to Fig. 7 linear half width CP = x = θ . D Total linear width of central maximum = PP′ = 2 x = 2 θ . D = 2 × 3 × 10 − 3 × 200 = 1.2 cm Example 8: Calculate the angle at which the first dark band and the next bright band are formed in the Fraunhofer diffraction pattern of a slit 0.3 mm wide (λ = 5890 Å ) . [UPTU, B.Tech. Special C.O. Exam. Aug 2008, B.Tech. I Sem. (C.O.) 2007]

Solution: In Fraunhofer diffraction pattern due to single slit of width a, the directions of minima are given by a sin θ = nλ , where n = 1, 2 , 3, … For first dark band, n = 1, and therefore a sin θ = λ or

or

sin θ =

−8 λ 5890 × 10 = = 0 .00196 a 0 .03

θ = sin−1 (0 . 00196 ) = 0 .112 °

212

The angle of diffraction θ′ corresponding to the first bright band on either side of the central maximum is approximately given by a sin θ ~ 3 λ / 2 ~1.5 λ

or

sin θ′ = 1.5

λ a

λ = 0 .00196, therefore a

but

or

sin θ′ = 1.5 × 0 .00196 = 0 .00294

θ′ = sin−1(0.00294) = 0.168 °

Example 9: In a single slit diffraction pattern the distance between the first minimum on the right and the first minimum on the left is 5.2 mm. The screen on which the pattern is displayed is 80 cm from the slit and the wavelength is 5460 Å. Calculate the slit width. Solution: The central maximum is exactly midway between the first minimum on its left and first minimum on its right as shown in Fig. 7. Therefore, the distance of first minimum from the central maximum is x=

2x 5.2 = = 2 .6 mm 2 2

From Fig.7, θ = x / D and from the expression of direction of minima in a single slit diffraction pattern a sin θ = ± nλ For

n = 1, sin θ = θ =

λ a

x λ = D a

∴ θ=

or

a=

λD x

Here, x = 2 .6 mm = 0 .26 cm, D = 80 cm and λ = 5460 Å = 5460 × 10 − 8 cm ∴

Slit width,

a=

5460 × 10 − 8 × 80 = 168 × 10 − 4 cm 0 .26

Example 10: A single slit of width 0.14 mm is illuminated normally by monochromatic light and diffraction bands are observed on a screen 2 m away. If the centre of second dark band is 1.6 cm from the middle of the central bright band, deduce the wavelength of light.

[U.P.T.U., B.Tech. II Sem. 2005]

Solution: For single slit diffraction pattern, the direction of nth minima is given by

for small θ,

a sin θ = nλ

or

sin θ =

sin θ ~ θ,

∴

θ=

nλ a

nλ a

If D is the distance of the screen from the slit and x is the distance of centre of second dark band from the middle of the central bright band, then θ=

x D

∴

nλ x = D a

or

λ=

xa , nD

213

Here, x = 1.6 cm = 1.6 × 10 − 2 m , a = 0.14 mm = 0 .14 × 10 − 3 m, n = 2 and D = 2 m ∴

λ=

1.6 × 10 − 2 × 0 .14 × 10 − 3 = 5.6 × 10 − 7 m = 5600 Å 2 ×2

Example 11: Calculate the wavelength of light whose first diffraction maximum in the diffraction pattern due to a single slit falls at θ = 30 ° and coincides with the first minimum for red light of wavelength 6500Å. [U.P.T.U., B.Tech. I.Q. Bank, 2000]

Solution: As shown in Fig. 8. the central maximum is exactly at midway between the first minimum on its either side. Therefore, the first diffraction maximum is about half way between the first and second minima. If λ1 is the wavelength of light, then a sin θ ~ 3 λ1 / 2 ~1.5 λ1

Ist minimum Slit

x

θ

...(1)

2x

Central maximum

Ist minimum

For red light of wavelength λ (= 6500 Å) a sin θ = 1 λ (∵ n = 1 for first maximum)

...(2)

Fig. 8

Dividing eqn. (1) by equation (2), we get 1.5 λ1 = λ or λ1 =

λ 6500 = = 4333.3 Å 1.5 1.5

Example 12: A single slit is illuminated by light composed of two wavelengths λ1 and λ 2 . One observes that due to Fraunhofer diffraction, the first-minima obtained for λ1 coincides with the second diffraction minima of λ 2 . What is the relation between λ1 and λ 2 ? [U.P.T.U., B.Tech. I Sem. (C.O.) 2006, I Sem. 2003, II Sem. 2001]

Solution: In the Fraunhofer diffraction pattern due to single slit of width a, the directions of minima are; a sin θ = ± nλ For wavelength λ1, the direction of first minima (n = 1) is, a sin θ1 = λ1

...(1)

Similarly, for wavelength λ 2 , the direction of second minima (n = 2) is, a sin θ2 = 2 λ 2

...(2)

According to the given problem, the direction of first minima (θ1) due to λ1 and the second minima due to λ 2 coincides that is, θ1 = θ2 = θ . Therefore, from eqns. (1) and (2), we have a sin θ = λ1 = 2 λ 2 ∴

λ 2 = λ1 / 2

214

Example 13: A lens of focal length 100cm forms Fraunhofer diffraction pattern of a single slit of width 0 .04 cm in its focal plane. The incident light contains two wavelengths λ1 and λ 2 . It is found that the fourth minimum corresponding to λ1 and the fifth minimum corresponding to λ 2 occur at the same point 0 . 5 cm from the central maximum. Compute λ1 and λ 2 .

[M.T.U., B.Tech I Sem. 2012]

Solution: In the Fraunhofer diffraction pattern due to a single slit of width a, the directions of minima are given by a sin θ = ± nλ For wavelength λ1, the direction of fourth minimum (n = 4) is given by ...(1)

a sin θ1 = 4 λ1 Similarly, for wavelength λ 2 , the direction of fifth minimum (n = 5) is given by

...(2)

a sin θ2 = 5 λ 2

According to the problem, the direction of fourth minimum (θ1) due to λ1 and the fifth minimum (θ2 ) due to λ 2 occur at the same point, that is, θ1 = θ2 = θ Therefore,

[from eqn. (1) and (2)]...(3)

a sin θ = 4 λ1 = 5 λ 2

If x is the linear distance of fourth minimum for λ1 or fifth minimum for λ 2 , we have θ = x / f (as θ is small), where f is the focal length of the lens. Here,

x = 0 .5 cm and f = 100 cm

∴

θ = 0 .5 /100 = 0 .005 rad

From eqn. (3), we have, a sin θ = 4 λ1 or

λ1 =

a sin θ a θ = 4 4

λ1 =

0 .04 × 0 .005 = 5 × 10 − 5 cm 4

(as θ is small)

Here a = 0 .04 cm and θ = 0 .005 rad ∴

Substituting value of λ1 in eqn. (3), we have λ2 =

4 λ1 4 × 5 × 10 − 5 = = 4 × 10 − 5 cm 5 5

Example 14: A lens whose focal length is 50 cm form a Fraunhofer diffraction pattern of a single slit of 0.3 mm width. Calculate the distances of the first dark band and of the next bright band on either side of the central maximum. The wavelength of light used is 5890 Å.

[U.P.T.U., B.Tech. I Sem. (old) 2009]

215

Solution: The direction of the mth minima in the Fraunhofer diffraction pattern due to a single slit of width ‘ a’ is given by a sin θ = mλ , where m = 1, 2, 3,...... For the first dark band or for the first minima, m = 1 or

sin θ =

λ , a

∴

a sin θ = λ

Here λ = 5890 Å = 5890 × 10 −8 cm, and a = 0 . 3 mm = 0 . 03 cm sin θ =

∴

5890 × 10 −8 = 1. 963 × 10 −3 rad 0 . 03

As θ is small, sin θ ≈ θ = 1.963 × 10 −3 rad

First minima

If y is the linear distance of first dark band from the axis, we have θ=

y or y = f θ, where f is the focal length of the lens f

Screen y

θ

a

d

Here f = 50 cm and θ = 1. 963 × 10 −3 rad ∴

Central maximum

Fig. 8(a)

y = 50 × 1. 963 × 10 −3 = 0.09815 cm

The angular position θ1 of the bright band or of the secondary maxima of either side is approximately (2 m + 1) λ given by a sin θ1 = 2 For first secondary maxima of first bright band, m = 1 sin θ1 =

∴ As θ1 is small, we have

∴ a sin θ1 =

3 λ 3 × 5890 × 10 = 2a 2 × 0 . 03

−8

3λ 2

= 2 . 945 × 10 −3 rad

θ1 = sin θ1 = 2 . 945 × 10 −3 rad

If the linear distance of first bright band on either side of the central maximum is y1, we have y1 = f θ1 = 50 × 2 . 945 × 10 −3 = 0.1472 cm Example 15: In obtaining the diffraction of a single slit the focal length of the lens used is 100cm, wavelength of light 4500 Å and width of slit 10 −

Solution: The width of central maximum, 2 r =

2

cm. Find the width of central maximum.

2fλ a

Here f = 100 cm, λ = 4500 Å = 4500 × 10 − 8 cm and a = 10 − 2 cm ∴

width of central maximum, 2 r =

2 × 100 × 4500 × 10 − 8 10 − 2

= 0.9 cm

216

Fraunhofer’s Diffraction at a Double Slit (Double Slit Diffraction) Let us suppose two parallel slits AB and DE, each of width a separated by an opaque distance b. Let a plane wavefront of monochromatic light of wavelength λ be incident normally upon the slits (Fig. 9). Suppose the light diffracted by the slits be focused by a convex lens L on the screen XY situated in the focal plane of the lens. The diffraction pattern obtained on the screen is characterised by a number of equally spaced interference maxima and minima in the region normally occupied by the central maximum in the single slit diffraction. The central interference maximum of this double slit pattern is four times more intense than the central maximum in the single slit diffraction pattern. The maxima on either side of the central maximum are of gradually decreasing intensity. This characteristic of diffraction pattern may be theoretically explained as follows: Explanation: According to the principle of Huygen, every point within the slit becomes the source of secondary wavelets and sends out secondary wavelets in all directions. All the secondary waves travelling normal to the slits or in the direction of incident light are brought to focus at C. The secondary waves travelling in a direction making an angle θ with the initial direction come to focus at P. According to the theory of diffraction at a single slit, the resultant amplitude due to all the wavelets diffracted from each slit is given by R=A

sin α α

and the resultant phase in this direction is, α =

π a sin θ λ

...(1) ...(2)

where A is the resultant amplitude in the direction, θ = 0 due to each single slit. We can therefore, replace all the secondary wavelets originating from the slit AB by a single wave of amplitude R and phase α originating from its middle point S1. Similarly all the secondary wavelets in the second slit DE can also be replaced by the similar wave originating from its middle point S2 . Hence the resultant amplitude at a point P on the screen will be due to the interference between the two waves of same amplitude R and a phase difference φ originating from S1 and S2 . Let us drop a perpendicular S1 M on S2 M . The path difference between the two wave originating from S1 and S2 and reaching at P after travelling in the direction θ will be, S2 M = (a + b) sin θ

...(3)

Therefore, the phase difference between them is φ = (2 π / λ ) (a + b) sin θ

...(4)

217

The resultant amplitude at P can be obtained by the vector amplitude diagram shown in Fig. 10, where QR and RS represent the amplitude of the two waves originating from S1 and S2 and angle φ as phase difference between them. According to Fig. 10 QS2 = QR 2 + RS2 + 2 (QR ) .(RS) cos φ R ′2 = R 2 + R 2 + 2 R . R . cos φ

or

= 2 R 2 (1 + cos φ) R ′2 = 4 R 2 cos2 φ / 2

or

...(5)

Substituting the values of R and φ from equations (1) and (4) in equation (5), we get R ′2 = 4 A2 where

β=

sin2 α α2

cos2 β

...(6)

φ π = (a + b) sin θ 2 λ

...(7)

Therefore, the resultant intensity at P is I = R ′2 = 4A 2

sin2 α α2

cos2 β

...(8)

This is the required expression for the intensity distribution due to a Fraunhofer diffraction at a double slit. It is clear from eqn. (8) that the resultant intensity at any point on the screen depends on two factors : 1.

sin2 α α2

2.

which gives diffraction pattern due to each single slit and

cos2 β which gives the interference pattern due to two waves of same amplitude (R ) originating from the mid–points of their respective slits, that is, from S1 and S2 .

Hence, the resultant intensity, due to double slit of equal width, at any point on the screen is given by the product of sin2 α / α 2 and cos2 β . If either of these factor is zero, the resultant intensity will also be zero. Let us now consider the effect of each factor. (i)

As we have seen in the case or single slit diffraction, the factor

A2 sin2 α α2

gives a principal maximum

at C in the direction, θ = 0 . On either side of this principal maximum alternate minima and secondary maxima of diminishing intensity are observed as shown in Fig. 11(a). The angular positions of minima are given by sin α = 0, but α ≠ 0

218

Y Y=

–3π

–π

–2π

π

O

–2

–1

–4π

–3π

–2π

–π

2π

3π

Y = cos2 β

Y –3

α2

α

(a)

–4

A2 Sin2 α

O

1

2

3

4

π

2π

3π

4π

β

(b)

I

4A2

O

2 Sin2 α α2

COS2 β

θ

(c)

Fig. 11

The values of α satisfying the above equation are

or or

α = ± mπ , where m = 1, 2, 3, … π a sin θ =± mπ λ sin θ = ± (mλ / a), m = 1, 2, 3,

The positions of secondary maxima due to this term approach to 3π 5π 7π α =± ,± ,± , . . . etc 2 2 2 (ii)

(From equation 2). ...(9)

...(10)

According to the interference term, cos2 β , the intensity will be maximum when cos2 β = 1

...(11)

219

that is, bright fringes are obtained in the directions given by β = ± nπ , where n = 0 , 1, 2, 3, … From equation (7), or

...(12)

π (a + b) sin θ = ± nπ λ (a + b ) sin θ = ± nλ

...(13)

When n = 0, θ = 0 , that is, the central maximum of interference pattern is observed along the direction of incident light. The various maxima corresponding to n = 0, 1, 2 ,… are the zero-order, first–order, second order ... maxima. Since the central maximum of diffraction pattern (due to individual slit) also lies along this direction, therefore at this point all waves arrive in the same phase. Hence, the intensity of this central maximum is highest. The intensity in the interference pattern will be minimum when cos2 β = 0 , that is, β = ± (2 n + 1) π / 2 or or

π (a + b) sin θ = ± (2 n + 1) π / 2 λ (a + b ) sin θ = ± (2 n + 1) λ / 2

...(14)

where n = 0, 1, 2 ,3, ... This equation represents that for interference minima the path difference between the parallel diffracted rays originating from any pair of corresponding points within the two slits be odd multiple of λ / 2 . It can be easily shown that for small values of θ , the maxima and minima are equally spaced. Thus, cos2 β term in eqn. (8) is responsible for interference fringes. If the slit width a is kept constant and b is varied, the position of maxima and minima due to diffraction remain unaffected while those due to interference undergo a change. Fig. 11(a) represents the intensity distribution due to diffraction term, A2 sin2 α / α 2 and Fig. 11(b) that due to interference term cos2 β . Fig. 11(c) represents the resultant of curves (a) and (b), which is the intensity distribution curve due to double slit. Fig. 11(c) is obtained by multiplying the ordinates of first two curves [Figs. 11(a) and (b)] at every point. Thus, the entire pattern due to double slit may be considered as consisting of interference fringes due to light from both slits. The intensities of these fringes being governed by diffraction occurring at the individual slit.

Effect of Increasing the Slit Width ‘a’ If the width of the slit ‘a’ is increased, the envelope of the fringe–pattern changes and central peak becomes sharper. The fringe spacing remains unaffected because it depends on slit separation ‘b’ which is fixed in this case. Hence, the number of interference maxima falling within the central diffraction maximum decreases.

220

Effect of Increasing the Slit Separation ‘b’ If the slit separation, b, is increased and slit width a is kept constant, then the fringe–spacing decreases and the fringes become closer together. Hence more interference maxima fall within the central maximum.

Effect of Increasing the Wavelength ‘λ’ When the wavelength of the monochromatic light falling on the slits increases, the envelope becomes broader, and the fringes move farther apart.

Missing Orders in Double Slit Fraunhofer Diffraction Pattern In the double slit diffraction (discussed earlier) we have taken slit width of each slit as ‘a’ and the separation between them as ‘b’. If the slit width is kept constant and the separation between them is varied, it is observed that the distance between two successive maxima changes. A close study of the resultant diffraction pattern reveals that certain orders of interference maxima are missing depending upon the relative values of the transparency a and opacity b. The directions of interference maxima are given by the equation (a + b) sin θ = ± nλ , where n = 0, 1, 2 , 3, …

...(1)

and the direction of diffraction minima due to each single slit are given by a sin θ = ± mλ , where m = 1, 2 , 3, …

...(2)

Let us suppose that for the same value of θ, a and b are such that both the eqns. (1) and (2) are simultaneously satisfied. In this situation the certain interference maxima corresponding to the diffraction minima for which θ is common will be absent. Dividing equation (1) by (2), we get (a + b) n = a m (i)

...(3)

If b = a , That is, the width of transparencies a and opacities, b are equal, then n = 2 m = 2 , 4, 6, … (Since m = 1, 2 , 3, ...) nd

th

th

The 2 order, 4 order, 6 order or even ordered interference maxima will be absent from diffraction pattern because these maxima will coincide with 1st , 2 nd , 3 rd ... order diffraction minima due to single slit. For m = 1, n = 2, that is, there should be five interference maxima (n = 0, ± 1, ± 2) in the central maximum but only three interference maxima will appear in the pattern, because 2 nd order (± 2) interference maxima are missing from both the sides of the centre. It is due to the fact for 2nd order maxima the condition of minima is also satisfied simultaneously with the condition of maxima. (ii)

If b = 2 a, then equation (3) becomes n = 3 m that is, whenm = 1, 2 , 3, 4, … etc. then n = 3, 6, 9, 12, … etc. It means that 3 rd , 6 th , 9 th … order interference maxima will coincide with 1st , 2 nd , 3 rd … order diffraction minima. Hence, third, sixth, ninth ... order interference maxima will be absent from the diffraction pattern. For m = 1, n = 3, that is, there should be seven interference maxima

221

(n = 0, ± 1, ± 2, ± 3) in the central diffraction maximum, but only five interference maxima are observed in the pattern because for third order interference maxima both the condition of interference maxima and diffraction minima are simultaneously satisfied and hence third order maxima is absent from both sides of the centre. (iii) If a + b = a, that is, b = 0 then double slit reduces into single slit and all orders of the maxima of interference will be missing. System is working as single slit diffraction pattern. From above discussion we conclude that, as the slit separation, b increases, the number of interference maxima appearing in the region of central diffraction maximum increases. Example 16: Two parallel slits having widths 0.15 mm each and separation between them is 0.30 mm. They are illuminated normally by light of λ = 6000 Å and the emergent light is focused by a convergent lens of 100 cm focal length. Deduce the positions of the first four interference maxima on one side in the focal plane of the lens. Solution: The positions of the interference maxima in the double slit diffraction are given by (a + b) sin θ = nλ or

 nλ   θ = sin−1   a + b

...(1)

Here a = 0 .15 mm = 0 .015 cm , b = 0 .30 mm = 0 .03 cm and λ = 6000 Å = 6000 × 10 −8 cm Since b = 2 a, n = 3 m therefore, 3 rd , 6 th , 9 th … order interference maxima will coincide with 1st , 2 nd , 3 rd … order diffraction minima. Hence, 3 rd , 6 th , 9 th order interference maxima will be absent in the pattern. If θ1, θ2 , θ3 and θ4 be the positions of first four maxima on one side, then According equation (1)  n × 6000 × 10 − 8  −3 θ = sin −1   = 1. 33 × 10 n rad. where n = 1, 2, 4, 5 ( 0 . 015 + 0 . 03 )   ∴

and

θ1 = 1.33 × 10− 3 × 1 = 1. 33 ×10 −

3

rad

θ 2 = 1.33 × 10− 3 × 2 = 2.66 ×10 −

3

rad

θ 3 = 1.33 × 10− 3 × 4 = 5 .32 ×10 −

3

rad

θ 4 = 1.33 × 10− 3 × 5 = 6.65 × 10 −

3

rad

The corresponding positions on the one side of the focal plane of the lens is given as x = f . θ Here f = 100 cm ∴

x = f θ1, f θ2 , f θ3 and f θ4 x = 1.33 mm, 266 mm, 5.32 mm and 6.65 mm.

222

Example 17: In a double slit Fraunhofer diffraction pattern the screen is 170 cm away from the slits. The slit widths are 0.08 mm and they are 0.4 mm apart. Calculate the wavelength of light if the fringe–spacing is 0.25 cm. Also deduce the missing order.

[U.P.T.U., B.Tech. I Sem. 2004]

Solution: The fringe–spacing in the case of interference pattern is given as ω=

Dλ 2b

or

λ=

ω .2b D

where 2b is the separation between the slits Here ∴

2 b = 0 .4 mm = 0 .04 cm, ω = 0 .25 cm and D = 160 cm λ=

0 .04 × 0 .25 170

= 5 .88 × 10 − 5 cm = 5880 Å . If a be the slit width and b the slit separation, then the condition of missing order is represented as a+ b n = a m Here b = 0 .4 mm and a = 0 .08, that is,

0 .08 + 0 .4 n = . 0 .08 m

or

n = 6 m, where m = 1, 2 , 3, …

or

n = 6, 12 , 18, …

Hence, 6th, 12th and 18th . . . orders will be missing.

A Plane Transmission Diffraction Grating (N-Slits Diffraction) A plane diffraction grating is an arrangement consisting of a large number of close, parallel, straight, transparent and equidistant slits, each of equal width a, with neighbouring slits being separated by an opaque region of width b. A grating is made by drawing a series of very fine, equidistant and parallel lines on an optically plane glass plate by means of a fine diamond pen. The light cannot pass through the lines drawn by the diamond; while the spacing between the lines is transparent to the light. There are about 15,000 lines per inch in such a grating to produce a diffraction of visible light. The spacing (a + b) between adjacent slits is called the diffraction element or grating element. If the lines are drawn on a silvered surface of the mirror (plane or concave) then light is reflected from the positions of mirrors in between any two lines and it forms a plane or concave reflection grating. Since the original gratings are quite expensive, for practical purposes their photographic reproduction are generally used. The commercial gratings are produced by taking the cast of an actual grating on a transparent film such as cellulose acetate. A thin layer of collodion solution (celluloid dissolved in a volatile solvent) is poured on the surface of ruled grating and allowed to dry. Thin collodion film is stripped off from grating surface. This film, which retains the impressions of the original grating, is preserved by mounting the film between two glass sheets. Now–a–days holographic gratings are also produced. Holographic gratings have a much large number of lines per cm than a ruled grating.

223

Theory of Grating X

Suppose a plane diffraction grating, consisting of large separation b, is illuminated normally by a plane wavefront of monochromatic light of wavelength λ as

S1

α

S2

K1

θ

shown in Fig. 12. The light diffracted through N slits is focused by a convex lens on the screen XY placed in the

(a+b)

θ

Sn–1

Kn–2

Sn

Kn–1

consists of extremely sharp principal interference maximum; while the intensity of secondary maxima becomes negligibly small so that these are not visible in the diffraction pattern.

P

θ θ

focal plane of the lens L. The diffraction pattern obtained on the screen with very large number of slits

L

C

O

Screen

number of N parallel slits each of width a and

A sin α

θ θ A sin α

Y

α

According to Huygen–Fresnel principle, all points

Fig. 12

within each slit become the sources of secondary wavelets, which spread out in all directions. From the theory of Fraunhofer diffraction at a single slit, the resultant amplitude R due to all waves diffracted from A sin α π a sin θ each slit in the direction θ is given by R = , where A is the constant and α = . Thus the α λ resultant of all the secondary waves may be treated as a single wave of amplitude R and phase α starting from the middle point of each slit and travelling at an angle θ with the normal. Thus the waves diffracted from all the N slits in direction θ are equivalent to N parallel waves, each starting from the middle points S1, S2 , S3 , ... Sn−1, Sn of slits. Draw perpendicular S1 Kn −1 from S1 on Sn Kn−1, then the path difference between the waves originating from the slits S1 and S2 is S2 K1 = S1S2 sin θ = (a + b) sin θ Similarly, between rays from S2 and S3 will be S3 K2 = S2 S3 sin θ = (a + b) sin θ Thus the path difference between the waves from any two consecutive slits is (a + b) sin θ . Therefore the 2π corresponding phase difference is (a + b) sin θ = 2 β . Therefore, as we pass from one vibration to λ another, the path goes on increasing by an amount (a + b) sin θ and corresponding phase goes on 2π increasing by an amount (a + b) sin θ . Thus the phase increases in arithmetical progression. Therefore, λ the problem reduces to find the resultant amplitude of N waves of equal amplitude ( A sin α / α ) and period but the phase increasing in the arithmetic progression. In this case the resultant amplitude in the direction θ is given by R′ = R

sin N β A sin α sin N β = . sin β α sin β

224

The corresponding intensity at P is given by A 2sin2 α

I = R ′2 =

The first factor of eqn. (1), that is,

α2

A2 sin2 α α2

single slit; whereas the second factor, that is,

.

sin2 N β sin2 β

...(1)

represents intensity distribution due to diffraction at a

sin2 N β sin2 β

gives the distribution of intensity in the diffraction

pattern due to the interference in the waves from all the N slits. Principal Maxima: The condition for principal maxima is sin β = 0, that is, β = ± nπ , where n = 0, 1, 2 ,3, … we have, sin N β = 0, so

sin N β 0 = becomes indeterminate. But it may be evaluated by applying the usual sin β 0

method of differentiating the numerator and denominator and applying its limit as β tends to ± nπ,

Thus

lim

β → ± nπ

sin N β = lim sin β β → ± nπ

d (sin N β) dβ N cos N β = lim =N d cos β β → ± nπ (sin β) dβ

Therefore, the intensity of principal maxima is proportional to N 2 and increases as the number of slits increase. sin N β Substituting the value of = N in eqn. (1), we get sin β I =

A 2 sin2α α2

N2

...(2)

Thus, in order to find the resultant intensity of any principal maxima, we have to multiply N 2 by a factor A2 sin2 /α 2 . Thus, if we increase the number of slits, (N ) the intensity of principal maxima increases. The directions of principal maxima are given by

sin β = 0, that is, β = ± nπ , where n = 0, 1, 2,…

or

π (a + b) sin θ = ± nπ λ

or

(a + b ) sin θ = ± nλ

...(3)

If we put n = 0 in eqn. (3), we get θ = 0 and eqn. (3) gives the direction of zero order principal maximum. The first, second, third, ..., order principal maxima may be obtained by putting n = 1, 2, 3, … in eqn. (3). Minima: The intensity is minimum when Therefore or

sin N β = 0 ; but sin β ≠ 0

N β = ± mπ N

π (a + b) sin θ = ± mπ λ

225

or

...(4)

N(a + b)sin θ = ± m λ

Here m can have all integral values except 0, N , 2 N , 3 N , … because for these values of m, sin β = 0 which gives the positions of principal maxima. Positive and negative signs show that the minima lie symmetrically on both sides of the central principal maximum. It is clear from eqn. (4) that for m = 0, we get zero order principal maximum, m = 1, 2 , 3, ..., N − 1 give minima governed by eqn. (4) and then at m = N , we get principal maximum of first order. This indicates that, there are (N − 1) equispaced minima between zero and first orders maxima. Similarly, between first (m = N ) and second (m = 2 N ) order principal maxima, there are (N − 1) equispaced minima. Thus, there are ( N − 1) minima between two successive principal maxima.

Secondary Maxima The above study reveals that there are (N − 1) minima between two successive principal maxima. Hence there are (N − 2) other maxima coming alternatively with the minima between two successive principal maxima. These maxima are called secondary maxima. To find the positions of the secondary maxima, we first differentiate eqn. (1) with respect to β and equating it to zero 2 2 sin N β  N cos N β sin β − sin N β cos β dI A sin α = .2  =0  2 dβ α sin2 β  sin β 

or or

N cos Nβ sin β − sin Nβ cos β = 0 ...(5)

tan N β = N tan β

To find the intensity of secondary maxima, we make the use of the triangle shown in Fig.13. We have

Therefore,

or

sin N β = sin2 Nβ sin2 β sin2 Nβ sin2 β

=

=

=

N tan β (1 + N 2 tan2 β) (N 2 tan2 β) / (1 + N 2 tan2 β) sin2 β N 2 tan2 β (1 + N 2 tan2 β) sin2 β N2 1 + (N 2 − 1) sin2 β

Putting this value of (sin2 Nβ) / sin2 β in eqn. (1), we get the intensity of secondary maxima as I' =

A 2 sin2 α α2

.

N2 [1 + (N 2 – 1)sin2 β ]

...(6)

This indicates that the intensity of secondary maxima is proportional to N2 / [1 + (N2 − 1) sin2 β ] whereas the intensity of principal maxima is proportional to N2 .

226

The ratio of the intensity of these secondary maxima to the intensity of principal maxima is obtained by dividing eqn. (6) by eqn. (2). Thus Intensity of secondary maxima Intensity of primary maxima =

I′ 1 = I 1 + (N 2 − 1) sin2 β

Hence, greater the value of N, the weaker are secondary maxima. In an actual grating N is extremely large. Hence secondary maxima are not visible in the grating spectrum. The intensity distribution curve for plane diffraction grating is depicted in Fig. 14.

Angular Half-Width or Width of the Principal Maximum The angular separation between the first two minima lying on either side of the principal maximum is called the angular width of principal maximum of any order. As shown in Fig. 15, 2 d θ n represents the angular width of nth order principal maximum. The nth order principal maximum lies in the direction given by equation (a + b) sin θ n = nλ

...(1)

If (θ n + d θ n) and (θ n − d θ n) represent the directions of first outer and inner sided minima adjacent to the nth order principal maximum, then dθ n will give the angular half width of nth principal maximum. The directions of minima are represented by an equation. N (a + b) sin θ = mλ

...(2)

where m is an integer and takes all integral values except 0, N , 2 N , … nN , because for these values of m eqn. (2) give the directions of zero, first, second, ... nth order principal maxima. As the first outer and inner sided minima adjacent to nth order principal maximum lie in the direction (θ n ± d θ n), this is corresponding to the value of m = (nN ± 1) . Hence eqn. (2) gives N (a + b) sin (θ n ± dθ n) = (nN ± 1) λ or

N (a + b) (sin θ n cos d θ n ± cos θ n sin dθ n) = (nN ± 1) λ

As dθ n is small, then cos dθ n ≈ 1 and sin dθ n ≈ dθ n Thus,

N (a + b) (sin θ n ± cos θ n dθ n) = (nN ± 1) λ

...(3)

227

or

N (a + b) sin θ n ± N (a + b) cos θ n dθ n = nNλ ± λ

...(4)

Multiplying eqn. (1) by N on both sides, we get ...(5)

N (a + b) sin θ n = Nnλ Substituting this value of N (a + b) sin θ n in eqn. (4), we get N (a + b) cos θ n dθ n = λ λ or d θn = N (a + b) cos θ n

...(6)

This is the angular half width of principal maxima. Substituting the value of (a + b) from eqn. (1) in eqn. (6), we get, 1 d θn = N n cot θ n The width of nth order principal maximum = 2dθ n =

2λ N(a + b)cos θ n

...(7) ...(8)

Thus, the angular width of principal maximum is inversely proportional to number of slits (N), that is, as N increases width of the maxima decreases. It is clear from expressions (7) and (8) that the angular half width or angular width of principal maximum in a plane transmission grating does not depend on the number of rulings per unit length, but it does depend on the total number of lines present on the grating.

Effect of Closeness of Rulings If a + b is small, that is, the lines on the grating are close together, the dispersive power will be large. That is, the angular spacing between maxima become large.

Effect of Width of the Ruled Surface If the width of the ruled surface, that is, N (a + b) increases the width of principal maxima decreases, that is, the maxima become sharper.

Advantage of Increasing the Number of Rulings on a Grating If the number of rulings in a grating increases, the principal maxima become intense and sharp. Moreover, the secondary maxima become weaker. Example 18: Two plane diffraction gratings A and B have same width of ruled surface but A has greater number of lines than B. Compare intensity of fringes, width of principal maximum and dispersive power in two cases. Solution: The intensity of principal maxima of the diffraction pattern in plane diffraction grating is proportional to N 2 , that is, to the square of number of lines on the grating. Thus, the intensity of fringes is greater for grating which has great number of lines, that is, grating A. The angular width of principal maxima is inversely proportional to the total width of ruled surface, that is, on N (a + b) . Since gratings A and B have same width of ruled surface, hence it is same for A and B. The dispersive power is inversely proportional to the grating element (a + b) . Since grating A has greater number of lines on the same width of ruled surface as B, it has a small grating element and hence has a large dispersive power than B.

228

Formation of Multiple Spectra with Grating We have seen that when a plane diffraction grating is illuminated normally by a monochromatic beam of light of wavelength λ, the principal maxima are formed in the direction given by (a + b) sin θ = ± nλ

...(1)

where (a + b) is the grating element, n the order of the maxima and λ the wavelength of light used. From the above equation we inferred that (i)

For a particular wavelength λ, the directions of principal maxima of different orders are different.

(ii)

For a particular order n, the light of different wavelengths will be diffracted in different directions; longer the wavelength greater is the angle of diffraction θ . Hence if white light is incident normally on a diffraction grating, each order of spectrum will contain principal maxima of different wavelengths in different directions.

Eqn. (1) shows that, for n = 0, θ = 0 for all values of λ . Thus zero order principal maximum for all values of λ lie in same direction. Hence the zero order (n = 0) principal maximum will be white. The first order (n = 1) principal maxima of all visible wavelengths form the first order spectrum and the second order principal maxima of all wavelengths form second order spectrum and so on. As λ r > λ v , the angle of diffraction for red colour is greater than that for violet colour and hence in every spectrum (as shown in Fig. 16), the violet colour being in the innermost position and red colour in the outermost position except the central maximum. Thus, the spectrum consists of white principal maximum of zero order having on either side of it the first order spectra, the second order spectra and so on. The spectra of each order consists of spectral colours in the order from violet to red. Most of the light is concentrated in the zero order principal maximum and the intensity goes on decreasing gradually as we move to the higher order spectrum.

Condition for Missing Order or Absent Spectra with a Diffraction Grating A study of the resultant diffraction pattern reveals that sometimes it happens that for certain values of angle of diffraction there is no spectrum or certain orders of spectra are absent. It can be understood in the following manner: The fundamental equation for the directions of principal maxima with a plane diffraction grating for normal incident of light is given by, (a + b) sin θ = nλ , where n = 0, 1, 2 , … where (a + b) is the grating element and n the order of principal maximum. In case of a single slit, the directions of minima in the diffraction pattern can be represented as

...(1)

229

a sin θ = mλ , where m = 1, 2 , 3, …

...(2)

As we have seen that the resultant intensity at a point of observation is a particular direction θ for  A2 sin2 α  sin2 Nβ  and diffraction grating is the product of two factors  , Therefore, if in any direction  sin2 β α2   sin2 Nβ sin2 β

is maximum and

A2 sin2 α α2

is zero, then the principal maxima will be absent in that particular

direction. Thus, if two eqns. (1) and (2) are simultaneously satisfied, the principal maxima of order n will not be present in the grating spectrum. Dividing eqns. (1) and (2), we get,

a+ b n a+b or n = = m a m a

This is the required condition of missing order spectra in the diffraction pattern. The ratio (a + b ) / a determines the orders of absent spectra. (i)

When b = a, that is, the width of transparencies (a) and opacities (b) of the grating are equal, then a+ b n= m = 2 m, a Therefore, when m = 1, 2 , 3, 4, ..., missing orders are 2 , 4, 6, 8, … Hence, the second order, 4 th order, 6 th order or even ordered interference maxima will be absent from diffraction pattern because these maxima will coincide with Ist , 2 nd , 3 rd , order

diffraction minima due to single slit. (ii) When b = 2 a then n = a + b m = 3 m, Therefore, when m = 1, 2, 3, 4, ..., missing orders are 3, 6, 9, 12, … a Hence, when width of the opacities (b) of the grating is doubled to that of transparencies (a), then 3 rd , 6 th , 9 th … order interference maxima will coincide with 1st , 2 nd , 3 rd … order diffraction minima. Hence, third, sixth, ninth... order interference maxima will be absent from the diffraction pattern. Similarly, when b = 3 a, n = 4 m, where m = 1, 2, 3, … Hence, 4 th , 8th , 12 th , 16 th ... order diffraction maxima will be absent from diffraction pattern of the grating.

Maximum Number of Orders with a Diffraction Grating The maximum number of spectra available with a diffraction grating in the visible region can be evaluated by using the grating equation for normal incidence as (a + b) sin θ n (a + b) sin θ n = nλ or n = λ where (a + b) is the grating element, θ n the angle of diffraction and n the order of maximum. The maximum possible value of angle of diffraction (θ n) is 90°. Therefore, the maximum possible order available with grating is given by

230

nmax =

(a + b) sin 90 ° (a + b) = λ λ

Thus, the grating element determines the maximum possible order. If the grating element (a + b) lies between λ and 2λ or grating element (a + b) < 2 λ then 2λ nmax < 1 polarised beam of unpolarised light is reflected from the surface of a transparent dielectric medium such as glass or water, the reflected light is either partially polarised (Fig. 5a) or Unpolarised completely plane polarised (Fig. 5b), Beam depending upon the angle of incidence. If µ the angle of incidence is 0° (normal incidence) or 90° (the grazing incidence), Partially the reflected beam is unpolarised. The (a) (b) polarised vibrations of the reflected beam are parallel to the reflecting surface. The percentage of Fig. 5 polarisation varies with the angle of incidence. At a particular angle of incidence the reflected beam is completely plane polarised as shown in

Completely plane polarised N

fig. 5(b). The angle of incidence at which maximum

ip

ip

90° µ θ Partially polarised

D No light

polarisation occurred is called angle of polarisation. The above conclusions can be verified by passing the reflected light through a rotating tourmaline crystal or polaroid. The examination of transmitted light shows the variaton in intensity which indicates that the reflected light is partially polarised. Keeping the tourmaline crystal or polaroid fixed in the position of minimum intensity and vary the angle of incidence till the intensity of reflected beam becomes almost zero (as shown in fig. 6). The angle of incidence at which this condition occurs is called the polarising angle. The polarising angle is different for different reflecting surfaces, it is 57° for air-glass reflection, 33° for glass-air reflection and 53° for air-water (at 0°C) reflection.

C Completely plane polarise Incident unpolarised beam

F

A Polarising angle

B E

Fig. 6

Reflecting surface

273

Brewster’s Law In 1881, Sir Brewster, on the basis of his experimental observations, discovered a simple relation between the angle of incidence at which the maximum polarisation occurs and the refractive index of the medium. He observed that for a particular angle of incidence, called angle of polarisation, the reflected light is completely plane polarised with the plane of vibration perpendicular to the plane of incidence. It was shown by him that

Unpolarised N Completely plane beam polarised with vibrations A perpendicular to the plane of paper C ip ip Air

the tangent of polarising angle is equal to the refractive index

Glass

B

90°

µ>1

of the medium, that is, µ = tan ip N

This is called Brewster’s law. Brewster also concluded that at polarising angle the reflected and refracted rays are at right

D Fig. 7

angles to each other. Let a beam AB of an ordinary light be incident on the glass surface at the polarising angle ip . A part of it is reflected along BC and a part is refracted along BD as shown in Fig. 7. According to Brewster’s law µ = tan ip =

∴

or

or

µ= sin ip sin r

=

sin ip sin r

=

sin ip cos ip

,

but from Snell’s law, µ =

sin ip sin r

sin ip cos ip

sin ip π sin  − ip  2 

ip + r = π /2, ip + ∠CBD + r = π

or

π − ip = r 2

As

∠NBN ′ = π, we have

or

∠CBD = π − (ip + r ) = π /2

When the light is incident at polarising angle, the reflected ray is at right angles to the refracted ray.

Production and Detection of Plane Polarised Light by Reflection Plane polarised light can be produced and detected in the laboratory by a simple instrument called Biot’s polariscope. Biot’s polariscope consist of two polished glass plates M1 and M2 [Fig. 8(a)] mounted at the end of a cylindrical tube. To avoid internal reflections and also to absorb refracted light the plates are painted black on their back surface. The plates M1 and M2 can be rotated about the axis as well as about the diameter of the cross-section of the cylindrical tube. A beam of unpolarised light AB is incident at the polarising angle (57 .5° ) on plate M1 at B and is reflected along BC. As the planes of M1 and M2 are parallel, this light is again reflected at an angle 57.5° by the second glass plate M2 . The glass plate M1 is known as polariser and M2 as analyser.

274 When the upper glass plate M2 is rotated about the vertical axis BC, the intensity of the beam reflected along CD gradually decrease and becomes zero for 90° rotation of M2 . This happens when plane of incidence of M1 and M2 are perpendicular to each other (Fig. 8(b)]. On further rotation of M2 the intensity gradually increases and finally becomes maximum at 180°. On further continuing the rotation of M2 about BC the intensity begins to decrease till at 270° and then increase and becomes maximum again at 360° of rotation. The variation in intensity by the rotation of M2 may be understood as follows :

M2

D Maximum light

C

57.5

M2 C

No light zero intensity)

57.5

M1

M1

B

A

B

A (a)

(b) Fig. 8

When the planes of M1 and M2 are parallel and the beam AB is incident on M1 at polarising angle, the reflected ray BC is completely plane polarised with its vibrations normal to the plane of incidence of M1. Since the plane of incidence of M1 and M2 are parallel, the vibrations in the beam BC are perpendicular to the plane of incidence of M2 . Since only those vibrations are reflected which are perpendicular to the plane of incidence hence the intensity of reflected ray CD is maximum. As M2 is rotated about CD, at 90° of rotation of M2 the vibrations in the ray BC become parallel to the plane of incidence of M2 . Since the vibrations which are parallel to the plane of incidence are never reflected, no light is reflected from M2 when it is perpendicular to M1. Hence, on rotating M2 , the intensity of beam CD varies with zero minimum. This signifies that the beam BC is completely plane polarised.

Law of Malus

Transmission plane of analyser

AE = a cos θ

and

a

pl an eo f

P

Tr po ans la mi ris ss er ion

a cos θ

In 1809, Malus experimentally discovered that when a beam of completely E(-ray) plane polarised light is incident on an analyser, the intensity of the emergent beam varies as the square of cosine of the angle between the planes of transmission of the analyser and the polariser. This law is known as law of Malus and holds good only for a completely plane polarised light. To prove the law, let the angle between the planes of θ transmission of the analyser and the polariser be θ, and a be the amplitude of the incident plane polarised light emerging from the polariser. The A plane polarised light of amplitude a may be resolved in two components along and perpendicular to the plane of transmission of the analyser as shown in Fig. 9.

a sin θ O(-ray)

Fig. 9

AO = a sin θ

The perpendicular component (a sin θ) is eliminated in the analyser while the parallel component (a cos θ) is freely transmitted through it. Therefore, the intensity of light emerging from the analyser is given by I0 = (a cos θ)2 = a2 cos2 θ = I cos2 θ where I = a2 is the intensity of the completely plane polarised incident light. Thus, the sum of the intensities of the transmitted rays is equal to the intensity of the incident polarised ray.

275 Example 1: Calculate the polarising angle for crown glass (µ = 1. 520 ), flint glass (µ = 1. 650 ) and water (µ = 1. 33 ).

[UPTU, B.Tech.I, Q.Bank, 2000]

Solution: According to Brewster’s law, µ tan ip

ip = tan −1(µ)

or

For crown glass : µ = 1.520 ip = tan −1 (1.520) = 56 °40 ′

∴ For flint glass : µ = 1.650

ip = tan −1 (1.650) = 58 °47′

∴ For water : µ = 1.33

ip = tan −1 (1.33) = 53 °03′

∴

Example 2: When the angle of incidence on a certain material is 60°, the reflected light is completely polarised. Find the refractive index of the material and also the angle of refraction. Solution: From Brewster’s law, the refractive index of the second medium with respect to the first is given by tan ip = µ, where ip is the angle of polarisation. In the given problem ip = 60 °, ∴

µ = tan 60 ° = √ 3

or

µ = 1. 732, angle of the refraction r = π /2 − ip = 90 ° − 60 ° = 30 °

Example 3: A ray of light is incident on the surface of a glass plate of refractive index 1.732 at the polarising angle. Calculate the angle of refraction of the ray.

[UPTU, B.Tech. I, Q.Bank, 2000]

Solution: According to Brewster’s law the refractive index µ is related with polarising angle ip as µ = tan ip , ∴

1.732 = tan ip

Here

µ = 1.732

or

ip = tan −1 (1.732) = 60 °

If r is the angle of refraction, then we know that π r = − ip = 90 ° − 60 ° = 30 ° 2 Example 4: A ray of light is incident on a surface of benzene of refractive index 1.50. If the reflected light is linearly polarised, calculate the angle of refraction. Solution: From Brewster’s law we know that µ = tan ip Here µ = 1.5 ∴

the angle of polarisation ip = tan −1 (µ) = tan −1 (1.5) = 56 .3 °

The angle of refraction r is given by r = π /2 − ip = 90 ° − 56 .3 ° = 33 . 7 °

276 Example 5: The refractive indices of glass and water are 1.54 and 1.33. Which will be greater—the polarising angle for a beam incident from water to glass or that for a beam incident from glass to water ? Solution: From Brewster’s law, the polarising angle ip is given by tan ip = µ For water-glass interface,

µ=

For glass to water interface,

µ=

µ glass

1.54 = 1.158 1.33

∴

1.158 = tan ip

or

µ water 1.33 = = 0 .864 µ glass 1.54

∴

0 .864 = tan ip

or ip = 40 . 8 °

µ water

=

ip = 49 . 2 °

The polarising angle for water to glass interface is larger. Example 6: If the polarising angle for a piece of glass for green light is 60° find the angle of minimum deviation for green light for its passage through a 60° prism, made of this same glass. [UPTU, B.Tech. II Sem. 2001]

Solution: According to be Brewster’s law, the polarising angle ip is related with the refractive index as, µ = tan ip , Here ip = 60 ° ∴ µ = tan 60 ° = 1.732 = √ 3 If δ m is the angle of minimum deviation for the given wavelength of light, then  A + δm  sin    2  µ= , Here A = 60 ° sin ( A /2) where A is the angle for prism Since, the prism is made of the same glass its refractive index µ should also be 1.732 60 ° + δ m sin 60 ° + δ m 2 or sin ∴ 1.732 = = 1.732 sin 30 ° sin 30 ° 2 or ∴

60 + δ m = 0 .866 = sin 69 ° 2 60 ° + δ m = 60 ° or δ m = 60 °. 2

sin

Example 7: Calculate the angle of polarisation for a beam of light from (a) air to water, (b) glass to water. Given µ for water = 1. 33 and µ for glass = 1. 54. Solution: (a) When light is incident from air to water refractive index of water 1.33 µ= = = 1.33 refractive index of air 1 According to Brewster’s law µ = tan ip or polarising angle ip = tan −1 (µ) ∴

ip = tan −1 (1.33) = 53 °3′

(b) When light is incident from glass of water

[UPTU, B.Tech. I Sem. 2000]

277 µ=

refractive index of water 1.33 = = 0 .8636 refractive index of glass 1.54

ip = tan −1(0 .8636) = 40 °49′

∴

Example 8: A beam of light travelling in water strikes a glass plate which is also immersed in water. When the angle of incidence is 51°; the refracted beam is plane polarised. Calculate the refractive index of glass. [UPTU B.Tech. I, Q.Bank 2000]

Solution: According to Brewster’s law, µ and ip are related as µ = tan ip Here ip = 51° and light is travelling from water to glass ∴

ωµ g

= tan ip = tan 51° = 1.2349

Thus, the refractive index of glass with respect to water = 1.2349 The refractive index of water with respect of air,

Hence,

aµ g

aµ ω

= 1.33,

aµ g

= aµ ω × ω µ g = 1.333 × 1.2349 =1. 646

∴

ωµ g

=

aµω

Example 9: Two polarising sheets have polarising directions parallel so that the intensity of the transmitted light is maximum. Through what angle must either sheet be turned if the intensity is to drop by one-third ? Solution: From Malus law, we have, I = I0 cos2 θ, where θ is the angle between the two polarising sheets. Here I0 = I m and I = I m /3,

∴

(1/3) I m = I m cos2 θ

∴

cos2 θ = (1/3)

or

θ = cos −1(1/ √ 3) = ± 55 °18 ° , ± 124 °42 °

Example 10: A polariser and an analyser are oriented so that the maximum amount of light is transmitted. To what fraction of its maximum value is the intensity of transmitted light reduced when the analyser is rotated through (a) 30°, (b) 45° and (c) 60° ? Solution: Let I m be the maximum intensity of the transmitted light and θ be the angle between the plane of polariser and the plane of analyser, then from Malus law, I = I m cos2 θ (a) When θ = 30 °, I = I m cos2 30 ° = I m (3 /4)2 ,

∴

I / I m = 0 . 75

I = I m cos2 45 ° = I m (1/ √2)2 = I m /2,

∴

I / I m = 0 . 50

I = I m cos2 60 ° = I m (1/2)2 = I m /4,

∴

I / I m = 0 . 25

(b) When θ = 45 °

(c) When θ = 60 °

278 Example 11: Analysing Nicol examines two adjacent plane-polarised beams A and B whose planes of polarisation are mutaully perpendicular. In one positon of the analyser, beam B shows zero intensity. From this position a rotation of 30° shows the two beams as matched (i.e., of equal intensity). Deduce the intensity ratio I A / I B of the two beams. Solution: According to Malus law, when a beam of plane polarised light of intensity I0 be incident on an analyser, the intensity I of the emerging light is given by, I = I0 cos2 θ, where θ is the angle between the plane of transmission of the analyser and the plane of vibration of the incident light. In one postion of the analyser, beam B shows zero intensity. This means that θ = 90 ° for beam B Obviously θ will be 0° for beam A. When the analyser is rotated through 30°, then we have θ = 30 ° for beam A and θ = 60 ° for beam B. If I′ be the intensity of beam A and also of beam B as transmitted by the analyser, then we have I ′ = I A cos2 30 ° = I B cos2 60 ° 2

 1    2

2

I A cos 60 ° 1 = = = I B cos2 30 °  3  2 3    2 

∴

Example 12: A polariser and an analyser are oriented so that the amount of light transmitted is maximum. How will you orient the analyser so that the transmitted light is reduced to (a) 0.5, (b) 0.25, (c) 0.75, (d) 0.125, (e) 0 of its maximum value.

[UPTU, B.Tech. I, Q. Bank, 2000]

Solution: According to the law of Malus, I = I0 cos2 θ, where I0 be the intensity of incident light, I the intensity of transmitted light and θ the angle between the plane of polariser and plane of analyser. Therefore, (a) When I = 0 .5 I0 , then 0 .5 I0 = I0 cos2 θ

or

cosθ = √ (0 .5)

or

θ = cos −1 √ (0 .5)

= 45 °

or

cos θ = √ (0 .25)

or

θ = cos −1 (0 .25)1 /2

= 60 °

or

cos θ = √ (0 .75)

or

θ = cos −1 (0 .75)1 /2

= 30 °

or

cos θ = √ (0 .125)

or

θ = cos −1 (0 .125)1 /2 = 69 °30 °

or

cos θ = 0

or

θ = 90 °

(b) When I = 0 .25 I0 , then 0 .25 I0 = I0 cos2 θ (c) When I = 0 .75 I0 , then 0 .75 I0 = I0 cos2 θ (d) When I = 0 .125 I0 , then 0 .125 I0 = I0 cos2 θ (e) When I = 0, then 0 = I0 cos2 θ

279 Example 13: At a certain temperature the critical angle of incidence of water for total internal reflection is 48° for a certain wavelength. What is the polarising angle and the angle of refraction for light incident on water at an angle that gives maximum polarisation of the reflected light ? (Given sin 48 ° = 0 . 7431). Solution: The relation between the critical angle and the index of refraction is 1 1 1 µ= = = ∴ µ = 1.345 sin C sin 48° 0 .7431 For maximum polarisation, the light must be incident at polarising angle ip . From Brewster’s law, we have, or µ = tan ip ip = tan −1 (µ) ∴

ip = tan −1(1.345) = 53 °22'

If r is the angle of refraction, then we have ip + r = π /2

or

r = π /2 − ip = 90 ° − 53 °22' = 36 °38'

Plane Polarised Light by Refraction (Pile of Plates) It is observed that when a beam of unpolarised Reflected plane polarised light falls on any thin transparent surface (glass) at polarising angle, only a small fraction of the incident light is reflected. For instant, in case of glass (µ =1.5) plate at polarising angle, all the vibrations (100%) parallel to the plane Unpolarised light of incidence are transmitted, whereas some of 32.5 the vibrations (about 15%) perpendicular to the plane of incidence are reflected and rest (about 85%) is transmitted or refracted. Refracted plane Pile of plates Hence, the refleted light is completely plane polarised light polarised having vibrations perpendicular to Fig. 10 the plane of incidence, while the refracted light is partially polarised as it has vibrations both in the plane of incidence (about 100%) as well as perpendicular to the plane of incidence (85%). Now if the refracted light is again allowed to pass through another identical glass plate placed in contact and exactly parallel to the previous plate so that light falling on new plate is again at polarising angle. Again a fraction of light having vibrations perpendicular to the plane of incidence will be reflected and rest is transmitted. Therefore, the light refracted from new plate has a small portion of light having vibration perpendicular to the plane of incidence (about 60% = 85% − 15%) and the entire portion of the light having vibrations (100%) parallel to the plane of incidence. Therefore, if we use a pile of parallel plates and the beam of unpolarised light is incident at the polarising angle on the system piles of plates, then in the final emergent refracted ray, light is having vibrations only in the plane of incidence. The light having vibrations at right angle to the plane of incidence is totally removed by repeated reflection. Hence, we get completely plane polarised light by refraction through piles of plates with vibrations in the plane of incidence as shown in fig. 10. Thus, piles of plates may be supposed to act as polariser.

280 If I p and I s be the intensities of the parallel and perpendicular components refracted light, then the degree of polarisation or proportion of polarisation is given by P=

Ip − Is Ip + Is

n

=

 2µ   n +  2 1 − µ 

2

where n is the number of plates and µ the refractive index of the plates. The pile of plates consists of a number of thin glass plates such as microscope cover slips arranged in a tube in contact with each other. The plates are inclined at an angle of 32.5°(90° − 57.5°) to the axis of the tube. Example 14: Calculate the proportion of polarisation for a beam of light passing through a pile consisting of 8 plates of glass. Given refractive index of glass = 1. 5. Solution: According to the expression for proportion of polarisation, n P= 2  2µ    n+  2 1 − µ  Here, n = 8 and µ = 1.5 ∴

8

P=

 2 × 1.5   8 +  2 1 − (1.5) 

2

= 0 . 581

Doubly Refracting Crystals There are certain crystals which exhibit the property of splitting of a light ray, incident on them, into two refracted rays. Such crystals are called doubly refracting crystals. They are of two types (i) Uniaxial crystals and (ii) Biaxial crystals. In uniaxial crystals, there is only a single direction known as optic axis along which two refracted rays are transmitted with the same velocity. Of two refracted rays only one follows the ordinary laws of refraction. The examples of these crystals are calcite, tourmaline and quartz. In biaxial crystals, there are two directions or two optic axis along which the velocities of refracted rays are same. None of the refracted rays obeys the laws of refraction. The examples of biaxial crystals are topaz, argonite, copper sulphate, cane sugar and mica.

281

Axis

Axis

C

B

A 102°

102°

H 102°

F

78°

102°

E

G

78°

G

C

H

F (a)

Optic

A

B

102°

Optic

102°

D

D

E

Optic

The most commonly used uniaxial crystal is Iceland spar or calcite. Calcite, also known as Iceland spar, is colourless transparent crystal. It is chemically crystallised calcium carbonate (CaCO3 ) and occurs in nature in di fferent form s, all of whi ch give rhombohedron on cleavage as shown in Fig. 11. Each of the six faces of the crystal is a parallelogram having angles of 78° and 102° approximately. At two diametrically opposite corners, A and H, all the angles of three faces meet there, are obtuse (> 90 ° ); while at the rest of six corners two angles are acute and the remaining one is obtuse. The two corners A and H are termed as blunt corners.

Axis

Geometry of Calcite Crystal

(b)

Fig. 11

Optic Axis: The optic axis in a doubly refracting crystals is a direction along which all the plane waves are transmitted with a single velocity without showing the effect of double refraction. Thus the optic axis is a direction along a line passing through any one of the blunt corners and making equal angles with each of the three edges which meet at the corner. A ray of unpolarised light travelling along the optic axis or in a direction parallel to the optic axis does not split up into ordinary (O) and extraordinary ( E) rays. Ordinarily, optic axis is not obtained by joining the two blunt corners. In a special situation when the rhombohedron is so cut that all of its edges are equal then the line joining the blunt corners coincides with the optic axis as shown in Fig. 11(b). Principal Section: A plane containing the optic axis of Optic axis the crystal and perpendicular to its two opposite A A F refracting faces is called the principal section of the D 109° B crystal for that pair of faces (Fig. 12). As crystal has six C 71° faces there are three principal sections corresponding to H each pair of opposite faces. A principal section of a F C E G Optic axis calcite crystal is a parallelogram having angles of 71° and H 109° as illustrated in Fig. 12(b). The principal plane or Principal section (b) (a) principal section of the ordinary ray (O-ray) is defined as a plane which contains the ordinary ray and the optic Fig. 12 axis, whereas the principal plane of the extraordinary ray ( E-ray) is a plane which contains the optic axis and the extraordinary ray. Ordinarily, the ordinary and extraordinary planes do not coincides except in a case when the plane of incidence and principal section are the same.

Ordinary and Extraordinary Rays When a narrow beam of unpolarised light be incident normally on a double refracting crystal, such as calcite, it splits up into two refracted rays. The two refracted rays emerge parallel to the incident beam but slightly relatively displaced by a distance. The relative displacement depends on the thickness of the crystal [Fig. 7(b)]. Of the two refracted beams one which obeys the ordinary laws of refraction is called

282 the ordinary ray or O-ray, and the other which behaves extra ordinarily and does not follow laws of refraction is called extraordinary ray or E-ray. Since sin i / sin r represents the ratio of the velocity of light in vacuum to that in refracting medium, the velocity of O-ray is the same in all directions within the crystal. For extraordinary ray this ratio (sin i / sin r ) varies with angle of incidence, its velocity varies with the angle of incidence and is different in different directions within the crystal. Apart from it, O-ray is always in the plane of incidence, but in general it is not true for the E-rays. Special Cases: It is an observed fact that, 1.

a ray of unpolarised light incident along the optic axis or in a direction parallel to the optic axis of the crystal is not split up into ordinary and extraordinary components, but both the rays travel along the same direction with the same velocity.

2.

a ray of unpolarised light incident normally to the optic axis of the crystal is not split up into ordinary and extraordinary rays, but both the rays travel with different velocities along the same direction.

Phenomenon of Double Refraction (In a Calcite Crystal) In 1669, Erasmus Bartholinus discovered that when Stationary

an unpolarised beam of light enters an anistropic

E O 78°

crystal such as calcite or quartz, it splits up into two plane polarised refracted beams travelling in different directions, one of which always obeys ordinary laws of refraction and the other, in general, does not obey them. This phenomenon of splitting of unpolarised light into two polarised refracted rays is known as double refraction.

102°

A

D

109°

i B r1

Rotatory (a)

N

E O

71°

r2

C N’ (b)

Fig. 13

The phenomenon of double refraction is easily investigated by placing a calcite crystal over a black mark on a piece of paper. The two images of black mark are observed as shown in Fig.13(a) by O and E. If, now, the crystal is rotated slowly about the vertical axis as shown in Fig.13(a), one of the two images remains fixed and other moves round the first (O). The stationary image behaves as if the crystal were an isotropic media such as water or glass and is called ordinary image and the ray which produces this image is called ordinary ray (O-ray). The second image is called extraordinary image and the refracted ray corresponding to this image is called extraordinary ray (E-ray). In Fig. 13(b), an incident ray AB is split up inside the crystal into O and E rays. The O-ray travelling along BC makes an angle r1 while the E–ray travelling along BD makes an angle of refraction r2 with the normal NN′ . The rays emerging from the crystal along CO and DE are parallel to each other as well as to the incident beam. The refractive index of ordinary ray, µ0 =

sin i sin i and that of extraordinary ray, µ e = . sin r1 sin r2

283 In calcite crystal r1 < r2 , therefore µ0 > µ e . Hence v0 < ve , that is, inside the calcite crystal the E-ray travels faster as compared to O-ray. Such crystals are called uniaxial negative crystals. There are other types of crystals in which µ e > µ0 or v0 > ve , that is, inside the crystals O-ray travels faster as compared to E-ray. Such crystals are called uniaxial positive crystals. The examples of negative crystals are tourmaline, ruby etc. and that of positive crystals are quartz, iron oxide etc.

Polarisation by Double Refraction We have seen that when a beam of unpolarised light is incident on a doubly refracting crystal, such as calcite, it splits up into two plane polarised beams, both the refracted rays, ordinary (O) and extraordinary (E), are polarised with vibrations at right angles to each other. The polarisation by double refraction was successfully demonstrated by Hugen's in 1678 by using two calcite crystal. 1.

When a beam of unpolarised light is incident normally on a surface of a calcite crystal, the ordinary (O) ray transmitts undeviated, whereas extraordinary (E) rays travels parallel to ordinary ray, but slightly displaced. If another identical calcite crystal is brought in the path of two emergent rays and if the principal sections of the two crystals are parallel as shown in Fig. 14(a) (extreme right) two images O1 and E1 are seen. The ordinary ray from the first crystal traverses the second crystal without any deviation, while the extraordinary ray passes the second crystal along a path which is parallel to the path inside the first crystal. Thus, if the two crystals are of equal thickness, the two rays represented by O1 and E1 emerge exactly parallel, but with a separation twice as much as with the first crystal alone [Fig. 14(a)].

2.

If the second crystal is rotated with A C C A E1 respected to the first about the axis E1 O1 along the incident ray, then each of the two rays, O and E, from first crystal D B O1 B D (a) Calcite Crystal θ=0° undergoes double refraction in the A A A E1 second crystal, thus giving rise to four O2 C D O2 O1 E1 rays and correspondingly four images. E2 O1 E2 C D E2 O2 D The old images O1 and E1 become C B B B fainter and in between them two more θ=45° θ=135° θ=90° (b) new faint images O2 and E2 are formed. (d) (c) A D If the second crystal is further rotated, E A D then it is observed that the images O1 O1 O1 E1 and O2 remain unmoved, while E1 and O B C E2 rotate around O1 and O2 in the E1 Calcite Crystal B C (e) circular path. It is also observed that θ=180° the intensity of old images, O1 and Fig. 14 E1, goes on decreasing while that of new images, O2 and E2 , goes on increasing until a rotation of (θ = ) 45° . That is, at the rotation of 45° all the four images become equally intense [Fig. 14(b)].

284 3.

On further rotation the intensity of old images O1 and E1 decreases while that of new images O2 and E2 increases and at 90° rotation the old images totally disappear and the new images, O2 and E2 , become highly intense [Fig. 14(c)].

4.

On continuing the rotation after passing 90°, the images O1 and E1 start reappearing and gaining intensity while images O2 and E2 correspondingly begin to decrease in intensity and at θ = 135 ° four images again become equally intense as shown in Fig. 14(d).

5.

At 180° rotation, the principal sections of the two crystals are again become parallel, but now, their optic axes are oriented in opposite directions [Fig. 14(e)] as a result of which the images O2 and E2 disappear while O1 and E1 after graining initial intensities superimpose and form a single image after emerging from the second crystal. It is possible only when the two crystals are of equal thickness [Fig. 14(e)].

6.

The similar changes in the reverse order occur if the rotation of second crystal is continued from 180° to 360° .

Thus, the experiment with two identical calcite crystals demonstrates the phenomenon of polarisation. Here first stationary crystal, called polariser, produces plane polarised vibrations in the O and E rays while the second one which is rotated, called analyser, analysis them. The physical explanation of Huygen's experiment may be understood as follows:

a sin

In Fig. 15, AB and CD are the principal sections of the A C Principal section a cos θ first and second crystals respectively, and θ the angle Principal section of first crystal between them. On entering the first crystal the of second crystal E E1 unpolarised light breaks into two plane polarised rays, O a θ E and E. The ordinary rays has vibrations perpendicular to 2 O2 as the principal section AB of the first crystal, while O in P a cθo a extraordinary ray has vibrations along the principal s O1 section AB. Let the amplitude of each ray 'a' be D B represented by PO and PE. Due to the phenomenon of Fig. 15 double refraction, the ordinary ray, on entering the second crystal, breaks into two components O1 and E2 . The amplitude of the O1 ray, a cos θ, is represented by PO1 and that of the E2 ray, a sin θ, is represented by PE2 in Fig. 9. Similarly, the extraordinary ray E on passing through second crystal breaks into ordinary ray O2 and extraordinary ray E1. The amplitude PO2 (= a sin θ) of O2 is perpendicular to the principal section CD of the second crystal and that PE1 (= a cos θ) of E1 ray is along the principal section CD of the second crystal. Therefore, the intensities of O1 and E1 are a2 cos2 θ while those of O2 and E2 are a2 sin2 θ . Therefore, at θ = 0 ° or 180 ° , sin2 θ = 0 and cos2 θ = 1. Hence, the intensities of O1 and E1 rays are maximum equal to a2 and those of O2 and E2 are minimum zero. At θ = 45 ° or 135° , we have cos2 θ = 0.5, and sin2 θ = 0 .5, therefore, intensities of O1, E1, O2 and E2 are equal and equal to a2 / 2 and hence four images appear equally bright.

285 At θ = 90 ° , we have cos2 θ = 0 and sin2 θ = 1, that is, the intensities of O1 and E1 images are zero while those of O2 and E2 are maximum and equal to a2 . Hence, all the experimentally observed affects occurred in different situations are theoretically explained. From the above study it is clear that in each situation the intensity of emergent light is same as that of the incident light.

Nicol Prism In 1828 William Nicol designed an ingenious optical device by specially cutting calcite crystal for producing and analysing plane polarised light and is known as nicol prism. Its principle is based on the phenomenon of double refraction. Actually Nicol prism is not a prism but a parallelopiped. Principle: It is well known that when an unpolarised beam of light enters the calcite crystal, it splits up into two plane polarised rays, as O-ray, and E-ray, with vibrations in two mutually perpendicular planes. If by some optical means, we eliminate one of the two beams, then we would obtain only one plane polarised beam. The nicol prism is designed in such a way so as to eliminate the ordinary ray by total internal reflection. Hence only the extraordinary ray is transmitted through the prism. Construction: Nicol prism is constructed C Y by taking a long rhomb of calcite so that Optic axis D its length AE is three times its breadth B Canada Balsam A A AD as illustrated in Fig. 16(a). The end A E S 22° E faces ABCD and EFGH of the rhomb are 90° 48° 14° ground such that the angles C and E S 14° P E-ray reduce from 71° to 68° . The crystal is then S 90° 68° 71° Y G 0 cut into two halves along the diagonal G C G H O-ray F A′ G′ and the two halves are cemented (b) E (a) together by a thin layer of Canada balsam. The Canada balsam is a transparent Fig. 16 substance whose refractive index lies midway between the indices of ordinary and extraordinary rays for calcite. The refractive index of the crystal for the ordinary ray is 1.66 and is 1.49 for the extraordinary ray; the refractive index of the Canada balsam is about 1.55 for both rays. Action: A ray of unpolarised light SP incident on one of the end faces A′ C of the nicol prism and nearly parallel to CG is split up into E and O rays with vibrations perpendicular and parallel to the principal section of the nicol prism as shown in Fig. 16(b). The ordinary ray in passing from calcite to Canada balsam travels from denser (µ0 = 1.66) to rarer (µ = 1.55) medium and is, therefore, totally reflected only when the angle of incidence at the Canada balsam layer is greater than a certain critical angle for two media (calcite and Canada balsam), that is, the critical angle  µ C = sin −1   = sin −1  µ0 

1.55    = sin −1 (0 .933) = 69° 1.66 

286 This totally internally reflected ray is finally absorbed by the blackend side CG′ of the prism and only the extraordinary ray is transmitted through the prism as it is travelling from rarer (µ e = 149 . ) to denser (µ = 1.55) medium. Hence, in the nicol prism the ordinary ray is eliminated by total internal reflection and only extraordinary ray with vibrations parallel to the principal section passes through, and the light beam emerging from crystal is plane-polarised. Thus, nicol prism can be used as polariser. Limitations: For the effective use of nicol prism as a polariser, the angle of incidence on the face A′ C, should be limited to a very narrow range of about 14° on either side of SP. For the angle of incidence greater than this value, the O-ray will strike the Canada balsam layer at an angle less than the critical angle and therefore will be transmitted along with E-ray. It can be understood as follows: Polariser (P) Analyser (A) When the angle of incidence on the E-ray E-ray crystal surface A′ C increases, the angle of incidence of O-ray on calcite-balsam (a) layer decreases. When ∠S0 PS becomes O-ray Analyser (A) Polariser (P) greater than 14°, the angle of incidence E-ray No for O-ray on calcite balsam layer light becomes less than the critical angle O (69°), hence O-ray also transmitted with O-ray (b) E-ray. Therefore, the light emerging Polariser (P) Analyser (A) from nicol prism will not be plane E-ray E-ray polarised. When the angle of incidence on the crystal surface is decreased or (c) O-ray when ∠SE PS decreases, the angle made by E-ray with optic axis also decreases, Fig. 17 as a result of which its refractive index i n c r e a s e s . W h e n ∠SE PS b e c o m e s greater than 14°, then µ e becomes greater then µ for balsam layer, that is, Canada balsam becomes optically less dense than calcite and E-ray is totally internally reflected by the balsam. Therefore no light emerges from the Nicol. Thus to avoid the unwanted transmission of O-ray and the total internal reflection of E–ray; the incident light should be confined within an angle of 14° on either side of SP.

Nicol as a polariser and analyser: The nicol prism is widely used to analyse the plane polarised light. Its analysing action may be easily understood from Fig. 17. When two nicols are arranged coaxially, the first nicol prism P which produces plane polarised light is known as polariser whereas; the second nicol A which analyses the polarised light coming from P is termed as analyser. Such a combination of two nicols is called a polariscope. When the principal sections of the two nicols are parallel as shown in Figs. 17(a) and (c), then the vibrations in the E-ray, which are in principal section of the polariser, are also in the principal section of the analyser. Consequently the E-ray from the polariser is freely transmitted by the analyser. In this position of parallel nicols, the intensity of the emergent light is maximum. When the analyser A is rotated about its axis from the position of parallel nicols, the intensity of emergent E-ray decreases. When the principal sections of two nicols become mutually perpendicular, then no light emerges from the analyser as shown in Fig. 17(b). In this situation the vibrations in E-ray, which are in the principal section of the polariser, become perpendicular to the principal section of the

287 analyser and have no component in the principal section of analyser and thus travel as an ordinary ray in it. Hence it is totally internally reflected at the balsam layer and therefore no light is transmitted through the analyser. The above facts can be used for detecting the plane polarised light. If the given light, viewed through a rotating nicol, shows variation in intensity with minimum zero, then the given light is plane polarised.

Huygen’s Theory of Double Refraction Huygen explained the phenomenon of double refraction in uniaxial crystals on the basis of wave theory of light using the principle of secondary wavelets. He made the following assumption: 1.

When the light wave strikes the surface of a doubly refracting crystal, every point of the crystal disturbed by the incident wavefront becomes the source of two secondary wavelets, ordinary and extraordinary, which spread out into the crystal.

2.

For ordinary ray (O-ray) for which the velocity is the same in all directions, the wavefront is spherical.

3.

For extraordinary ray (E-ray) for which the velocity is different in different directions, the wavefront is an ellipsoid of revolution. (An ellipsoid of revolution is a surface obtained by rotating an ellipse about either the major axis or the minor axis, this axis of rotation is along the optic axis as shown in Fig. 18).

Calcite Quartz Optic

θ axis O

Optic axis

θ E

O

4.

Positive crystal E When a ray of light is incident along the optic axis, it is Negative crystal not split up into E and O components. In this case, both the rays travel along the same direction with the same Fig. 18 velocity, therefore, the spherical wavefront corresponding to O-ray and the ellipsoid of revolution corresponding to E-ray touches each other at points where the optic axis through the origin of wavelets intersects them.

5.

In negative uniaxial crystals the sphere lies inside the ellipsoid, while in positive uniaxial crystal the ellipsoid lies inside the sphere as illustrated in Fig. 18.

The principles of Huygen's theory will now be applied to find the O and E-rays, and refracted wavefronts in which the crystal plane is so cut that the optic axis occupies special direction.

Case (I): Optic axis inclined to the refracting face and lies in the plane of incidence (a)

Oblique incidence: Let AB represents a section of the plane wavefront incident obliquely on the crystal surface XX′ . The optic axis of the crystal lies in the plane of B incidence (plane of the paper) and inclined as shown by dotted line A in Fig. 19. The point A of the wavefront AB, where it strikes the C Air X X´ crystal surface first, becomes the source of two secondary wavelets P Crystal G inside the crystal, one ordinary and the other extraordinary. According to Huygen's postulate the ordinary wave surface is Optic D H spherical; whereas the extraordinary wave surface is ellipsoid of axis revolution. Since calcite is a negative crystal, the sphere lies inside O-ray the ellipsoid and both touches each other at point P on the optic E-ray axis. During the time t (= BC / v, where v is the velocity of light in Fig. 19 air), in which the disturbance from B reaches C, the ordinary

288 spherical wavelet has travelled a distance equal to AG = v0 t = v0 (BC / v) = BC / µ0 , where v0 is velocity and µ0 the refractive index of the crystal for ordinary ray. The distance travelled by extraordinary ray AH during the time t is equal to AH = ve t = ve (BC / v) = BC / µ e , where ve is velocity and µ e the refractive index of the crystal for extraordinary ray. Therefore, during

the the the the

time t the incident wavefront reaches from B to C. The ordinary spherical wave surface acquires a radius BC / µ0 ; whereas the extraordinary ellipsoidal wave surface has semiminor axis BC / µ0 along the optic axis and semimajor axis BC / µ E at right angle to the optic axis, where µ E is the minimum refractive index for E-ray at right angle to the optic axis. For negative uniaxial crystal like calcite µ E < µ e < µ0 . To find the position of ordinary spherical wave surface in the crystal, draw a circle with A as centre and radius equal to BC / µ0 . This circle meet the optic axis at P. From C draw CG a tangent plane touching the circle, the CG represents the position of the ordinary wavefront and AG represents the direction of the O-ray. Similarly an ellipse with BC / µ0 as semiminor axis and BC / µ E as semimajor axis gives the extraordinary wave surface. A tangent plane CH to the ellipse indicates the position of the extraordinary wavefront and AH gives the direction of E-ray. Thus in a direction other than the optic axis both the E and O-rays travel with different velocities and are plane polarised in mutually perpendicular planes. (b) Normal incidence: Fig. 20 illustrates the similar construction of ordinary spherical wave surface and extraordinary ellipsoidal wave surface for normal incidence of the wavefront on the crystal surface XX′ . As incident wavefront is parallel to XX′ , all points on the surface are disturbed simultaneously. Hence at any time the secondary wavelets have the same size. The tangent planes CD and GH are parallel and represent the po si tions of the or di nary spher i cal wave sur face and extraordinary ellipsoidal wave surface respectively. AO and AE represent the ordinary and extraordinary rays, which travel along different directions.

X

Air

Crystal

A

B

D P

C

X´ P

G

H Optic axis

E-ray

E-ray O-ray

O-ray

Fig. 20

Case (II): Optic axis is parallel to the refracting surface and lies in the plane of incidence (a)

Oblique incidence: Let AB represents a section of plane wavefront incident obliquely on the crystal surface XX′ . The B Optic axis optic axis is in the plane of incidence and parallel to the A Air refracting surface as shown in Fig. 21. As the wavefront P C X X´ G touches the point A, it becomes the sources of ordinary and Crystal H ex traor di nary sec ond ary wave lets. These wave lets are represented by the semi-circle (for O) and the semi-ellipse (for E) within the crystal, and touches each other along XX′ . During the time in which the disturbance from B reaches C, E-ray O-ray the spherical ordinary wavefront occupies the position CG O-ray E-ray such that AG = BC / µ0 ; whereas extraordinary ellipsoidal Fig. 21 wavefront occupies the position CH such that AH = BC / µ e , where µ e is the refractive index for extraordinary ray along AE. If µ E is the principal refractive index for E-ray along semimajor axis, the semimajor axis of the ellipse is BC / µ E (where µ E < µ e < µ0 for calcite crystal); whereas the semiminor axis is BC / µ0 .

289 AO and AE represent the ordinary and extraordinary rays. The ordinary and extraordinary rays travel along different directions with different velocities inside the crystal. (b) Normal incidence: Fig. 22 illustrates the similar construction of ordinary and extraordinary wave surfaces for normal incidence of the wavefront on the crystal surface XX′. As incident wavefront is parallel to XX′ , all points on the re fracting surface are disturbed simultaneously. Therefore, at any time the size of the secondary wavelets are the same. The tangent planes CD and GH are the positions of the ordinary spherical wave surface and extraordinary ellipsoidal wave surface respectively. The ordinary and extraordinary wavefronts are parallel to each other and also to the crystal surface. AO and AE represent the ordinary and extraordinary rays. Although the O and E-rays are not separated and they travel along the same direction, yet there is double refraction. As they travel with different velocities a path difference is introduced between them. This property is utilized in the construction of quarter and half wave plates.

Case (III): Optic axis is perpendicular to the refracting surface and lies in the plane of incidence (a)

Oblique incidence: Let AB represents the section of the plane wavefront incident obliquely on the crystal surface XX′ . The optic axis is in the plane of incidence and perpendicular to the refracting surface as shown in Fig. 23. As the wavefront touches the point A it becomes the sources of ordinary and extraordinary secondary wavelets. These wavelets are represented by the semi-circle (for-O) and the semi-ellipse (for–E ray) within the crystal and touch each other at P in the direction of the optic axis. During the time in which the disturbance from B reaches C, the spherical ordinary wavefront occupies the position CG such that AG = BC / µ0 , where µ0 is the refractive index for ordinary ray; whereas extraordinary wavefront occupies the position CH such that AH = BC / µ e , where µ e is the refractive index for E-ray along AE. If µ E is the principal refractive index for E-ray along semimajor axis, the semimajor axis of the ellipse is BC / µ E ; whereas the semiminor axis is BC / µ0 . AO and AE are ordinary and extraordinary rays and travelling along different directions with different velocities.

(b)

Normal incidence: Fig. 24 illustrates the similar construction of ordinary spherical wave surface and extraordinary ellipsoidal wave surface for normal incidence of the wavefront on the crystal surface XX′ . In this case ordinary wavefront CD and extraordinary wavefront GH coincide at all instants. There is no separation between ordinary and extraordinary rays AO and AE. Both travel in the same direction with same velocity. Thus the phenomenon of double refraction does not occur in this case.

290

Case (IV): Optic axis is parallel to the refracting surface and perpendicular to the plane of incidence (a)

Oblique incidence: Let AB represents the section of the plane wavefront incident obliquely on the crystal surface XX′ . The optic axis is parallel to the refracting surface but per pen dic u lar to the plane of in ci dence, that is, perpendicular to the plane of paper as shown in Fig. 25 by fine dots. As the wave front touches the point A, it becomes the source of ordinary spherical wave surface and extraordinary ellipsoidal wave surface, but since O and E-wave surfaces are figures of revolution about the optic axis, their sections in the plane of incidence are circular. During the time in which the disturbance from B reaches C, the ordinary wavefront occupies the position CG such that AG = BC / µ0 , where µ0 is the refractive index for ordinary ray, whereas extraordinary wavefront occupies the position CH such that AH = BC / µ E , where µ E is the principal refractive index for E-ray along semimajor axis. Therefore BC / µ0 is the radius of the ordinary wave surface and BC / µ E is that of extraordinary wave surface. AO and AE represent ordinary and extraordinary rays and travelling along different directions with different velocities.

(b) Normal incidence: Fig. 26 illustrates the similar construction of ordinary and extraordinary wave surface for normal incidence of the wavefront on the crystal surface XX′ . CD and GH represent the positions of ordinary and extraordinary refracted wavefronts at the same instant. AO and AE are ordinary and extraordinary rays. The two rays (O and E-rays) are not separated but travel with different velocities along the same direction and thus double refraction occurs.

Mathematical Treatment for the Production of Plane, Elliptically and Circularly Polarised Lights A light having vibrations only along a single direction perpendicular to the direction of propagation of light is said to be plane polarised. Plane polarised light represents the simple type of polarisation. For such light the light vector vibrates simple harmonically along a fixed straight line perpendicular to the direction of propagation. When two plane polarised beams of monochromatic light are superimposed, then under suitable conditions the resultant light vector may rotate in a plane perpendicular to the direction of propagation and if its magnitude remains constant while the orientation varies regularly, then the tip of the vector traces a circle and the resultant light is said to be circularly polarised. If, however the magnitude of light vector varies periodically between it maximum and minimum values, the tip of the vector traces an ellipse and the resultant light is said to be elliptically polarised.

291

Superposition of Two Linearly Polarised Waves with Their Optical Vectors Parallel Let us consider two linearly polarised sinusoidal light waves of same frequency, are travelling along the same direction (say X-axis) and having their optical vectors parallel to each other but perpendicular to the direction of propagation (say along Z-axis). The magnitude of light vector, associated with these waves can be written in the form,

and

x1 = a1 sin (ωt + φ1)

...(1)

x2 = a2 sin (ωt + φ2)

...(2)

where a1 and a2 are the amplitudes and φ1 and φ2 the initial phases of the two motions. The resultant of these two waves would be given by ...(3)

x = x1 + x2 or

x = a1 sin (ωt + φ1) + a2 sin (ωt + φ2) = a1 (sin ωt cos φ1 + cos ωt sin φ1) + a2 (sin ωt cos φ2 + cos ωt sin φ2) [∴ sin ( A + B) = sin A cos B + cos A sin B]

or

x = sin ωt (a1 cos φ1 + a2 cos φ2) + cos ωt (a1 sin φ1 + a2 sin φ2)

...(4)

Let us take

a1 cos φ1 + a 2 cos φ2 = A cos φ

...(5)

and

a1 sin φ1 + a2 sin φ2 = A sin φ

...(6)

where A and φ are new constants. Thus, equation (4) becomes x = A sin ωt cos φ + A sin φ cos ωt or

x = A sin (ωt + φ)

...(7)

Equation (7) is similar as equations (1) and (2). Hence, the resultant wave is also a linearly polarised wave with its electric vector oscillating along the same axis. The amplitude of the resultant wave is obtained by squaring equations (5) and (6), and then adding. That is,

A2 cos φ + A2 sin2 φ = (a1 cos φ1 + a2 cos φ2)2 + (a1 sin φ1 + a2 sin φ2)2 A2 = a12 + a22 + 2 a1 a2 (cos φ1 cos φ2 + sin φ1 sin φ2)

or

A 2 = a12 + a22 + 2a1 a2 cos (φ1 − φ 2)

and phase by dividing equation (6) by equation (5) as, tan φ =

or

a1 sin φ1 + a2 sin φ2 a1 cos φ1 + a2 cos φ2

φ = tan−

1

 a1 sin φ1 + a2 sin φ2     a1 cos φ1 + a2 cos φ2 

292

Superposition of two linearly polarised waves with their optical vectors mutually perpendicular Let a beam of plane polarised light coming

Ordinary light

Plane polarised light

Optic axis

from a polariser be incident normally on a thin plate of calcite crystal cut with faces parallel to the optic axis and oriented in such

O, E

a way that the vibrations in plane polarised

Plate of Calcite crystal

light make an angle θ with the optic axis as shown in Fig. 27, then the amplitude A of the

Fig. 27

vibrations in the incident plane polarised light be resolved into two components ( E and O components) by O

having vibrations perpendicular to the optic axis of amplitude A sin θ.

A sin θ

doubly refracting calcite plate as shown in Fig. 28. E-wave having

As both the vibrations travel with different veloci ties in same

P

vibrations parallel to the optic axis of amplitude A cos θ and O-wave

direction, a phase difference δ is introduced between E-wave along PE

Optic axis

Y N A

θ

X

A cos θ E

and O wave along PO in traversing a distance equal to the thickness of the plate. In calcite plate ve > v0 , therefore, E-wave leads the O-wave. The displacements of E and O waves along and perpendicular to the

Fig. 28

optic axis may be expressed by the equations and

x = A cos θ sin (ωt + δ)

...(1)

y = A sin θ sin ωt

...(2)

Substituting A cos θ = a and A sin θ = b in eqns. (1) and (2), we get ∴

x = a sin (ωt + δ)

...(3)

and

y = b sin ωt

...(4)

x = sin ωt cos δ + cos ωt sin δ a

...(5)

From eqn. (3), we have,

From eqn. (4), we have

y = sin ωt and b

2  y 1 −   = cos ωt  b

Substituting these values of sin ωt and cos ωt in eqn. (5), we get y x = cos δ + a b

 y2  1 −  . sin δ  b2  

or

 y y2  x − cos δ = 1 − 2  . sin δ a b b  

2  y y2  x  Squaring both sides, we get  − cos δ = 1 − 2  sin2 δ a b  b  

Solving further, we get

x2 a

2

+

y2 2

b

−

2xy cos δ = sin2 δ ab

...(6)

This is the equation of an oblique ellipse. Thus, in general the emergent light is elliptically polarised. The exact nature of the resultant motion or the light emerging from calcite plate depends upon the value of δ .

293

Special Cases Plane Polarised Light Case I: When the thickness of the plate is such that δ = 0, 2 π , 4 π , … 2 nπ , then sin δ = 0 and cos δ = 1, therefore, eqn. (6) reduces to or

y 2 x  −   a b

x2 2

a = 0 or ± y = ± (b / a) x

+

y2 b2

−

2 xy =0 ab ...(7)

Thus the motion is described by a straight line, both x and y have the same sign. Thus, when two mutually perpendicular plane polarised waves are in phase then the emergent light is plane polarised with vibrations in the same plane as in the incident light as shown in Fig. 29. Case II: When the thickness of plate is such that δ = π , 3 π , 5 π , … (2 n + 1) π , then sin δ = 0 and cos δ = − 1, hence eqn. (6) reduces to y2 2 xy x2 + + =0 ab a2 b2 y 2 b x or  +  = 0 or y = − x  a b a

...(8)

This is again the equation of a straight line passing through the origin P. Thus the emergent light is again plane polarised having vibrations making an angle 2 tan−1 (b / a) with that of incident light as shown in Fig. 30.

Elliptically Polarised Light Case III: When the thickness of plate is such that δ = π / 2, 3 π / 2, … (2 n + 1) π / 2, where n = 0, 1, 2, … then sin δ = 1 and cos δ = 0, hence eqn. (6) reduces to x2 a2

+

y2 b2

=1

...(9)

This is the equation of a simple ellipse. Thus when two mutually perpendicular plane polarised waves are in π / 2 or (2 n + 1) π / 2 phase then the emergent light is elliptically polarised (provided a ≠ b). The plane of the ellipse being normal to the direction of propagation as shown in Fig. 31.

Circularly Polarised Light Case IV: When a = b and δ = (2 n + 1) π / 2, n = 0, 1, 2, ..., then sin δ = 1 and cos δ = − 1, hence eqn. (6) reduces to x 2 + y2 = a2 This is the equation of a circle of radius a. Thus the emergent light is circularly polarised as shown in Fig. 32.

294 Thus from the above discussion, we conclude that, when a plane polarised light enters into a thin plate of doubly refracting crystal, such as calcite, cut with faces parallel to the optic axis but oriented in such a way that the incident linear vibrations make an angle θ with the optic axis. The amplitude of incident vibration is split up into two ordinary and extraordinary components which are travelling along the same direction with different velocities in the crystal and finally emerge as a single component. The shape of the resultant transmitted vibrations depend on the phase difference and hence on the thickness of the doubly refracting crystal plate. The light emerging from the crystal plate is generally elliptically polarised, for 0 < δ < π the emergent light is right handed elliptically polarised because in this case the rotation of light vector is anticlockwise. Similarly for π < δ < 2 π , the emergent light is left handed elliptically polarised as light vector rotates in clockwise direction. When the thickness of the plate is such that the phase difference between the two components is even multiple of π (δ = 2nπ ), the emergent light is plane polarised with vibrations in the plane of incidence, whereas if δ is odd multiple of π [δ = (2 n + 1) π ], the emergent light is again plane polarised, but having vibrations in a direction making an angle 2θ [= 2 tan−1 (b / a)] with the plane of incidence. When δ = (2 n + 1) π / 2, where n = 1, 2, 3, ..., the emergent light is elliptically polarised and its axis (major and minor) coincide with the axis of the component linear vibrations. When δ = (2 n + 1) π / 2 and θ = 45 ° (that is, a = b), the emergent light is circularly polarised. Thus, the plane polarised and circularly polarised light are the special cases of elliptically polarised light. Example 15: Discuss the stage of polarisations of the following light waves: (i)

E = $j A cos (kx − ωt ) + k$ 2 A cos (kx − ωt + π / 4 )

(ii)

E = $j A cos (kx − ωt ) + k$ A sin (kx − ωt )

Solution: (i) The given wave equation is the resultant of superposition of two mutually perpendicular electric waves with unequal amplitudes. The general equation of resultant disturbance is expressed as, x2 2

a

+

y2

−

2

2x y cos δ = sin2 δ ab

b a = A, b = 2 A and δ = π / 4 y2 2 xy x2 π π + − cos   = sin2 2 2  4 4 A 4A 2 A2

Here ∴

x2

or

2

A

+

y2 2

4A

−

xy 2

2 A

=

1 2

...(1)

...(2)

Equation (2) is the equation of an oblique ellipse described in anticlockwise direction. Thus, the emergent light is elliptically polarised with plane normal to the direction of propagation. (ii)

The given wave equation is again the resultant of two mutually perpendicular electric waves with equal amplitude and with a phase difference of π / 2 .

In this case,

a = b = A and δ = π / 2

Therefore, from equation (1), we have

295 x2 A2

+

y2 A2

= 1 or x2 + y2 = A2

...(3)

This is the equation of circle with radius A. Thus, the emergent light is circularly polarised.

Phase Retardation Plates Retardation plates are used for the introduction of a phase difference between ordinary and extraordinary rays during their normal transmission through doubly refracting crystals. It is constructed by cutting a plate from a doubly refracting crystal by sections parallel to the optic axis. The phase difference between O-ray and E-ray produced by these plates may be deduced as under: If t be the thickness of a retardation plate in the direction of propagation, µ0 and µ E be the refractive indices for ordinary and extraordinary rays respectively, then the path difference introduced between O and E-rays is given by ∆ = (µ0 − µ E ) t µ0 > µ E the corresponding phase difference, δ = (2 π / λ ) (µ0 − µ E ) t The two most useful retardation plates are: (i) Quarter wave plate and (ii) Half wave plate 1.

Quarter Wave Plate: A quarter wave plate, also called phase retardation plate, is widely used for the production and the detection of elliptically and circularly polarised light. It is a plate of uniaxial doubly refracting crystal (like calcite) cut with optic axis parallel to the refracting faces and whose thickness is such as to produce a path difference of λ / 4 (or phase difference of π / 2) between the ordinary and extraordinary waves. When a plane polarised beam of light is incident normally on such a plate, the vibrations in the beam breaks into two types of vibrations; one type of vibrations (0) along the optic axis and the other type ( E) p e r p e n d i c u l a r t o i t . T h e s e t w o components travel in the same direction along the normal to the faces but with different velocities (Fig. 33). If t is the thickness of a quarter wave plate, µ0 and µ E be the refractive indices for the ordinary and extraordinary beams respectively, then the path difference introduced by the quarter wave plate between O and E-rays is given by ∆ = (µ0 ~ µ E ) t In negative uniaxial crystals like calcite, E-ray travels faster than the O-ray so that µ0 > µ E ∴

∆ = (µ0 − µ E ) t

In positive uniaxial crystals like quartz, O-ray travels faster than E-ray so that µ E > µ0

296 ∴

∆ = (µ E − µ0 ) t

For a quarter wave plate, the path difference between the two waves on emergence must be equal to λ /4 . ∴ ∴ For positive quartz crystal

[For calcite plate]

∆ = λ / 4 = (µ0 − µ E ) t λ t= 4 (µ0 − µ E ) t=

...(1)

λ 4 (µ E − µ0 )

...(2)

Practically the quarter wave plate has to be supported between glass plates. If the vibrations of incident plane polarised beam make an angle of 45° with the optic axis of the quarter wave plate, then the emergent light is circularly polarised. If the vibration of the incident plane polarised light do not make an angle of 45 ° , 0 ° or 90° with the optic axis, then the emergent light is elliptically polarised. 2.

Half Wave Plate: A half wave plate, also called retardation plate, is used in the Laurentz half-shade polarimeter for the measurement of optical rotation. It is a plate of doubly refracting crystal cut (like calcite) with optic axis parallel to the refracting faces and whose thickness is such as to produce a path difference of λ / 2 (or phase difference of π) between the ordinary and extraordinary waves. If t is the thickness of the half wave plate, µ0 and µ E be the refractive indices for the ordinary and extraordinary beams of light respectively, then the path difference introduced by the half wave plate between O and E-rays is given by ∆ = (µ0 − µ E ) t For negative uniaxial calcite crystal, µ0 > µ E

∴

∆ = (µ0 − µ E ) t

For positive uniaxial quartz crystal, µ E > µ0

∴

∆ = (µ E − µ0 ) t

For a half wave plate, the path difference between the two waves on emergence must be equal to λ / 2 ∴

∆ = λ / 2 = (µ0 − µ E ) t

∴

t=

λ 2 (µ 0 − µ E )

[For calcite crystal]

and

t=

λ 2 (µ E − µ 0 )

[For quartz crystal]

If the incident plane polarised beam makes an angle θ with the optic axis of the plate then the emergent light is also plane polarised with vibrations inclined at an angle of 2 θ with the optic axis. A major disadvantage with these plates is that they are true only for one particular wavelength because their thickness depends upon wavelength of light used.

Distinction Between Quarter Wave Plate and Half Wave Plate To identify quarter wave plate and half wave plate, a plane polarised light is allowed to pass normally through each of them and the emergent light is detected by a rotating nicol [Fig. 34].

297 1.

Quartz wave In case of a quarter wave plate, the Rotating Nicol plate light emerging from the quarter wave (a) Variation in intensity from max. to min. but not zero. plate is generally elliptically polarised. (a) no variation in intensity. It will be circularly polarised when (c) Variation in intensity Analyser Plane polarised (a) from max. to zero min. light the vibrations of the plane ploarised Half wave Rotating Nicol plate light falling on it make an angle 45° Variation in intensity from with the direction of the optic axis max. to zero min. of the plate and it will be plane Plane polarised Plane Analyser polarised if the vibrations of the light polarised (b) light incident plane polarised light make an angle 0° or 90° with the direction Fig. 34 of the optic axis. Thus, when a light emerging from quarter wave plate is examined by a rotating nicol the following may occurred: (i) When the emergent light is elliptically polarised, the intensity varies from maximum to minimum but the minimum intensity is not zero [Fig. 34 (a)].

(ii)

When the emergent light is circularly polarised, no variation in intensity is observed.

(iii) When the emergent light is plane polarised, the intensity varies from maximum to zero minimum. 2.

In case of a half wave plate, the light emerging from half wave plate will always be plane polarised whatever the orientations of the plate with respect to the plane of vibration of the incident light may be. Therefore, when the emergent light is examined through a rotating nicol, the intensity varies from maximum to zero minimum [Fig. 34(b)].

Example 16: A pair of crossed polarisers with axes at angles θ = 0 ° and 90 ° is placed in a beam of unpolarised light with intensity I0 , so that the light emerges from the first with I1 = I0 / 2 and from the second with I2 = 0 . A third polariser is placed between the two at an angle of 45°. (i)

What is I2 ?

(ii) If the third polariser rotates at angular frequency ω, show that, I2 = I0 (1 − cos 4 ωt ) /16 . Solution: (i) Let us suppose that the amplitude of unpolarised light is A0 . Suppose a1, a2 and a3 are the amplitudes of polarised light after passing through first, second and third polarisers respectively. Then I  a1 =  0  , 2 ∴

Now a3 = a1 cos 45 ° and a2 = a3 cos 45 °

a2 = (a1 cos 45 ° ) (cos 45 ° ) =

a1 2

or a2 =

1 2

 I0    2

I0 8 (ii) If the third polariser rotates at angular frequency ω, then a3 = a1 cos ωt and a2 = a3 cos (90 − ωt) = (a1 cos ωt) sin ωt a 1  I0  or a2 = 1 sin 2 ωt =   sin 2 ωt 2 2 2

∴

I2 = a22 =

∴

I2 = a22 =

I0 sin2 (2 ωt) 8

298 or

I2 =

1 − cos A   2 ∵ sin A =    2

I0 (1 − cos 4 ωt) 16

Example 17: Calculate the thickness of a doubly refracting crystal plane required to introduce a path difference of λ /2 between ordinary and extraordinary rays when λ = 6000 Å, µ0 = 1. 55 and µ e =1. 54. [GBTU, B.Tech. I Sem. (Old) 2010]

Solution: The thickness of a doubly refracting crystal plate required to introduce a path difference of λ /2 is,

∴

t=

λ , Here λ = 6000 Å = 6000 × 10 − 8 cm, µ0 = 1.55 and µ e = 1.54 2 (µ0 − µ e )

t=

600 × 10 −8 6000 × 10 −8 = = 3 × 10 −3 cm 2 (1.55 − 1.54) 2 × 0 .01

Example 18: Find the thickness of a quarter wave plate when the wavelength of light is equal to 5890 Å and µ0 = 1. 55, µ E = 1. 54 . Or Find the thickness of a quarter wave plate for the wavelength of light 5890 Å. The refractive index of O-and E rays are 1.55 and 1.54 respectively. [GBTU, B.Tech. I. Sem. (C.O.) 2012] Solution: The thickness of a quarter wave plate for a crystal for which µ0 > µ E is given by

∴

t=

λ , Here λ = 5890 Å, µ0 = 1.55 and µ E = 1.54 4 (µ0 − µ E )

t=

5890 × 10 − 8 5890 × 10 − 8 = =1 . 47 ×10 − 3 cm 4 (1.55 − 1.54) 4 × 0 .01

Example 19: Calculate the thickness of a quarter wave plate of quartz for sodium light of wavelength 5893 Å. The ordinary and extraordinary refractive indices for sodium are 1. 54425 and 1. 55336 respectively.

[UPTU, B. Tech. I Sem, Q. Bank, 2000]

Solution: The thickness of a quarter wave plate for positive crystal like quartz is

t=

Here µ0 = 1. 54425, µ E = 1. 55336 and for sodium light λ = 5893 Å = 5 . 893 × 10 − 5 cm ∴

t=

λ 4 (µ E − µ0 )

5 . 893 × 10 − 5 5 .893 × 10 − 5 = = 1.62 ×10 − 3 cm 4 (1. 55336 − 1. 54425) 4 × 911 × 10 − 5

Example 20: Calculate the thickness of a half wave plate of quartz for a wavelength of 5000 Å. Here µ e = 1. 553 and µ0 = 1. 554. [GBTU, B.Tech I Sem. 2010; UPTU, B.Tech. I Sem. (Old) 09] Solution: The thickness of a half wave plate of quartz is, t=

λ 2 (µ e − µ0 )

Here µ e = 1.553, µ0 = 1.544 and λ = 5000 Å = 5000 × 10 −8 cm ∴

t=

5000 × 10 −8 5000 × 10 −8 = = 2 . 78 × 10 −3 cm 2 (1.553 − 1.544) 2 × 0 .009

299 Example 21: Calculate the thickness of a calcite plate which would convert plane polarised light into circularly polarised light. The principal refractive indices are µ0 = 1.658 and µ E = 1.486 at the wavelength of light used as 5890 Å . Solution: The plane polarised light will be converted into circularly polarised if the thickness of a calcite plate creates a path difference of λ / 4 or phase difference of π / 2 between O and E-rays. Thus, the thickness of the plate is given by nλ nλ or , n = 1, 2, 5 … (µ0 − µ E ) t = t= 4 4 (µ0 − µ E ) Here ∴

µ0 = 1.658, µ E = 1.486 and λ = 5890 Å = 5890 × 10 − 8 cm t=

n × 5890 × 10 − 8 n × 5890 × 10 − 8 = = 8 .56 × 10 − 5 n cm 4 (1.658 − 1.486) 4 × 0 .172

Thus, the thickness of the calcite plate is given as t = 8.56 × 10 − 5 × 1, = 8.56 × 10 − 5 cm,

8 .56 × 10 − 5 × 3,

8 .56 × 10 − 5 × 5 cm

25 .68 × 10 − 5 cm,

42. 80 × 10 − 5 cm

The minimum thickness of the plate = 8 .56 × 10 − 5 cm Example 22: Calculate the thickness of a doubly refracting plate capable of producing a path difference of λ / 4 between ordinary and extraordinary rays with light of wavelength 5890 Å . The refractive indices for ordinary and extraordinary rays are 1. 53 and 1. 54 respectively. [UPTU, B.Tech., I Sem. (C.O.) 2006; II Sem. 05]

Solution: The thickness of a doubly refracting plate capable of producing a path different of λ / 4 or of positive quarter wave plate is given by t=

λ 4 (µ e − µ0 )

Here

λ = 5890 Å = 5890 × 10 −10 m, µ e = 1. 54 and µ0 = 1. 53

∴

t=

5890 × 10 −10 4 (1.54 − 1.53)

=

5890 × 10 −10 4 × 0 .01

= 1.47 × 10 − 5 m

Example 23: Find the thickness of a quarter or wave plate for the wavelength of light of 589 nm and or µ0 = 1. 55, µ e = 1. 54 . Find the thickness of a quarter wave plate for the wavelength of light of 5890 Å. The refractive indices for O- and E-rays are 1. 55 and 1. 54 respectively.

[GBTU, B.Tech I Sem. 2011; UPTU, B.Tech. II Sem. 2004]

Solution: The thickness of a quarter wave plate for negative crystal is, t=

λ 4 (µ0 − µ e )

Here,

λ = 589 nm, µ0 = 1. 55 and µ e = 1. 54

∴

t=

589 589 = = 147. 25 × 10 2 nm 4 × (1. 55 − 1. 54) 4 × 0 . 01

300 Example 24: Find the thickness of a quarter and half wave plates for the wavelength of light 589 nm and µ0 = 1. 55 and µ e = 1. 54. [UPTU, B.Tech I Sem. (CO) 2010] Solution: Here λ = 589 nm, µ0 = 1.55 and µ e = 1.54 The thickness of the quarter wave plate, tπ /4 =

λ 4 (µ0 − µ e )

∴

tλ /4 =

589 589 = = 147 . 25 × 10 2 nm 4 × (1.55 − 1.54) 4 × 0 .01

The thickness of the half wave plate,

tλ /2 =

λ 2 (µ0 − µ e )

∴

tλ /2 =

147 .25 1 .t = × 102 nm = 73 . 625 × 10 2 nm 2 λ /4 2

Example 25: A quarter wave plate is meant for λ 0 = 5 .893 × 10 − 5 cm. What phase retardation φ will show for λ = 4 .358 × 10 − 5 cm ? (Neglect changes of µ0 and µ E with λ ). Solution: If the quarter wave plate is of the smallest thickness then the thickness is given by t = or

λ 4 (µ0 − µ E )

path difference ∆ = (µ0 − µ E ) t = λ 0 / 4 The phase retardation for λ = 2 π / λ (path difference)

∴

phase retardation for λ = (2 π / λ ) .(λ 0 / 4) In the given problem, λ 0 = 5 . 893 × 10 − 5 cm and λ = 4 . 358 × 10 − 5 cm

∴

phase retardation, φ =

2π (4 . 358 × 10 − 5 )

×

5 .893 × 10 − 5 = 0 . 67 π rad 4

Example 26: A beam of linearly polarised light is changed into circularly polarised light by passing it through a slice of crystal 0 . 003 cm thick. Calculate the difference in the refractive index of the two rays in the crystal assuming this to be minimum thickness that will produce the effect and that the wavelength is 6 × 10 − 5 cm. Solution: We know that the least phase difference between O and E-waves on emergence from the doubly refracting plate must be π / 2 if the incident linear polarised light is changed into circularly polarised light. Thus the slice of crystal must be a quarter wave plate of thickness t =

λ 4 (µ0 ~ µ E )

where µ0 ~ µ E represents the difference in refractive index, therefore (µ0 ~ µ E ) = λ / 4 t, Here ∴

λ = 6 × 10 − 5 cm and t = 0 .003 cm (µ0 ~ µ E ) =

6 × 10 − 5 = 5 × 10 − 4 × 0 .003

3

301 Example 27: A sheet of cellophane is a half plate for light of wavelength 4000 Å. If the refractive index do not change with wavelength explain how the sheet would behave with respect to light of wavelength 8000 Å . Solution: The path difference, ∆ produced by a sheet is given by λ ∆ = (µ0 − µ E ) t = 1 = constant 2 Let the path difference produce by another light of wavelength λ 2 is (λ 2 / n) . 2 λ 2 2 × 8000 λ λ or ∴ ∆= 1 = 2 n= = =4 2 n λ1 4000 ∴

Path difference produced by second wavelength, ∆ =

λ2 4

Hence, the given cellophane sheet will behave like a quarter wave plate with light of wavelength 8000 Å . Example 28: Plane polarised light, is incident on a plate of quartz cut with faces parallel to optic axis. Calculate the thickness for which the phase difference between the two rays is 60 °, where µ0 = 1. 5442, µ e = 1. 5583 and λ = 5000 Å. Solution: When a plane polarised light is incident on a plate of quartz cut with faces parallel to optic axis, the beam splits up into ordinary and extraordinary rays. For a quartz of thickness t, the path difference between O-and E-rays is given by ∆ = (µ e − µ0 ) t 2π 2π The corresponding phase difference, δ = ∆= (µ e − µ0 ) t λ λ π π Here, δ = 60 ° = 60 ° × = , µ e = 1. 5583, µ0 = 1. 5442 180 ° 3 and λ = 5000 Å = 5000 × 10 − 8 cm λ .δ 5000 × 10 − 8 × π / 3 = 2 π (µ e − µ0 ) 2 π (1. 5583 − 1. 5442)

∴

t=

or

t = 0.00091cm

Example 29: The value of µ E and µ0 for quartz and 1. 5508 and 1. 5418 respectively. Calculate the phase retardation for λ = 5000 Å when the plate thickness is 0 .032 mm.

[UPTU, B. Tech. I Sem. 2008]

Solution: For a quartz crystal of thickness t, the path difference between the O and E-rays is given by 2π The phase retardation = (∆) λ

∆ = (µ E − µ0 ) t

=

2π (µ E − µ0 ) t λ

Here µ E = 1.5508, µ0 = 1.5418, t = 0 .032 mm = 0 .0032 cm and λ = 5000 Å = 5000 × 10 − 8 cm ∴

phase retardation =

2 ×3.14 × (1.5508 − 1.5418) × 0.0032 5000 × 10 − 8

= 3.617 rad

302 Example 30: Calculate the thickness of doubly refracting crystal required to introduce a path difference of λ / 2 between the O and E-rays when λ = 6000 Å, µ0 = 1.65 and µ E = 1.48 . Solution: The required thickness of doubly refracting crystal to introduce a path difference of λ / 2 is λ given by t= , 2 (µ0 − µ E ) Here ∴

µ0 = 1.65, µ E = 1.48 and λ = 6000 Å = 6000 × 10 − 8 cm t=

6000 × 10 − 8 = 0 .000176 cm = 1.76 × 10 − 2 (1.65 − 1.48)

4

cm

Example 31: A plate of thickness 0 .020 mm is cut from calcite with optic axis parallel to the face. Given µ0 = 1.648 and µ E = 1.481 (ignoring variations with wavelength), find out those wavelength, in the range 4000 Å to 7800 Å for which the plate behaves as a half wave plate and also those for which the plate behaves as a quarter wave plate.

[UPTU, B.Tech. I. Sem. Q. Bank, 2000]

Solution: The plate behaves as a half wave plate for λ, if (2 n − 1) λ (µ0 − µ E ) t = , where n = 1, 2, 3, … 2 2 (µ0 − µ E ) t 2 (1.648 − 1.481) × 0.0020 or λ= = cm (2 n − 1) (2 n −1) or

λ=

66800 66800 × 10 − 8 cm or λ = Å (2 n −1) 2n − 1

Substituting the different values of n, it is found that for n = 5, 6, 7, 8, we get λ = 7422.2 Å, 6072.7 Å, 5138.5 Å, 4453.3 Å in the given region (4000 Å − 7800 Å) for half wave plate. The plate will behave as quarter wave plate if (2 n − 1) λ 4 (µ0 − µ E ) t or λ = (µ0 − µ E ) t = , where n = 1, 2, 3, … 4 2n − 1 Here again µ0 = 1.648, µ E = 1.481 and t = 0 .0020 cm 4 (1.648 − 1.481) × 0 .0020 133600 ∴ λ= = Å (2 n − 1) 2n − 1 Substituting the different values of n, it is found that for n = 10, 11, 12, 13, 14, 15, 16, 17 respectively, we get λ = 7031.6 Å, 6361.9 Å, 5808 .7 Å, 5344 Å, 4948 .1 Å, 4606 .9 Å, 4309 .7 Å, 4048 . 5 Å in the given region (4000 Å − 7800 Å) for quarter wave plate. Example 32: A phase retardation plate of quartz has thickness 0 .1436 mm. For what wavelength in the visible region (4500 Å − 8000 Å) will it act as (i) quarter wave plate, (ii) half wave plate ? Given: µ0 = 1. 5443, µ E = 1. 5533 .

[UPTU, B.Tech I Sem. (C.O.) 2005]

Solution: (i) The thickness of a quarter wave plate is given by t = that is,

t=

λ or any odd multiple, 4 (µ E − µ0 )

(2 n + 1) λ where n = 0, 1, 2, 3, … 4 (µ E − µ0 )

303 ∴ Here Thermefore,

λ=

4 (µ E − µ0 ) t , (2 n + 1)

µ E = 1. 5533, µ0 = 1. 5443 and t = 0 .1436 mm = 0 .01436 cm 4 × (1. 5533 − 1. 5443) × 0 . 01436 51700 λ= = (2 n + 1) (2 n + 1)

Substituting n = 0, 1, 2, 3, ..., we get λ = 51700 Å, 17233 Å, 1034 Å, 7386 Å, 5744 Å, 4700 Å, … Hence in the visible region, for quarter wave plate λ = 7386 Å, 5744 Å and 4700 Å 2 (µ E − µ0 ) t 2 × 0 .009 × 0 .01436 25850 Å = = (2 n + 1) (2 n + 1) (2 n + 1)

(ii) For half wave plate, we shall have

λ=

Substituting n = 0, 1, 2, 3, ..., we get

λ = 25850 Å, 8617 Å, 5170 Å, 3693 Å, …

Hence in the visible region, for half wave plate λ = 5170 Å and 3693 Å

Sheet Polariser Sheet polariser is a device used for the production of linearly polarised light. The linearly polarised light may be produced in the number of ways and such a device or mechanism to produce linearly polarised light is called linear polariser. There are several types of linear polarisers based on different optical processes, but here we will concentrate only on sheet polarisers, which is an important member of dichroic polariser based on asymmetry of absorption. The simplest member of this family is the wire-grid polariser whose chemical version came in the form of H-sheet polariser invented by Land in 1938. The first sheet polariser is J-sheet polariser. The other world’s most popular polarisers are : H-sheet polariser and K-sheet polariser. Let us begin with wire-grid polariser.

1. Wire Grid Polariser Wire-grid polariser consists of a large number very thin copper wires arranged parallel to each other as shown in Fig. 35. When unpolarised light is incident on it then the component of electric vector along the length of the wire or parallel to the wires is absorbed or reflected. The electric field along the wire produce electric currents in the wire and the field energy is converted into the energy of the current and finally dissipated in the form of heat. For the horizontal component of electric vector along the X-axis (or for the component in the direction perpendicular to the wires) the electrons can not move very far across the width of each wire, therefore little energy is reflected and horizontal component is able to pass through the grid without much attenuation of energy. Thus, the transmitted light has an electric vector purely in the direction perpendicular to the length of the wires and hence linearly polarised with electric vector along the X-axis. However, for the system to be effective, the spacing between the wires

304 should be ≤ λ. Since the light wave is associated with a very small wavelength (~ 5 × 10 −5 cm), the fabrication of wire grid polariser with wire spacing of the order of 5 × 10 −5 cm is extremely difficult. But Bird & Parrish in 1950 were able to form a wire-grid polariser with 30,000 wires in about one inch. In fact it is extremely difficult to fabricate a wire-grid polariser to polarise visible light. The performance of a wire-grid polariser depends on the thickness of the wires and on the spacing between them. A wire-grid polariser in which thick copper wires are arranged with large spacing perform very poorly. If the wires are randomly oriented, both the components of electric vector are absorbed and device would not be able to transmit any light at all.

2. J-Sheet Polariser To remove the difficulties encounter in the fabrication of wire-grid polariser, E.H. Land in 1928 replaces the copper wires by a long chain of polymer molecules that contains atoms like iodine that provides high conductivity along the length of chain.These long chain molecules are aligred so that they are almost parallel to each other. In 1852, Herapath, discovered a synthetic dichroic crystal, iodoquinine sulphate, called Herapathite.The crystal transmits linearly polarised light of all colours. However, crystal can not be grown to sufficient bigger size. To over come the scarcity of a large single crystal, Land suggested that it should be possible to duplicate the effect of a single crystal with a statistical array of micro-crystals of the appropriate type. If an array of microcrystals could, by some means, be suspended in a transparent matrix and aligned in such a manner that their absorption axis are parallel, a sheet of this material would behave like a thin single crystal of dichroic material. To fabricate such an efficient polarising sheet, Land uses many microscopic crystals of iodoquinine sulphate (Herapathite) embedded in a transparent nitrocellulose polymer film. The needle like crystals are aligned during manufacture of the film by stretching. With the crystals aligned the sheet is dichroic. Thus, a large sheet of plastic containing alignment of millions herapathite crystals acts like one huge crystal. This is called J-sheet polariser. This sheet tends to absorb light which polarised parallel to the direction of crystal alignment, but transmits light which is polarised perpendicular to it. In this way Land was able to fabricate first dichroic sheet polariser, termed as J-sheet polariser. Thus, the aligned conducting molecules are similar to the wires in the wire-grid polariser and since the spacing between two successive chain molecules is small compared to optical wavelength, the sheet polariser is a better device than wire-grid polariser.The main difficulty with J-sheet polariser is its haziness.

H-Sheet-Polariser Now-a-days, J-sheet micro crystal polarisers have largely been superseded by molecular H-sheet polariser. The durability and practicality of these polariser makes them most popular. This is considered as the chemical version of wire-grid polariser.The improved H-sheet polariser was invented in 1838 by H.Land. H-sheet form is made from polyvinyl alcohol (PVA) polymer impregnated with iodine. Stretching of sheet during manufacture causes PVA chains to align in one particular direction. Valence electrons from the iodine dopant attached to the PVA molecules are able to move linearly along the polymer chains and make them conducting along the length of the chain.So incident light polarised parallel to the chains is absorbed by the sheet, but light polarised perpendicularly, is transmitted with very little absorption.

305 Thus, parallel, and equidistant long thin iodine atoms in the chains act like copper wires of the wire-grid polariser. They absorb light vibrations parallel to the alignment axis, but freely transmitted transverse vibrations. Hence, the emergent light is linearly polarised. For the alignment of conducting molecules, a large transparent sheet of easily stretched and chemically reactive PVA polymer is taken.The sheet is warmed and very quickly stretched to many times its original length. With the increase in linear dimension, the sheet becomes also narrower and thinner. During stretching, most of the long polymeric molecules, initially at random orientation, come to align in the direction of the stretching force. As soon as stretching is completed, the sheet is cemented to a rigid sheet of cellulose acetate, so that no unstretching can occur. The sheet is then dipped into a iodine-rich liquid solution. Immediately after dipping iodine diffuses into the polyvinyl alcohol layer and gets attached to the PVA molecules.The sheet so prepared is washed, dried and cut into pieces of any disered shape and size. The sheet is covered with glass plate to protect it from scratches. The H-sheet polarisers are designated as HN-32, means 32% transmission of incident light, HN-22 transmits 22%.

K-Sheet Polariser K-sheet polarisers are superior to H-sheet polariser as they are highly stable and their material is particularly resistant to humidity and heat. It is also made of polyvinyl alcohol, but they are not doped like H-sheet polarisers. In making H-sheet, we added atoms to the sheet, but here instead of adding atoms we takes atom away. To fabricate K-sheet polariser an oriented polyvinyl alcohols (PVA) polymer film is heated in a high temperature oven in the presence dehydrating HCl catalyst, where it undergo dehydration. In this way Land and Ragers removed 2N hydrogen atoms and N-oxygen atoms to make film highly dichroic.This produces a different type of polymeric molecule called polyvinylene. Polyvinylene molecules are aligned by streching the sheet in one particular direction. K-type sheet have their transmission axis perpendicular to the direction of stretch. A modern type of absorptive polariser is made of elongated silver nano particles embedded in thin (≤ 0 . 5 mm) glass plates.These polariser are more durable and can polarise light much better than plastic polaroid film. Such glass polarisers perform best for short wavelength infrared light and widely used in optical fibre communication.

Applications of Sheet Polarisers Polarising sheets are used in optical microscopes and sunglasses. They are also used in liquid crystal displays and to examine for chain orientation in transparent plastic products made from polystyrene or polycarbonate. Since polaroid sheet is dichroic, it will absorb incident light of one plane of polarisation. So sunglasses will reduce the partially polarised light reflected from level surfaces such as windows and sheets of water. Low cost, large acceptance angle (~30°) and linear aperture (5 cm or larger) are the most preferred advantage of these sheet polarisers. Sheet polarisers are inexpensive and can be made in large sizes. They are used as polarisers and analysers to produce polarise light and to examine the state of polarisation.

306 Sheet polarisers are not recommended for high power applications because the unwanted polarisation component is actually absorbed by polarising film where it is ultimately converted into heat, which must be dissipated. In addition, most of the polariser that employ glass protective plates will not transmit light below about 330 nm.

Production of Plane, Circularly and Elliptically Polarised Lights 1. Plane Polarised Light Though there are number of ways for the production of plane polarised light, the method using nicol prism is usually employed. When a beam of unpolarised light enters into a nicol, it splits up into ordinary and extraordinary components. The ordinary component is totally internally reflected at the Canada balsam layer and is absorbed, while the extraordinary component passes through them and emerges as plane polarised light. On rotating the analysing nicol the emergent light varies with zero minimum.

2. Circularly Polarised Light Circularly polarised light is produced by recombining two waves of equal amplitude and vibrations in mutually perpendicular directions and having a phase difference of π /2 or path difference of λ /4 between them. For obtaining circularly polarised light, the plane polarised light must fall normally on quarter wave plate in such a way that the plane of vibration of incident light makes an angle of 45° with the direction of optic axis. The experimental arrangement for producing circularly and elliptically polarised light is depicted in Fig. 36. The unpolarised light is allowed to pass through a nicol P (called polariser) to get plane polarised beam of light. Another nicol A (called analyser) is placed at some distance from P such that the two nicols are in crossed position. In this position of nicols no light is transmitted from A and hence field of view appears dark. A quarter wave plate Q mounted on a tube C is introduced between the two nicols in the path of plane polarised beam (without disturbing the positions of P and A) such that the polarised beam falls normally on a quarter wave plate. The amplitude of plane polarised beam on entering the quarter wave plate is split up into two mutually perpendicular components with a phase change of π /2, that is, into ordinary and extraordinary components. The field of view may now be bright. The quarter wave plate Q is rotated about the outer fixed tube B until the field of view is again appear dark. This happens when vibrations of incident light are along the optic axis of quarter wave plate. Keeping Q fixed, tube C is rotated so that the mark M on Q exactly coincides with zero mark on C. Now by rotating quarter wave plate Q the mark M is made to coincide with 45° mark on C. Now the position of quarter wave plate Q is such that the vibrations in the incident plane polarised light make an angle 45° with the

307 optic axis of the plate. In this situation the plane polarised light on entering the quarter wave plate Q is split up into O and E components of equal amplitudes. In this way the light emerging from quarter wave plate is circularly polarised. If the analysing nicol A is rotated, no change in intensity is observed in the field of view.

3. Elliptically polarised light Elliptically polarised light is produced by recombining two wave of unequal amplitudes and vibrations in mutually perpendicular directions and having a phase difference of π /2 or path difference of λ /4 between them. For obtaining elliptically polarised light, the plane polarised light must fall normally on quarter wave plate in such a way that the plane of vibration of incident light makes an angle other than 45° with direction of the optic axis. The experimental arrangement for producing elliptically polarised light is same as depicted in Fig. 36. The unpolarised light is allowed to pass through a polarising nicol P to get plane polarised beam of light. Another nicol A is placed at some distance from P such that the two nicols are in crossed position so that no light is transmitted from A . A quarter wave plate Q is introduced between the two nicols as shown in Fig. 36 so that the field of view appears to be bright. The quarter wave plate Q is rotated about the fixed outer tube B until the field of view is again appear dark. This happens when vibrations of incident light are along the optic axis. Again the quarter wave plate is rotated to such a position that vibrations in the plane polarised incident beam make an angle θ other than 0°, 45° and 90° with the optic axis of the plate (because at these angles the emergent light is either circularly polarised or plane polarised). In this position the plane polarised light entering the quarter wave plate is split up into ordinary O and extraordinary E components of unequal amplitudes. In this way the light emerging from quarter wave plate is elliptically polarised. If the analysing nicol prism A is rotated the change in intensity is observed between a maximum to a minimum value in the field of view, that is, the minimum intensity is not zero.

Detection of Plane, Circularly and Elliptically Polarised Lights 1. Plane Polarised Light To detect the plane polarised light, it is allowed to pass through a nicol prism rotating about the direction of propagation of light. If the intensity of the emergent light from the rotating nicol varies from maximum to zero minimum twice in each revolution, then the emergent light is plane polarised (Fig. 37).

Rotating Nicol

Plane polarised light

Variation in intensity with min. zero Analyser

Fig. 37

2. Circularly Polarised Light or Distinction between Circularly and Unpolarised Lights When a beam of circularly polarised light is allowed to pass through a rotating nicol, no change in the intensity of transmitted light is observed. Similarly, constant uniform intensity is also shown by an unpolarised light when viewed through a rotating nicol prism. Therefore, circularly polarised and ordinary unpolarised light behave in the same manner when viewed through a rotating nicol and hence cannot be distinguished by a nicol prism alone.

308 In order to detect circularly polarised light, the beam is first allowed to pass through a quarter wave plate and then viewed through a rotating nicol. If the original beam is circularly polarised, then it is converted into plane polarised beam by quarter wave plate. Hence, if this emergent beam is examined by a rotating nicol, it shows a variation in intensity from maximum to zero minimum twice during each rotation [Fig. 38(a)].

Quartz wave plate

Rotating Nicol Variation in intensity with minimum zero

Circularly polarised light

Plane Analyser polarised (a) light Quartz wave Rotating Nicol plate No change in intensity

Unpolarised light

Unpolarised light

Analyser

(b)

Fig. 38

On the other hand if the original beam is unpolarised, it will remain unchanged after passing through the quarter wave plate. Hence, on passing through the rotating nicol, it will show no variation in intensity [Fig. 38(b)].

3. Elliptically Polarised Light or Distinction between Elliptically and Partially Plane Polarised Lights When a beam of elliptically Quartz wave plate Rotating Nicol polarised light is allowed to pass through a rotating nicol, Variation in intensity with minimum zero the intensity of transmitted Elliptically Plane light varies from maximum to Analyser polarised polarised light minimum but the minimum (a) light intensity is not zero. The Quartz wave Rotating Nicol plate similar type of variation in Variation in intensity with intensity occurred when minimum not zero partially plane polarised light Partially plane Partially plane Analyser examined through a rotating polarised polarised nicol. Therefore, elliptically light (b) light polarised and partially Fig. 39 polarised light behave in the same manner when viewed through a rotating nicol and hence cannot be distinguished by a nicol prism alone. In order to detect the elliptically polarised light, the beam is first allowed to pass through a rotating nicol. If the original beam is elliptically polarised, then it is converted into plane polarised beam by quarter wave plate. Hence, if this emergent beam is examined by a rotating nicol, it again shows the variation in intensity from maximum to zero minimum twice during each rotation [Fig. 39 (a)]. However, if the original beam is partially polarised, it will remain unaltered after passing through the quarter wave plate. Hence, on passing through rotating nicol, it will show no variation in intensity [Fig. 39 (b)].

309

Conversion of Elliptically Polarised Light into Circularly Polarised Light To convert elliptically polarised light into circularly polarised light, the elliptically polarised light is first converted into plane polarised light and then into circularly polarised light. To do this the elliptically polarised light is allowed to fall normally on a quarter wave plate. When elliptically polarised light enters into a quarter wave plate, it breaks up into two unequal components vibrating mutually perpendicular and thus having a phase difference of π /2. The quarter wave plate introduces an additional phase difference of π /2 between the two components ( E and O-components) as a result, the total phase difference now becomes either π or 0. Hence the elliptically polarised beam after transmission through the quarter wave plate is converted into plane polarised beam. This plane polarised beam of light is then allowed to fall normally on another quarter wave plate in such a way that the vibrations in the incident plane polarised beam makes an angle of 45° with the optic axis of the plate. On entering into the quarter wave plate this polarised beam splits up into O and E-components of equal amplitude and period. The light emerging from second quarter wave plate is thus circularly polarised.

Analysis of Polarisation in a given Beam of Light Let us consider that we have a beam of light and it is desired to ascertain its nature from the following possibilities: 1. unpolarised light

2. plane polarised light

4. elliptically polarised light and

5. circularly polarised light.

3. partially plane polarised light

To determine the state of polarisation of a given beam of light, we introduce a nicol prism in the path of light beam and rotate it about the direction of propagation and the change in intensity is observed. Either of the following three cases may occur: (i)

If there is a variation in intensity with two maxima and two minima with minimum intensity zero in one complete rotation, then the incident beam is plane polarised.

(ii)

If there is no variation in intensity, then the incident beam is either unpolarised or circularly polarised.

(iii) If there is a variation in intensity with two maxima and two minima in one complete rotation, but with minimum intensity non–zero, then the given beam of light is either elliptically polarised or partially plane polarised. Thus the rotating nicol prism fails to distinguish between (i) circularly polarised and unpolarised light, and (ii) partially plane polarised and elliptically polarised light. For this purpose we use a quarter wave plate in front of a rotating nicol. To distinguish between circularly polarised and unpolarised beam, we put a quarter wave plate in the path of the beam followed by the rotating nicol and the change in intensity is observed. Either of the following may arise: (i)

If there is no variation in intensity then the incident beam is unpolarised.

(ii)

If there is a variation in intensity with minimum intensity zero, then the incident beam is circularly polarised. This is due to the fact that the quarter wave plate converted circularly polarised light into plane polarised light.

310 (iii) To distinguish between partially plane polarised light and elliptically polarised light, we put a quarter wave plate in front of the nicol prism with its optic axis parallel to the pass-axis of the analysing nicol at the position of maximum intensity and the change in intensity is observed. Either of the following may arises: (a)

If there is a variation in intensity with two maxima and two minima with minimum intensity zero in one complete rotation then the given beam of light is elliptically polarised.

(b)

If there is a variation in intensity with minimum intensity non–zero, then the incident light is partially plane polarised.

We may summarise the scheme of analysis in the following form as shown below:

Scheme of Analysis of a Given Beam of Light Given beam of light Incident on a rotating nicol

Variation in intensity with minimum non zero Conclusion: Given light is either elliptically polarised or partially polarised

Variation in intensity with minimum zero Conclusion: Given light is plane polarised

No variation in intensity Conclusion: Given light is either circularly polarised or unpolarised

Incident on a quarter wave plate with optic axis parallel to the pass-axis of the analysing nicol at the position of maximum intensity and then examined by a rotating nicol

Incident on a quarter wave plate in any position and then examined by a rotating nicol

Variation in intensity with minimum zero

Variation in intensity with minimum non zero

Variation in intensity with minimum zero

Conclusion: Partially polarised

Conclusion: Circularly polarised

Conclusion: Ellipitically polarised

No Variation in intensity Conclusion: Unpolarised

Example 33: Plane polarised light falls normally on a quarter wave plate. Explain what will be the nature of emergent light if the plane of polarisation of the incident light makes the following angles with the principal plane of the quarter wave plate: 0 ° , 30 ° , 45 ° , 90 ° . Solution: When a plane polarised light enters the quarter wave plate cut with faces parallel to the optic axis and oriented in such a way that the vibrations in plane polarised light make an angle θ with the optic axis then the amplitude A of the incident vibrations be resolved into two components E and O with amplitudes A cos θ and A sin θ respectively. On emerging from quarter wave plate a phase difference of π /2 is introduced between E & O-vibrations:

311 (i)

For θ = 0, a = A cos θ = A and b = A sin θ = 0 . That is, in this case O-vibrations are absorbed by the quarter wave plate and only E–vibrations are transmitted. Therefore the emergent light is plane polarised.

(ii)

For θ = 30 ° , a = A cos θ = A cos 30 ° = ( A √ 3 /2) and b = A sin 30 ° = A /2, that is, the two vibrations of amplitudes A √ 3 /2 and A /2 having phase difference of π /2 combine to form elliptical vibrations of ecentricity  b e = 1−    a

2

 ( A / 2)2   = 1 − ( A 3 / 2)2  

or

e = 1−

1 = 3

 2  ,  3

Therefore the emergent light is elliptically polarised. (iii) For θ = 45 ° , a = A cos 45 ° = A / √ 2 and b = A sin 45 ° = A / √ 2, that is, the E and O-components are of equal amplitude. Thus the emergent light is circularly polarised. (iv) For θ = 90 ° , a = A cos 90 ° = 0 and b = A sin 90 ° = A, that is, now E-vibrations are absorbed in the plate only O-vibrations are transmitted. Therefore the emergent light is plane polarised. Example 34: A calcite plate 0 .0031 mm thick is cut and polished such that its optic axis is parallel to the surface. A plane polarised white light from a polariser is incident normally on the plate. Estimate the type of polarisation in the emergent beam for wavelengths 7068 Å, 5893 Å and 5425 Å from the following data: (i) µ0 = 1.6521,

µ E = 1.4811

for

λ = 7068 Å

(ii) µ0 = 1.6580 ,

µ E = 1.4860

for

λ = 5893 Å

(iii) µ0 = 1.6628,

µ E = 1.4878

for

λ = 5425 Å

Solution: For a negative uniaxial crystal like calcite µ0 > µ E . Therefore, the path difference between O and E-rays for calcite plate of thickness t is, ∆ = (µ0 − µ E ) t and corresponding phase difference, δ= (i)

For λ = 7068 Å,

δ=

2π 2π ∆= (µ0 − µ E ) t ; λ λ

Here

2 π (1.6521 − 1.4811) × 0 .00031 7068 × 10 − 8

t = 0 .00031 cm =

3π 2

As the phase difference between O and E-rays is an odd multiple of π / 2, therefore plane polarised beam is converted into elliptically polarised light in general and circularly polarised light, if the vibrations in the incident plane polarised light makes an angle of 45° with the optic of plate. (ii)

For λ = 5893 Å, δ =

2 π × (16580 . − 14860 . ) × 0.00031 2π (µ − µ E ) t or δ = = 1.81π = 3.62 (π /2) λ 0 5893 × 10 − 8

As δ = 3 .62 (π / 2) between O and E-rays, which is not integral multiple of π / 2, therefore, the plane polarised light is converted into elliptically polarised light with axis making an angle with the components of linear vibrations.

312 (iii) For λ = 5425 Å, δ =

2 π × (1.6628 − 1.4878) × 0 .00031 2π (µ0 − µ E ) t or δ = = 2π λ 5425 × 10 − 8

δ = 2 π between E and O components, therefore the state of emergent light remains the same as incident light. Hence the emergent light is plane polarised. Example 35: Plane polarised light of wavelength 6000 Å is incident on a thin quartz plate cut with faces parallel to the optic axis. Calculate: (i)

the ratio of the intensities of the ordinary and extraordinary light if the plane of vibration of two incident light makes an angle of 30 ° with the optic axis.

(ii) the minimum thickness of the plate which introduces a phase difference of 60 ° between the ordinary and extraordinary rays. (iii) the minimum thickness of the plate for which the ordinary and extraordinary waves will combine to produce plane polarised light. Given µ0 = 1. 544 and µ E = 1. 553 . Solution: (i)

When a plane polarised light enters the quarter wave plate cut with faces parallel to the optic axis and oriented in such a way that the vibrations in plane polarised light make an angle θ with the optic axis then amplitude A of the incident vibration be resolved in two components: Amplitude of E-wave = A cos θ = A cos 30 ° = A √ 3 / 2, amplitude of O-wave = A sin θ = A sin 30 ° = A / 2 ∴

Intensity of E-wave ∝ ( A √ 3 / 2)2 = 3 A2 / 4

Intensity of O-wave ∝ ( A / 2)2 = A2 / 4 Therefore the ratio of the two intensities I E / I0 = 3 /1 (ii)

or

3 :1

Give, δ = 60 ° = π / 3 . Therefore, corresponding path difference between E and O-rays ∆ = ∴ ∴ (µ E − µ0 ) t = λ / 6 or t =

∴

∆=

λ (δ) 2π

λ π λ . = 2π 3 6

λ Here λ = 6000 Å = 6000 × 10 −8 cm, µ0 = 1. 544, µ E = 1.553 6 (µ E − µ0 ) t=

6000 × 10 − 8 = 0 .00111cm 6 (1 .553 − 1 .544)

(iii) The ordinary and extraordinary components will give plane polarised light on combination only when the thickness of the plate is such that a phase change of π or path difference of λ /2 occur λ between the two component on emergence, that is, t (µ E − µ0 ) = λ /2 or t = 2 (µ E − µ0 ) or

t=

6000 × 10 − 8 = 0.00333 cm 2 (1.553 − 1.544)

313

Basic Concept of Optical Activity It is found that when a beam of plane polarised light propagates through certain substances or crystals the plane of vibration (or plane of polarisation) of the emergent beam is not the same as that of the incident polarised beam but has been rotated through a certain angle. The phenomenon of rotation of the plane of vibration (or plane of polarisation) is called rotatory polarisation and this property of crystals and other substances is called optical activity and the substances which show this property are called optically active substances. The amount of optical rotation depends upon the thickness and density of the crystal or concentration in case of solutions, the temperature and the wavelength of light used. It is found that the action of turning the plane of vibration occurs within the optically active medium and not at surfaces. There are two types of optically active substances: 1.

Right-handed or dextro-rotatory: The substances that rotate the plane of vibration (or plane of polarisation) in the clockwise direction as seen by an observer facing the emergent light is said to be right handed or dextro-rotatory.

2.

Left-handed or laevo-rotatory: The substances that rotate the plane of vibration (or plane of polarisation) in the anticlockwise direction as seen by an observer facing the emergent light is said to be left-handed or laevo-rotatory. It is found that some quartz crystals are dextro-rotatory; while others are laevo-rotatory. These are denoted by d-quartz and l-quartz respectively, according to the sense of rotation which they produce. Fused quartz is an optically inactive substance.

Experimental Demonstration of Rotatory Polarisation The experimental arrangement for the demonstration of rotatory polarisation is depicted in Fig. 40. Here P and A are the two nicols mounted at a small distance apart. If plane polarised light from a polarising nicol P is examined by an observer facing the emergent light through analysing nicol A, it is completely cut off when the two nicols are in crossed positions as shown in Fig. 40(a). Thus no light emerges out of the analyser. If now a thin plate of quartz cut with faces perpendicular to the optic axis is inserted between the crossed nicols [Fig. 40(b)], the field of view appears brighten, that is, some light is now transmitted by the combination.

314 If now the analysing nicol, A, is rotated gradually about the direction of propagation of light, it is found that there is again complete extinction of light (no light emerges out of the analyser) for some different position of the analysing nicol. This indicates that it is still plane polarised but the plane of vibration of the incident polarised light has been rotated in passage through the quartz by an angle through which A has been rotated, to restore the condition of complete extinction of light. This property of rotation of plane of vibration (or plane of polarisation) of plane polarised light is not only possessed by quartz, but also by all organic compounds whose molecules are asymmetric, by crystals like sodium chlorate, cinna bar, sugar crystals and solutions like turpentine, sugar solution, quinine sulphate solution etc.

Specific Rotation The optical activity of a substance or optical rotation is measured by its specific rotation or specific rotatory power. The specific rotation of an optically active substance at a given temperature for a given wavelength of light is defined as the rotation (in degrees) produced by a path of one decimeter length in a substance of unit density. If θ is the rotation produced by l decimetre length of an optically active substance, the concentration of its solution is C gm/cc, then the specific rotation or specific rotatory power α tλ at a given temperature t for a given wavelength of light λ is expressed by αt = λ

10 θ θ or α t = (If l in cm) λ lC lC

The unit of specific rotation is deg. (decimetre) −1 (gm / cc) −1. The molecular rotation is given by the product of the specific rotation (α t ) and molecular weight of the λ

substance.

Polarimeters A device designed for the accurate measurement of the angle of rotation of the plane of polarisation of a plane polarised light by an optically active medium is said to be polarimeter. By measuring this angle specific rotation can be evaluated provided the concentration and path length are known. This measurement can be used to identify a substance and to study the structure of the molecules in chemistry. A polarimeter calibrated to read directly the percentage of cane-sugar in the solution is called saccharimeter. For medical purposes the concentration of sugar in the urine is determined in connection with the onset of diabetes. In the commercial field, the adulteration and general quality of samples of sugar can be tested. There are the following two types of polarimeters in common use: 1. Laurent's half-shade polarimeter.

2. Bi-quartz polarimeter.

315

Laurent’s Half Shade Polarimeter Apparatus Fig. 41 depicts the schematic optical arrangement of a Laurent’s half shade polarimeter. It consists of two separate nicols P (called polariser) and A (called analyser) mounted in brass tubes placed some distance apart and capable of rotation about a common axis. A glass tube T having a larger diameter in middle contains the active solution under examination. Two ends of the tube are covered by flat and parallel glass plates. It is mounted between the polariser P and analyser A on a rigid iron base. Monochromatic light of wavelength λ from a source S, rendered parallel by a convex lens L, falls on a nicol prism P. After passing through P the light becomes plane polarised. This plane polarised light now passes through a half shade device H and then through the solution whose specific rotation is to be determined and filled in the tube T. The transmitted light passes through analysing nicol A which can be rotated about the direction of propagation of light. The emergent light from nicol A is viewed through a Galilean Telescope G by an observer facing the emergent light.

Action of Laurent's half shade plate The Laurent system consists of a half shade plate in two halves, one of quartz cut parallel to its optic axis, and the other a matching plate of glass [Fig. 42(a)], so chosen as to absorb and reflect the same amount of light as the quartz plate. The quartz is a half wave plate, it introduces a path difference of λ /2 (or a phase difference of π) between the ordinary and extraordinary rays in transmission normal through it. Let the plane of vibration of the plane polarised light from the polariser P falls normally on half shade plate along CP [Fig. 42(b)]. The light passing through the glass plate remains uneffected, while that falling on the quartz plate is broken up into two components −one E-component, CX parallel to optic axis XY and the other O-component perpendicular to the optic axis, that is, along CB. As in quartz, O-component travels faster hence on emergence, O-component will gain a phase of π over the E-component. Hence on emergence from the quartz plate, O-component has vibrations along CD and E-component has vibrations still along CX. Therefore the emergent wave CQ is the resultant of vibrations along CD and CX . Here ∠ PCX = ∠ QCX = θ . Thus the angle between vibration planes of light emerging from quartz, CQ and that of light emerging from glass, CP is 2 θ .

316 Thus, there are two plane polarised beams–one emerging from glass with vibrations in the plane CP while other emerging from quartz

Right half or Glass half

with vibrations in the plane CQ. If the principal plane of the analysing nicol A is parallel to QCQ′ , the light from the quartz plate will pass unobstructed while light from glass will be partly obstructed. Thus the quartz half will be brighter than the glass half

(a)

Left half or Quartz half

[Fig. 43(a)]. If the principal plane of the analysing nicol A is parallel

(b)

(c)

Fig. 43

to PCP′, then the light from glass plate will pass unobstructed while the light from quartz plate will be partly obstructed. Thus right half

will appear brighter as compared to the left half [Fig. 43(b)]. But when the principal plane of the analysing Nicol A is parallel to the optic axis XCY , the two halves appear equally illuminated [Fig. 43(c)].

To find the specific rotation of cane sugar or of an optically active substance, the glass tube T is first filled with clean water and the analyser A is set in the position of equal brightness of the two halves of the field of view (Fig. 41). The reading of the verniers are noted. Now the sugar solution of known concentration is filled in the tube and tube is placed again in the same place. Due to the optical activity of the solution, the vibrations on passing from the quartz half and the glass half is rotated, therefore, on the introduction of the tube containing the sugar solution the

Angle of rotation

Determination of Specific Rotation of Sugar Solution Y

θ O

C

X

(Concentration)

Fig. 44

field of view is not appear equally bright. The analyser is rotated in the clockwise direction with respect to an observer facing the emergent light and is brought to a position so that the whole field of view is again appear equally bright. The new positions of verniers on the circular scale are noted. The difference in the two readings of the analyser gives the angle of rotation θ produced by the solution. In actual experiment, the angles of rotation for the solutions of various concentrations are measured. A graph is plotted (Fig. 44) between concentration C and the angle of rotation θ and the ratio θ /C is determined. The specific rotation of the cane sugar is then determined from the following relation α tλ =

10 θ lC

where l is the length of the tube in cm, C the concentration of the solution in gm/ c. c. and θ is the angle of rotation in degrees.

317

Biquartz Polarimeter A simple and accurate device for the measurement of optical rotation of an optically active substance is biquartz polarimeter. It consists of two semi–circular plates of quartz fitted together to make a disc (Fig. 45), one piece of the disc produces right handed rotation and the other left handed. This biquartz plate is placed just in front of the polarising nicol as shown in Fig. 45. (like Laurent's half shade plate). When plane polarised white light is passed through each half of this plate, its different components will be rotated through different angles by each half but in the opposite sense. Each rotates the plane of polarisation of yellow light through 90° thus, rotatory dispersion occurs in each plate. The left half plate rotates the plane of vibrations in anticlockwise direction and right half in the clockwise direction with respect to an observer facing the emergent light. The red rays are rotated least; while the violet rays are most. The intermediate rays are rotated through angles lying between least of red and maximum of violet. For yellow ray the rotation is 90°. If the principal plane of the analysing nicol be parallel to XX′ , the yellow light will be completely quinched and the other colours will be present in the same proportion in each half. In this position the field of view appears grey (blue + red). This is called the tint of passage or transition tint. A slight rotation of the analysing nicol to one side from this position will change the colour of one half blue, while that of other half red. If the analysing nicol is rotated to other side with respect to an observer facing the emergent light, the colours of two halves are interchanged.

Determination of Specific Rotation To determine the specific rotation of an optically active substance, the glass tube is first field with clean water and the analyser is set in the position tint of passage. The reading of verniers are noted. Now the solution of an optically active substance of known concentration is filled in the tube and analyser is rotated to obtain the position of tint of passage again. The new positions of verniers on the circular scale are noted. The difference between the old and new positions of the analyser gives the angle of rotation produced by the solution. In actul experiment, the angles of rotation for the solutions of various concentrations are measured, A grapoh s plotted between the concentration C and the angle of rotation θ and the ratio θ /C is determined. The specific rotation of sugar is then calculated from the following relation α tλ =

10 θ lC

where l is the length of the tube in cm, C the concentration of the solution in gm / c. c. and θ is the rotation in degrees.

318

Strength of Sugar Solution To determine the strength of sugar solution, the experiment is performed in the similar way as it was performed for the determination of specific rotation, that is, θ and l are found by experiment. The specific rotation at the temperature of a given solution is noted from the table of constants (Given in the Engineering Physics Practical of the same author). The concentration or strength of the solution is then calculated by using the formula C=

θ αl

Example 36: A 20 cm long tube containing sugar solution is placed between crossed nicols and is illuminated by light of wavelength 6 × 10 − 5 cm. If the specific rotation is 60 ° and optical rotation produced is 12° , what is the strength of the solution ? Solution: The specific rotation of a solution is given by α=

θ l×C

where θ is the rotation, C the concentration in gm / cm 3 and l is its length in decimetres. Given α = 60 ° , l = 20 cm = 2.0 decimetres, θ = 12 ° and λ = 6 × 10 − 5 cm ∴

C=

q 12 ° 1 = = = 0.1 gcm − l × α 2 × 60 ° 10

3

Therefore, it is a 10% solution of sugar, that is 1 gm of sugar is dissolved in 10 cm 3 of solution. Example 37: A 20 cm long tube containing 48 cm3 of sugar solution rotates the plane of polarisation by 11° . If the specific rotation of sugar is 66° , calculate the mass of sugar in the solution. [UPTU, B.Tech. I Sem. 2005, II Sem. (C.O.) 2004]

Solution: The specific rotation, α of a sugar is given by, α=

θ where θ is the rotation, C the concentration in gm / cm 3 and l the length in decimetres. l. c,

Here, α = 66 ° , l = 20 cm = 2 .0 decimetres, and θ = 11° ∴

C=

θ 11 1 = = gm / cm 3 l. α 2 × 66 12

1 cm 3 of sugar solution contains = 1/12 gm of sugar ∴

48 cm 3 of sugar solution contain = (1/12) × 48 = 4 gm

Thus, the mass of sugar in solution is 4 gm

319 Example 38: Calculate the specific rotation, which rotates the plane of polarisation 15 . 2 ° in 20% sugar solution of 25 cm length.

[UPTU, B.Tech. II Sem. 2006]

Solution: The specific rotation α of a sugar solution of concentration C is given by α=

θ l×C

Here θ = 15 . 2 ° , l = 25 cm = 2 . 5 decimetre and C = 20% = α=

∴

20 = 0 . 2 gm /cc 100

15 . 2 = 30.4 ° (dm)− 1(gm / cc)−1 2 .5 × 0 .2

Example 39: Determine the specific rotation of the given sample of the sugar solution if the plane of polarisation is turned through 13 . 2 ° . The length of the tube containing 10% sugar solution is 20 cm. [GBTU, B.Tech. I Sem. (C.O.) 2010, UPTU, B.Tech. I Sem. (C.O.) 09; II Sem. 2008]

Solution: The specific rotation α of a sugar solution of concentration C is given by α=

Here

θ l×C

θ = 13 . 2 ° , l = 20 cm = 2 .0 decimetre and C = 10% =

∴

α=

10 = 0 .1 gm /cc 100

13 . 2 = 66 ° (dm)−1 (gm / cc)−1 2 . 0 × 0 .1

Example 40: The plane of polarisation of plane polarised light is rotated through 6 . 5 ° in passing through a length of 2 .0 decimetre of sugar solution of 5% concentration. Calculate the specific rotation of the sugar solution. Solution: The specific rotation α at a given temperature t and for a given wavelength of light λ is given θ by α = , where θ is the optical rotation in degrees, l the length of the tube in decimetre and C the l×C strength of the solution in gm /cc Here θ = 6 .5 ° , l = 2 .0 decimetre and C = 5% = 5 /100 = 0 .05 gm / cc ∴

α=

6 .5 = 65 ° (dm)−1 (gm /cc)−1 2 × 0 .05

Example 41: If 1 mm of the quartz rotates the plane of polarisation of the light of wavelength λ by 18°, find the thickness of quartz which if placed between two parallel nicols then no light of this wavelength λ transmits through the system.

320 Solution: When two nicols are parallel, then light transmitted by it is maximum. For no light to be transmitted through second nicol, the thickness of quartz introduced between the two nicols must be such that it rotates the plane of polarisation of incident light (falling on it after passing through first nicol) by 90° because, then it will fall on second nicol at 90° and will be extinguished. As in the given problem, 1 mm thick quartz produce a rotation of 18° and rotation is proportional to the thickness of 90 ° quartz, hence the thickness of the required quartz = = 5 mm 18 ° Example 42: 80 gm of impure sugar is dissolved in a litre of water. The solution gives an optical rotation of 9 .9 ° when placed in a tube of length 20 cm. If the specific rotation of pure sugar solution is 66° dm−1 ( gm / cc )−1, find the percentage purity of the sugar sample. [UPTU, B.Tech. II Sem. (Old) 09; II Sem. 01]

Solution: The specific rotation α of a solution of concentration C is given by α=

θ , where θ is the optical rotation and l is the length of the tube in decimeters. l×C

or

C=

θ l ×α

Here θ = 9 .9 ° , α = 66 ° dm − 1 (gm / cc)− 1 and l = 20 cm = 2 .0 decimeter C=

∴

9 .9 ° = 0.075 gm / cc = 75 gm / litre 2 .0 × 66 °

In one litre of water 80 gm of impure sugar is dissolved which contains 75 gm of pure sugar. Therefore, the percentage (%) purity of the sugar sample =

75 × 100 = 93.75% 80

Example 43: A sugar solution in a tube of length 20 cm produces an optical rotation of 13°. The solution is diluted to one-fourth of its previous concentration. Find the optical rotation produced by 30 cm long tube containing the dilute solution. [UPTU, B.Tech. I Sem. (C.O.) 2007] Solution: The specific rotation,

α=

θ lC

...(1)

where θ is optical rotation, C the concentration and l the length of solution in decimeter. If θ′ is the optical rotation produced by the same diluted sugar solution of length l′ . When its concentration changes to C′ . Then α= From eqns. (1) and (2), we have

θ θ′ = l C l′ C′

θ′ l′ C′ or θ′ =

...(2) θ l′ C′ lC

321 Here l = 20 cm = 2 .0 decimeter, θ = 13 ° , C′ = C /4, l′ = 3 .0 decimeter ∴

θ′ =

13 × 3 .0 × (C /4) 13 × 3 .0 = = 4 .87 ° 2 .0 × C 2 .0 × 4

Example 44: Calculate the specific rotation if the plane of polarisation is turned through 26.4°, traversing 20 cm length of 20 percent sugar solution. [GBTU, B.Tech I Sem. 2010] Solution: If the plane of polarisation is turned through θ; the specific rotation α is given by, α=

θ lC

Here θ = 26 .4 °, l = 20 cm = 2 decimeter and C = ∴

α=

20 100

26 .4 × 100 = 56.0 ° (dm)−1 (gm / cc)−1 2 .0 × 20

Example 45: A sugar solution in a tube of length 20 cm produces optical rotation of 13°. The solution is then diluted to one-third of its previous concentration. Find optical rotation produced by 30 cm long tube containing the diluted solution.

[UPTU, B.Tech II Sem., 2006, I Sem. (C.O.) 04]

Solution: The specific rotation α is given by the relation α =

θ , where θ is the optical rotation, C the l×C

concentration and l is the length of the tube in decimetre. If θ′ is the optical rotation produced by solution of length l′ decimetre when its concentration is changed θ′ to C′ , Then α= l′ × C′ Thus

α=

l′ × C′ × θ θ θ′ or θ′ = = l × C l′ × C′ l×C

Given l = 20 cm = 2 .0 decimetre and θ = 13 ° , C′ = C /3, where C is the previous concentration of the solution θ′ = ?, l′ = 30 cm = 3 .0 decimetre. ∴

θ′ =

3 .0 × (C /3) × 13 ° 3 × 13 = = 6.5 ° 2×C 2 ×3

Example 46: Calculate the specific rotation of sugar solution from the following data: Length of the tube containing the solution = 22 cm, volume of the solution = 88 cc, amount of sugar in the solution = 6 gm, and angle of rotation 9 °54′ . Solution: The specific rotation of sugar solution is given by θ α= lC

[UPTU, B.Tech. I Sem. 2007]

322 Here l = 22 cm = 2 . 2 decimeter, C = ∴

6 gm 88 cm 3 α=

and θ = 9 °54′ = 9 .9 ° 9 . 9 × 88 = 66 ° 2 .2 × 6

Example 47: A certain length of 5% solution causes the optical rotation of 20 °. How much length of 10% solution of the same substance will cause 35° rotation. [UPTU, B.Tech. I Sem. 2006] Solution: If l2 length of the solution of concentration C2 of the same substance causes θ2 rotation, then from the formula of specific rotation. S=

θ1 θ = 2 l1 C1 l2 C2

or l2 =

θ2 l1 C1 θ 1 C2

Here, θ2 = 35 ° , l1 = l, C1 = 5%, θ1 = 20 ° and C2 = 10% ∴

l2 =

35 ° × l × 5% 7 = l 20 ° × 10% 8

Example 48: A 5% solution of cane sugar placed in a tube of length 40 cm, causes the optical rotation of 20 ° . How much length of 10% solution of the same substance will cause 35° rotation ? [UPTU, B. Tech. II Sem. 2007]

Solution: If l2 length of the solution of concentration C2 of the same substance causes θ2 rotations, then specific rotations is S=

θ1 θ = 2 l1 C1 l2 C2

or

l2 =

θ2 l1 C1 θ1 C2

Here θ2 = 35%, l1 = 40, C1 = 5%, θ 1 = 20 ° and C2 = 10% ∴

l2 =

35 ° × 40 × 5% 7 = × 40 = 35 cm 20 ° × 10% 8 mmm

323

U nit-IV

S ection

B L aser

Introduction

T

h e light emitted from an ordinary light source is incoherent because the radiation emitted from different atoms has no definite phase relationship with each other. In interference we defined

coherence between two sources of light as the existence of constant phase relation between them. The two independent sources cannot be coherent. For experimental purposes, two virtual coherent sources are obtained from a single parent source. In recent years, some sources are developed which are highly coherent, that is, the light emitted from them has a very high degree of coherence and is almost perfectly parallel. These coherent sources are called Lasers. Because of the high degree of coherence associated with a laser beam, it finds many important applications in science, industry, engineering, technology, medicine, holography, spectroscopy, fibre optical communication etc. The word 'LASER' is an acronym for Light Amplification by Stimulated Emission of Radiation. It is a process by which we get a strong, intense, monochromatic, collimated, unidirectional and highly coherent beam of light. It is also used for the light beam produced by this process. The most significant feature of light emitted by a laser is that the microscopic mechanism of emission in effect synchronizes the emission of many atoms, so that the emitted light maintains its definite phase relationship for time intervals much longer than that corresponding to the emission of a single atom. The principle involved in the laser is the phenomenon of stimulated emission which was predicted by Einstein in 1917. When an atom is in the excited state it can make a transition to a lower energy state through the emission of electromagnetic radiation, the emission can occur in two different ways. (1) The first is termed as Spontaneous Emission in which an atom in the excited state emits radiation even in the absence of any incident radiation. The rate of spontaneous emission is proportional to the number of atoms in the excited state. (2) The second is

324 termed as Stimulated Emission in which an incident signal of appropriate frequency triggers an atom in an excited state to emit radiation. The rate of stimulated radiation emission depends on the intensity of the external field and also on the number of atoms in the upper state. When stimulated emission occurs below the infrared region (frequencies below 1011c /s) of the electromagnetic spectrum, the term Maser will be employed and when stimulated emission occurs in the infrared, visible and ultraviolet regions (frequencies above 1011c /s) of the spectrum the term Laser is used. The phenomenon of stimulated emission was first used by Townes, in 1954, in the development of microwave amplifier using ammonia called MASER.Which is an acronym for Microwave Amplification by Stimulated Emission of Radiation. In 1958 Schawlow and Townes showed that the maser principle could be extended into the visible region and in 1960, Maiman built the first laser using ruby as the active medium. Since then, laser action has been obtained with atoms, ions and molecules in various materials or solids including liquids, glasses, flames, plastics, ionized gases, dyes, semiconductors at wavelength ranging from the ultraviolet to radio frequency regions, with power ranging from a few milliwatts to megawatts. Some lasers emit only in pulses and other emits a continuous wave.

Spontaneous and Stimulated (Induced) Emission of Radiation An atom may undergo a transition between two energy states E1 and E2 if it emits or absorbs a photon of the appropriate energy given by the relation E1 − E2 = ± hυ, where the plus sign indicates absorption of quantum of energy and the minus that of emission. Consider a sample of large number of free atoms some of which are in the ground state with energy E1 and some in the excited state with energy E2 . If photons of energy hυ = E2 − E1 are incident on the sample, basically three transition process can take place: absorption of radiation, spontaneous emission and stimulated emission of radiation.

Absorption of Radiation When the photon of light having energy hυ = ( E2 − E1) is incident on an atom in the lower energy state, the atom in the ground state E1may absorb the photon and jump to higher energy state or excited state E2 [Fig. 1(a)]. This process is called stimulated or induced absorption of photon. In fact the incident photon has stimulated the atom to absorb the energy.

325

Spontaneously Emission of Radiation Let us suppose that the atom is in excited state E2 . If we leave the atom as such there it automatically decays to the ground state by emitting a photon of energy hυ = ( E2 − E1).This process is called spontaneous emission and has been shown in [Fig. 1(b)]. Usually an atom in excited state can stay for about 10 −8 sec. The spontaneous emission has following characteristics: 1.

The emitted photon of energy hυ can move in any random direction.

2.

There is no phase relationship between the photons emitted from various atoms of the system.

3.

The rate of fall of electrons from excited state E2 to the ground state E1, at every instant, is proportional to the number of electrons left in the excited state E2 .

4.

The transition probabilities depend only on two energy states. Hence, the radiation coming out in spontaneous emission are incoherent.

Stimulated or induced Emission of Radiation Suppose the atom is in excited energy state E2 and a photon of energy exactly equal to E2 − E1 is incident on it [Fig. 1(c)]. The incident photon interacts with the atom and then it induces (or stimulates) the atom to come down to the ground energy state E1by emitting a new photon. Thus, when an atom ejects a photon due to its interaction with a photon incident on it, the process is called stimulated or induced emission. The emitted photon has exactly the same phase, direction and energy as the incident photon. The stimulated or induced emission has following characteristics: 1.

For each incident photon, there are two out going photons moving in the same direction.

2.

The direction of emitted photon is same as the direction of incident photon. Since the emitted photon has exactly the same energy, phase and direction as the incident photon, we can achieve an amplified and unidirectional coherent beam.

3.

The rate of stimulated emission or rate of transition to lower energy state is directly proportional to the number of atoms left in the excited state E2 and the energy density of the incident radiation.

Difference between Spontaneous and Stimulated Emissions Spontaneous Emission

Stimulated Emission

1. In this emission process the photons emitted 1. In this type of emission the emitted photons from various atoms have no phase relationship have same frequency and are in phase with between them. incident photons. 2. Achieved radiations are incoherence.

2. Achieved radiations are coherent and are unidirectional.

3. Emitted photon can move in any directions.

3. For every incident photon, there are two outgoing photons moving in the same direction.

4. In this emission process the rate of transition 4. In this process, the rate of emission is process, the rate of transition from excited state proportional to the number of atoms in the excited state and the energy density of the E2 to lower energy state E1 is proportional to the number of excited electrons remaining into incident radiation. excited state E2 . 5. Spontaneous emission disfavours laser action.

5.

Stimulated emission favours laser action.

326

Spontaneous emission and Einstein’s Coefficient of Spontaneous Emission

energy E1 and state 2 energy E2 respectively as shown in Fig. 2. An atom in the lower energy state can absorb radiation and get excited to the higher energy level E2 . The rate of absorption depends on the density of radiation at the particular frequency corresponding to the energy separation of the two levels, that is υ=

E1

Emission

having two states 1 and 2. Let N1 and N2 be the number of atoms in state 1 of

N2

E2

Absorption

An atom has a number of possible quantised energy states. Consider an atom

E2 > E1

N1

Fig. 2

( E2 − E1) h

The absorption process depends on the energy density of radiation at the frequency υ, this energy density is denoted by u(υ) and is defined such that u(υ) dυ represents the radiation energy per unit volume within the frequency interval υ and υ + dυ. The rate of absorption or rate of transition 1 → 2 is proportional to N1 and also to u(υ). Thus, the number of absorptions per unit time per unit volume can be written as P12 = N1B12 u(υ) the proportionality constant B12 is known as Einstein's coefficient of absorption of radiation and depends on the properties of states 1 and 2. Let us now consider the reverse process in which the emission of radiation takes place at a frequency υ when the atom de–excites from the energy level E2 to E1. During this process the atom de–excites either through spontaneous emission or through stimulated emission. In spontaneous emission, the atom, of its own accord, jumps to the lower energy state 1, emitting a photon of frequency υ. The probability of spontaneous emission 2 → 1 is independent of the energy density of the radiation field and depends only on the properties of states involved in the transition. Einstein denoted this probability per unit time by (P21)spontaneous = N2 A21 the probability constant A21 is known as Einstein's coefficient of spontaneous emission of radiation.

Stimulated emission and Einstein’s Coefficient of Stimulated emission In the case of stimulated emission, an atom in an excited energy state, under the influence of electromagnetic field of incident photon, jumps to a lower energy state, emitting an additional photon of same frequency (Fig. 2). Thus now two photons, one original and other additional emitted photon move in the same direction. The probability of stimulated emission transition is directly proportional to the number of atoms in the upper energy level as well as to the energy density of radiation and is written as (P21)stimulated = N2 B21u (υ) where B21 is the Einstein's coefficient of stimulated emission of radiation. The total emission probability or the probability of transition 2 → 1 is the sum of spontaneous and stimulated emission probabilities, that is, (P21)spontaneous + (P21)stimulated = N2 P21 = N2 [ A21 + B21u (υ)]

327

Transition Probabilities There are three transition probabilities associated with any pair of energy levels. 1. Spontaneous emission probability, 2. Absorption probability and 3. Stimulated–emission probability. The spontaneous–emission probability is the chance per unit time that an atom at the higher energy level will spontaneously emits a photon and jump to the lower energy level. The absorption probability is the chance that an atom at the lower energy level illuminated by photons of appropriate energy and intensity will absorb one of the photons and jump to the higher energy level. If an atom at the excited energy level is illuminated by photons of an appropriate energy, it may be caused to emit a photon identical to one already present and jump down to the ground energy level. The chance that it is does so is called the stimulated–emission probability. A relation between transition probabilities for spontaneous and stimulated emission or radiation may be obtained as follows:

Einstein's Relation between Spontaneous and Stimulated Emissions or Transition Probabilities Consider an assembly of atoms in thermal equilibrium at temperature T with radiation of frequency υ and energy density u(υ) . Let N1 and N2 be the number of atoms at any instant in the states 1 and 2 respectively. The probability of absorption that the number of atoms in state 1 that absorb a photon and rise to state 2 per unit time is given by N1P12 = N1B12 u(υ)

...(1)

The total probability of emission that the number of atoms in higher state 2 to drop to lower state 1, either spontaneously or under stimulation by, emitting a photon, per unit time is the sum of two probabilities, that is N2 P 21 = N2 [ A 21 + B 21 u(υ)]

...(2)

At thermal equilibrium, the absorption probability is equal to the total emission probability, that is N1P12 = N2 P 21 or or

N1B12 u(υ) = N2 [ A 21 + B 21 u (υ)] u (υ) =

A 21 (N1 /N2 ) B12 − B 21

u (υ) =

 A21  1   B 21  (N1 /N2 ) (B12 /B 21) − 1

...(3)

According to Einstein, the probability of stimulated absorption is equal to the probability of stimulated emission, that is B12 = B 21 Thus, from eqn. (3)

328   1   ( N / N ) − 1  1 2 

A  u (υ) =  21   B21 

From Boltzmann's law, the ratio of the populations of two levels at temperature T is expressed as N1 = e(E2 N2

− E1 ) / kT

or

N1 = e(hυ / kT ) N2

where k is the Boltzmann's constant and h is the Planck's constant. Therefore,

A  u (υ) =  21   B21 

  1    e(hυ / kT ) − 1  

...(4)

Now, according to Planck's law the energy density of radiation u(υ) is given by u (υ) =

 8πhυ3  1    3 (hυ / kT ) c − 1 e

...(5)

where c is the velocity of light Comparing eqns. (4) and (5), we get A 21 8 π h υ3 = B21 c3

...(6)

From eqn. (6) it is clear that the ratio between spontaneous emission coefficient and stimulated emission coefficient is proportional to υ3 . It means that at thermal equilibrium, the probability of spontaneous emission increases rapidly with the energy difference between two states.

Condition Necessary to Achieve Laser Action The transitions between two energy levels in an atom can occur by induced absorption, spontaneous emission and stimulated emission. To achieve laser action following condition must be satisfy. 1.

The rate of emission must be greater than the rate of absorption: Under normal condition, the number of atoms in the upper energy level is always smaller than that in lower energy level. If by some means, the number of atoms in higher energy state is made greater than that in lower energy state, the emission rate will become greater than the absorption rate.There are several mechanism to invert the normal population of energy levels the population inversion can be achieved by the continuous pumping of atoms in higher energy state.

2.

The probability of spontaneous emission (which produces incoherent radiation) must be negligible in comparison to the probability of stimulated emission (which causes coherent radiation): The condition can be achieved by taking working substance (active medium) such that its atoms have metastable states which have a lifetime 10 −3 sec or more instead of the usual 10 −8 sec. If certain atoms are excited to metastable state the probability of spontaneous emission will be quite negligible.

329 3.

The coherent beam of light must be sufficiently amplified: For this, active medium must be placed between two reflecting mirrors. The one of the mirror is fully reflecting while the other is partially transmitting. The photons emitted by stimulated emission are reflected back and forth in the laser medium by these mirrors so that they confined within the system long enough to allow them to stimulate further emission from other excited state. In this way reflected photons, travelling through the medium stimulate further emission as well as produce amplified coherent and intense beam of light which is emitted by partially transmitting mirror.

Various Pumping Methods The raising of atoms from the lower energy level to the upper one involves a large number of energy levels with complex excitation process. Several techniques of excitation are used in laser devices such as; excitation by a strong source of light, say flash lamp or arc lamp (optical pumping), excitation by electron impact (electrical pumping), excitation by chemical reaction (chemical pumping) etc. The commonly used pumping methods are as follows:

1. Optical Pumping In this most suitable pumping technique, the atoms in the ground state are excited to higher energy state by means of optical photons. The energy is supplied continuously in the form of short flashes of light. The excited atoms from the upper most level (in which the atom can reside only for about 10 −8 sec) go to the metastable state ( a state in which atom can reside for about 10 −3 sec, much longer than short lived uppermost level) to create a state of population inversion. In this case the frequency of pumping photons must be higher than emitted photons so that the atoms can be pumped to the higher energy level from the lower energy level. Optical pumping is suitable for those media which are transparent to light. Maiman employed optical pumping method for the production of laser beam in Ruby laser.

2. Electrical Pumping Electrical pumping is employed in gaseous medium where electricity can be conducted without effecting laser activity. A strong electric field is applied to accelerate electrons emitted by cathode of the tube. During the motion towards the anode the accelerated electrons collide with the atoms of the medium and give up their energy to excite them to the upper most level. From this level, the atoms jump to metastable state, creating a state of population inversion. Electrical pumping is used for population inversion in Argon–ion laser. Electrical pumping is also used in case of semiconductor laser medium. In this case instead of atoms, the charge carriers (electrons and holes) are excited and a state of population inversion is achieved in the junction region. In the junction region electrons combine with holes and produce laser beam. In this way in a semi–conductor laser, electrical energy is directly converted into light energy.

3. Inelastic atom–atom Collisions In this method of pumping, the accelerated electrons produced by electric discharge of the gaseous medium (a mixture of two gases) collide with the atoms of one kind of gas (which are responsible for pumping). The excitation energy of these atoms are readily transferred to the atoms of other kind of gas (which are responsible for laser transition) in their inelastic collisions with them. In this way population inversion is achieved by inelastic atom–atom collisions in Helium–Neon gas laser. In He–Ne laser, helium is the pumping medium and neon is the leasing medium.

330

4. Chemical Pumping In chemical pumping the energy necessary for pumping is generated by a chemical reaction. As an example, the heat energy evolved, when hydrogen combines with fluorine to form hydrogen fluoride, is used for pumping the atoms in a CO2 laser.

Metastable States When an atom gets sufficient energy, by any means, its electron jumps from inner to outer orbits. This state of atom is called excited state. The atom remains in an excited state for a period of 10 −8 sec. After that short period it comes back to the ground state by releasing excess energy spontaneously. For stimulated emission, the atom should remain for a longer time. As the atoms are continuously going to excited state by pumping process they should remain in the higher energy state until the population in the higher state (N2 ) becomes greater than that in lower state (N1) , that is, N2 > N1. A long lived energy state (~10 −3 sec ) from where the excited atom do not return to the lower level instantaneously is called metastable state. In fact metastable states are those energy levels of the atoms from which the transitions to the ground states are not allowed by selection rules. However, the atom can fall from the metastable state to the ground state either by giving up the appropriate amount of energy to another atom during a collision process or it may absorb radiation and go to a higher energy state which is allowed by selection rule and from there it may return to normal state by emission of radiation. Thus, if certain atoms are excited to the metastable state, the probability of spontaneous emission will be quite negligible.

Principle of Laser Action [Active System, Population Inversion (Inverted Population) and Pumping] To understand the laser action, we must understand such terms as population inversion, pumping and active system. These are also called main components of laser. A system in which population inversion is to be achieved is called an active system (for laser). The method of raising the molecules or atoms from lower energy state to the higher energy state is called pumping. In a system of atoms in thermal equilibrium, the number of atoms in the ground state is much greater than that in the higher energy state. This is called a normal population of atom among the available energy states. A state in which the number of atoms in higher energy state is greater than that in the lower energy state is called a state of population inversion. To maintain population in the higher energy levels, energy is to be supplied continuously to the medium. Now we shall discuss active system, population inversion, and most commonly used optical pumping in detail, as follows:

Active System and Cavity Resonator The active medium or working substance in a resonance cavity consists of a collection of atoms, molecules, or ions (in solid, liquid or gaseous form), and is able to amplify light waves. The working substance must have metastable states (life time ~10 −3 sec) so that the population inversion may achieve. In laser, active medium or system is placed in a specially designed cylindrical tube, called resonance cavity. The cavity resonator consists of a pair of mirrors which are facing each other. The

331 mirrors could be a plane or curved. One of the mirror is made fully reflecting while the other is partially transmitting so that an intense beam can emerge out of it. This cavity resonator makes the energy density large, whence stimulated emission dominates over spontaneous emission. The multiple reflection also makes the stimulated emission more coherent. Hence, the cavity resonator is able to convert the active medium into a light generator.

Population Inversion (or Inverted Population) In a sample of large number of atoms N0 , at temperature T in thermal equilibrium the distribution of atoms in different energy states follows Maxwell–Boltzmann distribution. At an absolute temperature TK the instantaneous population or the number of atoms in energy states E1 and E2 are given by N1 = N0 e − E1 / kT and N2 = N0 e − E2 / kT where k is to the Boltzmann's constant and N1 and N2 be the number of atoms in energy states E1 and E2 respectively. The relative population in two states is N2 /N1 = e −(E2

− E1 ) / kT

...(1)

For an ordinary conventional source of light, E2 > E1 and N1 > N2 , that is, the number of atoms in higher energy state E2 is less than the number of atoms in lower energy state E1. In these circumstances, the probability of induced or stimulated emission is much less than the probability of spontaneous emission. Since in spontaneous emission the emitted photons have random directions and random phase, a visible source of light emits incoherent radiation which is undesirable for laser action. The basic requirement of a laser is to have predominantly induced or stimulated transitions so that the radiation emitted should be in the same direction as the incident radiation and also they must possesses a definite phase relationship. This provides very intense, highly directional, high monochromatic and highly coherent amplified radiation. This can be achieved by population inversion, so that there are more atoms in higher energy state E2 than in the lower energy state E1. This condition is quite unnatural because for any equilibrium states N2 /N1 is less than unit. If by some technique, the number of atoms in the higher energy

hυ hυ

state E2 is made much larger than that in lower energy state E1, the predominantly stimulated emission is promoted. The situation in

hυ

which the number of atoms in the higher energy state is greater than that in the lower energy state is called population inversion. Under this condition the emitted photons which have same frequency and are in same phase interact with other atoms and produce large number of photons of same frequency and phase,

hυ hυ

hυ hυ hυ

hυ hυ hυ

goes on multiplying or amplifying by repeated induced emission. In unidirectional beam of light as shown in Fig. 3.

hυ hυ

hυ

and travel in the same direction. Hence, the number of photons this way one can get a highly intense, monochromatic, coherent and

hυ

Fig. 3

332 Optical Pumping: To maintain the population inversion we need an arrangement which insures the population inversion between E1 and E2 states. The process by which the population

(Higher energy state) E3

inversion is achieved or the process by which

(Metastable state) E2

(~10– 8 sec)

2

atoms are brought from lower energy states to higher energy states and maintained is called

population inversion 1 with ground state

rapidly decaying photon –3 (~10 sec) stimulated emission

3 stimulating pumping. There are various types of pumping (amplification) photon process but the most suitable and commonly used (ground state) E1 is optical pumping in which the atoms in the ground state are excited and raised to the higher Fig. 4 energy level by means of light photons. In optical pumping the population inversion is achieved by three energy levels scheme. To understand its working consider a material whose atom has three energy states and when it rises from ground energy state E1 to short lived excited energy level E3 (in

which the atom can reside only for 10 –8 sec) it does not come down to state E1, but transition from E3 is allowed to an intermediate metastable state with energy E2 (in which atom can reside for about 10 −3 sec, much longer than short lived state E3 ). Thus, when the atoms of the material are excited by a photon of energy hυ = ( E3 − E1), the atoms are excited to the state E3 by stimulated absorption, but the excited atoms do not decay back to E1 but some of the atoms decay spontaneously to metastable state E2 . Since the life time of metastable state E2 is much longer than the state E3 , the atoms reach the state E2 much faster than they leave state E2 . In this way the metastable state with energy E2 is all the time have larger number of atoms than the number in the lower energy state (Fig. 4) which means that population inversion is achieved. Now, if the photon of energy hυ = ( E2 − E1) is incident on the atoms of the metastable state stimulated emission will result.

Types of Lasers Now–a–days a number or kinds of lasers are used for producing light at frequencies from the far infrared to ultraviolet regions such as solid lasers, semiconductor lasers, gas lasers, chemical lasers, dye lasers, etc. Among them gaseous state lasers and semiconductor lasers are most commonly used. Some of the gaseous state lasers and semiconductor lasers are discussed below one by one.

Production of Laser A laser is a device that emits optical radiation in the form of an intense, highly monochromatic, highly directional and highly coherent beam of light. The laser beams can be produced in the laboratory in the following manner:

Construction and Working Principle of Ruby Laser A solid laser can be made by introducing impurity atoms into a crystal. Ruby was the first solid material, which was used in the production of laser and is still widely used. The ruby laser was first developed by Maiman in 1960. It consists of a single crystal of a pink ruby (Al2O3 ) doped with 0.05 per cent Cr3 + ions. The crystal is in the form of a cylindrical rod whose opposite

333 ends are ground flat and parallel, (one end is fully silvered and the other is partially silvered) so as to reflect a part of the light Ruby Glass Xenon flash tube incident normally on them and transmit rest part of it, as shown Rod tube in Fig. 5. The ruby rod is surrounded by a helical xenon flash tube which provides the pumping light to raise the chromium ions to upper energy level. In the xenon flash tube each flash lasts several milli seconds and in each flash a few thousand joules The Power source of energy is consumed. Only a small part of this energy is used in Laser Partially reflecting beam pumping the Cr3+ ions while the rest is wasted by heating up the end faces apparatus. For this purpose the cooling by water circulation in a Fully reflecting glass tube is provided (Fig. 5). In ruby laser the cromium atoms end face are the active atoms, which absorb energy in broad bands in green and yellow light. The solid state ruby laser (also called Cooling Cooling pulse laser) is a three level laser system. A simplified energy level Fig. 5 diagram of chromium ions in a ruby laser, indicating appropriate excitation and decay is shown in Fig. 5. In normal state, most of the chromium ions are in the ground state E1. When the ruby rod is irradiated by a flash of light, the 5500 Å radiation photons are absorbed by the chromium ions which are pumped to the excited state E3 . The excited ions give up part of its energy to the crystal lattice and decay radiationlessly to the metastable state E2 (transition 2). Since the state E2 has a much longer life time (10 − 3 sec), the number of ions in this state goes on increasing while, due to optical pumping the number of chromium ions in the ground state E1 goes on decreasing. Thus the population inversion is achieved between states E2 and E1. Perfect reflector Partial reflector E3 Radiation less Transition

2 E2

6943 Å 5500 Å Pumping 1 transition

Short-lived state

Metastable state

(i) Unexcited state

(iv) Stimulated emission of radiation along axis starts dominating

6943 Å 4

3

6943 Å Laser transition

(ii) Atoms start reaching the excited state

(v) Photons becoming further coherent

Ground state

E1

Laser

Fig. 6 (iii) Population inversion is reached with two coherent (stimulated) emision

(vi) Entire emission becomes coherent

Fig. 7

When an excited ion from the metastable state E2 drop down spontaneously to the ground state E1 (transition 3), it emits a photon of wavelength 6943 Å . This photon travels through the ruby rod and, if it is moving in a direction parallel to the axis of the crystal, is reflected back and forth by the silvered ends until it stimulates other excited ion and causes it to emit a fresh photon in phase with the

334 stimulating photon. Thus, the reflections will amount to additional stimulated emission – the so called amplification of stimulated emission. This stimulated transition 4 is the laser transition. The two photons will knock out two more photons and their total number will be four and so on. The process is repeated again and again, the photons thus multiply. Now the stimulated radiation along the axis starts dominating due to multiple reflections that is the photons travelling parallel to the axis of the rod will start a cascade of photons emission, while the photons which do not move axially escape through the sides of the crystal. The shaping of the beam with time is schematically shown in Fig. 7, it may only be taken as schematic representation, showing the dominant fact at each state. The hollow circles and black dots represent the unexcited and excited atoms respectively. When the photon beam becomes sufficiently intense, then a very powerful and narrow beam of red light of wavelength 6943 Å emerges through the partially silvered end of the crystal. The intensity of stimulated emission in the modern ruby laser is more than 106 Watt.

Construction and Working of Helium-Neon Laser (Gas Laser) To get a continuous and intense beam of laser, gas Fully silvered Power supply lasers are used. The spectral lines in a gas are narrow mirror or partial Laser and well-defined. A simplified diagram showing basic reflector beam features of a gas laser is shown in Fig. 8. The resonant system is a long and thin optical cavity with mirrors at He + Ne each end. Resonant system is also called a discharge tube. Helium-Neon (He-Ne) laser consists of a discharge (quartz) tube containing helium and neon in Helium-Neon discharge the ratio of 7 to 1 at a total pressure of about 1 torr (1 Partially silvered mirror mm of Hg). At one end of the tube, there is a perfect of partial reflector reflector while on the other end there is a partial Fig. 8 reflector. The spacing of the two reflectors is equal to an integral multiple of half wavelength of the laser light. A powerful radio – frequency generator is used to produce a discharge in the gas so that the helium are excited or pumped up to a higher energy level. He–Ne gas laser is a four level laser. The first few energy levels of He and Ne atoms are shown in Fig. 9. 20.61 eV When an electric discharge passes through the gas, the electron in the discharge tube collide with the He and Excitation by collision Ne atoms and excite or pumped them to the metastable with state 20 .61 eV and 20 . 66 eV respectively above the electrons ground state. Some of the excited He atoms transfer their energy to unexcited Ne atoms by collision. Thus, lighter He atoms help in achieving a population inversion in the heavier Ne atoms. When an excited O Helium (He) Ne atom drop down spont aneously from the metastable state at 20 . 66 eV to lower energy state at 18 . 70 eV, it emits a 6328 Å photon in the visible

Collision

Metastable state Laser 20.66 eV 6328 Å 6328 Å transition 18.70 eV

Spontaneous transition

E Radiation less transition Ground state

O Neon (Ne)

Fig. 9

region. This photon travels through the mixture of gas, and if it is moving parallel to the axis of the tube,

335 is reflected back and forth by the reflector-ends until it stimulates an excited Ne atom and causes it to emit a fresh 6328 Å photon in phase with the stimulating photon. The photons emitted spontaneously which do not move parallel to the axis of the tube escape through the side of the tube. The stimulated transition from 20 . 66 eV level to 18 . 70 eV level is the laser transition. The two photons will knock out two more photons and the process is repeated again and again, the photon thus multiply. When this beam becomes sufficiently intense, a portion of it escapes through the partially silvered end. The Ne atoms drop down from the 18 .70 eV level to lower metastable state, E, through spontaneous emission emitting incoherent light. From the level E, the Ne atoms are brought to the ground state through collision with the walls of the tube. Hence the final transition is radiation less. A gas laser has several advantages over the solid state lasers or Ruby laser. An important advantage is that the light is produced as a continuous beam rather than in ultra-short pulses as in the ruby laser. The remarkable spectral purity of the stimulated radiation, the narrow spectral width of the emission, the large power output per unit band width are the other features of the gas lasers. The gas laser beam is highly monochromatic and highly directional. This is because of the fact that in gas lasers the cyrstalline imperfection, thermal distortion and scattering are almost absent like solid state laser. The gas lasers are capable of operating continuously without any need for cooling. Therefore, He-Ne laser is superior than Ruby laser.

Characteristic (or Properties) of Laser Beam The laser beam has following main characteristics: 1.

High monochromaticity, high intensity, high degree of coherence and high directionality of laser radiation is of special significance in advanced research.

2.

A laser beam is very narrow and can travel to long distances without any spread. For a typical laser, the beam spreads less than 0 .01 mm for every meter. The beam spread on the surface of the moon, which is 384, 400 km away, is about a few kilometre.

3.

Output in a laser beam is many millions times more concentrated then the best search-light available.

4.

The laser gives out light into a narrow beam and its energy is concentrated in a extremely small region.

5.

Because of extremely high intensity of laser beam, it can produce temperatures of the order of 104 ° C at a focussed point.

6.

Because of its narrow band width, laser beams can be focussed on a very small area of the order of 10 −6 cm 2 . A good lens with a focal length of 1 cm can focus a laser beam to a spot a thousandth of a centimeter in diameter.

7.

An image of intensity 105 watt /cm 2 can be obtained by a laser of 10 mW power.

8.

Laser beam is extremely bright.

9.

The laser beam is completely coherent spatially as well as temporally. It is possible to observe interference effects from two independent laser beams.

336

Difference between Ordinary Light and Laser Radiation Basically light from a laser and from any ordinary conventional source, such as a flame or an incandescent lamp are both electromagnetic in nature but they differ enormously in coherency, monochromacity, directionality, intensity and other more detailed properties. Here we shall mention some important differences between ordinary light and laser beam on the basis of aforesaid characteristics. 1.

The light emitted by a conventional source is incoherent whereas those emitted by laser has a very high degree of coherence and is almost perfectly parallel.

2.

Spatially coherent beam of light from an ordinary source can be produced by passing it through a pin hole, whereas a laser beam is already highly spatially coherent. The high coherence of laser emission makes it possible to realize a tremendous spatial concentration of light power such as 1013 W in a space with linear dimension of only 1 µm. Radiation of such high intensity is used to cut metal and micro welding.

3.

The coherence time for a laser beam is usually much larger in comparison to ordinary light sources.

4.

The conventional light sources emit radiations in all directions, while the laser emit radiation only along one direction, that is, laser is highly directional.

5.

The light emitted by a laser is vastly more monochromatic than that of any conventional monochromatic source. The light emitted by ordinary source is never perfectly sharp but it spreads over a wide frequency range, whereas light from a laser has a very small spreading.

6.

Light from a lamp spread out more or less uniformly in all direction, whereas laser given out light into a narrow beam and its energy is concentrated in very small regions. For example, if we look at a 100 W lamp filament at a distance of 30 cm, the power entering the eye is less than a thousand of a watt. On the other hand, 1W laser would appear many thousand times more intense than 100W ordinary lamp.

Applications or Uses of Laser Radiations Because of high temporal coherence and large spatial coherence, laser beams have ingeneous applications in various branches of science, spectroscopy, engineering and computer. 1.

Industrial Applications: In industrial and technical fields the laser beam is used for drilling extremely fine holes in diamonds, teeth, paper clips, hard sheets and even in human hairs. It can melt and vaporise even the hardest metal. They are also used for cutting thick sheets of hard metals, welding on a microscopic scale, such as integrated circuits, is taken up by means of laser beams. The laser beam is used to vaporise unwanted material during the manufacture of electronic circuit on semiconductor chips. Laser cutting technology is widely used in the fabrication of space craft. Lasers have been used as light sources for telephoto pictures. Pulsed Q–switched lasers are suitable for technical motion picture photography. The spiky output from a normal ruby laser has been found to be useful in high speed photography.

337 It has been observed that finger-prints can be detected under laser light where the normal method of obtaining finger prints through dusting powder is ineffective. The laser technique can also be used for detection and analysis of older finger prints on, say, documents, currency bills, cloth, etc. This is possible because finger prints contain ultraminute quantities of stable amino acids, which when treated with certain chemicals become fluorescent. 2.

Medical Applications: In medicine, micro-surgery has become possible due to narrow angular spread of the laser beam. The laser beam can be focussed on harmful components to destroy them without seriously damaging the neighbouring regions. It can be used in the treatment of kidney stone, tumour, in cutting and sealing the small blood vessels in brain operations. Preliminary success has also been made to treat the human and animal cancer. The laser beams are frequently used in fibre–optic endoscopy. The time of completion of big surgical operations can be reduced considerably by the use of laser beam. It is also used in the delicate surgery of cornea grafting. Laser radiation is often used in controlling haemorrhage, for the treatment of liver and lungs, and for the elimination of moles and tumours developing on the skin tissues. Lasers are used in ophtalmology (scientific study of the eye and its diseases) to re-attach a detached retina and treating glaucoma (eye disease causing gradual loss of sight). Lasers are now increasingly used in therapy and also in stomatology. They can selectively destroy tissues of tooth affected by ovaries. New type of surgery with ultraviolet excimer laser is now used in the treatment of liver cancer. Laser technology has made many surgical procedures less painful for the patient.

3.

In Optical Communication and Holography: Laser has an important applications in optical communication and holography. Optic communication is a method of transmitting information from one place to another by sending highly intense laser light through an optical fibre. Recent and fast developments in the field of communication and frequent utilization of mobiles, T.V., computers, bar code scanners, automatic teller etc., necessitate such a system in which each information carrier has a capability of carrying simultaneously a large volume of communication traffic. Conventional telecommunication systems based on coaxial cables, radio and microwave links are able to transmit nearly one hundred pulse per second. But with laser through optical fibre it is possible to transmit 44 .7 million pulses per second. Lasers are mostly confined to long distance and single mode fibre applications. Highly intense laser beams are also used to illuminate hologram to reproduce three dimensional image of an object.

4.

In Communication: Narrow angular width or spread of the laser beam makes it a very useful tool for microwave communication. Communication with earth satellites, in rocketry, and in accurate range finders. The earth-moon distance has been measured with the help of lasers. In microwave communication the message in the form of signal is mounted on carrier waves through modulation. The position of a distant object can be determined with the help of range finder. Optical range finders not only afford accurate ranging but also make possible the determination of the size, shape and orientation of distant objects. Laser fluorescensors are used for remote environmental monitoring. The nature of the target is determined from the strength and spectral distribution of the fluorescence signal.

338 5.

Chemical Applications: Lasers have a wide chemical applications. It can initiate certain chemical reactions which could not be possible in the absence of photons. It is used in the laser Raman spectroscopy. (i)

During war, lasers are used in laser rifles, laser pistols and laser bombs which are used in night. Lasers are also used to detect and destroy enemy missiles.

(ii)

Laser torch is used to see objects at a very large distance.

(iii) Laser is also used in holography (three dimensional lensless photography) and non linear optics. (iv) In compact disc (CD) audio systems in place of the phonographic needle a laser beam is used. On a CD which is much smaller than a traditional record, a very large number of songs, speeches, serials, a huge amount of data in a small volume such as book of large volume like dictionaries and encyclopedias are available. Using lasers, video images can also be stored on discs and can be displayed on a T.V. with the help of a laser disc player. Now-a-days this technique is used in computers for data retrieval and storage. Lasers are widely used to send telephone signals over long distances through optical fibres. Laser communication lines based on fibre optics have been extremely useful in the modern computer network and in broad band communications. In fact laser techniques are so fast growing that everyday a new incredible application of laser appears. 6.

In Spectroscopy: The introduction of laser as a source of excitation has revolutionised the field of spectroscopy. The development of tunable dye lasers and the frequency doubling techniques have added a new dimension to the studies in Raman effect. The investigation of Raman scattering with the aid of lasers led the discovery of new Raman scattering phenomena of fundamental interest. The three new phenomena which arises from the non-linear interaction of a system with intense monochromatic radiation are: (i) the stimulated Raman effect, (ii) the hyper-Raman effect and (iii) coherent anti Stoke Raman scattering (CARS). The stimulated Raman effect has been observed in a very large number of Raman laser material, which provides us nearly hundreds of coherent sources of light from ultraviolet to infrared regions. Stimulated Raman scattering has been observed in optical fibre wave guides. The ability of laser beams to concentrate high power density of light at focal point has opened a new vista of micro Raman spectroscopic analysis of those biological and biomedical samples which are available in very small quantities. Coherent Anti-Stoke Raman Scattering is widely used in high resolution spectroscopy. This technique is frequently used as an effective spatial probe of concentration of gases in flames. This technique allows powerful fluorescence-free analysis of biological samples. Saturation absorption spectroscopy is used to study the fine details of the spectra of free atoms and molecules. Laser beams have also been used in the inertial confinement of plasma.

7.

Atmospheric Optics: Laser remote sensing is frequently used for the precise measurement of ozone in the atmosphere. Atmospheric optics uses lasers for the measurement of traces of pollutant gases, temperature, water vapour concentration, sometimes at ranges upto ten miles away. Laser radar (or lidar) has proved to be a powerful tool for the investigation of various atmospheric features. Lidar gives the distribution of atmospheric pollutants in different vertical sections as well as is useful in determining the concentration of matter at different height.

339 8.

Lasers in Astronomy: Radio astronomers have been able to amplify very faint radio signals from space with the uses of lasers. With the help of lasers it is possible to hear the bursts of light and radiation waves from start which emitted them over a millions of years ago.

Example 1: In a Ruby laser, total number of Cr +3 ions is 2 .8 × 1019 . If the laser emits radiation of wavelength 7000 Å . Calculate the energy of the laser pulse.

[UPTU, B. Tech. II Sem. 2006]

Solution: Energy of the laser pulse = total number of ions (n) × energy of one photon or

E = nhυ = n

hc λ

Here h = 6 .62 × 10 − 34 J – s, c = 3 × 108 ms −1, λ = 7000 Å = 7.0 × 10 − 7 m and n = 2.8 × 1019 ∴

E = 2.8 ×1019 ×

or

E = 7.94 J

6.62 ×10 − 34 × 3 ×108 7.0 × 10 −7

J (joule)

Example 2: Calculate the energy and momentum of a photon of a laser beam of wavelength 6328 Å [UPTU, B. Tech. I Sem. 2007]

Solution: The energy of the photon, that is, E = hυ =

hc λ

Here h = 6 .62 × 10 −34 J – s, c = 3 × 108 ms −1 and λ = 6328 Å = 6.328 × 10 −7 m ∴

Energy,

E=

6.62 × 10 −34 × 3 × 108 6 .328 × 10 −7 3 .14 × 10 −19

= 3 .14 × 10 − 19 J

or

E=

The momentum of photon, p =

− 34 h 6 .62 × 10 = = 1.05 × 10 −27 kg ms −1 λ 6 .328 × 10 −7

1.6 × 10 −19

eV = 1.96 eV

Example 3: In a CO2 laser, the energy difference between two levels is 0.121 eV. Calculate the frequency of radiation. Solution: We know that, E1 − E2 = hυ = ∆ E ∴ Here ∴

υ=

∆E h

∆ E = 0 .121 eV = 0 .121 × 1 .6 × 10 −19 J and h = 6.62 ×10 −34 J – s υ=

0 .121 × 1.6 × 10 −19 6 .62 × 10 − 34

= 0.029 × 10 −15 s −1

340

Example 4: Calculate the population ratio of two states in He-Ne laser that produces light of wavelength 6000 Å at 300 K . Solution: We know that population ratio

But,

E2 − E1 = hυ = =

Here

N2 = e − ( E2 N1

− E1 ) / kT

, where k is the Boltzmann’s constant

h c 6 .62 × 10 − 34 × 3 .0 × 108 = = 3 . 31 × 10 −19 J λ 6 .0 × 10 − 7

3 . 31 × 10 −19

= 2 .07 eV

1.6 × 10 −19

k = 8 .6 × 10 − 5 eV / K and T = 300 K,

So

kT = 8 .6 × 10 − 5 × 300 = 2 .58 × 10 − 2 eV

∴

N2 =e N1

  2 .07  − −2 2 . 58 × 10  

or

N2 = e− N1

80

Example 5: The Ruby laser has two states at 27°C and 227 °C. If it emits radiation of wavelength 7000 Å, then calculate relative population. Solution: The relative population of two states at temperature T K is represented as, N2 = e − ( E2 N1

− E1 ) / kT

E2 − E1 = hυ =

, where k is the Boltzmann’s constant

hc , λ

Here h = 6.62 × 10 −34 J – s, c = 3.0 × 108 m / s and λ = 7000 Å = 7 ×10 −7 m ∴

E2 − E1 = =

At

∴ At ∴

6 .62 ×10 −34 × 3.0 × 108 7 ×10 −7 2.84 ×10 −19 1.6 × 10 −19

J = 2.84 × 10 −19 J

eV = 1.77 eV

27° C = 27 + 273 = 300 K, Here k = 8 .6 ×10 − 5 eV / K N2 =e N1

  1 .77  − −5  8 .6 × 10 × 300 

=e

  1 .77  − −2  2 .58 × 10 

or

N2 = e− N1

69

227° C = 227 + 273 = 500 K N2 =e N1

  1 .77  − −5  8 .6 × 10 × 500 

=e

  1 .77  − −2  4 .3 × 10 

or

N2 = e− N1

41.1

341 Example 6: A laser beam can be focused on an area of 10 × 10 −14 m2 . If laser radiates energy at the rate 10 mW, find the intensity of focused beam. Solution: Intensity of a focused laser beam is given as I=

Power of laser beam P = Area A

Here P = 10 mW and A = 10 × 10 −14 m 2 I=

∴

10 × 10 − 3 W 10 × 10 − 14 m 2

= 1011 Wm −2

Example 7: Find the intensity of a laser beam of 1 mW power and having a diameter of 1.4 mm . Assume the intensity to be uniform across the beam. Solution: The intensity of a laser beam is I=

Power of the beam P = Area π r2

Here

P = 1 mW = 10 − 3 W (watt), d = 1.4 mm or r = 0.7 mm = 0.7 × 10 −3 m

∴

I=

10 − 3 3 .14 × (0 .7 × 10 − 3 )2

=

103 = 6.5 ×10 2 Wm − 3 .14 × 0 .7 × 0 .7

2

Example 8: A laser beam of wavelength 6328 Å and aperture 5 × 10 − 3 m from He-Ne laser can be focused on an area equal to the square of its wavelength. If the laser radiates energy at the rate of 10 mW , find (i) the angular spread of the beam, and (ii) the intensity of focused beam. Solution: (i) The angular spread of the beam is given by ∆θ ≈

λ d

Here λ = 6328 Å = 6 .328 × 10 − 7 m and the aperture, d = 5 × 10 − 3 m ∴ (ii)

∆θ ≈

6 .328 × 10 − 7 5 × 10

−3

≈ 1.265 ×10 −4 rad

If P is the energy radiated by laser and A the area on which the beam is focused, then the intensity of focused beam I is given by, I=

P A

Here P = 10 mW = 10 ×10 −3 W and A = λ 2 = (6 .328 × 10 − 7)2 = 40 .04 ×10 −14 m 2 ∴

I=

10 × 10 − 3 40 .04 × 10 −14

= 2.5 ×1016 Wm −2

342 Example 9: A laser beam a wavelength of 8 × 10 − 7 m and aperture 5 × 10 − 3 m. The laser beam is sent to moon. The distance of moon is 4 × 10 5 km from the earth. Calculate, (i)

the angular spread of the beam, (ii) the axial spread when it reaches the moon. [UPTU, B.Tech I, Sem. Q. Bank, 2000]

Solution: (i) If λ is the wavelength and d the diameter of aperture, then λ Angular width or angular spread of the beam, ∆θ ≈ rad d λ = 8 ×10 −7 m and d = 5 ×10 −3 m

Here ∴ (ii)

∆θ ≈

8 ×10 −7 5 ×10 −3

≈ 1.6 ×10 −4 rad

The axial spread of the beam when it reaches the moon W after traversing a large distance L is given as The axial spread, W ≈ ( L ∆θ)2 L = 4 ×105 km

Here

W ≈ (4 × 105 × 1.6 × 10 − 4 )2 ≈ 4 .096 × 10 3 km2

∴

Example 10: A laser beam has a power of 50 mW. It has an aperture of 5 × 10 − 3 m and it emits light of wavelength 7200 Å . The beam is focused with a lens of focal length 0 .1 m. Calculate the area and the intensity of the image.

[UPTU, B.Tech I, Sem. Q. Bank, 2000]

Solution: (i) If λ is the wavelength of laser light and d the aperture of lens (or mirror), then the angular spread of the beam is related to the aperture diameter and is given by, λ Here λ = 7200 Å = 7 .2 × 10 − 7 m and d = 5 × 10 − 3 m ∆θ ≈ d 7 . 2 × 10 − 7 ∴ ∆θ ≈ ≈ 1.44 × 10 − 4 rad 5 × 10 − 3 The area of the image or area I spread ≈ (∆θf )2 , Here f = 0 .1 m = (1.44 × 10 − 4 × 0 .1)2 or (ii)

A ≈ 2.074 ×10 −10 m2

If P be the power of the laser beam, then the intensity of image, I is given by I=

P 50 × 10 − 3 W = = 2.41 × 10 8 Wm − A 2 .074 × 10 − 10 m 2

2

Note : In some books, the angular spread (∆θ ) of laser beam is calculated by the formula, 1.22 λ (However, it is true for circular aperture) ∆θ = d Though, we have used following formula λ ∆θ = d Since the measurement of angular width is approximate, therefore for numerical calculations any one of the above formulae may be used.

343 Example 11: The coherence length of sodium light is 2 .945 × 10 − 2 m and its wavelength is 5890 Å . Calculate (i) the frequency (ii) the number of oscillations corresponding to the coherence length, and (iii) the coherence time. [UPTU, B.Tech., Special C.O. Exam. Aug 2008] Solution: (i) We know that υ = c / λ , Here c = 3 × 108 m and λ = 5890 Å = 5890 × 10 − 8 m ∴ (ii)

υ=

3 × 108 5890 × 10 − 8

= 5 .09 ×1012 Hz

The number of oscillations in any length l may be given as l n = c , Here l c = 2 .945 × 10 − 2 m and λ = 5 .89 × 10 − 7 m λ n=

∴

2.945 × 10 −2 5.89 ×10 −7

= 5 ×10 4

(iii) The coherence time τ c and coherence length l c of a laser light are related as τ c × c = l c or τ c = l c / c Here l c = 2 . 945 × 10 − 2 m and c = 3 × 108 m ∴

τc=

2 .945 × 10 − 2 3 × 108

= 9.816 ×10 −11s

Example 12: The coherence length for sodium D2 lines is 2 . 5 cm . Deduce (i) the coherence time, (ii) the spectral width of the line and (iii) the purity factor Q. Solution: (i) The coherence length l c and coherence time τ c of a laser light are related as l τ c × c = l c or τ c = c Here l c = 2 .5 cm = 2 . 5 × 10 − 2 m and c = 3 × 108 m /s c ∴ (ii)

τc =

2 . 5 × 10 − 2 3 × 108

= 0.83 ×10 −10 s

The spectral width, ∆λ of the beam is given by ∆λ = ∴

∆λ =

λ2 Here, λ for sodium D2 = 5896 Å = 5 .896 × 10 − 7 m lc (5 .896 × 10 − 7)2 2 .5 × 10 − 2

= 13.90 × 10 −

12

m

(iii) The purity factor Q is, Q=

5 .896 × 10 − 7 λ = = 4.24 ×10 4 ∆λ 13 .90 × 10 −12

Example 13: Assume that we chop a continuous perfectly monochromatic laser beam of wavelength λ = 6238 Å into 10 − 10 sec pulses using some sort of shutter. Calculate the resultant line-width, band-width and coherence length. Solution: The coherence length, l c is given by

344 l c = c × τ c = 3 × 108 × 10 −10 = 3 × 10 − 2 m Band-width,

∆υ =

1 1 = = 1010 Hz τ c 10 − 10

Line width,

∆λ =

(6238 × 10 λ2 ∆υ= c 3 × 108

−10 2

)

× 10 10 = 0.13 × 10 −10 m = 0.13 Å

Example 14: A laser source of wavelength λ = 6500 Å, coherence width 5 mm and power 50 mW shines on a surface 100 m away. Deduce (i) the angular spread (ii) areal spread, and (iii) illumination. Solution: (i) For the laser all power goes in an angle λ θ = , where a is the coherence width a 2λ Angular spread, 2 θ = ∴ a Here λ = 6500 Å = 6 . 5 × 10 − 7 m and a = 5 mm = 5 × 10 − 3 m Angular spread =

∴

2 × 6.5 × 10 −7 5 × 10 −3

= 2.6 ×10 −4 rad

The areal spread is, A = π ( Lθ)2 Here L = 102 m

(ii)

∴

A = 3 .14 × (102 × 1. 3 × 10 − 4 )2 = 5 . 31 × 10 − 4 m 2

Power of source P = Area A Here P = 50 mW = 50 ×10 −3 W and A = 5.31 ×10 −4 m 2

(iii) Illumination =

∴

Illumination =

50 × 10 −3 5.31 × 10 −4

= 94.162 Wm −2

Example 15: A laser beam (λ = 6000 Å) on earth is focused by lens of diameter 2 m on to a crater on the moon. The distance of the moon is 4 × 10 8 m. How big is the spot on the moon? Neglect the effect of a atmosphere.

[UPTU, B.Tech. I. Sem., Q. Bank, 2000]

Solution: If d is the diameter of the focusing lens, then the angular spread of the laser beam is, λ ∆θ ≈ d Here λ = 6000 Å = 6.0 × 10 −7 m and d = 2.0 m 6.0 × 10 −7 ≈ 3.0 ×10 −7 rad 2.0 If L is the distance of moon, then the area of the spot on the moon or the areal spread is given as

∴

∆θ ≈

A ≈ ( L∆θ)2 Here

L = 4 × 108 m

∴

A ≈ (4 ×108 × 3.0 ×10 −7)2 ≈ 1.44 × 10 4 m2

345

E xercises Section–A : Polarisation Question Bank 1. Light wave is transverse, but ordinary light is symmetric towards propagation direction, explain. Distinguish between polarised and unpolarised light. [UPTU, B.Tech. I Sem. (C.O.) 2007, II Sem 2004] 2. How would you obtain plane polarised light by reflection ? Explain the term plane of polarisation and plane of vibration.

[UPTU, B.Tech. I.Q.Bank 2000]

3. Explain Brewster’s law. Use this law to show that when light is incident on a transparent substance at the polarising angle the reflected and refracted rays are at right angles. [UPTU. B.Tech. I, Q. Bank 2000] 4. State and explain Malus law.

[UPTU, B.Tech. I, Q.Bank 2000]

5. Explain the phenomenon of double refraction in calcite or quartz and give the main reason for this phenomenon.

[UPTU, B.Tech. I Sem. 2008]

6. Discuss double refraction in calcite crystal. How can a phase retardation plate be obtained from it ? What is the quarter wave plate ?

[UPTU, B.Tech. II Sem. 2001]

7. Describe briefly, the double refraction in a calcite crystal. Define and discuss ordinary and extraordinary rays.

[UPTU, B.Tech. II Sem 2005, II Sem. 2003]

8. Explain the phenomenon of double refraction in calcite crystal. Describe the construction, working and use of Nicol prism.

[UPTU, B.Tech. II Sem. 2008, II Sem. 2002]

9. Explain the phenomenon of double refraction in a calcite crystal. Give the construction and theory of (i) quarter–wave plate and (ii) half wave plate. [GBTU, B.Tech. I Sem. 2010, I Sem. (old) 2009, I Sem. 2007]

10. Describe the construction of a nicol prism. Explain how it can be used as a polariser and as an analyser ?

[GBTU, B.Tech. I Sem. 2012; I Sem. (C.O.) 2012; UPTU, B.Tech. I Sem. (C.O.) 2006; I Sem. 2005, I Sem. 2004]

11. Describe the phenomenon of double refraction in uniaxial crystals. How is double refraction explained by Huygen’s theory ? Give the construction and theory of half wave plate. [UPTU, B.Tech. I Sem. 2005]

12. What do you understand by double refraction ? Explain Huygen's theory of double refraction in an uniaxial crystal. Give a diagram of Huygen construction of wavefronts in a negative uniaxial crystal when the optic axis is perpendicular to the plane of incidence. [GBTU, B.Tech. I Sem. (old) 2010; UPTU B.Tech. I Sem. (C.O.) 2009]

13. Explain Huygen's theory of double refraction in an uniaxial crystal. [UPTU, B.Tech. I Sem. (C.O.) 2007]

14. Explain phenomenon of double refraction. Describe Huygen's theory of double refraction when optic axis is parallel to upper face and lying in a plane of incidence.

[UPTU, B.Tech. II Sem. 2006]

346 15. How wire-gride polariser produces linearly polarised light ? 16. How will you obtain linear polarised light with the help of H-sheet polariser? 17. What do you mean by sheet polariser. What are the merits and demerits of sheet polariser? 18. Write short notes on : (a) J-sheet polariser, (b) K-sheet polariser. 19. What is meant by plane polarised, circularly polarised and elliptically polarised light ? Show that the plane polarised and circularly polarised light are the special cases of elliptically polarised light. [UPTU, B.Tech. I Sem. 2008, Special C.O. Exam. Aug 2008]

20. Show that the plane polarised and circularly polarised lights are the special cases of elliptically polarised light ?

[UPTU, B.Tech. I Sem. (C.O.) 2009]

21. Discuss theoretically the superposition of two linearly polarised sinusoidal light waves of the same frequency travelling in the same direction when the optical vectors are (a) parallel, and (b) mutually perpendicular. [UPTU, B.Tech. I Sem. (old) 2009, II Sem. 2005] 22. What is retardation plate ? Draw a ray diagram for extraordinary and ordinary rays before and after passing through a quarter wave plate.

[UPTU, B.Tech. I Sem. 2003]

23. Obtain an expression for the minimum thickness of a quarter wave plate. [UPTU, B.Tech. II Sem. 2004] 24. What is polarised light ? How will you produce and detect plane, elliptically and circularly polarised lights ?

[UPTU, B.Tech. II Sem 2005, I Sem. (C.O.) 2005, I Sem. 2001]

25. Describe the method to produce and detect circularly and elliptically polarized light. [MTU, B.Tech I Sem. 2012]

26. How would you produce and detect the following with the help of a nicol prism and a quarter wave plate (i) plane polarised (ii) circularly polarised and (iii) elliptically polarised light. [UPTU, B.Tech. I Sem. (C.O.) 2003]

27. Describe how a nicol prism can be used as polariser and analyser. [UPTU, B.Tech. II Sem. 2007; I Sem. 2005; II Sem. (C.O.) 2004]

28. Describe how, with the help of Nicol prism and quarter-wave plate, plane polarised, circularly polarised and elliptically polarised lights are produced and detected ? [UPTU, B.Tech. II Sem. 2006, II Sem. 2003]

29. Discuss the phenomenon of rotation of the plane of polarisation of light by optically active material. [UPTU, B.Tech. II Sem. 2008]

30. Discuss briefly the working principle of a polarimeter.

[UPTU, B.Tech. I Sem. 2003]

31. Describe the working principle of Half shade polarimeter. How can one use this experiment to find out the specific rotation of sugar. [MTU, B.Tech. I Sem. 2012] 32. Define specific rotation. Describe the construction and working of a Laurent's half shade polarimeter, explaining fully the action of the half shade device. How would you use it to determine the specific rotation of cane sugar solution. [GBTU, B.Tech. I Sem (C.O.). 2010, UPTU. B.Tech. I Sem 2009, I Sem. 2002]

347 33. Outline the principle of a half shade polarimeter and explain how will you use it to determine the specific rotation of a substance like Glucose.

[UPTU. B.Tech. II Sem (Old) 2009, I Q Bank 2001]

34. Describe a polarimeter and explain how it is used to measure the strength of sugar solution. Does it have any practical application ?

[UPTU, B.Tech. II Sem. 2003, I, Q. Bank, 2001]

35. Describe a polarimeter and explain how it is used to measure the strength of sugar solution. [UPTU, B.Tech. I Sem. 2001]

36. Describe the construction and working of a biquartz polarimeter. How will you use it to find the specific rotation of an optically active substance ?

[GBTU. B.Tech. I Sem. 2011, UPTU. B.Tech. I Sem 2006]

Short Answer Type Questions 1. What is polarisation ? 2. What do you mean by unpolarised light ? 3. Define plane of polarisation and plane of vibration. 4. What is plane polarised light ? Describe one method to produce it.

[MTU, B.Tech I Sem. 2012]

5. How will you produce plane polarised light by reflection ? 6. What is Brewster’s law ? 7. How will you produce plane polarized light by refraction ? 8. What is law of Malus ? 9. What are doubly refracting crystals ? 10. Define optic axis and principal section of the crystal. 11. What are ordinary and extraordinary rays ? 12. What is Nicol prism ? 13. How will you use Nicol as polariser and analyser ? 14. What are circularly and elliptically polarised light ? 15. What do you mean by phase retardation ? 16. What is quarter wave plate ? 17. What is half wave plate ? 18. How will you differentiate between quarter wave plate and half wave plate ? 19. How will you produce plane polarised light ? 20. What is H-sheet polariser ? 21. What is K-sheet polariser ? 22. What are the merits and demerits of sheet polariser. 23. How will you produce linear polarised light by wire-grid polariser ? 24. How will you produce circularly and elliptically polarised light ? 25. How will you detect plane polarised light using quarter wave plate ? 26. How will you detect circularly and elliptically polarised light with the help of quarter wave plate and rotating Nicol ? 27. How will you convert elliptically polarised light into circularly polarised light ? 28. What is optical activity ?

348 29. What are dextro–rotatory and laevo–rotatory substances ? 30. What is specific rotation ? 31. Distinguish between polariser and analyser. 32. What do you mean by optical rotation on what parameters does it depend ? [MTU, B.Tech I Sem. 2012] 33. Distinguish between dextro–rotatory and laevo–rotatory optically active substances. 34. What is the effect of wavelength of light on angle of rotation ? 35. Discuss the action of Laurent’s half shade plate. 36. What do you mean by strength of sugar solution ? 37. What is the principle of half shade polarimeter ? 38. What is the principle of biquartz polarimeter ? 39. Explain the phenomenon of optical rotation.

Unsolved Numerical Problems 1. A 60° quartz prism is cut with its faces parallel to the optic axis. Calculate the angle of minimum deviations for yellow light for each of the two polarised rays. The refractive index of quartz for the O–ray is 1.5422 and for the E–ray is 1.5533. 2. Two polarising sheets have there polarising directions parallel to that the intensity of the transmitted light is a maximum. Through what angle must either sheet be turned so that the intensity becomes one–half the initial value ? 3. Two Nicols have parallel polarising directions so that the intensity of transmitted light is maximum. Through what angle must either Nicol be turned if intensity is to be dropped by one–fourth of its maximum value. 4. For calcite, µ0 = 1. 658, µ E = 1. 486 for sodium yellow. Calculate the thickness of thinnest quarter wave plate of calcite for sodium yellow. 5. Calculate the thickness of a doubly–refracting crystal plate required to introduce a path difference of λ /2 between the ordinary and extraordinary rays when λ = 6000 Å, µ 0 = 1. 55 and µ E = 1.54. 6. Calculate the thickness of a quarter wave plate for light of wavelength 5896 Å, the refractive indices for ordinary and extraordinary rays being 1.54 and 1.55 respectively. 7. Plane polarised light is incident on a piece of quartz cut parallel to the axis. Find the least thickness for which the ordinary and extraordinary rays combine to form plane polarised light given that µ0 = 1. 5442, µ E = 1. 5633, λ = 5 × 10 − 5 cm. 8. A calcite plate is a half wave plate for light whose wavelength is λ. Assuming that the variation in the indices of refraction with wavelength can be neglected, how would this behave with respect to light of wavelength λ1 = 2 λ . 9. Elliptically polarised light falls normally on a quarter wave plate. Explain the nature of emergent light of the major axis of the ellipse makes the following angles with the principle plane of the quarter wave plate : 0°, 30°, 90°. 10. A left circularly polarised beam (λ = 5893 Å) is incident on a quartz crystal (with its optic axis cut parallel to the surface) of thickness 0.025 mm. Determine the state of polarisation of the emergent beam. Assuming µ0 = 1. 54425 and µ E = 1. 55336 .

349 11. Calculate the specific rotation of sugar solution if the plane polarisation is rotated by 13.2°. The length of the tube containing sugar solution is 20 cm and 5 gm of sugar is dissolved in 50 cc of water. 12. A tube of sugar solution 20 cm long is placed between crossed Nicols and illuminated with light of wavelength 6 × 10 − 5 cm. If the optical rotation produced is 13° and the specific rotation is 65°/dm / cm 3, determine the strength of the solution. 13. Find the specific rotation of cane sugar solution if the plane of polarisation is turned 26.4°.The length of the tube containing 20% sugar solution is 20 mm. 14. For a given wavelength 1 mm of quartz cut perpendicular to optic axis rotates the plane of polarisation by 18°. Find for what thickness will no light of this wavelength be transmitted when a quartz piece is interposed between the pair of parallel Nicols. 15. 20 cm length of a certain solution causes right handed rotation of 38°, and 30 cm of another solution causes left handed rotation of 24°. What optical rotation will be caused by 30 cm length of a mixture of the above solutions in the volume ratio 1:2 ? 16. For quartz the refractive indices for right handed and left handed vibrations are 1.55810 and 1.55821 respectively for λ = 4000 Å. Find the amount of optical rotation produced at λ = 4000Å by a plate of quartz 2.00 mm thick and with its faces perpendicular to the optic axis. 17. The rotation of the plane of polarisation of light (λ = 5893 Å) in a material is 10°/cm. What is the difference in the refractive indices of the material for clockwise and anticlockwise circularly polarised light ?

Answers Unsolved Numerical Problems 1.

40°54′, 41°54′

2.

±45 ° , ± 135 °

3.

± 30 °

4.

0.86 × 10 − 4 cm

5.

3 × 10 − 3 cm

6.

0.001474 cm

7.

0.00274 cm

8.

Quarter wave plate

9.

At 0°, plane polarised inclined at tan −1 (± b / a) with principal axis. At 30°, elliptically polarised. At 90° plane polarised inclined at tan −1(± a / b)

10.

0.77 π, right handed elliptically polarised

11.

66° (dm) −1 (gm/cc) −1

12.

0.1 gm/cc = 10%

13.

66°

14.

5.0 mm

15.

+ 3 ° (right handed)

16.

98°

17.

3 . 274 × 10 − 7

350

Section–B : Laser Question Bank 1. What do you mean by Laser? Distinguish between spontaneous and stimulated emission 2. What is LASER ? Explain its principle. Also explain the principle of optical pumping. [UPTU, B.Tech. I Sem. (C.O.) 2009]

3. Explain the terms 'spontaneous emission' and 'stimulated emission' of radiation. Obtain a relation between transition probabilities of spontaneous and stimulated emissions. [GBTU, B.Tech. I Sem. 2011] 4. Explain spontaneous and stimulated emission of radiation. How stimulated emission takes place with the exchange of energy between Helium and Neon atoms ?

[UPTU, B.Tech. II Sem. 2008]

5. What is spontaneous and stimulated emission of radiation ? Describe the method by which the electromagnetic radiation can interact with atomic energy levels.[UPTU, B.Tech. I. Sem. (C.O.) 2009] 6. Explain what do you understand by population inversion.

[UPTU, B.Tech. I Sem. 2007]

7. Explain the principle of optical pumping and stimulated emission of radiation. Discuss the properties of laser radiation and mention some of its applications.

[UPTU. B.Tech. I Sem. 2009; II Sem. 2003]

8. What are the differences between spontaneous and stimulated emissions ? Why is spontaneous radiation incoherent ?

[GBTU, B.Tech. I Sem. 2011; UPTU. B.Tech. I Sem. 2007]

9. What are Einstein’s coefficients ? Obtain a relation between them. Also discuss the essential conditions for laser action. [MTU, B.Tech I Sem. 2012] 10. What are Einstein's coefficients A and B ? Establish a relation between them. [UPTU, B.Tech. II Sem. 2007, I Sem. (C.O.) 2007, I Sem 2005, II Sem. (C.O.) 2004]

11. What are the differences between spontaneous and stimulated emissions. Which one, out of these two processes, is maximised in case of laser action, give suitable justification to your answer. [UPTU, B.Tech. I Sem. (C.O.) 2005, I Sem. (C.O.) 2003, I Sem. 2002]

12. Explain the action of Helium–neon Laser. How is it superior to a Ruby Laser ? [UPTU, B.Tech. I Sem. 2010, I Sem. (old) 2009, II Sem. 2007, I Sem.2001]

13. Describe the principle and working of three–level laser system.

[UPTU, B.Tech. I Sem. 2008]

14. What is a Laser ? What are its applications ? What is the difference between a continuous and a pulse laser ? Give one example of each type and discuss the basic principle and working of any one of them.

[UPTU, B.Tech. II Sem. 2004]

15. Explain the spontaneous and stimulated emission of radiation. Describe the working principle of a Ruby laser.

[UPTU, B.Tech. I Sem. (C.O.) 2010; I Sem. 2006, I Sem. 2003]

16. What is Laser ? Discuss the construction and working of a Ruby laser, explaining the principle of population inversion.

[UPTU, B.Tech. II Sem. 2006, II Sem. 2001]

351 17. Describe the construction and working of He-Ne laser.

[MTU, B.Tech. I Sem 2012]

18. Draw a neat diagram of He–Ne Laser and Describe its method of working.What are the characteristics of laser beam ? Discuss its important applications. [GBTU, B.Tech. I Sem. (C.O.) 2012, I Sem. 2012 UPTU, B.Tech. II Sem. (old) 2009, I Sem. 2004]

19. What are the characteristic properties of a laser beam ? Describe its important applications. [UPTU, B.Tech. Special C.O. Exam. Aug 2008]

20. Explain the principle of a laser and describe the working of a Ruby laser ? [UPTU, B.Tech. II Sem. 2006, I Sem. 2002]

21. Describe the construction and action of the ruby laser. [UPTU, B.Tech. I Sem. 2010, I Sem. 2005, II Sem. (C.O.) 2004]

22. Laser action is sometimes called "inverted absorption". Explain in which situation may A /B be small enough for laser action. [UPTU, B.Tech. I, Q. Bank, 2000] 23. Obtain relation between transition probabilities for spontaneous and stimulated emission of radiation. [UPTU, B.Tech. I, Q. Bank, 2000] 24. What are the requirements for producing laser action ? How are they achieved ? Explain the principle of laser by schematic diagram. [UPTU, B.Tech. I Sem. (old) 2010, I Sem. 2010, II Sem 2005] 25. Describe how laser radiation is different from ordinary light. Mention some important uses of laser in spectroscopy and industry.

[UPTU, B.Tech. I, Q. Bank, 2000]

26. Explain the principle of optical pumping and stimulated emission of radiation. Explain the action of a helium–neon laser. How is it superior to a ruby laser ?

[UPTU, B.Tech. II Sem. 2002]

27. Discuss the application of laser in holography and optical communication. [UPTU, B.Tech. I Sem. 2008]

Short Answer Type Questions 1. What is laser ? 2. What is spontaneous emission of radiation ? 3. What do you mean by stimulated emission of radiation ? 4. Distinguish between spontaneous and stimulated emissions. Which one is required for laser action ? [MTU, B.Tech. I Sem. 2012]

5. What are transition probabilities ? 6. What are the necessary conditions to achieve laser action ? 7. What are the components of laser ? Also describe spacial characteristic of laser. [MTU, B.Tech. I Sem. 2012] 8. Define population inversion. 9. What is pumping ?

352 10. Discuss in brief various pumping methods. 11. Define metastable states. 12. What is the principle of laser ? 13. What is optical pumping ? 14. How can you say that He-Ne laser is superior to Ruby laser ? [GBTU, B.Tech. I Sem. 2012, I Sem. (C.O.)2012]

15. What are the characteristic properties of laser beam ? 16. Distinguish between ordinary light and laser radiation. 17. Discuss some important applications of laser. ❍❍❍

Unit

V

F ibre O ptics & H olography

Section-A: Fibre Optics ...................................353–398 F undamental I deas about O ptical F ibres T ypes of F ibres A cceptance A ngle and C one N umerical A perture P ropagation M echanism and C ommunication in o ptical F ibre A ttenuation S ignal L oss in O ptical F ibre D ispersion

Section-B: Holography.....................................399–407 B asic P rinciple of H olography C onstruction of H ologram and W ave R econs truction A pplications of H olography (Qualitative)

Exercise .......................................................408–414 Q uestion B ank S hort A nswer T ype Q uestions U nsolved N umerical P roblems A nswers to Unsolved Numerical Problems

355

U nit-V

S ection

A F ibre O ptics

Introduction

T

h e advent of lasers sowed the seeds for the growth of entirely new crop of communication engineering the fibre optics or fibre communication optics. Prior to the introduction of fibre optics, in the conventional telecommunication systems the informations were transmitted through the electromagnetic waves which were propagated through transmission lines or waveguide structures having appropriate dimensions. All our communications depend on the transmission of signals through carrier waves. When we make a telephone call to a friend at a distant place, a sound wave carries the message from our vocal cords to the telephone. There, an electrical signal is generated which propagates along the copper wire. If the distance is too large, the electrical signal generated may be transformed into the electromagnetic waves and transmitted through the atmosphere, possibly by way of a communication satellite. The invention of optic fibre, as an alternative to electric and electronic, semi-conductor components, has wholly revolutionized the techniques of communication of information, even over very long distances, quite cheaply, but efficiently. The modern eagerness for telecommunications with carrier waves at optical frequencies of the electromagnetic system owes its origin to the discovery of laser. Earlier, before sixties, there were no suitable light sources available which could reliably used as the information carrier. On the other hand, around the same time telecommunication traffics were growing so rapidly that it was conceivable that conventional telecommunication system based on, say, coaxial cables, radio and microwave links could soon reach a saturation point.

356

Fast development in the field of communication in present time shows that the cost per unit information bandwidth is reduced considerably if a system is used in which each information carrier has a bandwidth capable of carrying simultaneously a large volume of communication traffic. This concept led to the development and introduction of microwave communication links. In fact, the amount of information carried by electromagnetic waves, having a frequency spectrum down to microwave region, is proportional to the frequency of the wave. A wave having a frequency of 100 can have 100 pulses at best, hence can transmit 100 pulse per second. Thus, the microwave communication and other traditional communications are unable to reduce such a heavy communication traffic. In this fast growing world, electronics and computers are absolutely essential for human beings because they touch every aspect of our daily lives in some way or others. Bar code scanners read the prices of our grocery items, computers dispense money from automatic tellers. We can use personal computer for banking, real estate, accessing news and entertainment, and a hundred and one other functions that are an integral part of our daily life. This increase in electronic methods for manipulating, interpreting, and communicating informations have revealed the limits of traditional communication technologies such as copper wire and radiowaves. Optical fibres have been developed to overcome these disadvantages. The invention of LASER and Light Emitting Diodes (LED) gave birth to the optical communication. A light beam acting as a carrier wave is capable of carrying far more informations than radiowaves and microwaves. In fact with optical communication in which light signals are used instead of electric signals, it is possible to transmit 44.7 million pulse per second. The optical waves provide such a large bandwidth that it becomes difficult to use it fully. Fibre optics is the overlap of applied science and engineering concerned with the design and application. Optical fibres are widely used in fibre optic communication which permits transmission over long distances. Fibres are used instead of metal wires because signal travel along them with low loss.

Fundamental Ideas about Optical Fibres History Guiding of light by the method of refraction, the principle that makes fibre optic possible, was first demonstrated by Daniel Colladon and Jacques Babinet in Paris 1840s, with Irish inventor John Tyndall. Fibre optic really developed in 1950s with the work of Hopkins and Narendra Singh Kapany in UK and Van Heel in Holland. These workers conducted experiments that led to the invention of optical fibre, based on Tyndall earlier studies. Modern optical fibres, where the glass fibre coated with a transparent cladding to offer a more suitable refractive index, appeared later in the decade. Development then focused on fibre bundles. First fibre optic semi-flexible gastroscope was patented by Basil Hirschowitz, C. Wilbur Peter, and Lawrence E Curtiss, researchers at the University of Michigan in 1956. In the process of developing gastroscope, Curtiss produced the first glass-clad fibres. Previous optical fibres had relied on air or impractical oils and waxes as the low index cladding material. In the right sense the actual evolution of fibre technology was taking place in the year 1960, after the successful invention of semiconductor laser and the light emitting diodes (LEDs). At that time the existing fibres had a loss even more than 1000 dB/km. In 1965, Charles K. Kao and George A. Hockhman of the British Company Standard telephones and cables were the first to suggest that the attenuation of contemporary fibres was caused by impurities, which could be removed, rather than fundamental physical

357

effect such as scattering. They speculated that optical fibre could be practical medium for communication, if the attenuation could be reduced below 20 dB/km. This attenuation level was first achieved in 1970 by researcher Robert D. Maurer et.al. at Corning Glass Works, America, now Corning Inc. they demonstrated a fibre with 17 dB attenuation per kilometre by doping silica glass with titanium. A few years later they produced a fibre with only 4 dB/km using germanium oxide as the core dopant. Now-a-days, attenuation in optical fibre cables are far less than those in electrical copper cables. The more strong optical fibre commonly used today was invented by Gerhard Bernsee in 1973 by Schott Glass in Germany, they utilizes glass for both core and clad and is therefore less prone to aging process. Since then fibre technology has advanced to point of fabricating fibre with a loss less than 0 . 5 dB/km. Fibre fabricated with recently developed technology are characterised by extremely low losses (~ 0 . 2 dB/km) as a consequence of which the distance between two successive repeaters could as large as 250 km. In a recently developed fibre optic system it has been possible to send 140 M bit/sec information through 220 km link of one optical fibre; this is equivalent to about 450000 voice channel-km. Now it is possible for a typical single optical fibre with great bandwidth at infrared wavelengths to carry 200 million telephone channels or twenty thousand TV channels or some combination of both with an attenuation of only below 0.2 dB/km. With this fast and amazing progress in the advancement of optical fibre technology, it has not very long since the cost of the system came down considerably, enabling people throughout the world to seriously begin the switch over from copper cable to optical fibres. Indeed the British Telecom stopped laying copper wires on its long distance routes where communication traffic is very high and started using optical fibres. Optical fibres are also being extensively used for local area networks that wire up telephones, televisions, computers or robots in office and cities. In 1991 the emerging field of photonic crystals led to the development of photonic crystals fibre, which guide light by means of diffraction from a periodic structure, rather than total internal reflection. The first photonic crystal fibre and their wavelength dependent properties can be manipulated to improve their performance in certain applications.

Optical Fibre Optical fibres serve as cables to carry huge amount of informations in the form of optical signals from one place to another over a wide bandwidth with negligible loss. An optical fibre is a hair-thin flexible transparent medium of cylindrical shape usually made of glass through which light can be propagated. The optical fibre has three principal sections, such as (i) the core, (ii) the cladding, and (iii) the jacket. Opaque Protective jacket The core is the innermost section of the fibre (Fig. 1) and has a remarkable property of conducting an optical beam. It is Fibre core made of glass or plastic. The core, the actual working structure of the fibre, is covered with another layer of glass with slightly different chemical composition or plastic, called cladding. The Cladding cladding has optical properties very different from those of the Fig.1 core. The optical fibre may have a abrupt boundary between the core and cladding or there may be a gradual change in the material between the two. The outermost section of the fibre is called, the jacket and is made of plastic or special kind of polymer and other materials. The opaque protective jacket protects the core from abrasion,

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interaction with environment, moisture, absorption, crushing and other vagaries of the terrestrial atmosphere and thus enhances its tensile strength. In fact the humidity of a normal atmosphere causes micro-cracks on the fibre surface to grow, which eventually degrades inherent high tensile strength of the fibre. The core acts like a continuous layer of two parallel mirrors.

Air

Losses

Air Cladding

Optical fibre

A message which is to be conveyed is first encoded into a light wave and then fed into the fibre, where it is propagated as a result of multiple internal reflections. A modern fibre consists Optical fibre with cladding of an optical rod of glass core coated with another type of Fig. 2 glass cladding. The refractive index of the core is higher than the refractive index of cladding material in order to utilize the phenomenon of total internal reflection for the propagation of light through the fibre. A bare fibre (without cladding) and a cladded fibre are shown in Fig. 2. The light signal travels through the fibre from the transmitter to the receiver and can be easily detected at the receiving end of the fibre (Fig. 3). The transmitter convert electrical signal to optical signals which is transmitted through the fibre. The information in the form of either voice, data or video is transmitted through the fibre. The receiver receives the optical signals from the fibre and converts the same to its electrical equivalent. Finally electrical signals are decoded to give the original information. The transmitters in optical fibre links are generally laser diode and receivers are semiconductor photodiodes. Infrared light rather than visible light is used more commonly for propagation, because optical fibres transmit infrared wavelengths with less attenuation and dispersion. To protect the fibre from chemical and mechanical attacks (as it routed, underground, overhead, through walls or under the ocean) several fibre cable designs are available. Each design has different properties such as water resistance, flame retardent, crush proof etc., but the components of all type of the fibre cables are roughly the same.

Receiver

Transmitter Optical fibre

LED or LASER

Photocell

Fig. 3

Fibre Fabrication Optical fibres are drawn from a furnace containing pure silica in the molted form and a small amount of dopant like GeO2, B2O3 and P2O5. The dopant produce the required change in refractive index. Most of the cable bulk is made up of strengthening and buffering materials which protect the fibre against chemical and mechanical attacks.

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The dimensions of the fibre (diameters of the core and cladding) vary, depending on the application of the fibre, but these dimensions must be kept within the tight specifications, for a particular type of a fibre. The type of glass used will also vary with the application, but the purity must also be kept very high. Advances in li ght sources, detectors and manufacturing techniques have led to new designs. Fibre design has concentrated on reducing the losses in a fibre in two ways. Decreasing attenuation losses is focused on bringing as much of the light originally launched in the fibre out the other end. Reducing dispers.ion limits the amount of distortion in the signal carried by the light through the fibre.

Principle of Operation of a Fibre An optical fibre is a cylindrical waveguide that propagates informations, coded in the form of light pulses, along its axis by the process of total internal reflection. Fibre consists of a glass core surrounded by another glass layer, called cladding. To confine the optical signal in the core, the refractive index of the core must be greater than that of the cladding. The boundary between the core and the cladding may either be abrupt or gradual depending upon the use of the fibre.

Types of Fibres Optical fibres, which form a passage way for transmitting optical signals, are designed in different ways to meet different requirements. On the basis of refractive index profile of the core and the way in which light signal propagates down the core, the fibres are broadly divided into three categories. (i) Step index multimode fibres (MMF), (ii) Step index single mode or monomode fibres (SMF), (iii) Graded index multimode fibres (GRIN)

(i) Step Index Multimode Fibres (MMF) In a step index multimode fibre a transparent glass core with a constant index of refraction is surrounded by another coaxial glass or plastic cladding of index of refraction lower than that of the core. The upper and lower interfaces between the core and cladding act as a cylindrical mirror at which the reflection of the transmitted light takes place. The multimode fibre has a larger core diameter of about 20–100 µm and the diameter of cladding is about 100–200 µm. The standard overall diameter of the MMF is 125 µm. These multimode fibres are referred to as step index fibres due to the step discontinuity of the refractive index profile at the core-cladding interface. The difference in the refractive index of the core and cladding materials is relatively large so that the critical angle (the minimum angle for total internal reflection) made µ by the ray with the line normal to the core-cladding boundary is small (θ c = clad ). Hence, there will be µ core

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Fig. 5

many paths

available for light signals to travel down the fibre.The light signals that meet the

core-cladding boundary at high angles, greater or equal to the critical angle for the boundary, are totally reflected into the core. Further, because of the cylindrical symmetry in the fibre structure, these totally reflected rays will further suffer complete reflections at the lower interface and therefore get guided through the core by repeated total internal reflections (Fig. 5). Rays that meet the interface at low angles (less than critical angle) will pass into and be absorbed by the cladding material. These rays do not convey light signals and hence the information along the fibre. The critical angle determines acceptance angle of the fibre, often reported as numerical aperture (NA) . A high numerical aperture allows light to propagate down the fibre in a set of many rays both close to the axis and at various angles with the axis, allowing efficient coupling of light into the fibre. Since the refractive of the core is constant all the rays making larger angles with the axis (or angles equal or greater than the critical angles with the normal to the core-cladding boundary) travel with same velocity in the core and take different times to reach the output end of the fibre, as their path lengths are different (Fig. 5). This spreading of arrival times of the transmitted rays at the exit end of the fibre cause a distortion or dispersion known as modal dispersion.Thus, the amount by which a pulse broadens as it passes through a multimode fibre is commonly known as dispersion. Due to this modal dispersion, a narrow pulse of light energy at the input end of the fibre, after travelling along the core, produces an attenuated output pulse dispersed over a wide period of time. That is, the light pulse containing the data transmitted is elongated or stretched out. If the output pulse spread too much, they begin to overlap on each others and may reach a point where one is not distinguishable from the next. In fact dispersion in optical fibre reduces the information carrying capacity of the fibre.

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In a stepped index multimode fibre, light propagates in different modes, each with a different path transit-time. If the diameter ‘d ’ of the core satisfies the condition d>

0 . 766 λ (NA)

where λ is the operating wavelength and NA the numerical aperture of the fibre equals to (µ2core − µ2clad ). A number of modes existing through the fibre become possible. If the number of modes in a step index fibre is larger than 50, then the number of modes in a multimode fibre is given by N =2.

π 2 a2 (NA)2 λ2

where a is the radius of the core. Step index multimode fibres generally have a large diameter core and are fabricated without sacrificing mechanical strength. Multimode fibres are used for short distance, shorter than 200 meters, communication links or for application where high power must be transmitted. Modal dispersion can be minimized by using single mode step or monomode step index fibres or graded index multimode fibres.

(ii) Step Index Single Mode or Monomode Fibres (SMF) In order to reduce dispersion upto almost zero level and to increase the information carrying capacity, a fibre with a core diameter less than about ten times the wavelength of the propagating light wave is fabricated. In the step index multimode fibres we had considered light propagation in the fibre as a set of many rays meeting at the core-cladding boundary at an angles (making with the normal to the boundary) greater or equal to the critical angle for that boundary. If the diameter of the core is reduced to a value about ten times of the wavelength of the propagating wave or if the refractive index difference between the

2

Fig. 6

core and the cladding is made very small, then only a single mode is propagated through the fibre (Fig. 6). Such a fibre is called step index single mode or monomode fibre. The most common type of single mode

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or monomode fibre has a core diameter of 8 to 10 µm and is designed for use in the infrared region. The mode structure depends on the wavelengths of the light used, so that the fibre actually supports a smaller number of additional modes at visible wavelength.Thus, in a single mode fibre, the transparent glass core is very thin and the refractive index of the cladding material is slightly lower so that the difference between the core and the cladding indices becomes very small. Thus, the single mode fibre is designed to eliminate modal dispersion by reducing the number of modes in the fibre to one. Since only a single ray path, parallel to the axis of the core, is possible in a single mode or monomode fibre, the dispersion caused due to the difference in the arrival times of different rays in multimode fibre would be completely absent. Thus, the information transmission capacity of a single mode fibre is much larger than that of a multimode fibre. In spite of negligible dispersion, very thin core of a single mode or monomode fibre creates mechanical difficulties in the manufacturing, handling and splicing the fibres. Hence, the fabrication of a single mode fibre is very expensive. Single mode fibres are used for most communication longer than 200 meters. Single mode fibres are frequently used under sea water. In order to get single mode, with all other modes cut-off, the diameter of the core must satisfy the relation 0 . 766 λ d< (NA) A single mode (or monomode) fibre does require a much more sophisticated light source in order to launch enough light into the tiny core.

(iii) Graded Index Multimode Fibres (GRIN) The graded index fibre offer a less expensive method of overcoming the modal dispersion or transit time dispersion. In step index fibres the refractive index of the core has a constant value. However, in a graded index multimode fibre, the index of refraction in the core decreases continuously in a nearly parabolic manner from maximum value at the core axis to a minimum constant value at the core-cladding interface (Fig. 7). Since the refractive index µ

µ2

gradually decreases as one moves away from the centre of the core, a ray entering the fibre is continuously bent towards the axis of the fibre because the ray encounters continuously a medium of lower refractive index and hence bends away from the normal, that is, towards the axis of the fibre. It is because of the

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fact that the ray moving towards the core-cladding boundary encounters continuously a rarer medium (medium of low refractive index). By gradually decreasing the index of refraction of the optical fibre core as a function of radius, the ray path becomes a smooth undulating curve which does not reach the core-cladding boundary so that the wave propagates as though in an unbounded medium. In such a parabolic index medium rays paths are smooth sinusoidal. The waves propagating at different angles of incidence in the fibre travel different distances from the fibre axis before suffering reflections to recross the axis. Hence, the rays making large angles with the axis pass more through the lower index periphery of the core, rather than high index centre. As we know, higher the refractive index lower the velocity of light (v = c /µ). The light rays travelling near the core axis move slowly than those passing near the core-cladding boundary. Hence, all the rays travelling in the core of the fibre, take same amount of time in traversing the fibre and reaching the receiving end of the fibre, in spite of different path lengths. Thus, the rays travelling in the graded index multimode fibre take equal time to reach the end of the fibre. Thus, the transit-time or modal dispersion is greatly minimized in graded index multimode fibres. The variation of refractive index of the fibre core of the graded index multimode fibre with radius, a, measured from the centre of the core is expressed as, p   x  µ ( x) = µ1 1 − 2∆     a  

2

where µ( x) is the refractive index of the core at a distance x from the centre of the core, µ1, the refractive index at the centre of the core and p the index profile. Index profile has a typical value of 2.0 for 850 nm. In the graded index multimode fibre, the number of modes is expressed (provided the number of modes  p  2 π 2 a2 (NA)2 is more than 50) as,  N =  p + 2 λ2 N is doubled to account for the two possible polarisations.

Acceptance Angle and Acceptance Cone of a Fibre We have seen that in the step index multimode fibre any light wave which is meeting core-cladding boundary at or above the critical angle will be totally reflected in the core and propagated. However, any light-wave meeting the boundary at angle below critical angle will pass into and be absorbed by the cladding. The external angle of incidence made by a ray with the axis of the fibre, corresponding to the critical angle of incidence at the core-cladding boundary, is termed as acceptance angle. Thus, the critical angle determines the acceptance angle of the fibre.

364

Hence, the maximum entrance angle (α ) subtended by a ray on the fibre axis at the entry point of the fibre for which the ray suffers total internal reflection on striking the core-cladding boundary, is called the acceptance angle (Fig. 8). Acceptance angle is different for different fibres and depends on the core material and the core diameter. Only light that enters the fibre a certain range of angles can travel down the fibre without leaking out. The range of angles at entry point which make angles at the interface ≥ the critical angle is called acceptance cone of the fibre. Acceptance cone is formed by rotating acceptance angle about the fibre axis and is also defined as the cone of light described at the entry end of the fibre with semi-angle less than or equal to the acceptance angle of the fibre (or the cone formed with acceptance angle as vertex angle). The half angle of the cone within which incident light is totally internally reflected is also defined as the acceptance angle. The size of the acceptance cone is a function of the refractive index difference between the fibre core and cladding materials. The light which is emerging from the Fig. 9 acceptance cone is trapped and guided in the fibre. The concept of acceptance angle and acceptance cone can be best understood with the following mathematical explanation and derivation: Let a ray of light OA, from an external medium of refractive index µ0 , is incident on entry end of the fibre at an angle θ e and refracted into the core along AB (Fig. 9). If θ is the angle made by the refracted ray AB with the axis inside the core of refractive index µ1, then according to Snell's law, sin θ µ0 = sin θ e µ1

or sin θ =

µ0 sin θ e µ1

...(1)

If θ i is the angle subtended by refracted ray AB on the core-cladding boundary, then from triangle ABC ...(2)

θ = 90 − θ i

As θ e decreases θ also decreases, so that θ i increases. When θ e is sufficiently small, θ i exceeds that of critical angle θ c (minimum angle for total internal reflection) for core-cladding boundary and the refracted ray AB is totally internally reflected at B and continue to propagate inside the core. The maximum angle of incidence (or entrance angle) subtended by a ray at the entry end of the fibre corresponding to which refracted ray in the core is totally reflected on striking the boundary, is called acceptance angle for that fibre. If the maximum value of θ e at which θ i = θ c (critical angle) or θ i ≥ θ c is α, then θe = α

at

θi = θc

....(3)

Substituting the value of θ from equation (2) in equation (1), we get µ sin (90 − θ i) = 0 sin θ e µ1 or

µ1 cos θ i = µ0 sin θ e

...(4)

Substituting θ e = α and θ i = θ c in equation (4), we get µ0 sin α = µ1 cos θ c or

µ0 sin α = µ1 (1 − sin2 θ c )

Applying Snell's law at a point B on core-cladding boundary when θ i = θ c

µ1 sin θ i = µ1 sin θ c = µ2 sin 90 °

...(5)

365

or

sin θ c =

µ2 µ1

...(6)

where µ2 is the refractive index of the cladding material. Substituting the value of sin θ c from equation (6) in equation (5), we get µ0 sin α = µ1 1 − (µ2 /µ1)2 or

µ0 sin α = µ12 − µ22

If the external medium around the fibre is air, that is, µ0 = 0 , then sin α = µ12 − µ22 α = sin−1 µ12 − µ22

or

This is the required expression for acceptance angle of a fibre with core and cladding of refractive indexes µ1 and µ2 respectively. The light rays which are emerging from a cone defined by the acceptance angle would be totally reflected from the core-cladding interface and continue to propagate down the core. On the other hand the rays which are emerging outside the acceptance cone on striking the interface totally absorbed by the cladding material and never propagate.

Numerical Aperture Another term related with the fibre is the numerical aperture (NA). Sometimes it is also called the figure of merit. Numerical aperture is a number which defines the light acceptance or gathering capacity of a fibre. In simple terms, there is a maximum angle made with the fibre axis at which the light may enter the fibre so that it is propagate in the core by several internal reflections. The sine of this maximum angle (acceptance angle) is a numerical aperture (NA). Mathematically numerical aperture (NA) is expressed as, 1 NA = sin α = (µ12 − µ22 ) µ0 If the external medium surrounding the fibre is air, then µ0 = 1 ∴

NA = µ12 − µ22

where µ1 is the refractive index of the core and µ2 that of cladding material. In terms of relative refractive index difference, ∆ = (µ1 − µ2 ) / µ1 numerical aperture may be evaluated as, ∆=

µ1 − µ2 (µ1 + µ2 ) (µ1 − µ2 ) = µ1 (µ1 + µ2 ) µ1

Since the difference in µ1 and µ2 is very small, therefore we can approximated µ1 + µ2 – 2µ1 ~ 2 2 2 2 µ − µ2 µ1 − µ2 ∆= 1 = ∴ 2µ1 ⋅ µ1 2 µ12 or or

∆ =

(NA)2 2 µ12

NA = µ1 2∆

or (NA)2 = 2 µ12 ∆

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This is the required expression for numerical aperture. Step index single mode fibres has small NA, whereas step index multimode fibre has large numerical aperture NA. A high NA allows light to propagate down the fibre in rays both close to the axis and at various angles with the axis, allowing efficient coupling of light into the fibre. However, this high numerical aperture increases the amount of dispersion. A low numerical aperture may therefore be desirable. The variation of numerical aperture (NA) with the acceptance angle α is shown in Fig. 10. The numerical apertures for the fibres used in short distance communication are in the range of 0.4 to 0.5, whereas for long distance communications numerical apertures are in the range of 0.1 to 0.3.

Numerical Aperture for Graded Index Fibres In the graded index fibre, the numerical aperture is a function of position across the core. To get the idea about the numerical aperture for graded index fibre, we are introducing the term local numerical aperture which is a function of radius of the core. From the geometrical optics it is obvious that the light falling on the fibre core at position r will propagate as the guided mode only if it is within the local numerical aperture NA(r ) at that point. Local numerical aperture at position r is expressed as,  r NA (r ) = [µ2 (r ) − µ22 ]1 /2 ~ NA (r = 0) 1 −    a

x

for r ≤ a

...(1)

= 0 for r > a where µ2 is the refractive index of the cladding material, a the radius of the core, x is the parameter describing the refractive index profile variation and NA(r = 0), numerical aperture at the centre of the fibre core. The axial numerical aperture NA (0) is, NA (0) = (µ2 (0) − µ22 ) = µ12 − µ22 where µ (0) = µ1 is the refractive at the centre of the core. NA (0) = µ1 2 ∆ , where ∆ =

µ1 − µ2 µ1

It is clear from equation (1) that or a graded index fibre numerical aperture decrease from axial numerical aperture NA (r = 0) to zero as r increases from zero to core radius ' a '.

Number of Modes and Cut-off Parameters of Fibres The number of modes supported by an optical fibre is obtained by an important parameter associated with the cut-off condition is called, cut-off parameter such as normalized frequency of cut-off, which is referred to V parameter or V–number. Mathematically V–number is expressed as, πd V= µ12 − µ22 λ0 πd or V≅ (NA) λ0 where λ 0 is the wavelength of a monochromatic light beam propagating in a multimode glass fibre of core diameter d, µ1 and µ2 are the refractive indices of core and cladding materials respectively, and NA = µ12 − µ22 is the numerical aperture, provided the environment of the fibre is air (for which µ0 = 0).

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If the external medium around the fibre has a refractive index µ0 , then V=

2 2  ∵ NA = µ1 − µ2    µ0  

πd µ (NA) λ0 0

The approximate total number of modes which the fibre will support is expressed as, Number of modes (N ) ≅ 12 V 2 provided the V number of V–parameter is considerably larger than unity. Each of the mode has a particular value of V–parameter, below that value the mode will cut-off. Out of all the modes only those modes will be propagated for which cut-off frequencies are less than the V–number. V –number can be reduced either by reducing the numerical aperture or by reducing the diameter of the fibre.

Comparison of Single Mode Index and Multimode Index Fibres Single Mode Index Fibre (SMF)

Multimode Index Fibre (MMF)

1.

In a single mode index fibre, the diameter of the core is very small and is of the same order as the wavelength of light to be propagated. It is in the range 5 µm–10 µm. The cladding diameter is about 125 µm.

1.

In a multimode fibre, the diameter of the core is large. It is in the range 30 µm − 100 µm. The clad ding di am e ter is in the range 125 µm–500 µm.

2.

The difference in the refractive indices of the core and the cladding materials is very small.

2.

The difference in the refractive indices of the core and the cladding materials is large.

3.

In SM fibre only a single mode is propagated.

3.

In MM fibre, a large number of modes can be propagated.

4.

SM fibre does requ ire a mu ch mo re sophisticated light source in order to launch enough light into the tiny core.

4.

MM fibre does not require any sophisticated light source.

5.

SM fibre is more expensive, but more efficient.

5.

MM fibre is less expensive.

6.

Acceptance angle and the size of the acceptance cone of a SM fibre is small.

6.

Acceptance angle and the size of the acceptance cone of a MM fibre is large.

7.

Numerical aperture of SM fibre is small.

7.

Numerical aperture of MM fibre is large.

8.

SM fibre has a very high information carrying capability.

8.

MM fibre has low information carrying capability.

9.

SM fibres are used when short distance communication is required.

9.

It is used for long distance communication.

10.

Modal dispersion in SM fibre is almost nil.

10.

Modal dispersion in MM fibre is the dominant source of dispersion.

11.

Material dispersion in SM fibre is low.

11.

Material dispersion in MM fibre is large.

12.

When a transmission has a very large band- 12. width, a single mode fibre is used e. g., undersea cables.

When the system bandwidth requirement is low, multimode fibres are used e. g., Data link.

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Comparison of Step Index and Graded Index Fibres Step Index Fibre (SI ) 1.

In a step index fibre, the refractive index of

Graded Index Fibre (GRIN) 1.

the core has a constant value.

In graded index fibre, the refractive index in the core decreases continuously in a nearly parabolic manner from a maximum value at the centre of the core to a constant value at core-cladding interface.

2.

For a SI fibre, the variation of refractive

2.

index is mathematically expressed as, µ(r ) = µ1 = µ2

Parabolic refractive index variation in GRIN fibre is mathematically expressed as,

 r 2 µ 2 (r) = µ12 1 −     α  

0 < r < a (core) r > a (Cladding)

where µ1 > µ2

3.

In SI fibre the propagating light rays reflect

= µ 22

3.

abruptly from the core-cladding boundary. 4.

For a given fibre diameter, the numerical

In SI fibre, there may be some irregularities at

r > a [CLADDING]

In GRIN fibre propagating light rays bend smoothly as they approach the cladding.

4.

aperture (NA) of SI fibre is large. 5.

0 < r < a [CORE ]

For the same fibre diameter, the numerical aperture (NA) of GRIN fibre is small.

5.

the interface between the core and cladding.

In GRIN fibre, there is no such irregularities at the interface.

6.

SI fibre has higher attenuation.

6.

GRIN fibre has lower attenuation.

7.

For a SI fibre of given physical size, with a

7.

For a GRIN fibre of same physical size,

8.

loss of power of the order of 12 dB/km, the

with an attenuation between 5 to 10

numerical aperture is of the order of 0.2 to

dB/km, the numerical aperture tend to run

0.35.

between 0.16 and 0.2.

In SI fibre, the time interval at the output

8.

end or pulse dispersion is expressed as, ∆τ =

end of pulse dispersion is expressed as,

 µ l µ1 l  µ1  − 1 = 1 ∆ c  µ2 c 

∆τ =

where l is the length of the fibre. 9.

Pulse dispersion in multimode step index

In GRIN fibre the time interval at the output

µ2 l  µ1 − µ2    2 c  µ2 

2

=

µ2 l 2 ∆ 2c

where l is the length of the fibre. 9.

Pulse dispersion in a GRIN fibre is small.

fibre is large. 10.

A good q uali ty SI fi bre may ha ve a bandwidth of 50 MHz km.

10.

The equivalent GRIN fibre can have 200, 400 or 600 MHz km bandwidth.

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Propagation Mechanism in Optical Fibres Whenever a ray of light travels from one material medium to another, it deviates from its original path. The refracted ray will bend toward the normal while going from a rarer to a denser medium or away from the normal while going from a denser to a rarer medium. The ratio of the sine of the angle of incidence to the sine of the angle of refraction is always constant. That is, sin θ i = constant = µ21 sin θ R

...(1)

This is known as Snell's law. The constant µ21 is called refractive index medium 2 with respect to medium 1, when a ray travels from medium 1 to medium 2. Whenever a light wave travels from one medium to another, the velocity of the wave changes. The velocity of light wave in vacuum is maximum and equal to c = 3 × 108 m/sec. The velocity of this wave in another medium will be less than c, say equal to v. The refractive index of the medium is, µ =c /v Suppose the velocity of light in medium 1 is v1 and that in medium 2 is v2 . The corresponding refractive indices µ1 and µ2 are µ1 = c / v1 and µ2 = c / v2 If we replace µ21 by µ2 / µ1 in eqn. (1) we get sin θ i sin θ R

=

µ2 µ1

...(2)

The speed of light wave may be different in different media of the fibre, but the frequency υ remains constant as light wave propagates from one medium to another. This means that for the relation, v = c /λ to hold good, its wavelength must change. In the fibre refractive index of the core µ1 is slightly greater than that of cladding µ2 , that is, µ1 > µ2 . From Snell's law, µ1 sin θ i = µ2 sin θ R . It is clear that sin θ R > sin θ i. In this case angle of refraction θ R is always greater than the angle of incidence θ i. As the angle of incidence increases the angle of refraction also increases. When the angle of refraction is 90°, the refracted ray just grazes (Fig. 11) along the surface, separating the two media. Any further increase in the angle of incidence will turn the refracted ray back into the same medium. This ray is said to be totally internally reflected. When light travelling in a dense medium hits the boundary at steep angle (greater than critical angle for the boundary) the light will be completely reflected.This effect is used in optical fibres to confine light in the core.

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Communication in Optical Fibre Let us consider a fibre having a core of uniform refractive index µ1 surrounded by a cladding of uniform refractive index µ2 such that µ1 > µ2 . Every light wave which travels along the core and strikes the core-cladding boundary at the angle greater than critical angle θ c (= sin −1 µ2 /µ1) will be totally reflected in the core region. Further, because of the cylindrical symmetry in the fibre structure, this ray will suffer total internal reflection at the lower interface also. Therefore, the light wave is propagated along (or guided through) the fibre core by a series of total internal reflections from the core-cladding boundaries. Even for a bent fibre, light guidance can occur through multiple total internal reflections. Since the refractive index changes abruptly at the core- cladding boundary this is a sort of step index fibre. In Fig. 12 the propagation path of one light wave is shown, it is possible only when a light wave is emitting from a very tiny point-source. In actual practice, several light rays emerge from a point source in different directions and have different colours with different frequencies or wavelengths. These rays strike the core-cladding boundary at different angles of incidence. Out of these, only the rays which are impinging on the interface at an angle greater than the critical angle are trapped inside the core through total internal reflection. Other rays either absorbed in the cladding or scattered.Thus, the multimode propagation is possible in fibre (Fig. 13). The rays follow zig-zag paths while travelling in an optical fibre by bouncing back and forth at the core–cladding boundaries and hence the rays will travel different total lengths of path and emerge at the far end at slightly different times. Thus, a distortion is produced which is called, transit time dispersion. As a result of this distortion the pulse of light containing the data transmitted is stretched out. If the pulses stretch too much, they begin to overlap each other and there–by cause distortion of the information being carried. This defect can be minimised by greatly reducing the diameter of the core and by reducing the difference in the indices between the core and cladding materials because the number of modes depends on the size of

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the core in a fibre. A fibre whose core diameter is of the same order as the wavelength of the light wave to be propagated is called monomode or single mode fibre because in such fibre only one mode exists. Since only a single ray path (Fig. 14) is possible in a single mode fibre, the dispersion caused due to the differences in the transit times of different rays in a multimode fibre would be completely absent. The dispersion in multimode propagation may also be reduced by considering Core propagation in graded index fibre. Since the refractive index of the core in a graded index Core fibre decreases continuously as one moves away from the Graded index multimode fibre centre of the core towards the Fig. 15 interface, a ray of light entering the fibre is continuously bent towards the axis of the fibre (Fig. 15), this follows from Snell’s law because the ray encounters continuously a medium of lower refractive index and hence bends away from the normal, that is, it bent towards the axis of the fibre. It is obvious that light waves with large angle of incidence travel longer path than those with smaller angles. As we know (v = c /µ) higher the refractive index slower is the velocity of light travelling through the fibre. Thus, the resulting fibre allows light in longer mode to travel faster than light in shorter modes.Thus, all propagating light waves will reach the end point of the fibre at the same time. Example 1: A silica glass optical fibre has a core refractive index of 1.5 and cladding refractive index of 1.450. Calculate the numerical aperture of the optical fibre. [U.P.T.U, B.Tech, I Sem (C.O.) 2010, I Sem 2009]

Solution: We know that the numerical aperture is, NA = µ core (2 ∆), where ∆ = ∴

∆=

So

µ core – µ clad µ core

1. 5 – 1. 45 = 0 . 033 1.5

NA = 1. 5 (2 × 0 . 033) = 1. 5 × 0 . 257 = 0 . 385

Example 2: Compute the numerical aperture and the acceptance angle of an optical fibre from the following data : µ1 (core) = 1.48 and µ2 (cladding) = 1.46. Solution: The numerical aperture of the fibre, NA = √ (µ12 − µ22 ) NA = √ [(1. 48)2 − (1. 46)2 ] = √ (2 .1904 − 2 .1316) = √ (0 . 0588)

or or Acceptance angle, α = sin

NA = 0.2425 −1

2 √ (µ1

− µ22 ) = sin −1 (0 . 2425) = 14°

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Example 3: An optical fibre has the following characteristics : Fibre index, µ1 = 1.36 and relative difference in index, ∆ = 0.025 . Find the value of the numerical aperture and the acceptance angle. Solution: The expression for numerical aperture in terms of relative refractive index difference ∆ is, NA = µ1√ (2 ∆) Here ∴ The acceptance angle,

µ1 = 1. 36 and ∆ = 0 . 025 NA = 1. 36 √ (2 × 0 . 025) = 1. 36 × 0 . 2236 = 0.304 α = sin −1(NA) = sin −1 (0 . 304) = 17.7 °

Example 4: Calculate a fractional difference between core and cladding refractive indices for a step index fibre having core and cladding refractive indices 1.55 and 1.50 respectively. Solution: We know that, the fraction difference ∆ is µ − µ2 ∆= 1 µ1 where µ1 and µ2 are the refractive indices of core and cladding materials respectively. Here µ1 = 1. 55 and µ2 = 1. 50 Hence,

∆=

1. 55 − 1. 50 = 0.032 1. 55

Example 5: Calculate the numerical aperture, acceptance angle, and the critical angle of the fibre from the following data : µ1(core refractive index) = 1.50 and µ2 (cladding refractive index) = 1.45 . [MTU,B.Tech I Sem 2012, UPTU, B.Tech I Sem 2009]

Solution: We know that, numerical aperture, NA = µ1 √ (2 ∆) where

∆=

µ1 − µ2 µ1

∴

∆=

1. 50 − 1. 45 = 0 . 033 1. 50

So, Acceptance angle,

NA = 1. 50 √(2 × 0 . 033) = 1. 50 × 0 . 257 = 0.385 α = sin −1(NA) = sin −1(0 . 385) = 22.63 °

According to Snell's law sin θ c = or

µ2 µ1

1. 45  µ   or θ c = sin −1  2  = sin −1   µ1  1. 50 

θ c = sin −1(0 . 967) = 75.3 °

Example 6: An optical fibre has the following data : µ1(core ) 1. 55 , µ2 (cladding) = 1. 50 , core diameter = 50 µm. Calculate the numerical aperture and acceptance angle. How many reflections per metre are suffered by the guided ray at steepest angle with respect to the fibre axis ?

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Solution: Numerical aperture NA = √ (µ12 − µ22 ) = √ [(1. 55)2 − (1. 50)2 ] = √ [(1. 55 + 1. 50) (1. 55 − 1. 50)] or Acceptance angle,

NA = √ (3 . 05 × 0 . 05) = 0.3905 α = sin −1(NA) = sin −1(0 . 3905) =23°

At the steepest angle, the angle of incidence at the core-cladding

θc

boundary exceeds the critical angle θ c . Thus, or

1. 50  µ   = sin −1 (0 . 9677) θ c = sin −1  2  = sin −1   µ1  1. 55 

d

θ c = 75 . 4°

If x the axial distance traversed by the ray between two

x Fig. 16

successive reflections suffered by the ray and d is the core diameter then From Fig. 16,

x = d tan θ c

or

x = 50 tan (75 . 4° ) = 50 × 3 . 8391 = 191. 95 µm

Number of reflections per metre is 1m 1m = = 5209 . 6 ~ 5209 x 191.95 × 10 −6 m Example 7: If the fractional difference between the core and cladding refractive indices of a fibre is 0.0135 and numerical aperture NA is 0.2425, calculate the refractive indices of the core and cladding materials. [UPTU, B.Tech, I Sem (C.O.) 2009] Solution: We have that, NA = µ1√ (2 ∆) and

∆=

µ1 − µ2 µ1

where µ1 and µ2 are the refractive indices of core and cladding materials respectively. Here ∴

and or or

NA = 0 . 2425 and ∆ = 0 . 0135 µ1 =

0 . 2425 0 . 2425 NA = = = 1.476 2∆ 2 × 0 . 0135 0 .1643

∆ = 0 . 0135 =

µ1 − µ2 1. 476 − µ2 = µ1 1. 476

1. 476 − µ2 = 0 . 0135 × 1. 476 µ2 = 1. 476 − 0 . 02 = 1.456

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Example 8: A step index fibre has core refractive index 1.466, cladding refractive index 1.46. If the operating wavelength of the rays is 0.85 µm, calculate the cut-off parameter and the number of modes which the fibre will support. The diameter of core = 50 µm. or A step index fibre has core and cladding refrective indices 1.466 and 1.460 reprectively. If the wavelength of light 0.85 µm. Find the normalized frequency and the number of mode supported by the fibre. [GBTU, B.Tech, I Sem (C.O.) 2012, UPTU, B.Tech I Sem 2011]

Solution: We know that, cut-off parameter or cut-off number or normalized frequency of cut- off is, V=

2πa 2 µ1 − µ22 λ

where a is the radius of the core, µ1 refractive index of the core, µ2 the refractive index of cladding and λ operating wavelength. Here a = 50 / 2 = 25 µm, λ = 0 . 85 µm, µ1 = 1. 466 and µ2 = 1. 46 2 × 3 .14 × 25 ∴ V= (1. 466)2 − (1. 46)2 0 . 85 = 184 . 70 √ (2 .149 − 2 .131) = 184 . 70 × 0 .134 V = 24.75

or

2

Number of modes,

N =

V 2 (24 . 75) ~ = 306 2 2

Example 9: The core diameter of a multimode fibre is 70 µm and the relative refractive index difference is of 1.5%. It operates at a wavelength of 0.85 µm. If the refractive index of the core is 1.46, calculate (i) the refractive index of the cladding, (ii) the normalised frequency V -number of the fibre, and (iii) the total number of guided modes in the fibre. Solution: In the given problem, radius, a = 70 / 2 = 35 µm, µ1 = 1. 46 and relative index difference, µ − µ2 ∆= 1 = 1. 5%, where µ1 is the refractive index of core and µ2 that of cladding material µ1 (i)

∆=

1. 5 µ1 − µ2 = 100 µ1

µ2 = (1 − 0 . 015) µ1 (ii)

or

µ1 − µ2 = 0 . 015 µ1

or

µ2 = 0 . 985 × 1. 46 = 1.438 2πa 2 The normalized frequency of cut-off or V–number = µ1 − µ22 λ Here λ = 0 . 85 µm

∴

V=

2 × 3 .14 × 35 (1. 46)2 − (1. 438)2 0 . 85

= 258 . 589 √ (2 .132 − 2 . 068) or (iii) The total number of guided mode in the step-index fibre is, 2

N=

V 2 (65 . 42) = = 2139 modes 2 2

V = 258 . 589 × 0 . 253 = 65.42

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Example 10: Compute the maximum value of ∆ and µ2 (cladding) of a single mode fibre of core diameter 10 µm and core refractive index 1.5. The fibre is coupled to a light source with a wavelength of 1.3 µm. V cut-off for single mode propagation is 2.405. Also calculate the acceptance angle. 2πa NA (cut-off) λ Vλ 2 . 405 × 1. 3 3 .1265 NA (cut-off) = = = = 0 . 09957 2 π a 2 × 3 .14 × (10 /2) 31. 4

Solution: We know that, V cut-off =

∴ Acceptance angle α = sin −1(NA) = sin −1 (0 . 0995) = 5.71° We know that,

NA = µ1√ (2 ∆) or ∆ =

or

∆=

or

∆=

We know that, or ∴

(NA)2 2 µ12

(0 . 09957)2 2 × (1. 5)2

0 . 0099 = 0.0022 4 .5 µ − µ2 or µ1 − µ2 = µ1∆ ∆= 1 µ1 µ2 = µ1 − ∆ . µ1 µ2 (cladding) = 1. 5 − 0 . 0022 × 1. 5 = 1.497

Example 11: A glass fibre has a core material of refractive index 1.47, cladding material of refractive index 1.45. If it is surrounded by air, compute the critical angle (i) at the core– cladding boundary, (ii) cladding-air boundary. Solution: If µ1 is the refractive index of core, µ2 that of cladding and µ0 that of surrounding air, then (i) Critical angle at the core-cladding boundary 1. 45  µ   θ c = sin −1  2  = sin −1   µ1   1. 47 or

θ c = sin −1(0 . 9864) = 80.5 °

(ii) Critical angle at cladding-air boundary  1  µ   θ c′ = sin −1  0  = sin −1   µ2  1. 45  or

θ c′ = sin −1(0 . 6896) = 43.6 °

Example 12: A step index fibre has core refractive index 1.466, cladding refractive index 1.46. Compute the maximum radius allowed for a fibre, if it supported only one mode at a wavelength 1300 nm. Solution: A fibre will support only one mode if its cut-off parameter V is less than 2 . 405.That is V < 2 . 405 or

2πa 2 µ1 − µ22 λ

< 2 . 405

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or

a