21,345 1,986 18MB
English Pages 573
Table of contents :
Cover......Page 1
Contents......Page 8
1. Ray Optics......Page 9
2. Wave Optics......Page 250
3. Dual Nature of Radiation & Matter......Page 367
4. Atomic & Nuclear Physics......Page 436
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ADVANCED
JEE PHYSICS for both Main & Advanced levels of JEE
OPTICS & MODERN PHYSICS
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ADVANCED JEE PHYSICS for both Main & Advanced levels of JEE
OPTICS & MODERN PHYSICS
RAHUL SARDANA M.Sc (Hons). Physics
7
. Ef·,1
=1~11; __I l
ELSEVIER A division ofRe¢ Elsevier India Pvt. Ltd.
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ADVANCED JEE PHYSICS, Optics & Modem Physics Rahul Sardana ELSEVIER A Division of Reed Elsevier India Private Limited. Copyright © 2014 Effective Testbooks India (ETI)
This book contains information obtained from authentic and highly regarded sources. Reprinted material is quoted with permission, and sources are indicated. Reasonable efforts have been made to publish reliable data and infonnation, but the author and the publisher cannot assume responSibility for the validity of all materials or for the consequences of their use. Neither this book nor any part may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, microfilming and recording, or by any information storage or retrieval system, without prior remission in writing ~om the publishers. The consent of ETI does not extend to copying for general distribution, for promotion, for creating new works, or for resale. Specific permission must be obtained in writing from ET/ for such copying. The export rights of this book are vested solely with the publisher. ISBN: 9789381269978
Published by Elsevier, a division of Reed Elsevier India Private Limited, under special arrangement with ETI. Registered Office: 305, Rohit House, 3 Tolstoy Marg, New Delhi110 001. Corporate Office: 14th Floor, Building No. JOB, DLF Cyber City, PhaseII, Gurgaon122002, Haryana, India .. Printed and bound at Rajkamal Electric Press, Kundli, Haryana.
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text overview, goals & focus
In the past few years, the UTJEE has evolved itself as an examination designed to check out true scientific skills. The examination pattern wants us to see those little details which, others fail to see. Those details which tell us how much in depth we should know to describe as much as possible. Keeping the present day scenario in mind, this book is written for students, to allow them not only to learn the tools that "Optics & Modem Physics" provides but also to see why they work so nicely in explaining the beauty of ideas behind the subject. The central goal of this text is to help the students develop a thorough understanding of the principles of "Optics & Modem Physics". This book stresses on building a rock solid technical knowledge based on firm foundation of the fundamental principles than on a large collection of formulae. The primary philosophy of this book is to act as a guide who creates a careful, detailed groundwork for strong conceptual understanding and development of problem solving skills like mature and experienced physicists. features of book theory with illustrations
"Optics & Modem Physics" are important topics, and in this book I have tried to make these topics lively, clear and precise to the greatest levels. I have generally seen students not stressing on the theoretical details. They always feel that doing more numerical problems will solve their purpose. But let me tell you here, that numerical problems are just the special cases of the theoretical concepts. The entire Physics is based on a simple program "IF~ THEN~ ELSE". Try to follow this and see how you get to your ultimate goal i.e., IIT~JEE. So, keeping this in mind, the entire theory part of all the chapters has been kept elaborative, simple to understand with supportive Illustrations at all the places. DO NOT TRY TO ATTEMPT ILLUSTRATIONS WITHOUT GOING THROUGH THE THEORY. conceptual notes, remarks, words of advice, misconception removal
Throughout the text, the Conceptual Notes and Remarks are highlighted which focus on the principal ideas and concepts that a student must take care of. Places where students commonly develop a misconception have been supported by Misconception Removal, highlighted in grey and supported by Words of Advice. Throughout my teaching career of 18 years I have always found my students getting benefitted from these Conceptual Notes, Remarks, Words of Advice, Misconception Removals. All these are actually used to provide warnings to the. students about common errors and ways to avoid them. problem solving techniques
These techniques, highlighted in grey, always ensure that the students become capable of solving a variety of problems in an easy way. Wherever necessary, the text is supplemented with them for having a thorough understanding to the application processes. in chapter exercises (ice) : topic wise
After you study the theory and apply it to the Illustrations, its time you practice something on your own and that too topic wise. For this purpose I have created In Chapter Exercises (ICE) (except for Mathematical Physics). Each ICE has the name of the topic(s) clearly mentioned on it. Please note that ICE are based on simple, single . concept classification technique. They are fully solved, so that if you come across only problem, then you just refer to the solutions.
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C:::::
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solved problems
After you have gone through the entire Theory (with Illustrations) and all the supplements (ICE, Conceptual Notes, Remarks, Words of Advice, Misconception Removal, Problem Solving Techniques), its high time to do problems that are a true mix of concepts studied. This section has problems that involve multiple concept usage so that your brain is exposed to the ultimate throttle required to extract the best from you at !ITJEE. practice exercise sets {fully solved)
Now comes the time when you are very much ready to do the practice as per the !ITJEE pattern. This section contains all the variety of questions that have been asked in the !ITJEE. In this section you will come across the following variety of questions. single correct choice type (SCCT)
Each question, in this section, has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. multi11le correct choice t}'pe (MCCT)
Each question, in this section, has four choices (A), (BJ, (C) and (D), out of which ONE OR MORE is/ are correct. reasoning based questions/ assertionreason type (ART)
This section contains Reasoning type questions, also called AssertionReason type question, each having four choices (A), (B), (C) and (D) out of which ONLY ONE is correct. E;ach question contains STATEMENT 1(Assertion) and STATEMENT 2 (Reason). You have to mark your answer as Bubble (A) If both statements are TRUE and STATEMENT 2 is the correct explanation of STATEMENT 1. Bubble (B) If both statements are TRUE but STATEMENT 2 is not the correct explanation of STATEMENT 1. Bubble (C) If STATEMENT 1 ls TRUE andSTATEMENT2 is FALSE. Bubble (DJ If STATEMENT 1 is FALSE butSTATEMENT2is TRUE. linked comprehension tyl)e (LCT) / paragraph type
This section contains Linked Comprehension Type Questions or Paragraph based Questions. Each set consists of a Paragraph followed by questions. Each question has four choices (A), (BJ, (C) and (D), out of which only one is correct. (F_or the sake of competitiveness there may be a few questions that may have more than one correct options). matrix match type (MMT) / column matching
Each question in this section contains statements given in two columns, which have to be matched. The statements in COLUMNI are labelled A, B, C and D, while the statements in COLUMNII are labelled p, q, r, s (and t). Any given statement in COLUMNI can have correct matching with ONE OR MORE statement(s) in COLUMNII. The appropriate bubbles corresponding to the answers to these questions have to be darkened as illustrated in the following examples:
If the correct matches are A > p, s and t; B > q and r; C > p and q; and D > s and t; then the correct darkening of bubbles will look like the following : P
q
r
s
t
A@@©©CD B@@©©CD c@@©©CD D@@©©CD
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t=:J
integer answer type questions (IATQ) / numerical type questions
In this section the answer to each of the question is a four digit integer, ranging from O to 9999. The appropriate bubbles below the respective question numbers in the ORS have to be darkened. For example, if the correct answer to question number X (say) is 6092, then the correct darkening of bubbles will look like the following: X.@e@@
CD CD
G)
CD
®@®• ®@®®
@@©@
®@®®
~~~~
®®®® @@e@ answers & solutions
Each· chapter contains answers followed by solutions to the problems. The solutions· are exhaustive with complete methods and reasons which will help you a lot to understand a particular concept. Short cuts are also included (wherever necessary) forenhancing you problem solving skills. This book, I hope, will nourish you with the concepts involved such that you get a great rank at !ITJEE. To conclude, I apologise in advance for the errors (if any) that may have inadvertently crept in the text. I would be grateful to the readers who bring errors of any kind to my attention. I truly welcome all comments, critiques and suggestions at [email protected]. PRAYING TO GOD FOR YOUR SUCCESS AT !ITJEE, GOD BLESS YOU!
The Author RAHUL SARDANA M.Sc.(Hons.) Physics
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CHAPTER 1
RayOptics (a) Reflection at Plane & Curved Surfaces _ _ _ _ _ _ _ _ _ _ 1.1 (b) Refraction at Plane Surfaces
1.27
(c) Refraction at Curved Surface
1.55
Solved Practice Problem
1.96
Practice Exercise Sets :> Single Correct Choice Type Questions _ _ _ _ _ _ _ _ J, 108
=
Matrix Match Type Questions (Column Matching Type), _ _ _..,,,40
:i
Integer Answer Type Questions
.43
Answers to In Chapter Exercises (ICE) & Practice Exercise Set
.45
Solutions to In Chapter Exercises (ICE)
.48
Solutions to Practice Exercise Sets
.52
CHAPTER4
Atomic & Nuclear Physics
:>
Multiple Correct Choice Type Questions
1. 135
AtomicPhysics ..... _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 4_7
:>
Reasoning Based Questions (Assertion Reason Type)
1. 141
Nuclear Physics....
4.21 4.48
:>
Linked Comprehension Type Questions (Paragraph Type) .......... 1. 144
Solved Practice Problem
:> :>
Matrix Match Type Questions (Column Matching Type) .............. 1.154
Practice Exercise Sets
Integer AnswerTypeQuestiohs ___________ 1,160
:i
Single Correct Choice Type Question~4.58
Answers to In Chapter Exercises (ICE) & Practice Exercise Sets ____ 1,163
:i
Multiple Correct Choice Type Question
Solutions to In Chapter Exercises (ICE) _ _ _ _ _ _ _ _ _ _ _ 1, 169
:i
Reasoning Based Questions (Ass'ertion Reason Type) _ _ _ _ 4.78
Solutions to Practice Exercise Set
:,
Linked Comprehension Type Questions (Paragraph Type) ............. 4.81
:,
Matrix Match Type Questions (Column Matching Type) ................ .4.89
:,
Integer Answer Type Questions _ _ _ _ _ _ _ _ _ _ _4.92
1. 193
CHAPTER2
Answers to In Chapter Exercises (ICE) & Practice Exercise Set,_ ____4.94
Wave Optics WaveOptiCS~1 Solved Practice Problems....
.41
Solutions to In Chapter Exercises (ICE) _ _ _ _ _ _ _ _ _ _ _4.99 Solutions to Practice Exercise Set
Practice Exercise Sets :i
Single Correct Choice Type Question,,_ _ _ _ _ _ _ _ _..,_.52
=
Multiple Correct Choice Type Question
.66
:i
Reasoning Based Questions (Assertion Reason Type)
.69
:i
Linked Comprehension Type Questions (Paragraph Type) ............. 2.71
:i
Matrix Match Type Questions (Column Matching Type) ................. 2.77
:i
Integer Answer Type Questions
.80
Answers to In Chapter Exercises (ICE) & Practice Exercise Set.~~,.83 Solutions to In Chapter Exercises (ICE)
.86
Solutions to Practice Exercise Sets
.92
CHAPTER3
Dual Nature of Radiation & Matter Dual Nature ofRadiqtion &Matter _ _ _ _ _ _ _ _ _ _ _ _ _..,,. I Solved Practice Problem
4J3
.1 S
Practice Exercise Sets :i
Single Cofrect Choice Type Question•s.._ _ _ _ _ _ _ _ __,_2 7
o
Multiple Correct Choice Type Question
.33
:,
Reasoning Based Questions (Assertion Reason Type)
.36
:,
Linked Comprehension Type Questions (Paragraph Type) .............3.38·
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I Ray Optics
Contents
 ··
 
  
      
'
(a) REFLECTION AT PLANE & CURVED SURFACES ..................................................... 1.1 (b) REFRACTION AT PLANE SURFACES ......................................................................1.27 (c) REFRACTION AT CURVED SURFACES ...................................................................1.55 Solved Practice Problems ..........••....•...•.•....•...•••••.••••...••....••...••...•••..•••....••...•....•.•..••....•••..1.96
Practice ·Exercise Sets
:> Single Correct Choice Type.Questions .•.....••..............•........................................1.108 :,
Multiple Correct Choice Type Ques1ions .............................................................1.135
:> Reasoning Based Questi?ns (Assertion Reason Type)...................................... 1.141 :,
Linked Comprehension Type Questions (Paragraph Type) ................................. 1.144
::> Matrix Match Type Questions (Column Matching Type) .........................••....•...... 1.154 ::> Integer Answer Type Questions ..................................................................•••.....1.160 Answers to In Chapter Exercises (ICE) & Practice Exe!cise Sets .................._................. 1.163 Solutions to In Chapter Exercises (lCE) ...........................................................................1.169 Solutions to Practice Exercise Sets ................................. ;...............................................1.193
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NATURE OF LIGHT: An Introduction
E
ight is a form of energy that makes object visible to our eyes or light is the form of energy that produces in us the sensation of sight. In Seventeenth century Newton and Descartes believed that light consisted of a stream of particles, called corpuscles. Huygens proposed wave theory of light and proposed that light is a disturbance in a medium called Ether. This theory could explain the
L
phenomena of interference, diffraction, etc. Thomas Young,
through . his double slit experiment, measured the wavelength of light. Maxwell suggested the electromagnetic theory of light. According to this theory, light consists of electric and magnetic fields,.in mutually perpendicular directions, and both are perpendicular to the direction of propagation. Heinrich Hertz produced in the laboratory the electromagnetic waves of short wavelengths. He showed
=c B In 1905, Albert Einstein revived the. old corpuscular theory using Plank's Quantum· Hypothesis and through his photoelectric effect experiment showed that light consists of discrete energy packets, called photons. The energy of each photon is
E=hf=hc A,
So, in view of these developments, light must be regarded to have a dual nahlre i.e., it exhibits the characteristics of a · particle in some situations and that of a wave in other sihlations. So the question'' Is light a particle or a wave?" is
purely inappropriate to be asked. At present, it is believed that light has dual nature, i.e., it has both the characters,
wavelike and particlelike.
that these electromagnetic waves possessed all the properties OPTICS : An Introduction
of light waves.
E
Optics is the study of the properties of light, its propagation through different media and its effects. In most of the sihlations, the light encounters objects of size much larger
than its wavelength. We can assume that light travels in straight lines called rays, disregarding its wave nature. This allows us to formulate the rules of optics in the language of geometry, as rays of light do not disturb each other on intersection. Such study is called geometrical (or ray) optics. It includes the working of mirrors, lenses, prisins, etc. When light passes through very narrow slits, or when it
Direction of
propagation
ii
passes around very small objects, we have to consider the wave nature of llght. This study is called wave (or physical) optics.
Light travels in vacuum with a velocity given by
1
. . 8'
c= ~=3x10 ms
Vµoto
1
·
where µo and Eo are the permeability and permittivity of free space (vacuum). The magnitudes of electric and magnetic fields are related to the velocity of light by the relation · ··
DOMAINS OF OPTICS
The study of light can be categorized into three broad domains. a) Geometrical Optics (Ray Optics) b) Physical Optics (Wave Optics) 1.1
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Optics & Modem Physics
c) Quantum Optics Please note that these domains are not strictly disjoint as the transitions between them are continuous and not sharp. However for convenience we consider them as distinct. These domains are distinguished as follows.
5.
The Laws of Refraction (discussed later). The Laws of Refraction govern the bending of light when the light goes from one medium to the other (rarer to denser or denser to rarer) medium.
BASIC TERMS & DEFINITIONS A.
GEOMETRICAL OPTICS (RAY OPTICS)
This branch involves the study of propagation of light based on the assumption that light travels in fixed straight line as it passes through a uniform medium and its direction is changed when met by a surface. of a different medium or if the optical properties of the medium are non uniform either in time or in space. The ray approximation is valid for the wavelength A. very small compared to the size of the obstacle (d) or the size of the opening through which the ray passes.
SOURCE
A body which emits light is called source. ·Tue source can be ·a point one or an extended one. A source is of two types. a) Self luminous : The source which possess light of its own. EXAMPLE : sun, electric arc, candle etc.
Nonluminous : It is a source of light which does not possess light of its own but acts as source of light by reflecting the light received by it.
b)
This approximation 1,. « d proves to be very good for the study of mirrors, lenses, prisms and associated optical instruments such as microscope, telescope, cameras etc. B.
PHYSICAL OPTICS (WAVE OPTICS)
This branch involves the study of propagation of light in the form of a wave and it deals with the phenomenon of interference, diffraction, polarization etc. This nature of light has to be taken when the light passes through very narrow slits or when it goes past very small objects. So this branch works effectively when 1,. » d . C.
QUANTUM OPTICS
This branch involves the study of propagation of light as a stream of particles called as Photons. This concept of light behaving as particles called photons is of utmost importance while studying the origin of spectra, photoelectric effect, concept of radiation pressure, Compton effect etc. FUNDAMENTAL LAWS OF GEOMETRICAL OPTICS
To a first approximation, we can consider the propagation of light disregarding its wave nature and assuming that light propagates in straight lines called rays. This allows us to formulate the laws of optics in the language of geometry. Thus, the bfanch of optics where the wave nature of light is neglected is called geometrical (or ray) optics. Geometrical optics is based ~m five fundamental laws. 1. Law of Rectilinear Propagation of Light. It states that light propagates in straight lines in homogenous media. 2. Law of Independence of Light Rays. It states that rays do not disturb each other upon intersection. 3. The Law of Reversibility of Light. According to this law, if a ray of light, after suffering a number of reflections and refractions, has its path reversed at any instant, then the ray retraces its path back to the source. 4. The Laws of Reflection. The Laws of Reflection govern the bouncing back of the incident ray after striking a surface to the medium from which it was coming.
=
EXAMPLE : moon, objects around us, book etc. REMARK(S) Sources are also classified as isotropic and nonisotropic. Isotropic sources give out light uniformly in all directions whereas nonisotropic sources do not give out light uniformly in 1~ll_dir~~!i?n~. _
RAY
The straight line path aiong which the light travels between two points ina homogeneous medium or in a pair of media is called a Ray: It is represented by an arrow head on a straight line, the arrow head represents the direction of propagation of light. A ray of light will always follow a path along which the time taken is the minimum.
    .  . ., 'Ray
I
REMARK(S)
_A ~ingle_!~~~ann_?~ be_~~ol~.t~~ !r ··::::>
sini = sinr i =r.
{The Law'of Reflection}
FERMAT'S PRINCIPLE OF LEAST TIME
According to this theorem, light will always follow the path taken between any two points by a ray of light is the path along which the time taken is the minimum. ·This principle is sometimes taken as the definition of a ray of light.
=
I PROBLEM SOLVING TRICK(S) I a)
Basic Problems in Optics : Most of the problems asked in
L_
optics. expect us fo find tl)e position and na!ure of the final imag·e formed bycertain optical systems for a given abject. The ..aptical_system_may b~ jul;;t a mirror, m~a lernr or a__
I I
1.4
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Ray Optics: Reflection at Plane & Curved'Surfaces combinatiOn of several reflecting and ~r9fractinQSurface5. b)
Basic Strategy for Solving the Problems : To handle these kinds of problems, first of all, we identify the sequence in which the reflection and refraction are taking place. The several events of reflection or refraction can be
named as Event 1, Event 2 and so on following the · sequence in which they occur. Now; the ihlage of Event t would be object for Event 2, image of Event 2 will be object of Event 3 and so on. This
way one can proceed to find the flnal Image.
I
ANGLE OF DEVIATION (6)
· Deviation (6) is defined as  the angle between the initial direction of the incident ray and the final direction of the reflected ray or the. emergent ray.
Illustration 1 Two plane mirrors are inclined to each other at an angle 0 . A ray of light is reflected first at one mirror and then at the other. Find the total deviation suffered by the ray.
Deviation produced in Reflection is 6 = 180° (i + r) Since r=i
=>
                CONCEPTUAL NDTE(S) The same ls fol!nd to'hold·good for three plane mirrors arranged mutually perpetldicular to each other thus fohning the comer of a cube such that the light incident on this arrangement suffers one reflection from each of the mirrors so as to emerge out antiparallel to the incident light. This arrangement of three mutually perpen_dicular plane mirrors fonning the comer of a cube is . called the CORNER REFLECTOR.
6=180°2i
The variation of deviation (6) with the angle of incidence (i) js shown in figure.
Solution a be the angle of in~idence for mirror M 1
p be the angle of incidence for mirror M, 61 be the deviation due to mirror M 1 and
8_= Jt
62 be thE! deviation due to mirror M 2 D
0
The deviation is minimum for grazing incidelice i.e., when
I I . I
i+i,then 6=6rr,;0 =0°.
1
: PROBLEM SOLVING TRICK($) • :he deviation is maximu~ for normal incidence i.e., when
I ,a) I· b)
[ c)
1=0 then, o=o= =180 .
!
While dealing with the case of multiple reflections suffered by a ray, the net deviation suffered by the incidentray is the : algebraiC; sy.m of deviation due to.each single reflection. So, 6total
= L 6!ndMdual
,_
""""""
' DO NOT FORGET TO TAKE INTO ACCOUNT THE SENSE OF I ROTATION WHILE SUMMING UP THE DEVIATIONS DUE TO SINGLE REFLECTION. 1'
TWO IDENTICAL PERPENDICULAR PLANE MIRRORS
If two plane mirrors are inclined to each other at 90° , the emergent ray is always antiparallel to the incident ray if it suffers one reflection from each (as shown in figure) whatever be the angle of incidence.
Fiom figure, we observe 61 =1t2a, 62 =1t2P Also ray is rotated in same sense i.e., anticlockwise, so Snet =~i'=Total deviation =31 +l\
=>
6=2it2(a+P)
Now in !J.OBC , LOBC + LBCO + LCOB = 180°
=>
(90'a)+(90°p)+0=1so 0
=> =>
a+P=0 6=,2it20=360°~20
1.5
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Optics & Modern Physics
Advanced JEE Physics Alternative Method: 5 = LBEC + LCEA + LAED
Illustration 2 Find the coordinates of the location of the image formed for an object kept at origin as shown in figure. y·
Now, LBEC = LAED (vertically opposite angle) =>
LBEC=180°2(o:+p)
=>
LBEC = 180° 28
Also, LCEA=2o:+2P
=>
{·: 8=o:+P)
=>
LCEA=2(0::1P)=28
5=(180°28°)+28+(180°28°) 5.=360°28
=>
REFLECTION FROM A PLANE SURFACE OR PLANE MIRROR 
When a real object is placed in front of a plane mirror, the image is always erect, virtual and of same size as the object. It is at same distance behind the mirror as the object is in front of it.
Solution The first thing we obsenre is that the object is virtual, because the ray of light is converging on plane mirror. Also, the coordinates of object are (0, 0, 0) and the image co
ordinates are the reflection of object coordinates in the mirror as shown in figure.
0
"4
d t+f d +t
14 8
dd
(a) Point Object
(b) Extended Object
LATERAL INVERSION
The image formed by a plane mirror suffers lateralinversion. That is, in the image the left is tu.med to the right _ and viceversa with respect to object. Howev.er, the plane mirror .does not turn up and down, as shown in figure.
0 cm +1
The image lies on norma~ of mirror at I . From MOP, we have
=> =>
sin(30°) = PO 8 PO=4cm
OI = 2(PO) = 8 cm So coordinates of I are X
= 8cos(60°) = 4 cm,
y=8sin(60°)=4.J3 cm and
Z=O Image
Objeet
Image Actually, the plane mirror reverses forward and back in threedimensions (and not left into right). If we keep a righthanded coordinate system in front of a plane mirror, only the zaxis is reversed. So, a plane mirror changes righthanded coordinate system (or screw) to lefthanded. y
Objeet
y
So, the coordinates of image are (4, 4../3, 0) ; PROBLEM SOLVING TRICK(S)
j For _finding, the :loC:Btion of an image of a point object placed in i front of. a plane· mirror, we must see the perpendicular distance··of
l the object f_rom the mirror.
I
·
,,.! 0
J•x 0
z Right Handed
=
System
'
¾
•' 
x'+to·
Coirect OM=Ml
z' Left Handed System
1.6
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Incorrect OM=Ml
I

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Ray Optics: Reflection at Plane & Curved Surfaces
=
FIELD OF VIEW OF AN OBJECT'
Suppose a point object O is placed in front of a mirror, then a question arises in mmd whether this mirror will form the image of this object or not. The answer is yes, it will form. A mirror, irrespective of its size, forms the ini.ages of all objects lying in front of it. But every object.has its own field of view for a given mirror. Field of vi_ew is the region where diverging rays· from object or image are present. If our eyes are present in field of view then only we can see the object or an image as the case may be. Field of view of image is decided by rays which get reflected or refracted from the extremeties or the extreme ends of the mirror or a lens and depends on the location of the object in front of mirror or lens.
MINIMUM SIZE OF A PLANE MIRROR TO SEE A COMPLETE IMAGE CASEI: To find the minimum size· bf mirror to see a full image we use the fact that light rays from extreme parts of object should reach eye after reflection from mirror. Let us consider following two situations a) The minimum size of mirror to see one~s full height is H where H is the height of man. To see full image 2 mirror is positioned in such a way so that rays from head and foot reach eye after reflection from mirror, as shown in the figure.
r(x+y)
y
_J __
f y
i
Field of view of image
E Man.
b)
Field of view of object
CONCEPTUAL NOTE(S) It _has been observed that a convex mirror gives a wider field of view than a plane mirror. Therefore, the convex mirrors are us~d as rear view mirrors in vehicles. Though they make the estimation of distances more difficult but still they are preferred because for a large movement of the object vehicle there Is only a small movement of the image.
A ray starting from head (A) after reflecting from upper end of the mirror (F) reaches the eye at C . Similarly the ray starting from the foot (E) after reflecting from the lower end (G) also reaches the eye at C . In similar triangles ABF and BFC AB=BC=x (say) Similarly in triangles CDG and DGE , we have CD=DE=y (say)
Now, we observe that height of the man is
2( x + y)
and that
the length of mirror is (x + y) , i.e., the length of the mirror is half the height of the man. Please note that the mirror can be placed anywhere between the centre line BF (of AC) and DG (of CE); CONCEPTUAL NOTE(S)
a) Field
Field
of
view of
convex , mirror
of
0
view of a
plane
b)
In· order to see full image of the man, the mirror is positioned such that the lower edge of mirror Is at height half the eye level from the ground. Minimum size is independent of the·distance between man and ·mirror.
mirror
CASEff: The minimum length of the mirror required to see the full image of a wall behind the man who is standing at the
middle of the mirror and the wall is ; , where H is the ·height of wall. The ray diagram for this situation is shown in figure. 1.7
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Optics & Modern Physics
Advanced JEE Physics A
·
f 2x
. 360°
Ca'.culate · · 8
t
i 1
(x+y)
.
360' is an INTEGER
(X;cY)
f
8
n·
. 360' is a· FRACTION 8
•.,
•
2y
:t Wall
C
Man I+ d
ltis an ODD• lnteg"ef i~~
IMsan EVEN! Integer
Mirror d+t
In mangles HBI and IBC let HI= IC= x . Now, in mangles HBI and ABF , we have
\
~
AF HI
FB BI
AF
2d
n is an Integral' Part of fraction
Object lies symmetrically ' on th~ angle
Object lies . UnsYmmetrically
bisector
;=a
~
AF=2x _Similarly if, CK= Kf = y·, then DG = 2y. Now, we o]:,serve
that height of the wall is 3(x+ y) while that of the mirror is
(x+y).
·
360' n='_(·   1')
: . a
,~~. ....:.:__,.' • CONCEPTUAL NOTE(S! . 
a)
If an object is placed between two•parallel mirrors (8 = 0°), .
·the number of im?Q_;s formed Wm,be infinite, ~b)
NUMBEROF IMAGES IN INCLINED MIRRORS
Let 0 be the angle between two plane mirrors and' n be the number of images formed:
360 if 360 is odd 8 ' · 8 Then n= (360 ) . 360. [ 1 , if 1s even
8
Further when
~=[(3:
1
_ ),
:o
3
,,''
w
'
M,
J.
J
360 is odd, then ·0
360
Further if
?1
8
0
a'
All th8·. images. lie on a,_clrcle with radius equal to the distanc~, "~etween ~e object_,;~ and th0 point· of intersection the mirrors c·. • '• The riu(nber ·of images formed may b8 different from the numberfOf Images see_n (which depends on .the· position ·of,. the observer). · · · ·
1

:
a
1~ 0=9no,.~,., l"
~6~
,.,.'I
.
"..~., M; "
.l12 I,. _______ ,! __ '.'° ___ ,
if.object lies symmemcally on the
liz
angle bisec:to! of tw~ mirrorS
if object lies tinsymmemcally ROTATION OF A.PLANE MIRROR
is a fraction, then the number of images
form~d will be integral part of the fr~ctlon e.g.
il
:o
3
is 4.8,
then, n = 4 . Following diagram shows _the process to calculate n . Net deviation produced by two plane mirrors.inclined at an angle 0 is 6=360°28 Clearly 6 is independent of the angle of incidence of the ray oflight. · · · ·
When a mirror is rotated by an angle 0 (say anticlockwise), keeping the incident ray fixed, then· the reflected ray rotates . by 28 along the same sense, i.e., anticlockwise. y(or N) N (ory)
'. ' ''' ' . '' ' j,•j
N' ..
R
t
R'.
'' '' \' :...._,.~ ,, . \
•· 'I
\ i0
,a•
,,
i0~
'
Initially
=
On Rotation of Mirror
1.8
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Ray Optics: Reflection at Plane & Curved Surfaces
Let I be the incident ray, N the normal and R the reflected ray, then on rotation, I remains as it is, N and R shift to N' and R'. From the two figures we can observe that the reflected ray earlier made an angle i with yaxis while after rotating the mirror it makes the angle (i20). So, we conclude that the reflected ray has been rotated by an angle 20.
d$ + 2 da =O dt dt d$ =~2 da dt dt So, the angular speed of the reflected ray is double the angular speed of the mirror. Since, y = xtan~ dy =xsec'$d$ dt dt
CONCEPTUAL NOTE(S) If a plane mirror rotates with angular velocity ro , then the
reflected ray rotates with angular velocity 2ro (excluding rotation
Since l~;l=2oo
of ~~~r ~ith n?r~al ~s_t~e ~is).
=>
Illustration 3
A plane mirror hinged at O is free to rotate in a vertical plane. The point O is at a distance x from a long screen placed in front of the mirror as shown in figure. A laser beam of light incident vertically downward is reflected by the mirror at O so that a bright spot is formed at the screen. At the instant shown, the angle of incidence is 8 and the mirror is rotating clockwise with constant angular velocity o, , Find the speed of the spot at this instant.
Normal (N)
,/ ,,
Screen
2
l'Jil=(xsec $)(2oo)
So, the speed of the spot is
1:1
= 2xoosec' $
VELOCITY OF IMAGE IN A PLANE MIRROR
To understand and interpret the moving images of moving objects in front of plane mirror, we must understand the following cases. CASE!: Object moving along the normal to the plane mirror which is at rest. All velocities measured w.r.t. grOund frame. y
8 ,'
0 ·
X
V V 0___.,_ ______ ,....._I
. Mirror ··::.._.;
Lx
00
Solution Let P be the bright spot, shown on the screen. Let the distance of point P from 0 1 be y at this instant shown in figure. Then according to the problem we need to calculate dy dt
(v.= 0)
Velocity of object with respect to mirror is Varn= vi Velocity of image with respect to mirror is
vhn =vi Velocity of object with respect to image is V01 =V0 V1 =(2v)i CASEII: Object moving parallel to the plane of mirror (at rest)
i
!: _____ J
From the figure 0+0+$=90°
... (1)
8+$+a=90°
... (2)
=> =>
a=0 $+20=90°
=>
$+2a=90°
Lx
I
l (v. = 0)
Velocity of object w.r.t. mirror is
Dom =vj 1.9
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Velocity of image w .r. t. mirror is
Solution
= vj
Vrm
Velocity of object w.r.t. image is iio1
=0
CASE III:
Object moving neither along the normal nor alohg the parallel to the plane mirror (at rest).
1
o~Vo,
~ 1 Vr,,,
The component of velocity of image perpendicular to mirror is
Lx
½=2Vm  \1
0
°" (v,L =2
J
={ijOm)along mirror
; Since, both the object and the image approach the mirror with i equal and ORposite speed, so we have
(Vimtormaltomlrror = (Vam)no,ma]tomirror
VIVm=(VaVm)
=>
V1 =2VmVo
=>
1
V, =./100+64 =./164 ms and 0=tan'(¾)
Illustration 5
(v,.)11 =(vem)u
=>
(v,): +(V,),:
Velocity of image(½)=
A point object is moving with a speed of 10 msI in front of a mirror moving With a speed Of 3 msI as shown in figure. Find the velocity of image of the object with respect to mirror, object and gro~nd.
t··
; Step 2 : Then the velocity of image w.r.t. mirror is
Solution
V,m =.(~1m\ +(V1m)1. , However, velocity' of image w.r.t. any· other observer, say A is
: then given by f
I
,
(v™)_,_ =(s./3131)=(5./3+3)1 ms1
For component of velocity parallel to mirror, we have
(iit~)1 (v0M)1 = v0
iiM
=
5}o =5}
= c:::================================= 1.10
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Ray Optics: Reflection at Plane & Curved Smfaces Since,
=>
=>
(vlM)=(v,Mt +(v1M)
2. 11
(v,M)=(5,J3+3)i5f
A ray of light travels from paint A to a point B afterbeing reflected from a plane mirror as shown in figure. From where should it strike the mirror?
B t
'' '' ''' '' '
(s,/3+3)1sJ=v,(31) A
=>
(s,/3+3)i+31s]=v,
=>
v, =[(s,/3 +6)1sJ] ms'
=>
v,0 =(5,/3+6)1sJ(s,131sJ)
=>
v10 =(5,/3 +6+5,J3)i +(55)]
=>
v10 = (10,ffi + 6)1 ms1
t
5cm
'
1+ 20
3.
20cm
cm +1
A plane mirror is inclined at an angle 8 =60° with horizontal surface. A particle is projected from point P on the ground (see figure) at t =0 with a velocity v at an angle ct with horizontal. The image of the particle is observed from the frame of the particle projected. Assuming the particle does not collide the mirror. Find the time when image will come momentarily at rest with respect to particle.
Illustration 6
A plane mirror in yz plane moves with a velocity
3i
as shown in figure. An object O starts moving with a
velocity 4/ + J 4k . Find the velocity of the image.
V
0
a p
GROUND
4.
Two plane mirrors are inclined to each other such that a ray of light incident on the first mirror and parallel to the second is reflected from the second mirror parallel to the first mirror. (a) Find the angle between the two mirrors. (b) Also calculate the total deviation produced in the incident ray due to the two reflections.
5.
Two plane mirrors M1 and M2 _ are inclined at angle 0 as
0•
Mirror (M)
Solution
e
Since the mirror is placed in yz plane, so the y and z components of the velocity of the image remain the same as that of the object. However, perpendicular to the mirror, the velocity of approach of object towards the mirror is always equal and opposite to the velocity of approach of the image towards the mirror, so, we have
shown in figure. A ray of light 1, which is parallel to M1 strikes M2 and after two reflections, the ray 2 becomes parallel to M2 • Find the angle
M,
e. 2
(voM); =(v,M), =>
(v 0 ),(vM), =(v,), +(vM),
=>
(v 1 ), =2(vM),(vo).
=>
(v,), =2(31l41 =lOi
So,
6.
v, =10i + ]4k
Calculate the deviation suffered by an incident ray in the situation shown in figure after it suffers three successive reflections.
M,
~
ICE I c;;l
1'
'
j
•.. 50°
· · ~~.
BASED ON REFLECTION AT PLANE SURFACES. (Solutions on page 1.169) :
1.
A ray of light travelling in the direction
i(i + .Jsj)
is incident
30°
M,
on a plane mirror. After refleciion, it travels along the direction
iU .Jsj).
Find the angle of incidence.
7.
Two plane mirrors are placed parallel to each other and 40 cm apart. An object is placed 1O cm from one mirror. 1.11
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Advanced JEE Physics Find th8 distance frolTl the object tO the iesp0ctive image for each of the five images that are closest to the object. 1
a.
13.
A ray of liQht is incident on 8.11 arrangement of two plafle: mirrors inclined at an angle 0 with each other. It suffers two ' reflections one from each mirror and finally moves in a , direction making angle u with the incident ray ( a is acute). , Find the angle a and show that it is independent of angle 1 of Incidence.
14.
A ray of light is incident at an angle of 30" with the horizontal. At what angle with horizontal must a plane mirror ! be placed in its path so that it becomes' vertically upwards after reflection? '
Find the number of images formed of an object O enclosed , by three mirrors AB , BC , AC having equal lengths ln ·
situation shown in figure.
A
6
!
0
, 9.
..
· ··c
s··
15.
A point l\>Ource of light S , placed at a distance L in front of , the centre of a mirror of width d, hangs vertically on a wall. A man Walks in front of the mirror along a line parallel to the mirror at a distance 2L from it as shown. Find the greatest distance over which he can see the image of the light · source in the mirror.
l }1
A ray of light is incident on a plane mirror along a vector j T+ J k . The normal at the point of incidence is along
Find the smallest size of a looking glass which a man with a : face 24 cm x 16 cm should purchase that will enable him to see his·whole face completely, lf the (a) man is one eyed.
(b) man is two eyed. Given that the separation between his eyes is 8 cm . In what direction should A beam of light is to be sent from point A (shown in figure) contained in a mirror box for it to fall onto point B after being reflected once from each of the four walls. If the points A and B are in one plane perpendicular to the walls of the box (i.e., in the plane of the , drawing) then in what direction should the beam be sent , fromBtoA?
A small curved reflecting surface can be considered to be a part of a sphere. Hence, such surfaces are called spherical mirrors. Depending upon the surface silvered, these are of two type&eoncave and convex, as shown in figure. Some important terms are described below. a) Pole or Vertex: Centre P of the surface of the mirror. b) Centre of Curvature : Centre C of the sphere. c) Radius of Curvature : Radius R of the sphere. d) Principal Axis : Line PC , joining the pole and the centre. e) Linear Aperture : Distance XY between the extremities of the mirror surface. Note that Sil).ce lenses are also made of spherical surfaces, the above terms also apply to lenses, except that the pole is replaced by a new term called as Optical Centre.
•A
Silvered
surface
•B
112.
ap surface
10 ms1
/ od3°~0bject
Mirror
Lx
axis
y , , ',
R
A
/ y
' , Concave' .. ____ ... _.,.,. Convex mirror mirror
1
 30° 
_f_ri_!l_g!J)_aJ
/I~
Silvered
The object and the mirror move with velocity shown in , figure. Calculate the velocity of the image. 5ms
I
REFLECTION FROM CURVED SURFACES
i+2L+I
11.
16.
i + j . Find a unit vector along the reflected ray.
i.L_..
' 10.
1
Two plane mirrors are inclined to each other at an angle of 70" . A ray is incident on one mirror at an angle 8 . The · ray reflected from this mirror falls on the second mirror from where it is reflected parallel to the first mirror. Find the value 9.
Important Terms and Definitions a) Centre of curvature : It is the centre of the sphere of which the mirror/lens is a part. b) Radius of curvature : It is the radius of the sphere of which the mirror/lens is a part. c) Pole : It is the geometrical centre of the spherical reflecting surface of which the mirror/lens is a part.
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· Ray Optics: Reflection at Plane & Curved Surfaces d) e)
f)
Principal axis (for a spherical mirror) : It is the straight line joining the centre of curvature to the pole. Focus : When a narrow beam of rays of light, parallel to the principal axis and close to it, is incident on the surface of a mirror (lens), the reflected (refracted) beam either converges to a point or appears to diverge from a point on the principal axis. Th.is point is called the focus (F) . Focal length (for a mirror) : It is the distance between pole and the principal focus (F) .
Real image : If reflected (or refracted) rays converge to a point (i.e. intersect there), then the point is a real image. h) Virtual image: If reflected (or refracted) rays appear to diverge from a point, then the point is a virtual image. i) Real object : If the incident rays diverge from a point, · then the point is a real object. j) Virtual object : If incident rays converge and appear to intersect at a point behind the mirror (or lens), then the point is a virtual object. g)
SIGN CONVENTIONS FOR MIRRORS
While solving problems, we must follow a set of sign conventions given for convenience. According to this sign convention a)
Origin is placed at the pole ( P) .,
b) c)
All distances are to be measured from the pole ( P) . Distances measured in the direction of incident rays are taken as positive. Distances measured in a direction opposite to that of the incident rays are taken as negative. Distances above the principal axis are taken as positive. Distances below the principal axis are taken as negative. This sign convention is used to find the position and nature (virtual or real, erect or inverted) of the image formed by the mirror (or lens). Object distance is denoted by u , image distance by v , focal length by f and radius of curvature by R.
d) e) f)
g)
h) i)
PARAXIAL RAYS
Paraxial rays are the rays which are either parallel to the principal axis or make small angles with it i.e., these rays are nearly parallel to the principal axis, Our treatment for the spherical mirrors has been restricted to these rays and due to this we shall be considering the curved mirrors that have smaller aperture. However, for the sake of convenience, comfort and clarity, we shall be drawing the diagrams of larger size.
Note that generally we keep the object to the left of the mirror (or lens), so that the ray of light starting from object must go from left to the right i.e., towards positive direction of xaxis. Now· since the distances have to be measured from the pole consequently, u must always be negative, v is positive (for a virtual image) and negative (for a real image). / is positive (for a convex mirror) and negative (for a concave mirror). For both the mirrors and lenses. Magnification for a real image is negative i.e., mreal = 0
FOCUS AND FOCAL LENGTH
Magnification for a virtual image is positive i.e.,
When a narrow beam of light, parallel to the principal axis and close to it, is incident on the surface of a mirror (lens), the reflected (refracted) beam is found to converge to or appears to diverge from a point on the principal axis. Titis point is the focus also called Principal Focus in case of mirror(s). The plane passing through the focus and perpendicular to the prigcipal axis is called focal plane.
mvmua1
=EB CONCEPTUAL NOTE(S)
The convention that all distances measured along the ray of light are ,positive and all distances measured opposite to the ray of light are negative matches exactly with the Cartesian coordinate system, where we can simply place the origin at the pole P and say that all distances to the left of the pole are negative, all distances to the right of the pole are positive, all distances above the pole are positive and all distances below the pole are negative.

C
+
+ '
Incident Ray:
' ''P
+.~+ Concave mirror
Focal length
(f)
Convex mirror
'' ''
is the distance of focus (F) from the pole
''
t'
(p) of the mirror or the optical centre for a lens. j)
For solving problems in which any of u, v, f(or R) is to be found, we must make ·sure that no convention should be applied on the quantity to be found. The
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Advanced JEE Physics tmknown quantity will automatically take up its sign from which we shall make obvious conclusion. The diagrams show the application of sign convention to curved mirrors.
k)
., 1(±)1
··c F
,
:'...t:'
\~_~:
• _)
b)
F
C
,i·
'
' ce~.J r
' :+t ~'
'' '' ''8''
,
' f=.B.:'
·...~)
2 ,'
'\
• _)
frincipal Axis
e'
p•
F
C
,____,;::>+\·_~...i,i],+
r,
'f  R '' '1
l~_i_j r
I
2 ,'
t§:}_~] '
'
:+ f ~
.,._

· Fc
P.is Pole, Fisfocus and C is·Centre of Curvature 1
 .......
', ..... _ rc·
I
...... :(£)1
(
',
p p
A ray of light passing through the focus (in case of concave mirror) or appearing to pass through the focus (in case convex mirror) is reflected parallel to the principle axis.
!+ R
:... t:
'._ ...
c)
',::~ ... ..,.... __
 '..F
C
A ray of light passing through the centre of curvature falls .normally on the mirror and is therefore reflected back along the same path i.e., retraces its path.
'8' ,_ _ !  .
R .:
  ........ __ _ F
RULES FOR OBTAINING IMAGE
These rules are based on the laws of reflection, i.e: the angle of incidence equals the angle of reflection, i = r and are used to find the location, nature (real or virtual, inverted or erect) and size of the image formed by a spherical mirror. Take any two rays coming from any given point on the object. Find out at which point these rays actually meet (or appear to meet) after reflection from the mirror. This point is the real (or virtual) image. In this way, taking one point after another on the object, the entire image can be constructed. a) A ray of light coming parallel to principal axis, after reflection passes through the focus (in case of co:ri.cave mirror) or appears to come from the focus (in case of convex mirror).
d)
C
Incident and reflected rays at the pole of a mirror are synunetrical about the principal axis. (Because for the pole principle axis acts as normal and by Laws of Reflection i = r ). So by observing the size of erect image in a mirror we can decide the nature of the mirror i.e., whether it is convex, concave or a plane mirror.
M
M
M"
M'
IMAGE FORMATION BY CONCAVE MIRROR
I
OBJECT POSITION_
At infinity
DIAGRAM
POSITION OF IMAGE
~
/ C
At the principal focus ( F)
,p
F~
or in the focal plane
~
=
NATURE OF IMAGE
1.14
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Real, inverted and extremely diminished
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Real, inverted and
Beyond C
Between F and C
At C
At C
Real, inverted and of same size as the object
Between F and C
Beyond C
Real, inverted and magnifi~d
At F or in the focal plane
At infinity
Between F and P
dirn)nished
Real, inverted and highly , magnified
Behind the mirror
Virtual, erect and magnified
I
IMAGE FORMATION BY CONVEX MIRROR
I
OBJECT POSITION
For all positions of object
DIAGRAM
0
F
C
POSITION OF IMAGE
NATURE AND SIZE OF IMAGE
Images formed between the Pole and the focus (F) .
AIWays forms a Virtual, Erect and Di_minished Image
p~
RELATION BETWEEN FOCAL LENGTH (!) AND RADIUS OF CURVATURE (R)
A ray parallel to the principal axis passes through the focus (as in concave mirror) or appears to pass through the focus (as in convex mirror). The normal to the mirror(s) •at the point of reflection i.e., A must pass through the centre of curvature. In triangle CAN , we have
tani= AN NC
For paraxial rays and_mirrors of small aperture, we have ~ieei=·AN NC In triangle FAN , we have
... (1)
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Optics & Modem Physics P=a+i
tan(2i) = AN NF
and y=r+P
CONCAVE MIRROR

f .........
.....  1C

..... ...i1 2i
A,
!~
,F
' ''

'' ' '''
 A f''.::....
CONVEX MIRROR
~
·,,1
pp p(
r
' ... , .................. _ 2·' I
 ....I
F,
'' '
'
'' ' t,t f...:
R
R
 ,c
... (2)
From (1) and (2), we get
2 (AN)=AN NC NF 2 1 =>    ... (3) NC NF Since, aperture is small, so N coincides with P, so we have NC e; PC and NF" PF For convex mirror, we have
=>
,1
' ''' ''' _ _ _ V +I ,
'' ' ''' '' '
Again for paraxial rays and mirror of small aperture, we have
tan(2i)" 2i = ~~
0
o+R
i+u>< Since by Laws of Reflection, we have i=r =>
a+y=2P
Applying paraxial ray approximation, we get AP AP AP tana=a=, tanP.,P= and tany=y=. PO PC PI => tana+tany=2tanp =>
AP+ AP = 2 (AP) PO PI PC
Using sign conventions, we have· PO=u, PI=v and PC=R
1 1 2 +=
PC=+R and PF=+f
=>
t=¾
R Since we know that / =  , so we get
(u) (v) (R)
2
For concave mirror, we have 1 1 2 1 +==u v R f
PC=R and PF=f
f=~ 2
So, for a curved mirror of small aperture, focal length is half the radius of curvature. MIRROR FORMULA
Consider a point object O placed on the principal axis of a concave mirror. A ray of light, incident on the point A at an angle of incidence i on the mirror. makes an angle r with the normal as sho\\ n in the figure. From the Laws of Reflection we know that i = r . Further to find the location of the image let us take another ray along the principal axis so that it hits the mirror normally at the point P to reverse its path and meet the other ray at I . This point of intersection of the two rays happens to be the place where the image is formed. Since from geometry we know that in a triangle, external angle equals sum of internal opposite angles, so for triangle CAO and triangle CAI , we have 1.16
For Convex Mirror Similarly we can drive a formula for a convex mirror. Since from geometry we know that in a triangle, external angle equals sum of internal opposite angles, so for triangle CAO and triangle CAI, we have i = a.+y
For Concave Mirror
=
(Mirror Formula}
and P=r+y
0
p
' ' 13"  ... _ ___c r·lr' v=60 cm According to the mirror formula, we have 1.18 ==================================
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1
1
When the object lies between F and P , then the image becomes virtual i.e., u and· f are negative while v is positive. So from mirror formula we get,
+=v u f =>
v1 + u1 =11 = constant
1 v
Differentiating with respect to time, we get
1 (u)
1
+=
(!)
v2 dv ~2 du =O
dt
~
dt
~
~:=(::)~; dudt
Here
... (1)
is the rate at which the object distance
u is
changing i.e., it is the object speed if the mirror is stationary. Similarly, dv is the rate at which v (distance between
dt
image and mirror) is changing i.e., it is image sPeed if the mirror is stationary. So if at a known values of v and u, the object speed is given, we can find the image speed from the above formula.
.1 =>
1
1
=u V f
u'(~;)v'(!:)=o (~:)=(::)(~:)
Now, when u is further decreased, v also decreases to keep 1 ·  constant. So, du.th  is e rate at w hichb. o ~ect 1s f di approaching towards mirror and ( ~:) is rate at which the image is approaching towards the mirror.
::;1'
+
C
,;,"
/
,
p
"'~,..,.,,,+ P
/
'' I
+
""'oa+,
'' ''' ' 1'
I
''t+ V +I
'>+u+
u . Therefore, image speed is more than the object speed. Thus, the above entire discussion can simply be concluded as follows. CONCLUSION
(du) = rate of decrease of u dt . =>
(
!~) = rate of increase of
{object speed) {image speed}
v
Further, when the object lies between oo and C, then v < u , {from equation (1)) Hence, when the object is moved towards the mirror, its image (which is real) will recede from the mirror with speed less than the speed of object. When the object is at C, image is also at C ~
When an object is moved from oo to F , the image (real) moves from F to oo and then when the object is further moved from F to P image (now virtual) moves from +«.> to P .
~CONCEPTUAL NOTE(S) ·When the object Is either at centre of curvature C ot at pole P , the two speeds are equal. When the object is at pole, then due to the small aperture of the mirror, it car) be assumed, as if the image is being formed by a plane mirror.
·
FINDING COORDINATES OF IMAGE OF A POINT
If the coordinates of a point object (x 0 , y0 ) with respect to the coordinate axes shown in figure are known to us and the coordinates of image be (x;, Yi) then for finding the
V=U
xcoordinate, we use the mirror formula, acc0rding to which 1
Hence, when the object is at C speed of image is equal to the speed of object. When the object lies between C and F then v > u
1
1
+=v u f 1
1
1
+=X; Xo f
so, the image speed is more than the object speed. 1.19
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Optics & Modern Physics
Advanced JEE Physics
=___Eg_
X
=>
So, image coordinates are (  ~O, 1) cm
Xof
i
7
I
y
I PROBLEM SOLVING TRICK(S)
i\
a)
(Xo, yo)•
_
II
___.,\,________.x
,
bj
JP
..!.+..!.=! V U f
For finding the ycoordinate, we apply the concept of magnification (m) , according to which, we have
=>
I EXAMPLE ! mirror of focal
'I
I A point object is placed at
Place the object to the left of the mirror (or lens), so that sign convention matches with the familiar sign convention in the coordinate geometry. Both for concave as well as convex mirrors, use ,he same mirror formula i.e .
(40, 1) cm in front of a concave
length 5 cm having Its pole at the origin (0, 0). Assuming the principal axts to be along xaxis, flrid
Substitute the numerical values of the given quantities with proper sign (+ve or ve) as per sign convention. I Though the SI unit of distance' is metre, it may be more ' d) convenient in some problems to take the given distances, in cm rather than in m. But then your answer too will be in cril. e) Do not ,give any sign to .the quantity to be determined. In your answer, the unknown quantity will be obtained with its proper sign. In addition to the above hints, if you remember the following facts, it will ·help you. a) Since the object is always placed to the left of the mirrc;>rso, u is always negative.
b) c) d)
the pOsitlon of th.e image form~d.
e)
SOLUTION The si!uation discussed in the problem is shown in figure. yaxis
f)
' g)
I
i h)
7cm \p 'l__.~xaxis
Ii ")
(0,0)
,/
I
l 1, ' , '' ..,__4ocm1>1 '
U
: c) I
1
(\
and m=~
For a concave mirror, f is negatiVe. For a convex mirror, f is positive. A real image is formed in front of the mirror, so for a real image v is neg·ative.
A virtu·a1 image is formed behind the mirror, so for a virtual : image v Is positive. 1 A real Image is always inverted, so for a real image h· IS negative. ·A virtual image is always erect, so for a virtual image h is positive. · For the real image of a real object and the virtual image 6f. a virtual object, m is negative. For the virtualimage of a real object and the real image of a virtual object, .m is positive.
GRAPH BETWEEN
!
VERSUS
V
1
1
40
v = ,. cm 7
. h1 V S1nce, m·==h0
h0
Hence, the mirror formula i.e.,
U
"'.~
1
40
1
I
1
1
1
v
u
I
=>
+=
But h0 =7 cm
=>
=+
0
=
1
v u
=>
h, =.1 cm
 ··· ·
1.20

u
I
=
~=! h 7 
! + ,! =.!. becomes V
(~o}
h,
U
Let us first take the case of a concave mirror. Here, two cases are possible. Case .1 : 'When the Image formed is Real. When the image is real, i.e., object lies between F and infinity. In such a situation u, v and f are negative.
1
+v ("40) (~5) =>
!
_____
_\
1 v
1 u
1
I
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Ray OpHcs: ReflecHon at Plane & Cnrved Surfaces Comparing with y = mx + c, the desired graph. )Viii be a straight line with slope 1 and intercep,t on yaxis is equal to 1
1
1
1
1
1 1
v+ Cu) =t ~
f
=,+v u I
Comparing with y = mx + c , the desired graph is a straight line_
Do not confuse here, the slope m with magnification. 1/v
ofslope m = ,1 and intercept on yaxis equal·to
i.
The graph i$
thus shown in figure.
1/1~.
1/v
[/
I~
4 ""~"1/u 1/f
/ 1/f
Case 2 : When the Image formed is Virtual. When the image is virtual, i.e., object lies between l' and P . Under such situation .u and / are negative whil~ v is positive. The'mirror formula thus becomes 1 1 v1 u f
,/
.. ~45° ~....cc=11/u PROBLEM SOLVING TRICK(S) a)
Comparing it with y = mx + c the desired graph is a straight . . .· .. . 1 line with slope m ': 1 andintercept on yaxis is equal to 
1
As focallength of a spherical mirror f
=~ dep·ends only on
the radius of mirror and is independent of wavelength of light and refractive index of medium so the 'focal length of a spherica! mir_ror in air or water' and for rec;J or blue light is same. This is also why the image _formed by mirrors do not show Chromatic aberration.
.
1/v b)
In case of'spherical mirror if R+ oo (i.e., it becomes plane), f=
...J._ _...,
1
t
1
1
;+ (20) = (15)
=> v=60 cm Negative sign with v means that it is formed to right of pole P1 at a distance of 60 cm from P1 , ( 10 cm· behind M, ). V
=>
.
(60)
m, =;;= 20 =3
So, image ( 11 ) formed is real, inverted and three times size
of object i.e., 6 mm. This image
(I1 )
formed now acts as object for the convex
mirror. Further, this image formed is 10 cm to the left ~f P, and the incident ray from the original object goes to the right
M, 20 cm +1 30 cm +1 50 cm       ; M i+60cm
1 1 +=v 10 20 v=20 cm
u=20 cm
'!, =15 cm
(20) u 10 So, image (I) formed is virtual, erect and two times the size of object (here 11 ). Hence the size of I is 12 mm . So, finally I is formed at 20 cm in front of convex mirror M2 , with size 12 mm ; virtual and erect.
Since, .!:_ + .!:_ = .!.
Illustration 12 Find the coordinates of image of point object P formed after two successive reflections in figure, considering the first reflection at concave mirror and then at convex mirror. y f.:,=20 cm
=>
=>
V
V
m2 ===2
f,=15cm P
1 1 1 +=v, (20) (15)
=>
v1 =60 cm
So, magnification m., = _.'.J. =  GO = 3 (Inverted) u 20
~OE\'_.X 2mm
M,
M, t+20 cm_.,.
A'P' =m, (AP)=3x2=6 mm
For reflection at convex mirror M2 , we have u=+l0cm
f, =+20 cm Since.!:.+.!:.=.!.
'
a
=>
V
u
1
1
I
1
+=v, 10 20 v 2 =20 cm
gnif' . v, (20) . 1cation, m2 =   =    = 2 Agam, ma u 10
=>
C'P'=m,(CP')=2x8=16rnrn
So, the coordinate of image of point object P as measured from the origin O is (30 cm, 14 mm)
1+SOcmiM
=
I
=>
~
+
u
1.24
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Ray Optics: Reflection at Plane & Cttrved Sttrfaces (b)
BASED ON REFLECTION AT CURVED SURFACES
(c)_
(Solutions on page 1.173)
1.
2.
3.
An object of height 2.5 cm is placed at a 1.5 f from a concave mirror, where f is the magnitude of the fcical length of the mirror. The height of the object is perpendicular to the principal axis. Find the height of the image. Is the image erect or inverted?
the. ball· at time t =
X=O
8.
0
•
A
6.
B
• I State whether it is a convex mirror or a concave mirror. (b) ,Draw a ray diagram to locate.the mirror.and its focus. Write down the steps of construction of the ray diagram. Consider the possible two cases: (i) When distance of I from AB is more than the distance of O from AB and (ii)
When ·distance ·of o· from AB is more than the distance of I from AB
9.
Convex and concave mirrors have the same radii of curvature R. The distance between the mirrors is 2R. At what point on the common optical axis of the mirrors should a point source of light A be placed for the rays to converge at the point A after being reflected first on the convex and then on.the concave mirror?
1D.
An object ABED is placed In front of a concave mirror beyond centre'of curvature C as shown in figure. State the shape of the image.
A concave mirror has a radius of curvature of 24 cm . How far is an object from the mirror when the image formed is (a) virtual and 3 times the size of the object. (b) real and 3 times the size of the object and
real and
An Image I is, formed· of a point object 0 by a mirror whose principal axis is AB as shownin figure.
(a)
Find the distance of Object from a concave mirror of focal length 1O cm so that image size Is four times the size ·o_t the object.
(c)
T is time period of
C x~axis+H
' \ 1 ~m __________ _
5.
, where
oscillation?
A mir/'or (in a laughing gallery) .forms an erect image fbur times enlarged, of a boy standing. 2.5 m away. Is the. mirror concave or convex? What is its·radiu~ of curvature?
A concave mirror forms the real Image .of a point source lying on the optical axis at a distance of 50 cm from the mirror. The focal length of the mirror is 25 cm . The mirror is cut' in two halves and these halves are drawn apart at a distance of 1 cm in a direction perpendicular to the optical axis. How will the images formed by the halves of the miriOr be arranged?
T
2
(! __+____________ _ 4.
to the image ofthe swinging bait At what point does the ball appear to coincide with Its image. What will be the lateral magnification of the .image of
¾the size of the object?
A thin flat glass plate is placed in front of a convex mirror. At what distance b' from the plate should a point source of light S be placed so that its image produced by the rays reflected from the front surface of the plate coincides with the,image formed by the rays reflected fi'om the mirror? the focal length of the mirror is f = 20.cm andthe distance from the plate to. the ·mirror a= 5 cm . How can the coinciderice of the images be established by direct observation?
1B_,_E_____J\P A
,,.
D
C
F
I
An_ object is 30 cm from a spherical mirror, along the central axis. The absolute value of lateral magnification of an inverted image is
i.
Find the focal length of the mirror?
ab+t 12. 7.
A ball swings back and forth In front of a concave mirror. The motion of the ball is described approximately by the equation x =fcos(rot), where f is the focal length of the mirror and x is measured along the axis of mirror. The origin is taken at the centre of curvature of the mirror. (a). Derive an exRression for the di_stance from the mirror
A thinrod of length
½is placed along the principal axis of a
concave mirror of focal length f such that its image just ,__ _,,toLJchesJhl;! rod. Calc_ulate magnifica_te;io,..n~._______...,
1.25
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REFRACTION OF LIGHT AT PLANE SURFACES
T
he phenomenon of the bending of light rays as they travel from one medium to the other is called . Refraction. The surface separating two media is called an Interface. In other words, the phenomenon of bending of light rays at the boundary between two media is called refraction. Incident Ray N
A
: Normal Medium 1 (µ,)
i'
,
Interface
, ,
,
Q1
1r
~
~
~
1,Medium 2 (µJ
:
!,
!, I t,
ABSOLUTE REFRACTIVE INDEX
j
t,
Refracted Ray
LAWS OF REFRACTION
a)
b)
REFRACTIVE INDEX (RI)
The refractive index of a medium is not determined by its density. It is,govemed by the velocity of light in the medium. The lesser the value of the velocity of light, the more is the refractive index of the medium, and the denser. is the medium. A medium having greater refractive index is called denser medium whereas the other medium is Called rarer medium.
1' t>
B ~
is simply called as the Absolute Refractive Index of medium 2, expressed as µ 2 or simply µ .
The incident ray, the refracted ray and normal at the point of incidence to the surface separating the Mo media all lie in the same plane. Snell's Law For two media, the ratio of sine of angle of incidence i to the sine of the angle of refraction r is constant (for a beam of particular wavelength). For a given set of media this constant is called the refractive index of the medium 2 with respect to medium 1 (represented as 1µ2) i.e., sini µ . =constant= 2 = 1µ 2 smr µ1
The absolute refractive index of a medium is defined as the ratio of the speed of light in vacuum to the speed of light in the medium, speed of light in vacuum C >1 µ speed of light in medium V Absolute refractive index is more than one because the speed of light is maximum in vacuum/ air. RELATIVE REFRACTIVE INDEX
The relative refractive index of medium 2 with respect to medium 1 is denoted by 1 µ 2 and is given by
1
_
µ, _(
µ, ;,( (SNELL'S LAW)
:J:J_ v, v,
The relative refractive index of medium 1 with respect to medium 2 is denoted by 2 µ 1 and is given by
OR µ 1 sini=µ 2 sinr where µ 1 and µ 2 are Absolute Refractive Indices of Medium 1 and 2 respectively and 1 µ 2 is the refractive index of medium 2 with respect to medium 1. If medium 1 happens to be the vacu~m, then the constant
2
:J:J
µ, ( v, µ,==(=µ, v,
===================================== = 1.27
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Optics & Modern Physics
Advanced JEE Physics CONCEPJ"UAL NOTE(S) a)
1
The velocity of light in air Is nof.much different from that in vacuum. Hence, while defining the refractive index of a
me_dium we often take velOcity of light in air rather, than that in vacuum Medium
Refractive Index'"'
Water
.±=1.33 3
=> =>
sini < sinr
i'< r
REFRACTION : IMPORTANT POINTS
' a)
Whenever light goes froni. one medium to another, the frequency of light (/) remains unchanged. Since µ=
b)
V
, !=1.50
Glass
2
Relative refractive index can be less than one. If we calculate the refractive index afwater with respect to glass, then
where A., and "modi= being wavelengths of light in air and medium respectively.
(34)
8 ' 9µ =·µw ==
R'=2RsecrRtani Also, we observe that R' =BC +CM =2Rtanr+Rtani From equations (1) and (2), we get 2secrtani = 2tanr + tani => secrtanr=tani
. . . (1)
space/vacuum, during the same time intervals for both the media.
Let me illustrate this thing to you. For that let me take_ two media, one rarer of length L1 , refractive index µ1 and other
... (2)
denser of length L2 and refractive index µ 2 , as shown. 2
RARER(µ,)
DENSER(µ')
V1={f;
V2= ~2
L,..
L,..
Using the concepts of trigonometry, we get
,tani
1sinr cosr
Time taken by light to travel a distance L, in rarer medium . h speed v = C 1s • wit 1
µ,
=>
tani
t,=h=(L')=µ,L, V1
... (1)
C
The time taken by light to travel a distance L2 in denser
=> =>
:,
2i =!:__!..
2 2
medium with a speed v2
... (3)
=_..:... µ,
is
... (2)
Also, from Snell's Law, we get
sini=v'3sinr Solving the above equations, we get
... (4) Now if both times are equal, as said above, then
R' =.fi.R So the new area of shadow is
=
tl = t2 ~
µ1L1
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= µ2L2
... (3)
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Ray Optics: Refraction at Plane Surfaces Since, Optical Path Length (OPL) is the distance travelled by light in vacuum/air/rarer medium during the same time it travels a distance L2 in medium. So, from (3), we get
=µairLair = µmediumLmedium =µ2L2 Since, µair =1 , so, we get OPL = Lrur = µmediumLinedium
10
d
and t, =
v: = (3x10')
6x1015 s
1.8
µlLl
So, tmin (b)
{for same time in air and medium} Since light always travels slower in denser medium, s6 the OPL (the distance in air corresponding to same time in both) is always longer than the actual thickness L of the medium.
= 2 X 10
15
s
The total number of wavelengths in a film of refractive index µ , thickness d is n = Optical Path Length Wavelength of Light ~
µd
n='/,.
So, total number of wavelengths, is
CONCEPTUAL NOTE(S)
n = µ,d, + µ,d, + µ,d,
Also, note that µ 1L1 is OPL in air/rarer medium and µ 2 L2 is OPL
'/,.
in denser medium. However for standard purposes OPL is the distance travelled by light in vacuum/air to travel a distance L in _ a medium during the same time in either air or medium. ______ _
So, from equation (3), we conclude that for a pair of media, Optical Path Length ) = (Optical Path Length) ( in Air/Rarer Medium in Denser Medium
1,,
'/,.
~
1 n=i:(µ 1d1 +µ 2 d2 +µ 3 d3 )
~
n=
~
n = 1000 (4.5) = 4500 = 75 600 600
10xlO_, ((1.2)(1)+ (l)(l.5)+(1.8)(1)) 600
=µ2L2
=>
µ1 L1
~
L2 µ, L1
µ,
Since µ 1 < µ 2 , so we get
L2 < L1 Due to this reason, L2 is also called the Reduced Thickness.
So, in general, we get R~duced) = ( µra= ( Thickness µdenser
)L,=, = OPL in air
Illustration 21 A light ray enters the atmosphere of a planet and descends vertically 20 km to the surface. The index of refraction where the light enters the atmosphere is 1 and it increases linearly to the surface where it has a value 1.005. How long does it take the ray to traverse this path. Solution Since variation is linear, so we have
x0 µ1
=
µdenser
Illustration 20 A light beam of wavelength 600 nm in air passes firstly through film 1 of thickness 1 µm and refractive index n1 =1.2 , then through an air film 2 of thickness
1.5 µm and finally through film 3 of thickness 1 µm and refractive index n3 = 1.8 . Which film does the light cross in the least time and what is that least time? (b) Calculate the total number of wavelengths (at any instant) across all three films together.
~
2xl0 4 0 1.0051
0.005x µ= 1 + 2xl0 4
Now, by definition, we have C
µ(x) = v(x)
~
v(x)
= µ[x)
(a)
where µ(x)=µ=l+ O.OOS: and c=3xl0 8 rns1 2x10 X=O, µ= 1
i
Solution
(a) Since, 11
Similarly,
d,
v,
f2
10' (3xl08) 1.2
d, C
x,µ
4x1015 s
1.5x10' 3xl08
2x1015 s
X, =
2 X 10' m, µ= 1.005
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=>
=> =>
3xl08
3xl0 8 0.005 1 +  4x 2xl0
C
v(x) = µ(x)
X
di
1 di = , [(1+2.5x10'x)dx] 3xl0 t 1 2> =>
4 )'
1 ..
1 .
C
V
=>
smz=smr
=>
sini C ==µ sinr v
[The Law of Refraction}
µ 1 sini =µ 2 sinr
0
1 =1[ 2 xl0' + (2xl0 3xl08
sini
VECTOR FORM OF SNELL'S LAW
fdl=, f (1+2.5x10'x)dx 0
dx
Jb 2 +(dx)'
3xl08 1+2.5x107 x
dx
~ini and
Using our knowledge of cross product of vectors, we have
(2.5x10')] 2
ixfi=(l)(l)sini, 0 outwards
5
1=6.68x10 s
i
LAWS OF REFRACTION USING FERMAT'S PRINCIPLE
Consider a refracting surface / interface separating medium 1 from medium 2. Let the incident light start from A , in medium 1, hit the surface at O and get refracted to a point B, in medium 2. Let the points A and B be at perpendicular distances a and b_from the interface. Further, let A and B be at a separation d as shown in figure. The time taken by the light to go from A to O to B is t=tA...,.o+to)oa
Medium 1 (µ 1) Medium
A
n
(µ,)
'
and ilxf=rxfi=(l)(l)sinr, 0 outwards So, from above we conclude that µ 1 (1 X fi) = µ 2 (f X fl)
=> =>
Ja' +x'
Jb' +(dx)'
C
V
I+~
Now, according to Fermat's Theorem, t is MINIMUM, so In Medium 1
..'!!_=0
In Medium2
dx
REFRACTION THROUGH A COMPOSITE SLAB
'
.',o ''
(dx)
X
Interface
I+
=>
=>
Medium2
1
'r
''
=>
Medium 1 (vacuum)
d
' :' r'
b
!
B +I
J:..i( Ja' +x' )+J:..i( Jb' +(dx)') C
di
1
V
2x
di
1 ( 2(dx)(1)) 0
2c Ja' +x' + 2v Jb +(dx)' 2
1
X
cJa'+x'
1
(dx)
vJb'+(dx)'
From the figure, we observe that
=
Consider the refraction of light ray through a series of media as·shown in figure. The ray AB is incident on interface X1Y1 at an angle i . The ray is deviated in medium 2 along BC towards the normal. Then it falls on interface X2Y2 and is again deviated towards normal along CD. If the last medium is again Medium 1, the ray emerges parallel to the incident ray. Let r1 and r2 be angles of refraction in Medium 2 and Medium 3 respectively. Then from Snell's Law, sini µ 1  .  = 2= µ2 ... (1) smr1 µ 1 sinr smr2
µ
2
 . 1= 3= µ3
... (2)
sinr2 µ 1 3  .. == µ1
... (3)
Slill
µ2 µ3
µ 1 = refractive index of medium 1 µ2
µ3
= refractive index of medium 2 = refractive index of medium 3
·= ==================================== 1.34
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Ray Optics: Refraction at Plane Sn,faces A
Solution 1
X,
f
Given, wµ 8 =2. and aµ,=~
8
Y,
2
t.
'
X,
3
t
~
~
I
X,
Y, Y, ~
Multiplying (1), (2) and (3), we get 1µ2x2µ3x3µ1 =1
=>
LATERAL SHIFT ON PASSING THROUGH A GLASS
tµ2x2µ3=1µ3
In general if a ray passes through a number of conposite parallel plate glass slabs, then 1 µ2 X 2µ3 X 3µ4 X 4µ5
= 1µ5
Illustration 22
A light beam passes from a parallel plate glass slab of refractive index µ 1 placed·in a medium of refractive index µ 1 • Show that the emerging beam is parallel to the incident beam.
SLAB
Consider a ray AO incident on the slab at an angle of incidence i through the glass slab EFGH of thickness t . After refraction the ray emerges parallel to the incident ray. Let PQ be perpendicular dropped from P on incident ray produced as OQ . The lateral displacement caused by plate, x = PQ = OPsin(ir)
OM . (·1r) x=sm
~
Applying Snell's Law at A, we get µ 1 sini =µ 2 sinr1
~
1:!.= sinr1
~
µ2
/._OP= OMl cosr
l
COST
Solution
COST
... (1)
sini
t . (.ir ) x=sm
t (smzcosrcosismr . . .. ) x=COST
~
x = t(sini cositanr) A
Air(µ= 1)
µ,
µ,
Applying Snell's Law at B, we get µ 2 sinr2 = µ 1 sine
=>
1:!. = sinr2 µ2
sine
Air(µ= 1)
,e
''
' N'
... (2)
B
'
sini . Smce µ=.
From equation (1) and (2), we get i=e i.e., the emergent ray is parallel to incident ray.
smr
=> ::::::>
Illustration 23
Refractive index of glass with respect to water is 9/8 . Refractive index of glass with respect to air is 3/2 . Find the refractive index of water with respect to air.
. sini smr=
µ
tanr
sini
~
1.35
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If i is very small then r is also very smalt hence, sin i + i , and cos i + 1, Then expression for lateral displacement takes the form.
b)
At near normal incidence (small" angle of incidence~ i )' apparent depth (d') is given by
Observer ;;;11. Medium 'I
:=n(1¾)
f I
RARER RARER
APPARENT DEPTH
An object O placed in a medium of 'refractive index µ is observed from air at a small angle a to the normal to the interface (in figure, angle a is shown exaggerated for clarity) i.e., for near normal incidence.
d
l
'' 'A '' ''
'
B "'
,, '
fo·•"~,'
6X
',p
io '
Observer Medium
µrerawu
,
tanp
d =distance of abject from the' interface
µ==,S=
surta'c8.'

·
\\,
·
..
Illustration 24 A fish rising vertically to the surface of water in a lake at a uniform speed of 3 ms1 • It observes that a bird diving vertically towards the water at a uniform speed of
_!, find the 3
Solution Let x be the depth of the fish F below the surface of water, and y be the height of the bird B above the surface
d
Apparent depth, d' = 
µ
=d(1¾)
at an instant.
To the fish, the bird will appear to be farther away, at an apparent height y' given by y
In case the object is seen through n number of slabs with different refractive indices, the total apparent shift is simply the sum of individual shifts, so Ax=Ax1 +Ax2 +&3 + .... +filn
=
Perp~ndicu1ar
actual speed of dive of the bird.
The apparent shift in normal direction (or the normal shift) in the position of the object is
a)
··,
v :;= v13locity of object perper:idicular to interface, relative to surface. v: = ve1acity at iffiage ta interface relative ta
9 ms1 • If the refractive index of water is
Therefore, from Snell's Law, we get, smt sina. tana. d sinr sinP tanp d'
=,,
= real depth.~ = apparent
d' = distance of image from the interface depth.  · ,..
d d'
dX= dd'
RI of medium of incidence obje_qt
Medium(µ)
P: ' ''' !'
Since angles a. and j3 are small, so sin a. ~ tan a. and sinp "tanp.
=:,
µrolaUvo
where µ ra· = ro live RI of medium of refraction observer
Air
H the object O is at a real depth d from the interface, its apparent depth d' can be calculated. From t,.s ABO and ABO', tancx
4
' d  iV d' =  and v' =·
~
.,'
'' d'
'~·
DENSER
DENSER
( µ""'.'m•H•m.) µfishmedium =:,
ax=ai(1 :J~a,(1:J+a,(i:J+ . +a"(i :J . CONCEPTUAL NOTE(S) If the medium in which the object is placed is rarer (µ,,< 1') and it is seen from the denser medium, the apparent shift calculated will be negative. If means· that the object apparently shifts away from the observer. If the shift comes out to be ,positive, the ima9e of the· object shifts towards the observer.
'
y
y = (;) =µy
The total apparent distance of the bird from the fish is s=x+y' =:>
s=x+µy
Differ~ntiating w.r.t. time t, we get C
ds dx dy dt ~ dt + µ dt
1.36
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!f
•B'
SHIFT OF POINT OF CONVERGENCE OR DIVERGENCE
r
•B
y
If a glass slab of thickness t., refractive index µ is placed in the path of a convergent (or divergent) beam of light, the point of convergence (or divergence) gets shifted by
s
Air Water
X
~={1¾)
i •F µ
Substituting the values, we get 9=3+(±) dy 3
dt
i+t ....
1+t+1
Therefore, the actual speed of dive of the bird is given by
(B) Divergent beam
(A) Convergent beam
dy dt
=(93) (~) = 4.5 ms' 4
,.
Illustration 25 A vessel is filled with a nonhomogeneous liquid whose refractive index varies with the depth y from the
free ·surface of liquid as
µ·= ( 1
+;}
Calculate the
apparent depth as·seen by an observer from above, if H is the height to which the liquid is filled in the vessel.
Illustration 26 A point object O is placed in front of a concave mirror of focal length 10 cm . A glass slab of refractive
index µ =
¾and thickness 6cm is inserted betweenobject
and. mirror, Find the position of final image when the distance x shown in figure is (a) 5 cm (b) 20cm 6cm
Solution Let us consider a thin layer of liquid of thickness dy at a distance y below the free surface of liquid. The apparent
•
0
depth of this layer having real depth dy is dH' = dy . µ Free Surface
r
Solution The normal shift produced by a glass slab is given by
y
H
l =>
dH;= dy
=>
dH'= dy ' H y) .' ' ·( , 1+
~=(1¾}=(1¾}(6)=2 cm i.e., for the mirror the object is placed at a distance (32 ~) = 30 cm from it. Applying mirror formula i.e. 1 1 1 +=, we get
µ
V
U
f
1 1 1 =
ToW _apparent depth is obtained. by integrO:ting this expreSsion within appiopriate ~imifs. So, H
Hlog,(H +y)
H'=Hlog,2
x
1432 cm+i
30 10 => v=15 cm (a) When x =5 cm The light falls on the slab after being reflected from the mirror as shown. But the slab will again shift it by a distance tu= 2 cm . Hence, 'the final real image is formed at a distance (15 + 2) = 17 cm from the mirror. v
================================== = 1.37
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Optics & Modern Physics 4
Further, =:>
3
sin(45°) sinr
r=32°
=tanr =tan(32°) hlO =0.62 h
Now EF GE
20
(b) When x = cm This time too the final image is at a distance 17 cm from the mirror but it i~ virtual as shown.
Solving this, we get
h=26.65 cm MULTISLABS
If a number of slabs (or immiscible liquids) of depth d1 , d,, d3 , •••• and refractive index µ 1 , µ 2 , µ 3 , •••• are placed one over the other, the real depth is d=d,+d,+d,+ .... The apparent depth is given as
Illustration 27
A cubical vessel with nontransparent walls is so located that the eye of an observer does not see its bottom but sees all of the wall CD. To what height should water be poured into the vessel for the observer to see an object F arranged at a distance of b =10 cm from comer D ? The face of the vessel is a = 40 cm and refractive index of water .
d' =E.!..+E1_+ d3 + ... µ1
µ3
Therefore, for the combination, the effective µ is
µ=I d'
4
IS,
µ2
d,+~+~+...
L~
r(iJ
(i:)+(i:)+(i:)+...
If there are only two slabs, of equal thickness, d, =d, =d, d+d
3
.
d, f,,===~==='~ __ t_ d2
!;}}jit;tir.1~!f~
+· d3
µJ
Solution
1 . __ _ _ _ _ _.J. __
Since, the vessel is cubical, GE=ED=h (say)then EF=EDFD
LGDE=45°
t_
and Illustration 28
Eye~A
The bottom of a tub has a black spot. A glass slab of thickness 4.5 cm is placed over it and then water is filled to the height of 8 cm above the glass slab. Looking from top, what shall be the apparent depth of the spot below the water surface? Also find the effective refractive index of the combination of glass slab and water layer. (Refractive index of glass is
But tan(45')=1=GE ED
cc,,
ED=GE=h
cc,,
EF=EDFD=h10
~ 2
and of water is
± ). 3
Solution
The apparent depth is given as d d 4.5 8
d,=µ:+µ:=rn)+(¾)=3+6=9cm
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I
6.
I.
4 µ2=3
,.
3 µ1=2
\
surface of water. Where will the image of this source in the flat mirrorlike bottom of a vessel be if the depth of the vessel full of water is d ? Refractive index of water is
!'"
n =~ . co·nsider only two steps.
4.5cm
•·
Spot
7.
µclfuctire
= !!,_ =
d, + d,
d,
d,
A plate with plane parallel faces having refractive index 1.8 rests on a plane mirror. A light ray is incident on the upper face of' the plate at 60° . How far from the entry point will the ray ·emerge after reflection by the mirror of the plate is 6 cm thick?
The effective refractive index is given as Real Depth µeffective Apparent Depth
=>
A point source of light is arranged at a height h above the
'''
= 4.5 + 8 = 12.5 = 1 _39
9
''' N:'
60~1
,M
9
IC:Eilli BASED ON GENERAL REFRACTION (Solutions on page 1•.176)
1.
An object lies 100 cm inside water. It is viewed from air nearly normally. Find the apparent depth of the object.
2.
A concave mirror is placed Inside water with its shining surface upwards and principal axis vertical as shown. Rays are jncident parallel to the principal axis of concave mirror. Find the position of final image.
!! water 4/3
~
R=40cm
3.
4.
5.
f
A coin Is placed at the bottom of the slab and is being observed from the air side along the normal to the surface. Find the apparent PC?Sition· of the coin from the surface. Air
3cm h,
Glass
t
A ray of light falls onto a planeparallel glass plate 1 cm thick at an angle of 60° . The refractive index of the glass Is
./3 ,
A 2 cm thick layer of water covers a 3 cm thick glass slab.
Water
~.
Some of the light is reflected and the rest, being refracted,, passes into the glass Is reflected from the bottom of the plate, refracted a ~econd time and emerges back into the air parallel to the first reflected ray. Determine the djstance £ between the rays .
The velocity of light in air is 3 x 108 ms1 • If yellow light of wavelength 6000 A is passedfrom air to glass of refractive index 1.5, determine the velocity, the wavelength and the colour .of light in glass.
f
happens to be 1 m above the water. Determine the length of the shadow of the pole at the bottom of the lake if the sunrays make an angle of 45° with the Water surface. The
9.
!
h,
A pole 4 m high is driven into the bottom of a lake and
refractive index of water is
A small object Is placed on the principal axis of a concave spherical mirror of radius 20 cm at a distance of 30 cm . By how .much will the position and size of the image alter, when a parallelsided slab of glass of thickness 6 cm and refractive index 1.5 is Introduced between the centre of curvature and the object? The parallel sides are perpendicular to the principal axis.
f 2cm
8.
Air
30cm ~
MirrOr
.
10. A ray of light ·is refracted through a sphere whose ·material has a refractive index µ in such a way that it passes through the extremities of two radil which make an angle .~ with each other. prove that if a;. is the deviation Of the ray caused by its passage through the sphere, cos(~;")= µcos(%}
11. A vertical beam of light of crosssectional radius r is incident symmetrically on the .curved surface of a glass hemisphere
(µ =%)
of radius 2r placed with its base on a
horizontal table. Find the radius of the luminous spot formed on the table.
12. A material having an index of refraction µ is surrounded by vacuum and i_s in the Shape of a quarter circle of radius R . A light ray parallel to the base of the material is incident from the left at a distance of 'L above the base .and emerges out of the material at an angle 8 • Determine an expression for a .
Coin
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Optics & Modern Physics CRITICAL ANGLE
According to Snell's Law, we have sini smr
d
.= µ,
sin C sin 90
j 13. A circular disc of diameter d lies horizontally in~ide a
I
metallic hemispherical bowl of radius a. The disC. iS just visible to an eye looking over the .edge. The bowl is now filled with a liquid of refractive ·index µ. Now, the whole of the c;lisc is ju$t visible to the eye· in the same positioli. Show
that
,
d=2a(µ'
V
µdensl'r
· C =µrarer 1 sm =µderner
µdenser
· 1 ) µ2 +1
Eye
= µrarer
where µdenser is ~he refractive index of the denser medium w.r.t. the rarer medium. The lesser the value of µdenser, the greater is the critical angle C . For a given pair of media, since µ depends on the wavelength of light the critical angle ·also depends on the wavelength. The greater the wavelength, the greater will be the critical angle.
............ _ _ _ _ _ _ _ _ _ __
,  .......... _
.............. _ ...........
Critical angle TOTAL INTERNAL REFLECTION (TIR)
Media Pili
When a ray of light goes from a denser to a rarer medium, it bends away from the normal. If the angle of incidence in the denser medium is increased the angle of refraction in the rarer medium also increases. At a particular angle of incidence in the denser medium (called as the Critical angle C ), the angle of refraction in the rarer medium is 90° (i.e., the refracted ray grazes the interface). This angle of refraction in the denser medium for which the refracted ray grazes the interface is called the critical angle for the pair of interface. ' Please note, that for small angles of incidence, both reflection and refraction occur, however we shall be neglecting the reflection at the interface as most of the light is refracted. However, when i > C, no part of light is refracted· and the entire light is reflected back to the denser medium itself. This phenomenon is called total internal reflection (TIR) and was first noted by Kepler in 1604.
0
Denser µ2
=
1.40
1(1..· ·,) µdense'(
WaterAir
I
GlassAir
I
GlassWater
µd· _µ._4(3_4 _____ _
µ,
1
3
µ, µ,
3/2
3.
49°
µd===
1 · 2
63°
EXAMPLES OF TOTAL INTERNAL REFLECTION
a)
Mirage : Mirage is an optical illusion ol;,served in deserts and roads on a hot day. When the ai_; near the ground is hotter o (and hence rarer) than the air above, E '........ Denser j > 8~ there occurs a Rarer continuous decrease of refractive index of air towards the ' 'Earth ground. A ray of light from a point 0 of a tree is, therefore, refracted more and more away from the normal. Ultimately it gets totally reflected to reach the eye E. To tf1e observer it appeai;s to come from I, which is the image of O . This image gives the impression of reflection from a pond of water. Looming : Similarly, in extremely cold regions (near polar regions), the refractive index decreases ·with height. Due to TIR (shown in figure), the image of a hut appears hanging in the air. This is called looming.
µ1
Images formed by TIR are much brighter than those formed by the mirrors (or lenses). Some loss of intensity always takes place, when light is reflected from a mirror (or refracted through a lens).
. C =sin
b)
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rm ' ' ~~.._,...... \:,.,/ Rarer
.......................... ! goo ~·r' 1:c:c:__4r d
Earth
c)
d)
e)
The µ of diamond is 2.5, for which C is only 24°. Diamonds are cut such that _i > C ; so TIR takes place again and again inside it The light coming out from few meticulously cut surfaces makes it sparkle. Air bubbles in water shine due to TIR The working of an optical fibre is due to multiple TIR inside it.
f)
Porro prisms used in periscopes or binoculars bend the ray due to TIR. Some examples are shown in figure. 45"
/
(a) Bending of rays by 180'
L~l
µ
Fish in Glass Tank Similarly, if a source of light is kept in ·a pond, its light will come out only through a circular region. For any incident angle i greater than C, the light will be totally reflected back into the water, making corresponding region on the surface of water appear dark. Illustration 29 Light is incident making an angle 0 with the axis of a
transparent cylindrical fiber of refractive index. n =
/
i =45°
90'
'' ''
45'
~
as
shown in figure. Determine the maximum value of 8 so that the light entering the cylinder does not c~me out of. the curved surface. ·
t)~  ()
(b) Bending of rays by 90'
Solution The ray of light is incident at A and it just gets reflected totally at B . Therefore incident angle at B is equal
A' + '
',, p
r,
i
r1 =Ar2 =AC
Using Snell's Law,
r,
lsini=µsinr1 =µsin(AC) ~
sini = µ(sinAcosCcosAsinC)
~ sini = µ[(sinA),/1sin 2 C (cosA)(sinC)]
B
But (r1 )max is possible when i=imax =90° i.e., incident ray grazes the interface AB . Now, applying Snell's Law at AB, lxsini=µsinr1 ~
1
sin(90°) = µsinr
r1 =sm
~
'1 =C
... (2)
From equations (1) and (2), we get r1 +r2 >2C Since, r1 + r2 = _A Therefore, the condition becomes A>2C ~
sm(1)>sinC
~
sm
~
µ > cosec(
sini=sinAJµ 2 l ~cosA
=>
i=sin1 (sinAJµ 2 1cosA)
The light will emerge out of the. prism only if the angle of incidence i is greater than the above value.
. 1(1) ·µ
~
=>
... (3)
MAXIMUM DEVIATION
The angle of deviation D is maximum when the angle maximum, i.e., i = 90°.
is
Dmu = (i + e)A = (90°+ e)A Under such conditions of grazing incidence, r1 = C A
. (ZA) >µ1
1J B.'' C
=
And at the. second surface, 1.46
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emergent ray for the same incident ray, values of au the angles.
µsinr2 = lsine
=>
sine= µsinr2
A
.,
Since r1 +r2 =A ~
sine=µsin(Ar,)=µsin(AC)
~
e = sin' (µsin(AC))
indicating the
,,,,,/
+i,;cr.'" /
Illustration 34 An isosceles glass prism has one of its faces coated . with silver. A ray of light is incident normally on the other face (which is equal to the silvered face). The ray of light is reflected twice on the same sized faces and then emerges through the base of the prism perpendicularly. Find angles of prism. Solution As the ray is incident normally at the face AB , so
90'
B
Solution
For total internal reflection to take place at surface AB, we have i>C
=>
'1 =0
.45'
C
sini>sinC
. . C =1 sm Smce, µ
E
~
sin45'>(t)
~
µ>Ji
~
µmin
=Ji
When the prism is immersed in water, the boundary AB now separates glass from water. B
0 C
Since, we know that r1 + r2 = A , so we get
... (1)
r2 =A=180'20 Now, LDfE=180°90°2r2 ~
LDfE=180°90°360°+40
{·: r2 =18020)
~ LDFE=40270' Since, r, = 90°  LDFE ~
... (2) ... (3)
r,=360'40
Again LBFG = 90°  0 = 90°  r, =>
T3
=8
... (4)
From equations (3) and (4), we get 50=360° ~ 0=72' and 180°20=36'

c =sm · •(. µ,=, · 1 (Ji) ·  ) =Sm 
~
C=70.12°
µ,_
1.33
Since i = 45° and also, we.observe that i
µsinr1 =µsinr2
=,.
r1 = r,
Since, rt' + r2
... (2)
::::>
sini sine Further,µ=.=.smr1 smr3
=A
2
..
{because i = e}
... (3)
Solving equations (1), (2) and (3), we get r2 = 45° and r1 = 30°
smi µ== sinr1
. (A+Dm;•) 2
Slil
. Sin
(A) 2
Note that if .the prism is equilateral or isosceles, then the ray inside the prism is parallel to its base.
Now, for TIR (total internal reflection) to take place at the face BC; we have
=
r (say)
A r=
=r2 15°
=>. r3 = r1
=
1.48
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Ray Optics: Refraction at Plane Surfaces A
A
s~~c Illustration 37 The angle of minimum deviation for a glass prism with refractive index ./3 equals the refracting angle of the prism. What is the angle of the prism? Solution Since we know that
./3
r1 = r3 = r (say)
=>.
6,+6Q+6, =120°
=>.
(ir)+(1802C)+(ir)=120°
=>.
2i2(60°C)+180°2C=120'
sin(1)
=>.
2i = 60°
=>.
cas(1)=
=>.
A=30° 2 A=60°
(say)
r=60°C Given, OTotaI = 120°
sinA
./3 = 2cos( 1)
=i
=>.
=()min =AI so we get
=>.
=>.
=>.
=60°
Also, r2 ::::; C
sin(1) Bminimum
Since, r1 + r2 = r2 + r3
Similarly by symmetry, we have i1 =i,
(A+6m) s. m µ= 2
Since
r1 = r3 = r i1=i2=i r2~c
=> i=30° Now, according to Snell's Law, we have sini sini µ = 1 = sinr1 sinr
~
But r1 +r2 =60° =>.
r+C=60° r=60'C
=>.
µ
=>.
µ
=>.
Illustration 38 The path of a ray of light passing through an equilateral glass prism ABC is shown in the figure.
sin(30°) sin(60°C) 0.5
./3 cosc1.sinC 2
A
2
But sin C = 1. µ
BL.__ _ _',L_ _ __:,.C
The ray of light is incident on face BC at an angle just greater than the critical angle for total internal reflection to take place. The total angle of deviation after the refraction at face AC is 120° . Calculate the refractive index of the glass. Solution The ray diagram is drawn for the sake of convenience.
=>.
./3Jµ'1 =2
=>.
3(µ'1)=4
=>.
3µ 2 =7
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=>
µ=~
=>
µ=1.52
Optics & Modern Physics B µ=A+,_ 2
'
WHITE LIGHT
White light consists of infinite number of continuous wavelengths (colours) ranging from 4000 A to 7800 A. For convenience it is divided into seven colours. Violet, Indigo, Blue, Green, Yellow, Orange, Red called as 'VIBGYOR' pattern. The Violet having least wavelength (maximum frequency) and Red having maximum wavelength (minimum frequency). VARIATION OF REFRACTIVE INDEX WITH COLOUR (CAUCHY'S FORMULA)
The refractive index
(µ)
of a medium varies with
wavelength (:\,) according to Cauchy's formula
where A and B are constants. The smaller the value of A, the larger is the value of µ . Thus, µ is maximum for violet colour and minimum for red. The deviation of a ray depends on µ it is larger for higher µ . Hence, violet suffers ~e maximum deviation and red the minimum. If light from sodium lamp falls on a prism then it disperses (breaks) into two lines called D1 ( 5890 A) and D, ( 5896 A) lines. Thus we observe that a prism causes deviation as well as dispersion. If Dv, DR and '!)y are the deviations caused by prism for violet, red and mean yellow rays, then for prism with small refracting angle (A) , we have Angular Dispersion D=Dv D, =(µv µ,)A DISPERSIVE POWER OF A PRISM
The ratio of angular dispersion to the mean deviation is called dispersive power, so Dispersive Power is
B C µ=A+,_,+,,,+ ...
where A , B and C are constants. From above we observe that refractive index decreases with increase of wavelength. It is maximum for violet and minimum for red colour and due to this variation of. the refractive index with the wavelength or the colour, a composite beam of light entering a prism splits into constituent colours.
0)
Angular Dispersion Mean Deviation
D Dy
Dv D, Dy
where Dy is the deviation of mean light i.e., yellow light, whose wavelength is considered as mean of all the wavelengths present. Further for a prism of small refracting angle A , we have
D=(µl)A So, we have
DISPERSION
Dv=(µv1)A, D,=(µ,1)A and Dr=(µr1)A
It has been observed that when a beam of composite light (consisting of several wavelengths) passes though a prism, it splits into its constituent colours. This phenomenon is called dispersion. The band of colours thus obtained on a screen is called the spectrum. If white light is used, seven colours are obtained as shown in the figure. The sequence of colours is VIBGYOR, from bottom to top.
So the dispersive power ro becomes o,
(µvµ,)A (µy1)A
(µvµ,) µy1
~
µ1
where dµ:=µvµ, and µ=µy The dispersive power co has no units and no dimensions. Its value depends on the material of the prism. COMBINATION OF TWO PRISMS
Red
Orange Yellow
Green Blue Indigo
Violet The dispersion of light takes place because the refractive index µ of the medium depends on the wavelength of light as given by Cauchy's formula, according to which
From a single prism, it is not possible to get deviation without dispersion, or to get dispersion P' without deviation. However, two small angled prisms may be p combined to produce Dispersion without Deviation or Deviation without Dispersion. The prism placement for both is shown here. The placement remains the same. It is just that we are to decide the relation between their refractive indices such that required condition may be achieved.
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This give~
DISPERSION WITHOUT DEVIATION
Two prisms can be combined in such a way that the deviation of the mean ray produced by one is equal and opposite to that produced by the other. Such a combination is called a direct vision pris'm.
A'=(µvµ')A µ~µ~
Also from (1) we get
(µv µ,)A=(µ~ µ~)A'
A
=>
'White
1""""'~~~R7c1~:....
ligh::,I
=>
V
or
A''=
(.!:.=!.)A. µ'1
The net dispersion produced is
=(Dv D,)(D~ D~)
Net Dispersion =(µv µ,)A(µ~ µ~)A' . Net_ ·)=(µvµ•J(µl)A(µ~µ~)(µ'l)A' ( D1spers10n µ 1 µ'  1 Net Dispersion = roD ro'D' where co and ro' are dispersive powers of prisms P and P'.
o
B.
µ1
DD' =(µl)A(µ'
So, in this arrangement of prisms, the mean deviation ·(o) caused by one prism is cancelled by the mean deviation (D') caused by the other prism i.e. DD'=0
(µl)A(µ' ~ l)A' =0
µ1.. coD =ro'D'
.
is the condition for Deviation without Dispersion. The net mean deviation is
Crown ~~A,'
=>
(µv µ,)(µl)A=(µ~ µ~)(µ' l)A'
i)A'
Illustration 39 The refractive indices of the crown glass for blue and red light are 1.51 and 1.49 respectively and those of the flint glass are 1.77 and.1.73 respectively. An isosceles prism of angle 6° is made of crown glass. A beam of white light is incident at a small angle on this prism. The other flint glass isosceles prism is combined with the crown glass prism such that there is no deviation of the incident prism. (a) Determine the angle of the flint glass prism. (b) Calculate the net dispersion of the combined system. Solution (a) When angle of prism is small and angle. of incidence is also small, the deviation is given by o= (µ 1 )A
· Net deviation by the two prismsis.zero, when deviation due to one cancels the deviation due to the other. So, 01 0,=0
DEVIATION WITHOUT DISPERSION (ACHROMATIC PRISM)
... (1)
It is possible to combine two prisms of different .materials in such a way that each cancels the dispersion due to the other, Thus, the net dispersion is zero but a deviation is produced. So, in this arrangement of prisms, the dispersion (Dv DR) caused by one. prism is cancelled by dispersion ( D~ 
Here, µ 1 and µ 2 are the refractive indices for crown . and flint glasses respectively, where
o;)
produced by the other prism. /
Crown
µ,
1.51 + 1.49 2
l 5 d · an µ,
1.77 + 1.73 2
l 75 ·
Angle of prism !or crown glass is A, = 6° Substituting this values in equation (1), we get (1.51)(6°)(1.75l)A, =0 This gives A2 =4°
i.e., (DvD,)(D~D~)=0 or
(µvµ,)A(µ~µ~)A'=0
Hence, angle of flint glass prism is 4°
... (1)
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COLOURS OF OBJECTS AND COLOUR TRIANGLE
Flint
The colours of objects are due to·a number of phenomena.
6,+62=0
(b) Net dispersion due to the two prisms is given by Net Dispersion=(µ~ µ,,)A,(µ,,
µ,,)A,
=> Net Dispersion = (1.511.49)(6°)(1.77 1.73)( 4°) =0.04° ~
Net dispersion =0.04°
Illustration 40 A ray of light is incident on a prism ABC of
refractive index
COLOUR TRIANGLE
If, Red (R), Green (G) and Blue (B) are primary colours. If
./3 as shown in figure. B
The colours of opaque bodies are due to Selective Reflection. For example grass appears green because when white light is incident on grass, it absorbs all colours except green which is reflected. Black appears black because it absorbs all colours falling an it an reflects nothing. Similarly white appears white because it reflects all colours falling on it and absorbs nothing. · The colours of transparent bodies are due to Selective Transmission. For example a glass appears blue, because it absorbs all colours except blue, which it transmits. The colours of sky, rising and setting sun are due to scattering while the colours of soap bubble and kerosene oil film are due to interference.
P denotes Peacock Blue also called Cyan, M denotes Magenta, Y denotes Yellow and W denotes White, then from colour triangle we observe that
D 60'
R
E Find the angle of incidence for which the deviation of light ray by the prism ABC is minimum. (b) By what angle the second prism must be rotated, so that the final ray ·suffer net minimum deviation. (a)
At minimum deviation, we have r1 = r2 = 30° According to Snell's Law, we have
sini
µ=.1 smrt ~
.Js
sini1 sin(30°)
. . =2 .rs
~
Stnl1
~
i1 =60°
RAYLEIGH LAW
(b) In the position shown, net deviation suffered by the ray of light should be minimum. Therefore, the second prism should be rotated by 60° (anticlockwise). B. Dl'CT=..,E 60° 60° 60°
60'
A
B
R+G+B=W R+G=Y G+B=P R+B=M B+Y=W R+P=W G+M=W
Solution (a)
G
6_0'
60'
According to Lord Rayleigh, intensity (I) of scattered light is inversely proportional to the fourth power of the wavelength 1,. • So, l Ioc1,. 4
It can also be concluded that the amplitude (a) of the scattered light is inversely proportional to the square of the wavelength. 1 So, aoc ..2 {·.Joca 2 } 1
C
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Ray Optics: Refraction at Plane Surfaces COLOUR OF THE SKY
When light from the sun travels through earth's atmosphere, it gets scattered by the large number of molecules of various
gases. It is found that the amount of scattering by molecules, called Rayleigh scattering, is inversely proportional to the fourth power of the wavelength. Thus light of shorter wavelength is scattered much more than the light of longer wavelength. Since blue colour has relatively shorter wavelength, it predominates the sky and hence sky appears bluish.
r
ii. the system is placed in wale. r what will be the net deviation? Refractive index of water .
A ray of light incident.on the face of a prism is refracted and escapes through an adjacent face. What ·is the maximum permissible angle of refraction of th~ prism, If it is made of glass with a refractive index of µ = 1.5 .
5.
In an isosceles prism of angle 45° , it is found that when the angle of incidence is same as the prism angle and the emergent ray grazes the emergent surface. (a) Find the refractive Index of the material of the prism. (b) For what angle of incidence the angle of deviation will be minimum? 
I
I
6.
A prism of fli_nt glass
(a)
BASED ON PRISM (Solutions on page 1.183) The path of a ray undergoing refraction in an equilateral prism is shown in figure. The ray suffers refraction at the faqe AB ,and the refracted ray is incident on the face AC
at an angle slightly greater than the critical angle and hence, totally reflected. After refraction at the face BC the emergent ray makes an angle of 30° with normal at BC at" the point of emergence. Find the A
(a) (b)
2.
3.
(µ =¾) with an angle of refraction 9
i) .
At what angle should a ray of light fall on the face of the prism so that inside the prism the ray is perpendicular to the bisector of, the angle of the prism.
(b)
Through what angle will the ray tum after passing through both face5: of the prism?
7,
Light rays from ~ source are incident on a glass prism of_ index of refraction µ and angle of _priSm q.. At near normal incidence, calculate the angle of deviation of the emerging rays.
8.
One face of a prism with a refractive angle of 30° is coated with silver. A ray incident on another face at an angle of 45° is refracted and reflected from the silver coated face and retraces its path. What is the refractive index of the prism?
9.
A ray of light is incident at an angle of 60° at one face of a prism having refracting angle 30° . The ray emerging out of the prism makes ah angle of 30° with the incident ray. Find the angle of emergence and calculate the refractive index of
the material of the prism.
corresponding angle of incidence i. refractive index of the prism.
10.
A beam of light enters .a glass prism at an angle a and emerges into the air at an angle J3 . Having passed through th_e prismj the beam is deflected from the original direction by an angle 6 . Find the refracting angle of the prism and the refractive index of the materiatof the prism. In a prism of refractive index µ = 1.5 and refracting angle 60°, the condition for minimum deviation is fulfilled. If face AC is polished
60° µ= 1.5
i)
in a direction perpendicular to the other face.
Prove ttiat the ray will fail to emerge from the other face if cotA < cotC1 , where C is critical angle for the material of prism.
11.
The index of refraction for violet light in silica flint glass is
1
I
1.66 and that for red light is 1.62. Find the angular dispersion of visible light passing through a prism of apex angle 60° , if the angle of incidence is 50° ·:
12.
A light ray is passing through a prism with refracting angle A= 90° .and· refractive index µ = 1.3 . Find the minimum and maximumangle of deviation.
13.
A ray of li9ht. is incident at an angle of 60° on the face of a prism h~ving refracting angle 30° . The ray emerging out of the prism makes an angle 30° with the incident ray. Find the angle of emergence of the ray.
s~~c Find the net deviation,~~~
A ray of light is incident upon one face of a prism (angle of prism
µ,(~~)+µ,(":)=(µ,µ,)(~)
=>
l1_ + ~ = µ2  µ1
PI PC Since PO=u, PI=+v, PC=+R so we get 12+µ2 =µ2µl u V R
If the object is in air, then µ 1 = 1 and µ 2 =µ,so we get 1 µ' µ1 +=u V R
image formed is real.
Consider a spherical surface of radius R separating the two media 1 and 2 (µ, > µ 1 ) . A point object O is placed on the principal axis to the left of the pole P · at a considerable distance from it. The incident ray from O falls on point A and is refracted according to
. ... (1) Since the rays are paraxial, so the angle a is small and hence the angles i and r will also be small. Thus, applying such
paraxial approximation, then sin i ~ i and sin r ;;; r , so we get from (1)
... (2)
PO
CASE 2 : When the object lies in the rarer medium and the
image formed is virtual. Consider a spherical surface of radius R separating the two media 1 and 2 (µ 2  µ 1 ) . A point object O is placed on the principal axis to the left of the pole P . The incident ray from 0 falls on point A and is refracted according to ... (1) Since the rays ~e paraxial, so the angle a is small and hence the angles i and r will also be small. Thus, applying such paraxial approximation, then sini i and sinr r, so from (1), we have
=
=
... (2)
µ,
0
I
''' '' y ,.__R_., '
(Real)
µ,
I
I
Denser
'' ''
,.__ U    ,    V +I
!+ u ~
Using the geometrical property that an exterior angle of a triangle is equal to the sum of the two internal opposite angles, we get from triangles AOC and AIC,
i=a+y and r=r+p
... (3) ... (4)
Substituting the value of i and r from Equations (3) and (4) in Equation (2), we get µ 1 (a+r)=µ 2 (rP)
=>
µ 1a+µ 2 P=(µ 2 µ 1 )r
I
'1
R +1
l+V+i
Using the geometrical property that an exterior angle of a triangle is equal to the sum of the two internal opposite angles, we get from triangles AOC and AIC,
i"=a+y
... (3)
and r=P+r
... (4)
Substituting the value of i and r from Equations (3) and (4) in Equation (2), we get µ, (a+y) = µ 2 (P+r)
... (5)
µ 1aµ 2P=(µ·2 µ 1 )r
Now, since the aperture of the refracting surface is small, so
=>
M and P are very close to each other and hence, We have
Now, since the aperture of the refracting surface is small, so M and P are very close to each other and hence we have
=
1.56
... (5)
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Ray Optics: Refraction at Curved S111faces AM AM a:=:tana=eeMO PO'
AM AM yeetany=eeMC PC
AM AM peetanP=ee and MI PI AM AM yaatany=eeMC PC
µ,( ,:)+ µ,( ,:)=(µ, µ,)( ~ )
.l:L+l:L= µ2µt PO PI PC Since PO =n, PI =+v, PC =R so we get
µ,( :)µ,( ,:)=(µ, µ,)( ~~)
12+12 = µ1 µ2
_&_~= µ2µ1
PO PI PC Since PO=u, PI=v, PC=+R so we get
11+ µ2 = µ2 µ1 u
V
R
CASE 3 : When the object lies in the denser medium and the image formed is real. Consider a spherical surface of radius R separating the two media 1 and 2 (µ, > µ,) . A point object O is placed on the
principal axis to the left of the pole P . The incident ray from 0 falls on point A and is refracted according to ... (1) Since the rays are paraxial, so the angle a is small and hence
the angles i and r will also be small. Thus, applying such paraxial approximation, then sin i =i and sin r =r , so from (1), we have ... (2)
u V R Simply replace subscript 2 by 1 and 1 by 2 in the formula derived in Case 1 or Case 2. CASE 4 : When the object lies in the denser medium and the image formed is virtual. Consider a spherical surface ·of radius R separating the two media 1 and 2 (µ 2 > µ,) . A point object O is placed on the principal axis to the left of the pole P . The incident ray from 0 falls on point A and is refracted according to ... (1) Since the rays are paraxial, so the angle a is small and hence
the angles i and r will also be small. Thus, applying such
paraxial approximation, then sini::: i and sinr =r, so from
(1), we have ... (2)
µ, Denser
C

R __,___ V+!
Using the geometrical property that an exterior angle of a
14u.a
triangle is equal to the sum of the two internal opposite
Using the geometrical property that an exterior angle of a triangle is equal to the sum of the two internal opposite angles, we get from triangles AOC and AIC y=a.+i ... (3) and r=P+r
... (4)
Substituting the value of i and r from Equations (3) and (4) in Equation (2), we get
µ,a.+ µ,p = (µ, µ,)y
... (5)
Now, since the aperture of the refracting surface is small, so M and P are very close to each other and hence we have
AM AM a:::tana.=aaMO PO' AM AM peetanP=aa and MI PI
angles, we get from triangles AOC and AIC a.=i+y
... (3)
and P=r+y ... (4) Substituting the value of i and r from Equations (3) and (4) in Equation (2), we get µ, ( ur) = µ1 (Pr) ~
µ, ( r a)=µ, (P +r) ~
p
µ,a.µ,P=(µ,µ,)y
... (5)
Now, since the aperture of the refracting surface is small, so M and P are. very close to each other and hence we have
AM AM a:=tana=aaMO PO' AM AM paatanP=ee and MI PI AM AM 1 "' tany= MC" PC
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µ,( :)µ,( ~)=(µ, µ,)( ~ )
...!:L_..!::1= µ2µ1
~
u
h+ µ2 = µ2 µ~
µ2
u
R
V
1:1_ _ ..e..t = µ1  µ2
PO PI PC Since PO=u, PI=v, PC=R .soweg~t
PO PI PC Since PO =u, PI =v, PC =R sowe get
11 +11 = µ1 
µ,( :)µ,( ~)=(µ, µ,)( ~ )
R
V
REFRACTION AT CONCAVE SURFACE
CASE 2: When the object lies in the denser medium.
For a concave. refracting surface the image formed is always
Consider a spherical surface of radius R separating the two media 1 and 2 (µ 2 > µ 1 ). A point object O is placed on the principal axis to the· left of the pole P . The incident ray from 0 falls on point A and is refracted according to ... (}) µ 2 sini = µ 1 sinr
virtual irrespective of the placement of the object. CASE 1: When the object lies in the rarer medium. Consider a spherical surface of radius R separating the two media 1 and 2 .(µ 2 > µ 1 ) • A point object O is placed on the
principal axis to the left of the pole P .' The incident ray from 0 falls on point A and is refracted according to
: .. (1) Since the rays are paraxial, so the angle a is small and hence the angles i and r will also be small. Thus, applying such paraxial approximation, then sin(:: i and sinr::: r, so from (1), we have .... (2) µ,
µ,
Rarer
Denser
Since the rays are paraxial, so the angle a. is small and hence the angles i and r will also be small. Thus, applying such paraxial approximation, then sini =i and sinr.:= r, so from,
(1), we have ... (2) A µ, Rarer
0
/ f;~; 0
l
C
·R
C
''B
Using the geometrical property that an exterior angle of a triangle. is equal to the sum of the two internal opposite angles, we get from triangles AOC and AIC
y=a+i
... (3)
and r=P+r
... (4)
Substituting the value of i and r from Equations (3) and (4) in Equation (2), we get µ,(ya)=µ,(yp) ~
µ,aµ,P=(µ, µ,)y
... (5)
Using the geometrical property that an exterior angle of a triangle is equal to the sum of the two' internal opposite angles, we get from triangles AOC and .AIC
i=a+y
... (3)
and r=P+y
... (4)
Substituting the value of i and r from Equations (3) and (4) in Equation (2), we get µ, (a +y) = µ, (P +y) ~
AM AM a=:tana=eeMO PO'
AM AM a=:tana=eeMO PO'
AM AM peetanP=ee· and MI PI
AM AM · peetanP=ee and MI PI AM AM yeetany=eeMC PC
AM AM yeetany=·MC ee PC
1.58
... (5)
Now, since the aperture of the refracting surface is small, so M and P are very close to each other and hence ~e have ·
Now, since the aperture of the refracting surface is smalt so M and P a_re very close to each other and hence we have
=
µ,aµ,p = (µ, µ,)y
~
µ,( :)µ,( ~)=(µ, µ,)( ~ )
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Ray Optics: Refraction at Curved Surfaces
=>
J:L_l:.!.= µ1µ2 PO
PI
Since PO=u, PI=v, PC=+R so we get
AB=utani=ui and A'B'=vtanr~vr
CONCEPTUAL NOTE(S)
b)
For both ·convex and concave spherical surfaces, the refraction formulae are same,, only proper signs of u , v and R are to be used. For refraction from rarer to denser medium, the refraction formula is
u
R
V
For refraction from denser to rarer medium, we Interchange µ 1 and µ 2 and obtain the refraction formula, J:g__+&=~µ2 u .v R
d)
Using Equations (1), (2) & (3), we get m= ~:·
(µ 1 = 1) and the ,denser medium
has refractive index µ (i.e., µ 2
=µ ), then
u
V
R
.2._+l'.= µ1 e)
u
µ2  µ1 R
The factor
u µ,
H m is positive, the image is erect and virtual.
, ·   CONCEPTUAL NOTE(S) 11+ µ2 = ~ µ1
is equally applicable to R plane refradting surfaces i.e., surfaces for which R i oo . Let us . Real Depth denve µ '''"''""'"~ using this. Apparent Depth
for refraction from medium:to air, we have v
m=~l::1
=>
The refraction formula
for refraction from air to medium, we have
_!___+!'.= µ1 (ii)
=:; =(:)(;,)
H m is negative, the image is inverted and real.
If the rarer medium is air
(i)
... (2) ... (3)
The magnification is defined as height of image A'B' m height of object AB
.&.+ µ2 = µ2µ, ci)
... (1)
µ,
Now, in MBP and M'B'P, we have
12+1:!.=µlµ2 u V R
a)
~=11
=>
PC
Applying
R is called power of the spherical
refracting surface. It gives a measure of the degree to which the refracting surface can converge or diyerge the rays of light passing through It. For airmedium interface, the power is
.Hi.+ µ 2
=µ
u
V
µ1 R with proper sign conventions and values, we get
u
2 
V
_µ_+.!.= 1µ =0 (d) V oo d V=µ
:::)
I
I
P= µ1
R
R>oo
i
TRANSVERSE MAGNIFICATION
Instead of a point object O let us now, keep an extended object AB at point O such that its image A'B' will be formed at point I. The distance x(=u) and y(=v) are related by the above formula.
~~
I
B
!~~ 0 i.e., image of object O is formed at a distance
,
app
A
~ µ
on same side.
::. dactuar µ Real Depth
s1 ,u
!
f+ve
d
So d
A"!
I
2
µ = Apparent Depth
V
A ray from point B of the object is incident at point P and is refracted, in accordance with Snell's Law, such that µ 1 sini=µ 2 sinr
Illustration 41
In figure, a fish watcher watches a fish through a 3 cm thick glass wall of a fish tank. The watcher is in level
{applying paraxial approximation)
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with the fish; .the index of refraction of the glass is .
that of the wateris
.
.
_ll_
5
For face CD , we have
and
,8
_2_  ~ =0
!.
AI,
3
To the fish, how far away does the watcher appear to be? · (b) To the watcher, how far aw'ay does the fish appear to be?
11.16
(a)
:..a cm
o
AI, =(11.16)(¾)=6.975 cm
o
FI, =8+6:975 •
o
Ff; = 14.975 cm
6.8cm
'' '' ·Observer
'
Illustration 42
There are two objects 0 1 and 0 2 at an identical distance of 20 ·cm on the two sides of the pole of a
Solution
spherical_ ·concave refracti.ng J,oundary of radius 60,cm. The indices ·of• refraction of the llledia on two sides of the I ,, ' I ,.
(a)
boundai; are 1 and
OA=3cm So,
AI; =(n, )(OA)
o
AI, =(¾)c3l=4.8 cm
the object (a) o, 'wheri see;,_ from (b)
E
C
r·
O, when see!' from· 0 1 •
The formula forrefraction from a curved boun.dary ls
:f.,
_&_.!:':!_= µ,µ, v u' R
4
.'
F
(a)
From the ray diagram drawri, we get,
G  + +ve
D
' . ' µ =1, µ =4 · u1 =20cm, R=60cm, 1 2 3
For refraction at EG(R> oo), using
n2 _!::!_= n2 11i V u R 4· 8. 5
µ!
=1
0

60 cm _____.._'
FI, =6.5+6.8=13.3 cm
For face Ef, we have
8·
v=24 cm
4
,.
(b)
BI, =(6.8>(¾)(¾)=8.16 c_m
C
'.
'
Keeping the object 0 2 on the left of the· pole P as shown, here, we get
E
··, i
.
• µ, (4/3)
_,:_'_________
.
, ~
O, I P D
G
4 3
u1 =20cm, R=+60cm, µ 1 ,=, µ2 =1
O•A . ''·',B.   "7 ~F
=
i'
Thus, theobject 0 1 , will appear at a distance of 24 cm from P towards C .
...2._  _3_ = 0 BI, 6.8
o
µ,= (4/3)
20cm:
(4.8+3) 1+
(b)
... ~·
.::=::_,:.:~c~ r o,_ P~~.
BI, =(7.8>(¼)(¾)=6.5 cm
So,
o, .
Solution
___ J..: __ B, 'f.
3 BI,
(¾} respecfo'.ely. Find the location of
  .. ____ _ O,
C
+ve....,_ 1 60 cm +1 20cm
1.60
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Ray Optics: Refraction at Curved Surfaces
1 (¾) 1(¾)
Solution
(a)
;  (20) =;;a
Applying the formula for the refraction at the curved boundary i.e., '
=>
v=16.36 cm Thus, the object 0 2 will appear at a distance of 16.36 cm from P towards 0 2 •
µ2 _l:!_ = µ2 µ1 V u R
For, refraction at the first surface, the pole is P1 and we observe that
Illustration 43
An object of height 1 cm is kept at a distance of 40 cm fr~m a concave spherical surface having radius of curvature R = 20 cm , separating air and glass having refractive index µ = 1.33 . Find the location, height and the nature of the image formed. 1
1 (¾) 1(¾) ;  (oo)
V=3R the left of pole P1 •
C
Titis image acts as an object for the refraction at the secon.d surface, with pole P2 • For this refraction, we have
' 20cmi I'
,..........
i+40
cm+:'
4
U=(3R+2R)=SR, R=R, µ1 =1, µ 2 =3
Solution According to Cartesian sign convention, u=40cm, R=20cm
v=~R 2
1.33 _ _ 1_= 1.331 V (40) (20)
Thus, the final image 12 is at a distance
v=32cm
Magnification, m _h2 h,
µ,v µ,u
±_1
±
l, __1_=_3_ V SR R
Applying the formulaµ,_&_=µ,µ, ,we get V u R
=>
(+R)
Thus, the first image 11 is formed at a distance of 3R to
µ,=1.33
emf
=
lx(32) 1.33x(40)
(b)
0.60
towards left. The ray diagram is shown in figure.
So, height of image, h, = mh., = 0.6 x 1 = 0.6 cm
_
The positive sign of magnification indicates that the image is virtual and erect.
¾) is
P, µ=1
µ=4/3 (a)
µ=413
Find the position of the image due to refraction at the first surface and the position of the final image. (b) Draw the ray diagram showing the position of the two images.
I
I
.
 ............... . .

o+3Ra>t+2R_.,
refracted by a spherical air bubble of radius R situated in water.
P,
I

....r~...  «:{_. ...... •!;  ...........12___ .......... P1 P2
Illustration 44
R
7
__,~=~?"" 
_!µ~~;'.'.4/'_:3 _ _________ ...

A parallel beam of light travelling in water ( µ =
SR from P2 2
5R/2M
Illustration 45 A solid glass with radius R and an index of refraction 1.5 is silvered over one lielllisphere. A small object is located on the axis of the sphere at a distance 2R to the left of the vertex of the unsilvered hemisphere. Find the position of final image after all refr~ctions and reflections have taken place. Solution The ray of light first gets refracted then reflected and then again refracted. For first refraction and then reflection
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the ray of light travels from left to right while for the last refraction it travels from right to left. Hence, the sign
C
conventio:h. will change accordingly.
A''~D
Solution
Applying Snell's Law at face AB, we get
First Refraction : µ2 _!:!_= µ2µl V u R conventions, we get 1.5 1 1.51 ;;; (2R) = ~
Using
=>
V1
appropri9:te
1
=>
with
appropriate
sign
R
on face CD, we get
R
1.514
../2
1.514../2
OE
oo
0.4
Illustration 47
A glass sphere of radius R =10 cm having refractive µ2 V
_
1:2_ = µ 2  µ 1 u R
index µ8 with reversed sign
convention, we get 1 1.5 11.5
=
=>
= µ,  µ,
Solving this equation, we get OE"6 m
2
Again using
v,
1
Now applying, µ,  µ, V u
=2 Third Reflection : Vz
.
smr=2
=> r=30° i.e., ray becomes parallel to AD inside the block.
+=Vi 00 R
=>
(1)sin45° = (../2)sinr
sign
~oo
Second Reflection : 1 1 1 2 Using +==v u f R conventions, we get
1
with
1.5R
=!
2
is kept inside water. A point object O is
placed at 20 cm from A as shown in figure. Find the position and nature of the image when seen from other side of the sphere. Also draw the ray diagram. Given refractive index of water is µw
R

=_! . 3
v,=2R A
0 M 20
0
B
cm.i
Solution R/2 1+1,SR~
i.e., final image is formed at the vertex of the silvered face i.e., at the pole of silvered/ curved surface.
A ray of light starting from O gets refracted twice. The ray of light is travelling in a direction from left to right. Hence, the distances measured in this direction are taken positive. Applying
11+&= µ2 µ1 , twice with appropriate signs at the Illustration 46 Figure shows an irregular block of material of refractive index ../2 . A ray of light strikes the face AB as shoWII. in tlie figure. After refraction it is incident on a spherical surface CD of radius of curvature 0.4 m and enters a medium of refractive index 1.514 to meet PQ at
u
E . Find the distance OE upto two places of decimal.
=
1.62
R
V
two refracting surfaces, we get
M
p 2
0
I
2
B + +ve
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Ray Optics : Ref.actio11 at Curved Sm/aces
µ
(¼) + (¾) (¾¼) (20) ~
10
AI1
Now, the first image 111 acts as an object for the second surface, so~ we have BI1 = u =(30+20)=50 cm
u
= µ, µ,
V
R
~
, we get
u, =(3Rv,)
10
~
i.e., the final image 12 is virtual and is formed at a distance 100 cm (towards left) from B . The ray diagram is as shown. N
::::::r,
o
A
C
8
"420 cm+1 1+ 30 cm +1 1+100 c m       + 1
u2 = (3R ( 2:
u, =(4µ9)R 2µ3
For refraction at curved surface 52 , we have 1
µ
(1µ)
v,
(4µ9)R 2µ3
(%)
v,
Illustration 48 A ·glass rod· has ends as shown in figure. The refractive index of glass is µ. The object O is at.a distance 2R from the surface of larger radius of curvature. The distance between apexes of ends is 3R. R
(4µ9) >0 (10µ9)(µ2)
... (4)
The equation (4) is satisfied when Origin for first refraction
Origin for refraction at second surface
A
B
'' S' :
C
I
D
S'2 ,._ 2R _ _ R.,_3R/2___.. 1
R/2
Case I: (4µ9)>0
R/2
9 µ>4
..2R  ;    3R ____..,
Find the distance of image formed of the point object from right hand vertex. (b) What is the condition to be satisfied if the image is to be real? Solution For refraction at curved surface 51 ,
... (3)
The image will be real if v 2 is positive, i.e.,
J;?'4~~~ r';l~
(a)
... (2)
(4µ9) (10µ9)(µ2)
0
air
3 ))
On solving the above expression for v 2 , we get
Following points should be kept in mind while drawing the ray diagram. (i) At P the ray travels from tarer to a denser medium. Hence, it will bend towards normal PC . At M, it travels from a denser to a rarer medium, hence, moves away from the normal MC. (ii) The ray PM when extended backwards meets the principal axis at I1 and the ray MN when extended meets the principal axis at I2 •
O
... (1)
2µ3
vertex/pole of surface 52 • Object distance for second refraction is
BI, =100 cm
r,
v,·= 2µR
The first image acts as object for refraction at second surface S2 • The origin of our Cartesian coordinate system is now at
(¾) + (¼) = ¼¾ (50) .BI,
(µ1)
µ 2µ3 =v, 2R
AI1 =30 cm
Again applying .l:L+ µ,
1
;, (2R) =R
and (10µ9)(µ2)>0 ~
0.9
1.64
R
F~r first re~actio~ at AP1B, ~~ have
µ,
=
=µ2 
1 v,
1
=
6
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Ray Optics: Refraction at Curved Surfaces

=>
V1
====
~~
sphere of refractive index 1.5 and of radius r is silvered to make the inner side reflecting. An object is placed at the axis of the sphere at a distance 3r from the centre of the
=6 mm A
sphere. The light from the object is refracted at the
P,
P,
unsilvered part, then reflected from the silvered part and
again refracted at the unsilvered part. Locate the final image formed.
~\
B
So, the first image will be formed at 6 mm towards left
•
of P,
0
For second refraction at AP2B, the distance of first
~
image I, from P, will be 6mm+4mm=10mm (towards left). So, we get
±_1 _Q_ __l_ _ 3_ v,
10
4 1 1 4 3v, 6 10 15
=>
Vi=5 mm
Consider the figure shown. A hemispherical cavity of radius R is carved out from a sphere (µ =1.5) of radius 2R such that principal axis of cavity coincides with that of sphere. One side of sphere is silvered as shown. Find the value of x for which the image of an object at O is formed at 0 itself.
2
=>
/
,.__ _ _ 3r _ _ _....,
3.
±
•l
(ii) The ray diagram is shown in figure· Q,/
..... I,
+
+ve
.··· ...~x::·..,·~.: .' ~
I,
C
2mm 2mm
i.
6 mm>...
Using newCartesian sign convention, we get
=:>
Using new Cartesian sign converttion,
Object d~tarice BO= u Image distance OB'= v Focallength OF=+/
Also M'B'F artd llMOf are sll?lllar,·
A'B'
... (2)
Dividing both sides by uvf , we get
1
1
1
=V U
!
Ca~e 2 : When a virtual image is formed When an object AB is placed between the optical centre 0 and the focus F of a conVex lens, the image A'B' formed by the convex lens is virtual, erect and magnified. as shown in figure. · ·since, triangles A'B'O and ABO are s~ar, so Wf: have A'B' OB' =... (1) AB OB Also, triangles A'B'F and MOF are similar, so we have A'B'
B'F
OM
OF
1
Since, M'B'O and MBO are similar A1B1 OB' So, AB= OB
... (1)
Also, M'B'F arid ~OF are similar
A'B' FB' So,=OM ·OF Since OM= AB, therefore
•
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Ray.Optics: Refraction at Curved Surfaces A'B' FB' =AB OF From (1) and (2), we get OB' FB' OFOB' OB OF OF Using new Cartesian sign convention, we get OB=u, OB'=v, OF=/
... (2)
= f +v f
=>
v u
=>
vf=ufuv
=>
uv=ufvf
Dividing both sides by uvf , we again get 1
1
1
I
V
u
IMAGE FORMATION BY CONVEX LENS OBJECT POSITION
At infinity
DIAGRAM
F
Between F and 2F
At F
2F
2F
F
Beyond 2F
At 2F
POSITION OF IMAGE
2F
2F
2F
2F
NATURE AND SIZE OF IMAGE
At the principal Focus (F) or in the focal plane
Real, inverted and extremely diminished
Between F and 2F
Real, inverted and diminished
At 2F
Real, inverted and of same size as the object
Beyond 2F
Real, .inverted and magnified
At infinity
Real, inverted and highly magnified
1.69
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Advanced JEE Physics OBJECT POSITION
DIAGRAM
Between F and optical centre
POSITION OF IMAGE
On the same side as the object
2F 2F
F
NATURE AND SIZE OF IMAGE
Virtual, erect and magnified
IMAGE FORMATION BY A CONCAVE LENS
I
OBJECT POSITION
DIAGRAM
For all positions of object
POSITION OF IMAGE
hnages formed between the optical centre of the lens and the focus (F)
2F 2F
NATURE AND SIZE OF IMAGE
Always forms a Virtual, Erect and Diminished Image
Illustration 52
'' ' o~~~~~~~.ri:~t'___________ : ____ ~!3__ _
A point object O is placed at a distance of 0.3 m from a convex lens of focal length 0.2 m . It then cut into two halves each of which is displaced by 0.0005 m as shown in figure. Find the position of the image. If more than one image is· formed, find their number and the distance between them.
'
'
''
,_,__ V = 0.6 m +I
Since the triangles OL,,L2 and 01112 are similar. So, we have
j_
0
2 X 0,0005
ffi
I,I, =OB= u+v OA u
L,L,
I,I, L,L, 1+
0.3 m +1
=,,
Solution Each part will work as a separate lens and will form its own image. For any part, we have u = ..:..Q.3 rn, / = +0.2 m .
1 1 1 Therefore, from lens formula,    =  , we have V
1 V
=,,
U
f
1 1 0.3' 0.2
v=0.6m
So, each part forms a real image. of the point object O at 0.6 m from the lens, as shown in figure.
= 0.3 + 0.6 0.3
0.9 =3 0.3
I,I, = 3(L,L,) = 3(2x 0.0005) = 0.003 m
Illustration 53
An object is placed 45 cm from a converging lens of focal length 30 cm . A mirror of radius 40 cm is to be placed on the·other si«:fe of lens so that the object coincides with its image. Find the position of the mirror if it is · (a) convex? (b) concave? Solution
The object and image will coincide only if the light ray retraces its path and it will happen only when the ray strikes the mirror normally. In other words the centre of the curvature of the mirror and the rays incident on the mirror are collinear.
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Ray Optics : RefractioH at C11rued S11rfaces (a)
The rays after refraction from lens must be directed towards the centre of curvature of mirror at C . If x is the separation, then for the lens u=45cm, v=x+40, /=30cm
r_ _ _ tc
0, I
1+ 45
=>
V1
=120 cm
In the second case, shift due to the glass slab is given by
fil=(1f;c}=(1/8 }=4 cm i.e., u will now become ( 404) = 36 cm, so now we have
1 1 1 +=v, 36 30 =>
v2 =180 cm
Therefore, the screen will have to be shifted 60 cm away from the lens.
cm+t x    4 0 cm+i
1 1 1 Using lens formula    =V
U
f
Illustration 55
Find the distance of an object from a convex lens of focal length 10 cm if the image formed is two times the size of object. Focal length of the lens is 10 cm .
1 1 1 x+40 45 30 45 3 X= ( 0) 40=50cm 4530
=
Solution
(b) In case of concave mirror, the refracted rays from lens meet at C, the centre of curvature (C) of the mirror.
A convex lens forms both type of images, real as well as virtual. Since, nature of the image is not mentioned here, so we will have to consider both the cases. CASE I : When image is real In this case v is positive and u is negative with
lvl =2lul, so if u=x then v=2x and f =10 cm
0, l
. . . 111 Sb u stitutmgm =,we get V
1+45cm+1+x~ Using lens formula
1 1 1 = where u=45cm, V
U
f
v=x40, /=30cm,weget 1 x,40
1 45
1 30
x 40 = 45x30 4530 x=90+40=130 cm
1 1 1 +=2x X 10 3 1 2x 10 => x=l5cm x = 15 cm, means object lies between F and 2F. CASE II : When image is virtual In this case v and u both are negative. So if u=y then v=2y and f=lOcm V
A lens with a focal length f = 30 cm placed at a
1 2y
Through what distance should the screen be shifted for the image of the object to remain distinct? The refractive index of the glass of the plate is µ = 1.8 .
1
1
1
U
f
1 1 1 +=2y y 10
distance of a= 40 cm from the object produces a sharp image of an object on the screen. A plane parallel plate with thickness of d =9 cm is placed between the lens and the object perpendicular to the optical axis of the lens.
+=v, 40 30
f
. . . 1 1 1 Su bstitutmg m,    =  , we get
Illustration 54
Solution In the first case,
U
1 10
y=5cm y = 5 cm, means object lies between F and P . LINEAR MAGNIFICATION (m)
The linear magnification (also called lateral or transverse magnification) m produced by a lens is defined as the ratio of the height of image to the height of the object. So, 1.71
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Advanced JEE Physics Height of Image Height of Object
m
I 0
{·: f
Since triangles ABC and A'B'C are similar, so
A'B'
CA'
AB
CA
( dv)=(~)du dt u' dt
AB=O CA=U
Substituting the values in equation (1 ), we get Magnitude of rate of change of position of image is
...CA'=+v
=>
I
V
0
u
 dv = 0.09 rns1 dt
=
  . u ,v 
I
V
0
u
V
Lateral magnification, m =u Rate of change of lateral magnification is given by dv du uv (0.4)(0_09)(1.2)(0.01) dm dt dt u' dt (0.4)2 
ni==
Please note that for both the lens and mirror we have
mreal = NEGATIVE i.e. mroal < 0 mvirtual
= POSITIVE
i.e. mv1rtua1
;> 0
dm
Linear/Transverse/Lateral Mcmn_lfication proauced by a lens'is' I 0
V f u f+u
where I. is size ·of image perpendicular to Principal Axis ~nd O is size of object perpendlcular to principal Axis. Axial Magnification :· Axial magnification iS the ratio of the size of image along the principal ·axis to the size of the
object along the ·pril1cipal axis. SO m axiai
·= Size of Image along Principal Axis Size of 2/. Indicate the nature of the combination (concave, convex or plane) in each case. Solution The formula
1
(d) When d = 2/: The incident parallel beam emerges out as a parallel beam but inverted. The combination behaves as a plane glass slab, which inverts the beam.
r
(e)
d
1
1
F
/1 /, f,f,
=+
When d > 2 /: The incident parallel beam emerges out as a convergent beam. The combination behaves as a convergent or convex lens.
r
is valid only for small values of d compared to / 1 and /,. Therefore, we cannot use this formula in the given cases. However, we can draw the ray diagram to decide the nature. of the combination. (a) When d < f : The ray diagram is shown in figure. The outcoming rays are convergent. Obviously, the combination is a convex lens with F < f.
Illustration 68
······...
·
F....., (b) When d = f : The incident parallel beam converges to a point and then passes without any more deviation. The combination behaves like a convex lens of F = f .
Two equiconvex lenses of focal lengths 30 cm and 70 cm , made of material of refractive index =1.5 , are held in contact coaxially by a rubber band round their edges. A liquid of refractive index 1.3 is introduced in the space between the lenses filling it completely. Find the position of the image of a luminous point object placed on the axis of the combination lens at a distance of 90 cm from it. Solution According to Lens Maker's Formula, we have
l_)
J_= (1.5l)(J_ __ 30
=>
R1
R1 =30cm
Similarly, radius of curvature of the second lens is 70 cm . Since 1 1 1 1 =++... (1)
F(c)
R1
F
When / < d < 2/: The incident parallel beam emerges
/1 f, f,
Here, / 1 =30 cm,/, =70 cm
out as a divergent beam. the combination behaves as a divergent or concave lens.
Now / 3 is calculated again using the Lens Maker's Formula, so we get ,.___F+1
1 .!:..=(1.31)(J_) /, 30 70 1.81
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Advanced JEE Physics
=>
I, =70 cm
=> F =30 cm {from equation (1)) According to Lens formula, applied on the combination of lenses, we have
1 1 1
=
v u
1
F
1
;  (90)
1
= 30
v=45 cm LENSES WITH ONE SILVERED SURFACE
P=2_=3_~ f /, Im where f, is focal jength of lens and Im is focal length of spherical'rnirror formed due to silvering of surface. To have · a fundamental understanding of this we Can understand the silvering of lenses using the following' arguments. A ray incident on a lens with its backside· silvered will be refracted through the lens twice and will be reflected from the mirror once, as shown. a) Light from object O passes through lens ·10 form image I,.
When one face of a lens is silvered as shown in figure it acts like a lensmirror combination.
b)
The image ! 1 acts as an object (virtual) for the curved mirror to form image 12 •
c)
The image I2 acts as an object (virtual) for the lens· to form the firtal image I •
=
Lenses with one face silvered act like lensmirror combination
It is obvious from ilie ray diagram as shown in figur~ that the incident ray of light is refracted through the lens twice (i.e., once when light is incident on the lens and second time when reflected by the mirror) and reflected from the m!rror once.
+
+
The silvered lens acts like a mirror with equivalent focal ' length F , given by The combination acts like a mirror whose effective power is given by P,,e1 =2Pi+Pm wheie Pi is the power of the lens and Pm is the power of the mirror. Since for a mirror we have
p =~
m Im
and for a lens, we have
P, =2_=(µ1)(~~) f, R, R2 So, the combination acts like a mirror having net focal length given by
1 Fnet=pnel
111121
+F /, Im f, f, Im where I, is focal length of lens and Im is focal length of spherical mirror formed due to silvering of surface. SIGN CONVENTION While using the above formula, we make use of tti_e following sign conventions.
a)
f is positive for converging (convex) lens and concave mirror.
b)
f Is negative for diverging (concave) lens and convex mirror.
For example, for a planoconvex lens, from Lens Maker's Formula we get
2_ = (µ1)(22) = µ1 f,
Rao
R
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Ray Optics: Refraction at Curved Surfaces R µ1
=>
/,=
a)
when plane surface is silvered, fm+
0.03 m and when viewed normally through the curved surface it appears to be 0.036 m . If the actual thickness is 0.045 m , find the
co
(a) (b) (c) (d)
Since we know that
1
2 f,
1 "'
2(µ1)
=
F
=> . b)
R
R
F
refractive index of the material of the lens. radius of curvature of lens. focal length if its plane surface is silvered. focal length when the curved surface is silvered.
Solution
2(µ1)
when convex surface is silvered, then in general we know the relation between radius of curvature and the focal length is given by
(a)
Since, µ =>
µ=
Real Depth Apparent Depth
d. dapp
R
= 0.045 =1.5 0.03
fm=z
Since we know that
1
2
2
=+F f, R
=>
2(µ1)
R
2 2µ +=
R
R

F=~ 2µ
0.045 m
Illustration 69
The plane surface of a plano•convex lens of focal length 60 cm is silver plated. A point object is placed at a distance 20 cm from the convex face of lens. Find the position and nature of the final image formed.
(b)
Using1 .!::,__b_=µ,µ, V U R
=>
Solution
1 1.5 (0.036) (0.045) R=0.09 m=9 cm
weget I
11.5 (R)
Since, P = 2Pi + Pm 1
=>
2
1
I'= t, Im where, Ji = +60 cm => =>
and fm + co
1 2 1 1 == F60oo30 F=30 cm
(c)
The problem is reduced to a simple case where a point object is placed in front of a concave (converging) mirror of focal
But f," > oo
length 30 cm . Using mirror. formula i.e.,
=>
1 1 1 +=v u f 1
1 1 v 20 30 v=60 cm
+=
1 F
R1 =+9 cm
2
I,
where, 2.=(µl)(J__l_) I, R, "'
where u=20cm and /=30cm =>
If the plane surface is silvered, then 1 2 1 =+F f, f.,
1+ 20
cm +f
=> The image is virtual and erect
=>
1
2(µ1)
F
R1
1 2(1.51) F +9 => F=9 cm The nature is given by applying negative sign to the final result. So, this will behave as a concave mirror. =>
Illustration 70 The greatest thickness of a planoconvex lens when
viewed normally through the plane surface appears to be 1.83
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Advanced JEE Physics (d) When curved surface is silvered then R, > oo , R, = 9 cm 1 F
2 f,
1
=
fm
=>
_! = 2(µ1)(~_!_)~
=>
.!=2(µ1)+3.
=>
_!=2µ
~
F
,oo99
9
F
F
9
9
1 1.5x2
=
+9 /=3cm F
=>
DEFECTS OF IMAGES : ABERRATIONS
The simple theory of image formation developed for mirrors and lenses suffers from various approximations. As a result, the actual images formed 'contain several defects. These defects can be broadly divided in two categories. a) Monochromatic Aberration The defects, which arise when light of a single colour, is used, are called monochromatic aberrations. b) Chromatic Aberration The index of refraction of a transparent medium differs for different wavelengths of the light used. The defects arising from such a variation of the refractive index are termed as chromatic aberrations.
shows spherical aberration for a concave mirror for an object at infinity. The rays parallel to the principal axis are incident on the spherical surface of the co_ncave mirror. The rays close to the principal axis (Paraxial Rays) are focused at the geometrical focus F of the mirror. The rays farthest from the principal axis (Marginal Rays) are focused at a point F' somewhat closer to the mirror. The intermediate rays focus at different points between F and F'. Also, the rays reflected from a small portion away from the pole meet at a point off the axis. Thus, a three dimensional blurred image is formed. The intersection of this image with the plane of figure is called the Caustic Curve. If a screen is placed perpendicular to the principal axis, a disc image is formed on the screen. As the screen is moved parallel to itself, the disc becomes smallest at one position. This disc is closest to the ideal image and its periphery is called the Circle of Least Confusion. The magnitude of spherical aberration may be measured from the distance FF' between the point where the paraxial rays converge and the point where. the marginal rays converge. The parallel rays may be brought to focus at one point if a parabolic mirror is used. Also, if a point source is placed at the focus of a parabolic mirror, the reflected rays will be very nearly parallel. The reflectors used in automobile headlights are made parabolic and the bulb is placed at the focus. The light beam is then nearly parallel and goes up to large distance.
MONOCHROMATIC ABERRATIONS F
A.
SPHERICAL ABERRATION
Throughout the discussion of lenses and mirrors with spherical surfaces, it has been assumed that the aperhlre of the lens or the mirror is small and the light rays of interest make small angles with the principal axis. Only then, it is possible to have a point image of a point.object.
The rays reflect or refract from points at different distances from the principal axis. In general, they meet each other at different points. Thus, the image of a point object is a blurred surface. Such a defect is called Spherical Aberration. Figure
=
A lens too produces a blurred disc type image of a point object (due to finite aperture of lens). Figure shows the sihlation for a convex and a concave lens for the rays coming parallei'to the principal axis. We see from the figure that the marginal rays deviate a bit strongly and hence, they meet at a point different from that given by geometrical optics formulae. Also, in the situation shown, the spherical aberration is opposite for convex and concave lens. The point FM, where the marginal rays meet, is to the left of the focus for convex lens and is to the right of the focus for the concave lens. M++.
M+\1
P+1...__·,
PA;.,r
P+~rl/
P,L
M+¥
M +f¾.
1.84
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Ray Optics: Refraction at Curoed Surfaces The magnitude of spherical aberration for a lens depends on the radii of curvahrre and the object distance. For minimum spherical aberration the ratio of radii of curvahrre of lens is R1 R1
further above P1 • The image seen on the screen. thus have a cometlike appearance. Image of P
2µ 2 µ4 µ(2µ+1)
However, it cannot be reduced to zero for a single lens which forms a real image of a real object. A simple method to reduce spherical aberration is to use a stop before and in front of the lens. A stop is an opaque sheet with a small circular opening in it. It only allows a narrow pencil of rays to go through the lens hence reducing the aberration. However, this method reduces the intensity of the image as most of the light is cut off. Otherwise, the spherical aberration is less if the total deviation of the rays is distributed over the two surfaces of the lens. Example for this is a planoconvex lens forming the image of a distant object. If the plane surface faces the incident rays, the spherical aberration is much larger than that in the case when the curved surface faces the incident rays. In the former case, the total deviation occurs at a single surface whereas it is distributed at both the surfaces in the latter case.
R
Coma can be reduced by properly designing the radii of curvature of lens surfaces. It can also be reduced by using appropriate stops placed at appropriate distance from the lens. C.
d=f,f, B.
COMA
CURVATURE
So far we have considered the image formed by a lens on a plane. However, it must be kept in mind that the best image may not be formed along a plane. For a point object placed off the axis, the image is spread both along and perpendicular to the principal axis. The best image is, in general, obtained not on a plane but on a curved surface. This defect is known as curvature. It is intrinsically related to astigmatism. The astigmatism or the curvature may be reduced by using proper stops placed at proper locations along the axis. E.
It has been observed that if a point cbject is placed on the principal axis of a lens and the image is received on a screen perpendicular to the principal axis, the image has a shape of a disc because of spherical aberration. The basic reason is that the rays passing through different regions of the lens meet the principal axis at different points. If the point object is placed away from the principal' axis and the image is received on a screen perpendicular to the axis, the shape of the image is like a comet. This defect is called Coma. the lens fails to converge all the rays passing at different distances from the axis at a single point. The paraxial rays form an image of P at P' . The rays passing through the shaded zone forms a circular image on the screen above P' . The rays through outer zones of the lens form bigger circles placed
ASTIGMATISM
Spherical aberration and coma refer to the spreading of the image of a point object in a plane perpendicular to the principal axis. The image is also spread along the principal axis. Consider a point object placed at a point off the axis of a converging lens. A screen is placed perpendicular to the axis and is moved along the axis. At a certain distance, an approximate line image is focused. If the screen is moved further away, the shape of the image changes but it remains on the screen for quite a distance moved by the screen. The spreading of image along the principal axis is known as Astigmatism (you must not confuse this with a defect of vision having the same name). D.
The spherical aberration can also be reduced by using a combination of convex and concave lenses. A suitable combination can reduce the spherical aberration by compensation of positive and negative aberrations. If two thin lenses are separated by a distance d, then condition for minimum spherical aberration is
~,
DISTORTION
It is the defect arising when extended objects are imaged. Different portions of the object are, in general, at different distances from the axis. The relation between the object distance and the image distance is not linear and hence, the magnification is not the same for all portions of the extended object. Hence a line object is not imaged into a line but into a curve and shown.
(b)
(c)
Object (a) and its distorted Images (b) & (c)
1.85
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Advanced JEE Physics CHROMATIC ABERRATION
B.
The inability of a lens to form the white image of a white object is called chromatic aberration. In this case the lens forms coloured images of a white object. The chromatic aberration arises .due to the fact that the focal length of a lens depends upo~ the refractive index of material of the lens. The lens has different refractive indices for· different colours or wavelengths in accordance with Cauchy's formula given by
This is the spread of images perpendicular to principal axis and is given by
B
µ=A+, ). Accordingly, the refractive index is maximum for violet
(). =4000A)
I,Iv
u
u
)0 =(rov' )0 u
f
u
ACHROMATISM & ACHROMATIC DOUBLET
ro,+ro'=O f, f, f, = ro, f, ro,
and minimum for red (). =7800A) . Since
1
focµ1
where ro1 and ro 2 are dispersive powers of materials of
Hence focal length of a lens is maximum for red and minimum for violet
=>
= v,O _ VvO =(v,vv
The lens system free from chromatic aberration is called achromatic combination. This is obtained by using two lenses of different materials and different focal lengths and process is called, to Achromatise which satisfies the relation
7=(µ1i(~ ~) =>
LATERAL CHROMATIC ABERRATION
lenses for focal length a)
freo > J,,,;,,~.
Figllre_ represents the chromatic aberration caused by a lens in the image of an object AB of size O . FR and Fv are second principal foci for red and violet
b)
colours respectively. The images of object AB are of different sizes and of different colours between AvBv and ARBR . The chromatic aberration is of two types. >+V,+< 14.Vv~
f,
and
f,
respectively.
and ro, are always positive, therefore f,/ f, must be negative. This means the combinatj.on mu~t have one lens convergent and other divergent. For the achromatic combination (also called Achromatic Doublet) to be convergent, the powCr of convex lens must be greater or the focal length of convex lens must be smaller than that of concave lens. As dispersive power for crown glass is less than that for flint glass, therefore the convex lens must be made of crown glass while concave lens must be made of flint glass. Condition for minimum chromatic aberration obtained by two thin lenses of same medium separated by a distance d is As
ro1
d~f.,+f, 2
DEFECTS OF EYE
A,
A normal eye has nearer point at D ( 25 cm) called distance
of distinct vision and far point at c.o.
Chromatic Aberration
A. A.
AXIAL CHROMATIC ABERRATION
This is the spread of images along the principal axis and is given by VRVv
SHORTSIGHTEDNESS OR MYOPIA
A shortsighted eye can see only nearer objects. It is due to elongation of eyeball. It is corrected by using a concave lens, whose focal length is ~qua! to the far point of defective eye.
rov' =1
~\_
where ro is dispersive power, v is distance of image from lens for mean (yellow) colour and f is mean focal length of lens. If object is at infinity, then axial chromatic aberration, f,fv=rof
1.86
Corrective DEFECTIVE EYE Image is not Created on the Retina
Lens DEFECT CORRECTED
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LONGSIGHTEDNESS OR HYPERMETROPIA
A long sighted eye can see only farther objects. It is due to contraction of eyeball. It is corrected by using convex lens. This lens brings the nearer point of defective eye at a distance which equals to the distance of distance of distinct vision D( = 25 cm).
~\_
MAGNIFYING POWER
~Tu I
l
Corrective DEFECTIVE EYE Image Js Created beyond the Retina
C.
When a small object is placed between optical centre and focus of a convex lens, its virtual, erect and magnified image is formed on the same side of the lens. The lens is held close to eye and the distance of the object is adjusted, till the image is formed at the least distance of distinct vision from the eye. For a normal eye, the least distance of distinct vision is 25cm.
Lens DEFECT CORRECTED
Case I : When image is formed at D It is defined as the ratio of the angle subtended by the image at the eye and the angle subtended by the object seen directly, when both lie at the least distance of distinct vision. It is also called angular magnification produced by the simple microscope. It is denoted b)' M. By definition, magnifying power of the simple microscope is given by
PRESBYOPIA
A presbyopic eye can see objects only within a definite range. This defect is corrected by using bifocal lenses. D.
M=Q_ a
In particle, angles a and p are small. Therefore, angles a and p can be replaced by their tangents i.e.
ASTIGMATISM
It arises due to distortion in spherical shape in cornea. This defect is corrected by using cylindrical lenses. OPTICAL INSTRUMENTS
An optical instrument is a device which is constructed by a suitable combination of mirrors, prisms and lenses. The principle of working of an optical instrument in based on the laws of reflection and refraction of light The common types of optical instrument are a) Projection instruments: These are used to project on the screen a real, inverted and magnified image of an opaque or transparent object so as to be viewed by a large audience. The object is, however, so fitted that its image is seen in erect form. An eye, a photographic camera, a projection lantern, an episcope, an epidiascope, an over~head projector, a film projector, etc., are examples of projection instruments. b) Microscopes: These are used to see very small objects in magnified form which otherwise cannot be seen distinctly when placed close to the naked eye.
·M = tanp =CA'=~ tana CA u If f is focal length of the lens acting as simple microscope, then 1 1 1 =v u f =>
~=1~ u f
=>
M=l~ f
B"
... ____________ _
Q
A'
F"
EXAMPLE A simple microscope and a compound microscope.
c)
u
Telescope: These are used to see astronomical and distant objects in magnified form which, otherwise cannot be seen clearly with the naked eye. EXAMPLE An astronomical telescope, a Galilean telescope, a terrestrial telescope, a reflecting telescope, etc.
4
1+f~    D +
D
M=l+/
1.87
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b)
a)
" CONCEPTUAL NOT!=(S) .· From above it follows that lesser'is the focal length of the convex. lens used as simple microscope, greater is the·
value of the magnifying power obtained. Further, the positive value of magnifying power of a simple microscope tells that .image formed is erect and hence virtuat.
c)
b)
Case II: When image is formed at infinity It is defined as the ratio of the angle formed by the image (situated at infinity) at the eye to the angle formed by object at the eye, when situated at least distance of distinct vision. It is denoted by M .
d) e) f)
In science laboratories, a magnifying glass is used to see slides and to read the vernier scales attached to the instruments. The use of magnifying glass enables us to place the object close to eye, making it appear bright and yet clearly visible. In position AB, object lies close to the eye. In absence of lens, the object will not be clearly · visible. It is also used by astrologers to read the fate lines of the hand. Used by Biology students to see slides. Used by detective department to match finger prints.
Illustration 71
A man with normal near point (25 cm) reads a book with small print using a magnifying glass (a thin convex lens) of focal length 5 cm. Find the (a) closest and farthest distance at which he can read the book when viewing through the magnifying glass. (b) maximum and minimum maguifying power possible using the above simple microscope.
I+ u = f+i i.D
Draw a line· A'B' = AB and perpendicular to principal axis at a distance CA'= D (least distance of distinct vision). Joint B' C . Then LB' CA' = a is the angle formed by object at the eye, when situated at distance D . The angle formed by the image situated at infinity at the eye is same as the angle formed by the object AB at the eye. Thus, LBCA = P is the angle formed by the image at the eye,
Solution (a) For a normal eye, far and near points are co and 25 cm, respectively. So, we have Vmm
=25 cm
, 1 1 1 Using lens formula,=
v
u
f
U=/
(f)1
By definition, M=f_
> 
Jii
=>
A,
V 60 3 . . M agnif1cation, m===
')' ' J.:!1:.=" 3 xrn, ,l2x9.Bf dt h
H
(
Jii
,c{3)
u 20 The image formed by first half lens is shown in Figure1
o
B
~5'
:
F _________ A,_
i
A
Hh
C,
:45°
Hh
'
r,
*
mirror. Mirror forms its virtual image A2 C2 . So, 2
B,
dt
So, speed of insect is
=>
=¼(~~)
'' :' '' ' •A,
A convex lens of focal length 15 cm is split into two halves and the two halves are placed at a separation of 120 cm . Between these two halves of the convex lens, a plane mirror is placed horizontally and at a distance of 4 mm below the principal axis of the lens halves. An object of length 2 mm is placed at a distance of 20 cm from one half lens as shown in figure. f=15cm
!. ~
2mm
1+
For the second half of the lens, using lens formula 1 1 1 =,we get V U f 1
1 v 60 v=+20
1 15
=
=>
V 20 1 m===u 60 3
f=15cm
~
½
So, length of final image A,B, = A2 B2 = 2 mm . However, point B2 is 2 mm below the optic :axis of
20 cm ,_.....,...__ ___._4 ~m
second half lens. Hence, its image B3 is formed
120 cm H
Find the position and size of the final image. (b) Trace the path of rays forming the image.
I mm 3
above the principal axis. Similarly, point A, .is 8 mm below the principal axis. Hence, its image is
1.98
4mm
Figure·2
Problem 4
·
2mm
C,
V=1.1x10 3x=1.8 m => x=0.6m Hence, u=0.6m and v=+1.2m .11111 1 Usmg  =    =     = ! V U 1.2 0.6 0.4 =>
+ '' ''
0
''
' 11 2 ',
I
.,..,. , .,.
a.1/ ,
IVI
h'',~//
'.,,
Similarly, y coordinate of image I,
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y = _:(lvlsincxhcoscx) On substituting the_values of lvl and h from (1) and (2),weget y=O.
=>
So, the coordinates of the final irhage·are
[
20
v2 =  cm 3
/;;:::J
6.67 cm
V 20/3 2 and m2===u 10 3 About Final Image Net magnification is given by
t(2coscx+l), o] cosa.+1
(b) Ray diagram is shown in figure
m=m1m2
=31
i.e., height of the image is 3 ~ ½ = 1 cm Since, the net magnification isnegative, so the final image is inverted.
Further y coordinate of a point of the image will be,
Y1 =myom2.6.
CONCEPTUAL NOTE(S)
... (1)
The Ycoordinate of 12 is zero is very obvious because a ray of
with respect to the principal axis of L1
light starting from 11 and passing ihrgugh O' will suffer no
So, y coordinate of image of A is
deviation. Hence, 12 must be forined an this line itself [e.,
y,. =(½) => (b)
0.9 m   ~ 0.8 m
+1
•
1 _45 + 1.so:10• (i..=i..,) A.o Solution
· 1 (1 1) , ApplyingLensMaker'sFormula, =(µ1) f R, R, we get
The given system happens to be a part of an equilateral prism of prism angle 60° as shown in figure.
__!_ =(~ 0.3
=>
2
1)(!R .!..) R
{·: R, =Rand R2 =R}
R=0.3
Now applying, !!:1__ µ 1 = µ 2 µ 1 at air glass surface, we get V
=>
(say}
2
surface. Therefore, applying !!:1__!:!_ = µ 2 µ 1 at glass water surface,
Since according to Snell's Law, we have
V
sini smr
u
R
we get
=.
4 3
sini=n'1 sin(30')
. Smee, ni
v1 =2.7 m
This image 11 will act as the virtual object for glass water
60'
f1=f2==30°=r
=>
R
So, the first image 11 will be formed at 2.7 m from the lens.
At minimum deviation, we have
n1
u
, =1.2 + 10.Sxl0' 2
i..,
where A. 0 = 600 run
=>
. ·{l.2 + 10.SxlO'}(l)3 Slilt 1.5  _ 
=>
, , 1(3) t=sm 
(600)1
2
2
v,
=>
3 2 2.7
(¼)(¾) 0.3
v, =1.2 m
So, the second image 12 is formed at 1.2 m from the lens1or
0.4 m from the plane mirror.
4
4
Problem 10 A thin equiconvex lens of glass of refractive index
This iITI4ge 12 will act as a virtual object for mirror. Therefore, third real image 13 will be formed at a distance of 0.4 m in front of the mirror after reflection from it. Now this image acts as a real object for waterglass interface. Hence applying,
&  1:2 = µ, R µ, , we get V
U
µ = ~ and of focal length 0.3 m in air is sealed into an 2
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(¾)(¼)
(0.80.4)
0.3
3 2
=>
Applying Snell's Law at face AB, we get sini sinr1
1 n=
v, =0.54 m
So, the fourth image is formed to the right of the lens at a distance of 0.54 m from it. Now finally applying the same formula for glassair surface, we get
1(I) l_ =  2=0.9 m I2
v, 0.54
=>
sini1 =nsinr1
=>
i1 =sin1 (nsinr1 )
Substituting value of r1 , we get i1 =sin1 {nsin(45°C)}
=>
0.3
i1 = sin1 [ n(sin45°cosC cos45°sinC)]
=>
Hence, the position of final image is 0.9 m relative to the
lens (rightwards) i.e., the image is formed 0.1 m behind the
=>
mirror.
. =sm. 1[ .J2n [ y1;,:; ~ n,J]
11
Problem 11
A right angle prism ( 45° 90° 45°) of refractive index n has a plane of refractive index n1 (n 1
r1 =Ar2 =45°0°=45°
Now applying Snell's Law at face AB, we get sini1 n=sinr1
=>
1.352
sini, sin(45')
=> · sini1 =(1.352)(1)
sinC =:i
n
=>
sin i1 = 0.956
=>
i, = sin1 (0.956) "73°
Therefore, required angle of incidence is
4 = 73°.
Problem 12
B
Now, it is given that r2
=>
C
=C
r1 =Ar, =(45'C)
A ray of light travelling in air is incident at grazing angle (incident angle = 90°) on a long rectangular slab of a transparent medium of thickness t = 1.0 m . The point of incidence is the origin A(O, 0). The medium has a variable index of refraction
n(y)
~ven by
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dy
The refractive index of air is 1.0 =,,
K~ .
,,..
"
Solution
... ~ =>
cm +t+ 20 cm +1
Rays coming from object AB first refract from the lens and then reflect from the mirror.
[ ·: Distance of object is 15 cm)
For refraction from the lens, we have
u=20 cm, [=+15 cm
R =15 cm 2
A
R=lO cm
r r0.6cm
_ of new lens system. Then,
:, =(µ, 1)(! ~)+(µ1)(_~ 
Optic Axis
B,
In the second case, let µ be the refractive index of the liquid filled between lens and mirror and, let f' be the focal length
3
!)
of Lens
r
B
Optic Ax.is of Mirror
cm
i A, ..,.._ 30 cm ao cm_.,.. 20 cm 111
1 =(' 1)'(2)_(µ1)= 1_µ1={2µ) F'2
R
F' =~ = ___!Q_ 2µ 2µ
RRR
1. 1 1 Applying lens formula, =,we get V U f
R
{·: R=lO cm}
Now, the image coincides with the object when it is placed at 25 cm distance.
=> '=>
2µ
=>
25µ=40
=> =>
v=+60 cm
V +60 m1 ===3 u (20) So, the first image formed by the lens will be 60 cm from it (or 30 cm from the mirror) towards left and 3 times magnified but inverted. Length of first image A 1B1 would be
___!Q_ = 25 5025µ=10
=>
and linear magnification is given by
F'=25
=>
1 1 1 :;; (20) = 15
40 µ==1.6 25 µ=1.6
A 1B1 = 1.2 x 3 = 3.6 cm (inver(ed). For reflection from mirror, we have
Image formed by lens (A1B1 ) will behave like a virtual
Problem 14
A convex lens of focal length 15 cm and a concave mirror of focal length 30 cm are kept with their optic axis PQ and RS parallel but separated in vertical direction by 0.6 cm as shown. The distance between the lens and mirror ls 30 cm. An upright object AB 'of height 1.2 cm is placed on tlie optic axis .PQ of the lens at a distance of 20 cm fr~m the lens. If A' B' is the image after refraction from the lens and the reflection from the mUTor, find the distance of A'B' from the pole of the mirror and obtain· its magnification. Also l~cate positions of A' and B' with respect to the optic axis RS .
object for the mirror at a distance of 30 cm from it as shown. Therefore u = +30 cm , f = 30 cm .
1 1 1 Applying mirror formula, +=,we get V
1
1
U
f
1
+=v 30 30
=>
V=15 cm
and linear magnification is given by V (15) 1 m ===+, u +30 2
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Advanced JEE Physics So, the final image A'B' will be located at a distance of 15 cm from the mirror (towards right) and since magnification is
+.!., 2
length of final image would be
(n2 > n1 ) at an angle of incidence O as shown in the figure.
A'B' = 1.8 cm
Since the p"oint B1 is 0.6 cm above the optic axis of mirror,
therefore, its image
B'
would be
(0.6)(½) =0.3 cm
above
optic axis. Similarly, point A 1 is 3 cm below the optic axis, therefore,
its image A' will be 3 x
Monochromatic light is incident on a plane interface AB between two media of refractive indices tt 1 and n2
A'B'=3.6x½=1.8 cm =>
Problem 15
½=1.5 cm below the optic axis as
The angle 0 1s infinitesimally greater than the critical angle for the two media so that total internal reflection takes place. Now if a transparent slab DEFG of uniform thickness and of refractive index n3 is introduced on the interface (as shown in the figure), show that for any value of n3 all light will ultimately be reflected back again into medium II. Consider separately the cases (a) n3 < n1 and (b)
n, > n1 Medium I (n,) D  E : Medium Ill :
shown.
I
G:
(nJ
e,' ''' '' '
Optic Axis of Lens
Optic Axis of Mirror
1.5 cm
A'B'= 1.8 m
15cmj
(n,)
Solution At interface AB, 0 is infinitesimally greater (slightly greater) than the critical angle for interface, so
Net magnification of the image is given by
m=m1
Medium II
'' '
A'
~
lF
A~~~~B
xm =(3i(+½)=%
1
2
0>sin
(::)
(a) When n3 < n1
A'B'=(m)(AB)=(%)8
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Medium Ill
Medium II
Medium I
,.'
q
p'
e,'
Medium Ill
i>9
I'
'
i (C)1>m
Hence, TIR will now take place on Medium I and Medium III interface and the ray will be reflected back to Medium IIL Case II: f1i < n2 < n3
equation only. But two cases are deliberately formed for b~tter understanding of refraction, Snell's Law and total internal reflection (TIR).
1bis time while moving from Medium II to Medium III, ray of light will bend towards normal. Again applying Snell's Law at P , we get n2 sin0·= n3 sini ~
sini = " 2 sin9
n,
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1bis section contains Single Correct Choice Type Questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct.
. 1.
.

A transparent hemisphere has 'a radius of curvature 8 cm and an index of refraction of 1.6. A small object 0 is placed on the axis halfway between the plane surface and the Spherical surface i.e. 4 cm from each. The distance between the two images when viewed along t,he axis from the two sides of the hemisphere is approximately
0
2mm
1
1+ 20
cm +1.
(A) the distance between the.images is 2 mm ./
(B)
µ= 1.6
the distance between the images is 4 mm
(C) the distance between the two images formed by
+
· suchalensis6min
·
(D) only one image will be formed by the lens (A) 7.5cm (C) 2.5cm 2.
3.
(B) 8.5 cm (D) 13.5 cm
A square wire of side 3.0 cm is placed 25 cm in front of a concave mirror of focal length 10 cm with its centre on the axis of the mirror and its plane normal to the axis. The area enclosed by the image of the wire is (A) 7.5 cm' (B) 6.0 cm2 (C) 4.0 cm' (D) 3.0 cm' An object is placed at a distance 2/ from the pole of a convex mirror of focal length f . The linear magnification is 1 2 (B) (A) 3 3
(C)
3 4
(D) 1
5.
Ah object is placed at 26 cm from a convex mirror of · focal length 20 cm. The distance of the image from the pole of the mirror is (A) infinity (B) 10 cm (C) 15 cm (D) 40 cm
6.
A point object is placed at a distance of 25 cm from a convex lens of focal length 20 cm . When a glass slab of thickness t and refractive index 1.5 is inserted between the lens and the object, the image is formed at infinity. The thickness t of the slab is, (A) 5 cm (B) 10 cm (C) 15 cm (D) 20 cm
7.
Light is incident normally on face AB of a prism as showri in figure. A liquid of refractive index µ is placed on face AC of the prism. The prism is made of glass of refractive index
4.
=
A convex lens of focal length 10 cm is painted black at the middle portion as shown in figure. An object is placed at a distance of 20 cm from the lens. Then
1.108
i. The limits of 2
µ for which
total internal reflection takes place at the face AC is
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I
I
Liquid
(C) concave, 15 cm (D) convex, 15 cm 14. A boy of height 1.5 m with his eye level at 1.4 m stands before a plane mirror of length 0.75 m fixed on the well. The height of the lower edge of the mirror above the floor is 0.8 m. Then (A) the boy will see his full image. (B) the boy cannot see his hair. (C) · the boy cannot see his feet. (D) the boy cannot see both his hair and feet.
~ c 90° B
J3
(A) µ 41°
(B)
'' 450,''
< 41°
(D) ;, 41°
192. When a ray of light is refracted by a prism such that ihe angle of deviation is minimum, then (A) the angle of emergence is equal to the angle of incidence. (B) the angle of emergence is greater than the angle of incidence. (C) the angle of emergence is smaller than the angle of incidence. (D) the sum of the angle of incidence and the angle of emergence is equal to 90° .
(A) 45' (C) 120°
(B)
90°
(D) 180°
196. A pianoconvex lens. has a thickness of 4 cm. When placed on a horizontal table with the curved surface in
contact with it, the apparent depth of the bottommost point of the lens is found to be 3 cm. If the lens is inverted, such that the plane face _is in contact with the table, the apparent depth of the centre of the plane face 25 cm. The focal length of the lens is is found to be 8 (A) 50 cm (B) 75 cm (C) 100cm (D) 150cm 197. If J 8 and JR are the focal lengths of a convex lens for
blue and red lights respectively and F, and FR are the respective values for a concave lens, then (A) J, > JR and F, >FR ,.(B) f, FR
(C) J, > JR and F, m, ) , and the distance between the two positions of the lens is x , the
focal length of the Jens is (A)
(C)
X 
(B)
m1 +m2 X
(D)
(m, +m,)'
x m1m2 X
(m, m,)'
231. A screen is placed 90 cm from an object. The image of the object on the screen is formed by a convex lens at
226. A lens forms a sharp image on a screen. On inserting a
parallel sided glass slab between the lens and the screen, it is _found necessary to move the screen a distance d away from the lens in order for the image to
be sharp again. If the refractive index of the material of the slab is n , the thickness of the slab is d (A) nd (B) n · nd (C) (n1)d (D) n1 n
two different locations separated by 20 cm. The focal · length of the Jens is (A) 10.7cm (B) 21.4cm (D) 32.1 cm (C) 15.8cm 232. A light ray travelling in glass medium is incident on glassair interference at an angle of i_J;i.cidence 8 . The
reflected (R) and transmitted (T) int~nsities, both as function of 0, are plotted. The.correct sketch is ,Intensity
227. A plano convex glass
Jens
(µ, =¾)
100o/o , ·
of
radius of curvature R =10 cm is placed at
T
,' ,,
(A)
___R..
0
~_,.,..'
o~9LO'+ 0
a distance of y from a
concave lens .of focal ,..._ X    Y ___,.. length 20 cm . The distance x of a point object O from the piano convex lens so that the position· of final image is independent of y'is (B) 30cm (A) 20cm (C) 40cm (D) 60cm
, Intensity 100%
T
(B)
228. A thin lens has focal length /, and its aperture has
diameter D .
Ii forms
an image of intensity
central part of the_ aperture, of diameter
i'.
If the
~ , is blocked
by an opaque paper, the focal. length of the lens and the intensity of image will become
(A) 1.128
fr,½
Intensity 100%
T
(C)
__ f:!_.
(B)
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T
(B)
,, 
a virtual, erect, samesized image can be obtained using a plane mirror.
(C) a virtual, erect, magnified image can be formed using a concave mirror.
(D)
(D) a real, inverted, samesized image can be formed using a convex mirror.
R 
O!=:.:::c__ _...:::::90=,=+. e 239. When an object is at distances x and y from a IellS, 'a
233. A lens is placed between the source of light and a wall. It forms images of area A1 and Ai on the wall for its two different positions. The area of the source of light is (A)
~A,A,
(C)
(B)
(D)
A,; A,
(C)
2.+2.)1
( A,
the curved surface, then the system behaves like a
(C) rµ
(B)
r µ1
(D)
r(µ1)
..{xy
(D)
x+y 2
240. If the central portion of a convex lens is
A,
234. The plane face of a planoconvex lens is silvered. If µ be the refractive index and r the radius of curvature of concave mirror of radius r (A) µ
real image and a virtual image is formed respectively
having· same magnification. The focal length of the lens is given by (A) xy (B) x+y
wrapped in black paper as shown in the figure, ~ (A) no image will be formed by the remaining portion of the lens. · (B) full image will be formed, but it will be less bright. (C) the central portion of the image will be missing. (D) there will be two images, each producecl by one of the exposed portions of the lens.
241. A plane mirror made of glass slab 235. A ray of light falls on the surface of a spherical ·paper
position of the final image is
deviation of the emergent ray from the direction of the incident ray is
(A)
(B)
2(ap)
(aP)
(D)
pa
(C)
2
is 2.5 cm
thick and silvered at the back. A point object is placed 5 cm in front of the unsilvered face of the mirror. The
weight making an angle a with the normal and is refracted in the medium at an angle p. The angle of
(A) (aP)
(µ, =1.5)
(B)
16 3
cm from unsilvered face
25
(C)
cm from unsilvered face 3 12 cm from unsilvered face
(D)
14 cm from unsilvered face
236. The magnification of an object placed in front of a
convex lens of focal length 20 cm is +2 . To obtain a magnification of 2, the object has. to be moved by a distance equal to (A) 40 cm
(C) 20cm
242. The distance between an object and the screen is 100 cm. A lens proQ.uces an image on the screen when
placed at either of two position~ 40 cm apart. The power of the lens is approximately (A) 4.25 D (B) 4.50 D (C) 4,75 D (D) 5.0 D
30 cm (D) 10cm (B)
. 237. A concave mirror is placed on a horizontal table with its axis directed vertically upward. Let ( 0) be the pole of the mirror and C its centre of curvature. A point object is placed at C . It has a real image also located at C . If the mirror is now·filled with water, the.image will be
(A) real and will remain at C . (B) real and located at a point between C and oo . (C) virtual ru,d located at a point between C and 0. (D) real and located at a point between C and 0.
243. A real image of an object is formed ·by a convex lens at
the bottom of an empty beaker. The beaker is now filled with a liquid of refractive index 1.4 to a depth of 7 cm. In order to get the image again at the bottom, the beaker should be moved 0
238. All of the following statements,are correct except
(A) the magnification produced by a convex mirror is always less than one.
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(A) 10
(B) 20
(C) 30
(D) 40
250. An object is placed at a distance of 12 cm from a convex lens on its principal axis and a virtual image of certain
244. The convex surface of a thin concavoconvex lens (refractive index 1.5) has a radius of curvature 20 cm. The concave surface has a radius of curvature 60 cm. The convex side is silvered and placed on a horizontal surface. At what distance from the lens should a pin be placed on the optic axis such that its image is formed at
size is formed. If the object is moved 8 cm away from the lens, a real image of the same size as that of the virtual image is formed. The focal length of the lens in cmis (A) 15 (B) 16 (D) 19 (C) 18
the same place?
(A) 15 cm (C) 22.5 cm
(B) 7.5 cm (D) 30 cm
245. In PROBLEM 244, if the concave part is filled with
water (refractive index
± ), the distance from the lens at 3,
251. A thin converging lens of refractive index 1.5 has a power of +0.5 D . When this lens is immersed in a liquid, it acts as a diverging lens of focal length 100 cm. The refractive index of the liquid is (A)
4
(B)
3
(C)
5 3
() D
2
which the pin should be placed to form the image at the same place is
90 cm 13 . 135 (C) cm 13
(A)
(B)
45
cm 13
2
252. The distance between an object and its real image
180
(D) cm 13
246. An equiconvex lens of glass
3
(µ, =1.5)
of focal length
10 cm, silvered on one side behaves like a
formed by a lens is D . If the magnification is m , the focal length of the lens is mD (A) (ml)D (B)
m
(C)
(A). convex mirror of focal length 5 cm (B) convex mirror of focal length 20 cm (C) concave mirror of focal length 2.5 cm (D) concave mirror of focal length 10 cm 247. A pianoconvex lens of focal length 30 cm has its plane surface silvered. An object is placed 40 cm from the lens
(ml)D
(D)
m'
m+l mD
(m+l)'
253. A piano convex lens of focal length 16 cm, is to be made of glass of refractive index 1.5. The radius of curvature of the curved surface should be (A) 8 cm (B) 12 cm (C) 16 cm (D) 24 cm
on the convex side. The distance of the image from the
lens is (A) 18 cm (C) 30 cm
254. A real image of a point object O was formed by an
(B) 24cm (D) 40 cm
248. Refraction takes place at a concave spherical boundary ' 3 separating glass air medium. If µg = , then for the
2
image to be real, the object distance (A) is independent of the radius of curvature of the refracting surface (B) should be greater than the radius of curvature of the refracting surface (C) should be greater than two times the radius of curvature of the refr~cting surface (D) should be greater than three times the radius of curvature of the refracting surface 249. A parallel beam of light incident on a concave lens of focal length 10 cm emerges as a parallel bearu from a convex lens placed coaxially~ the distance between the lenses being 10 cm. The focal length of the convex lens incm is.
equiconvex lens of focal length f and the magnification was found to be unity. Now the lens is cut into two symmetrical pieces as shown by the dotted line and the right part is removed. The position of the image formed by the remaining part is at
0
f
(A)
f
(C) i_ 2
(B)
2/
(D) Infinity
255. A convex lens, made of a material of refractive index 1.5 and having a focal length of 10 cm is immersed in a liquid of refractive index 3.0. The lens will behave as a
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' lens of focal length 10 cm. (A) converging (B) diverging lens of focal length 10 cm. (C) converging lens of focal length
10 3
(C) 12cm
cm.
(D) diverging lens of focal length 30 cm. 256. A point source S is placed at a height h from the bottom of a vessel of height H (< h). The vessel is . polished at the base. If the water is gradually filled in the vessel at a constant rat'e a m 3s1 , the distance d of image of the source from the bottom of the vessel varies with time. t as
S•
260. The maximum and minimum distances between a convex lens and an object, for the magnification of a real image to be greater than one are (A) 2/ and f (B) f and zero (C) oo and 2/ (D) 4/ and 2/ 261. A point object is placed on the optic axis of a convex lens of focal length f at a distance of 2f to the left of it. The diameter of the lens is d . An observer has his eye at a distance of 3 f to the right of the lens and a distance h below the optic axis. The maximum value of h to see the image is d (A) d (B)
r
i
• h
4
H
(C)
(A)
(q
t=, l=, 1=, l=, 0)
(D)
257. A convex lens of glass has power P in air. If it is immersed in water its power will be (A) more than P (B) less than P (C) p (D) more than P for some colours and less than P for others 258. A biconvex lens, made of a material of refractive index 1.5, has radius of curvature of each side equal to 0.5 m. The power of the lens is
(A) 0.5 D (C) 1.5 D
(D) 6cm
(B) 1.0 D
(D) · 2.0 D
259. A convex lens forms a real image 4 cm long on a screen. When the lens is shifted to a new position without disturbing the object or the screen, again real image is formed on the screen which is 16 cm long. The length of the object is (A) 8 cm (B) 10 cm
d
2
3
(D) d
262. A convex lens is immersed in a liquid of refractive index greater than that of glass. It will behave as a (A) convergent lens (B) divergent lens (C) plane glass (D) homogeneous liquid
263. If the top half of a convex lens is covered with black paper, (A) the bottom half of the image will disappear. (B) the top half of the image will disappear. (C) the magnification will be reduced to half. (D) the intensity will be reduced to half. 264. In displacement method, the lengths of images in the two positions of the lens between the object and the screen are 9 cm and 4 cm respectively. The length of the object must be · (A) 6.25 cm (B) 1.5 cm (C) 6 cm (D) 36 cm
265. A convex lens of focal length 15 cm is placed on a plane mirror. An object is placed 20 cm from the lens. The image is formed (A) 12 cm in front of the mirror (B) 60 cm behind the mirror (C) 60 cm in front of the mirror (D) 30 cm in front of the mirror 266. A convex lens of focal length 40 cm is held coaxially 12 cm above a concave mirror of focal length 18 cm. An object held x cm above the lens gives rise to an image coincident with it. The x is equal to (A) 12cm (B) 15 cm (C) 18cm (D) 30cm
0 xcm
12cm
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267. Two thin lenses of powers 2 D and 3 D are placed in contact. An object is placed at a distance of 30 cm from the combination. The distance in cm of the image from the combination is
(A) 30 (C) 50 .
(A) f_
(ll) 40 (D) 60
lenses is
(C)
lf,.f,I
(ll)
h + !,
f,f,
(D)
I,!,
It,  !,I
(D) dispersion
270. A plane mirror is Rlaced at the bottom of a tank containing a liquid ofrefractive index µ . A small object P lie~ at a height h above the mirror. An observer O , vertically above P , outside the liquid, observe P and its image in the mirror. The apparent distance between these two will be
o, ''' '' ' 'I ' Ph
(D)
3
(D)
f
3/ 2
274. An astronomical telescope has an angular magnification of magnitude 5 for distant objects. The separation between the objective and the eyepiece is 36 cm and the final image is formed at infinity. The focal length f, of
275. A piano convex lens .of radius of curvature R fits exactly into a plano concave lens such that their plane surfaces are parallel to each other. If the lenses are made of differen~ materials of refractive indices µ1 and µ 2 , then focal length of the combination is given by (A)
(ll)
R 2(µ, +µ,)
(C)
ZR
R 2(µ,µ,)
(D)
R
276. A compound microscope has an objective of focal length 2.0 cm and an eye piece of focal length 6.25 cm separated by 15 cm. If the final image is formed at the least distance of distinct vision (25 cm), the distance of the object from the objective is (A) 1.5 cm (B) 2.5 cm (C) 3.0 cm (D) 4.0 cm
:~
(B)
(C)
2/
the objective and f, of the eyepiece are (A) f 0 =45cm and f>,9cm (ll) f 0 =50cm and /,=10cm (C) f,=7.2cm and /,=5cm (D) / 0 =30cm and /,=6cm
h+f,
269. Chromatic aberration in a lens is caused by (A) reflection (ll) interference
(C) diffraction
(B)
2
268. Two convex lenses of focal lengths f, and f, are mounted coaxially separated by a distance. If the power of the combination is zero, the distance be~~en the
(A)
concave mirror of focal length f : To obtain a real image of same magnification, the object has' to moved 1;>y a distance
2h .µ1 2h µ
277. In PROBLEM 276, the magnifying power of the
271. A person can see clearly between 1 m and 2 m .. His corrective lenses should be (A) bifocals with pow~r 0.5 .D and additional +3.5 D (B) bifocals with power ~1.0 D and additional +3.0D (C) concave with power LO D (D) convex with power 0.5 D
microscope is
(A) 10 (C) 20
(B) 15 (D) 30
278. A point object is placed at a distance of 20 cm from a
glass slab (µ, =
¾) half immersed in water (µ. = ¾) as
shown in figure. The distance between two images when se~nfrom the other side of the slab is
272. Spherical aberration in a thin lens can be reduced by (A) using a monochromatic light. (B) using a doublet combination. (C) using a circular. annular mask over the lens. (D) increasing the size of the lens.
9cm
l++i
:
0
273. A virtual image of an object is formed with a magnification of 2, when the object is placed infront of
a
1+20 cm+i
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(A) 1 cm (C) 4cm
(B) 2cm (D). 6cm
279. A compound microscope has .a magnification of 30. The focal length of the eyepiece is 5 cm. If the final image is formed at the least distance of distinct vision ( 25 cm),
possible magnification, we choose the lenses of focal lengths (A) 100 cm, 0.3 cm (B) 10 cm, 0.3 cm (C) 10 cm, 4 cm (D). 100 cm, 4 cm 287. The angular magnification of a telescope which contains
an objective of focal length / 1 and eyepiece of focal length /, is
the magnification produced by the objective is (A) 5 (B) 7.5 (C) 10 (D) 15
(AJ 280. The least distance of distinct VisiOn is 25 cm. The focal length of a convex lens is 5 cm . It can act as a simple
microscope of magnifying power (AJ 4 (B) 5 (C) 6 (D) None of these 281. An astronomical telescope has an eye piece of focal length 5 cm. If the angular magnification of normal adjustment is 10, the distance between the objective and the eye.piece is (A) 45cm (B) 50 cm (C) 55cm (D) 110 cm 282. The focal lengths of the objective and the eyepiece of an
astronomical telescope are 100 cm and 20 cm respectively. Its magnifying power in normal adjustment is (A) 5 (B) 2 (C) 25 (D) 4 283. Two convex lenses of focal lengths 0.3 m and 5 cm are used to make a telescope. The distance kept between
them is equal to (A) 0.35 m (C) 5.3m
(C)
4 3
(DJ
/1 + /, I, f,f, !1 + I,
288. An achromatic combination is to be made using a convex and a concave lens. Tfle two lenses should have (AJ their power equal. (B) their refractive indices equal. (C) their dispersive powers equal. (DJ the product of their powers and dispersive powers equal. · 289. For a thin equiconvex lens, the optics axis coincides with the xaxis.and the optical centre coincides with the origin. The coordinates of a point object and its image
are (40, 1) cm and (50,2) cm respectively. Lens is located at (A) x=0 (B) X=lOcm (D) x~30cm (C) x=+20 cm 290. The near point of a person is 50 cm and the far point is for seeing distant objects are respectively
285. The angle of incidence for an equilateral prism is 60° . The refractive index of prism so that the ray inside the prism is parallel to the base of the prism is 9 .. 8
(BJ
1.5 m. The spectacles required for reading purpose and
(B) 5.3 cm (D) 0.15m
284. To have larger magnification by a telescope (A) the objective should be of large focal length and the eyepiece should be of small focal length (B) both the objective and the eyepiece should be of large focal lengths (C) , both the objective and the eyepiece should be of small focal lengths (D) the objective should be of small focal length and the eyepiece should be of large focal length
(A)
(C)
I, !1 /1 I,
(B). ..fi.
(D) ,/3
286. Four convergent lenses have focal lengths 100 cm, 10 cm, 4 cm and 0.3 cm. For a telescope with maxirnuffi
(A) +2 D, (¾) D
(B)
+(¾) D, 2 D
(C) 2 D, +(¾) D
(D) (¾) D, +2 D
291. Astigmatism for a human eye can be removed by using
(A) concave lens (C) cylindrical lens
(B) convex lens (D) prismatic lens
292. A hollow convex lens of glass behaves like a
(A) plane mirror (C) convex lens
(B) concave lens (D) glass plate
293. The far point of a myopic eye is 250 cm. The correcting lens should be a (A) diverging lens of focal length 250 cm. (B) converging lens of focal length 250 cm. (C) diverging lens of focal length.125 cm. (D) converging lens of focal length 125 cm. 294. A person cannot see clearly beyond 50 cm. The power Of the lens required to correct his vision is (A) 0.5 D (B) +0.5 D
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(C)
2 D
(D) +2 D
295. A ray travelling in negative xdirection is directed towards positive ydirection after being reflected from a surface at point P . The reflecting surface is represented by the equation x2 + y2 = a2 • Then coordinates of point
296. A person cannot see clearly objects at a distance less than 100 cm,. The power of the spectacles required to see clearly objects at 25 cm is (A) +1 D (B) +3 D
(C)
+4 D
(D) +2 D
297. An object is kept at a distance.of 16 cm from a thin lens and the image formed is real. If the object is kept at a
P are
distance of 6 cm from the same lens the hnage formed
is virtual. If the size of the images formed are equal, the focal length of the lens will be · (A) 8 cm (B) 5 cm (C) 11 cm (D) ../% cm 298. A person can see clearly objects lying between 25 cm and 2 m from his eye. His vision can be corrected by
(A) (a, 0)
=
(B)
(0.6a, O.Ba)
c;, Jz)
using spectacles of power
(A) +0.25 D (C) 0.25 D
(B)
+0.5D
(D)
0.5D
(C)
( 0.Ba, 0.6a)
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This section contains Multiple Correct Choice Type Questions. Each question has four choices (A), (B), (C) and (D), out of which ONE OR MORE is/ are.correct.
1.
The xy plane is the boundary between two transparent media. Medium 1 with z 2:: 0 has a refractive index ..fi. and medium 2 with z < 0 has a
·../3. A ray of light in medium 1 given A= 6../31 + s../3J10k is incident on the
refractive index by the vector
6, (C) =µ 61 4.
A ray of light from a denser medium strikes a rar~r medium at angle of incidence i . The reflected and the refracted rays make an angle of 90° with each other. The angles of reflection and refraction are r and r' respectively. The critical angle is (A) sin1 ( tanr) (B) sin1 ( tani) 1 (C) sin (tanr') (D) tan1 (sini)
5.
A single converging lens is used as a simple microscope. In the position of maximwn magnification.
plane of separation. The refracted ray makes angle r with +z axis and incident ray makes an angle i with z axis. Then, (A) i = 120° (B) i =60° (C) r=45° (D) r =135°
2.
A ray of light travels from a medium of refractive index µ to air. Its angle of incidence in the medium is i, measured from the normal to the boundary, and its angle of deviation is a. The curve that best represents the plot of deviation 6 (along yaxis) with angle of incidence i (along xaxis) is
Select the correct statement(s). (A) the object is placed at the focus of the lens. (B) the object is placed between the lens and its focus. (C) the image is formed at infinity. (D) the object and the image subtend the same angle at the eye. ·
il 6,
(A)
6.
o, 0'""':.._4_.,,__.. 0
'
'
,,
3.
(D)
'' 6 ~ il 1 I J!. 0
ilkut.
A light of wavelength 6000 A in air enters a medium of refractive index 1.5 . Inside the medium, its frequency is v and its wavelength is 'A. . (A) v=5x10 14 Hz (B) v=7.5x1014 Hz (C)
1. = 4000
A.
(D) ,_ = 9000
A.
il
il
(C)
(D)
0
In PROBLEM 2, (A) 9=sin'(¾)
2
i
'
il ,
.
0
'' '
''
0
7.
(A) may form a real image (B) must form a real image (C) may form a virtual image (D) may be a parallel beam
n
2
If a converging beam of light is incident on a concave mirror, the reflected light
i
8.
Two points P and Q lie on either side of an axis XY as shown. It is desired to produce an image of P at Q
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using a spherical mirror, with XY as the optic axis. The mirror must be p
•
(A)
b
C
(B)
b
C
(C)
(D)
a
ab C
X+•Y
14. A lens ofrocal length / is placed in between an object
• Q
and screen fixed at a distance D . The lens forms two real images of object on the screen for two of its different positions, a distance x apart. The two real images have magnifications m1 and m2 respectively
(A) converging (B) di verging (C) positioned to the left of P (D) positioned to the right of Q 9.
(m 1 >m,).
An object and a screen are fixed at a distance d apart. When a lens of focal length / is moved between the object and the screen, sharp images of the object are formed on the screen for two positions of the lens. The magnifications produced at these two positions are M 1 and M2 • (A)
d>2f
(B)
d>4f
(C)
MM =1
(D)
IM,1IM,1=7
1
2
d
10. Resolving power of an electron n,,.icroscope is R, and that of optical microscope is R0 .
(A) R, > R, (C) R, =R,
(B) (D)
R, < R0 Data Insufficient
(A)
X
!=m1 mi
(C)
D2x2
! =4D 
(B)
m1m2 =1
(D)
D?.4/
15. A planet is observed by , an astronomical refracting telescope having an objective of focal length 16 m and an eye piece of focal length 2 cm . (A) The distance between objective and eye piece is 16.02 m. (B) The angular magnification of the planet is 800. (C) The image of the planet is inverted. (D) The objective is larger than the eye piece.
16. A parallel beam of white light falls on a combination of
11. In PROBLEM 10, the correct argument for the correct selected option is that (A) electrons have greater wavelength than visible light. (B) electrons have lesser wavelength than visible light. (C) resolving power is inversely proportional to the wavelength of the wave used for detecting an object by the microscope. (D) resolving power is inversely proportional to the square of the wavelength of the wave used for detecting an object by the microscope. 12. The distance between two point objects P and Q is 32 cm . A convex lens of focal length 15 cm is placed between them so that the images of both the objects are formed at the same place. The distance of P from the lens could be (A) 20cm (B) 18 cm (C) 16 cm (D) 12cm
a concave and a convex lens, both of same material. Their focal lengths are 15 cm and 30 cm respectively for the mean wavelength in white light. On the other side of the lens system, one sees (A) a coloured pattern with violet at the outer edge. (B) a coloured pattern with red at the outer edge. (C) white light again. (D) that it is unable for the lens to converge the rays at a point. 17. Consider a ray of light going from A to B. Let the ray traverse, in going from A to B, distances S 11 s2 ,s 3 , ................... sm in media of indices n1 , n2 , n3 , ••...•..••.•. nm respectively.
(A) Total time offiight
1 t=Ln,s, m
C i=l
(B) Total time offiight t = _!
f s,
C i=l
(C) Optical path length is (O.P.L.) =
f n,s, i=l
13. The graph shows the variation of magnification m produced by a convex lens with the image distance v . The focal length of the lens is
B
f
m
(D) For inhomogeneous media the O.P.L. = n(s)ds A
and the ray travels along 'Stationary Pathways'.
L,.__
'+ V
_.L._ _ _ _
,.__ a,c
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(µ, =¾)
on the lens returns as a parallel beam from the arrangement. Select the correct statement(s).
of focal length 20 cm . For the final
(A) The beam diameters of the incident and reflected beains must be the· same.
image of object to be formed at infinity, which of the following is/ are correct? ' (A) A concave lens of focal length 60 cm is placed in
CBl a= 211,Hl,I
1
is/are (A) The mirror is concave
(B) The mirror can be convex or concave but it cannot
i
be plane (C) The object lies between pole and focus (D) The object lies·beyond focus
I
M,!
I= lo
(C)
V
' µ
= Vo µ
(B) (D)
M,
(A) The distances (in cm) of three nearest images mirror Mt are 5, 35 and 45 respectively. (BJ The distances (in cm) of three nearest images mirror M, are 5, 35 and 45 respectively. (C) The distances (in cm) of three nearest images mirror M 1 are 15, 25 and 55 respectively. (D)_ The distances (in cm) of three nearest images
21. A ray of light has speed v0 frequency lo and wavelength A.0 in vacuum. When this ray of lightenters in a medium of refractive index µ , corresponding values are v , I and " . Then (A)
.............. !······ 0 j
from from from from
mirror M 2 are 15, 25 and 55 respectively.
I= lo
22. For which of the pairs of u and I for curved mirror(s), the,image formed is smaller in size.
26.
In the case of hypermetropia (A) the image of a near object is formed behind' the retina.
(BJ the image of a distant object is formed in front of
(A) u=45cm, l=l0cm (B) u=10cm, l=20cm (C) u=60cm, l=30cm (D) u=20cm, l=:30cm
the retina. (C) a concave lens should be used for correction. (D) a convex lens should be used for correction. 27.
23. A di~erging lens of focal leng\h I, is placed in front of and coaxially with a concav~ mirror of focal length / 2 •
Their separation is d . A parallel beam of light incident
Which of the following produce a virtual image longer in size than the object?
(A) Concave lens (CJ Concave mirror
(BJ Convex lens (D) Convex mirror
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C
A concave mirror has focal length 15 cm. Where should
an object be placed in front of the mirror so that the image formed is three times the size of the object?
(A) 7.5 cm (C) 17.5 cm
(B) 10 cm (D) 20 cm
29. A concave mirror of focal length f forms an image 2 times the size of object. The object distance from the mirror is
(A)
j_
(B)
3/
(D)
4
(C)
2
30.
4/ 3
µ = 1.5
(A) appear to meet after extending the refracted rays backwards. (B) actually meet at some point. (C) meet (or appear to meet) at a distance of 60 cm from the spherical surface. (D) meet (or appear to meet) at a distance of 30 cm from the spherical surface.
j_
35. The focal length of a lens in air and refractive index are f and µ respectively. The focal length changes to / 1
2
A point object P moves towards a convex mirror with
when the lens is immersed i:r:t a liquid of refractive
a constant speed V, along its optic axis. The speed of
index
.the image
(A) is always less than V. may be less than, equal to or greater than V, depending on the pcisition of P . (C) increases as P comes closer to the mirror. (D) decrease as P comes closer to the mirror. · (B)
!:
and it becomes / 2 when the lens is immersed 2 ' in a liquid of refractive index 2µ . Then
w
h
2(µ1)
f
(B)
(D)
(C)
2{µ1)
h
f
µ1
!,=!
31. A bird flies down vertically towards a water surface. To a fish inside the water, vertically below the bird, the
36. Two
thin
lenses,
when
in
contact,
produce
a
combination of power +10 dioptre. When they are 0.25 m apart, the power is reduced to +6 dioptre. The respective powers of the lenses in dioptre, are (A) 1 and 9 (B) 2 and 8 (C) 4 and 6 (D) 5 each
bird will appear to (A) be closer than its actual distance. (B) be farther away than its actual distance. (C) move slower than its actual speed. (D) move faster than its actual speed. 32. There are three optical media 1, 2 and 3 with their refractive indices µ 1 > µ 2 > µ 3 • S~lect the correct statement(s) {A) When a ray of light travels from 3 to 1 no TIR will take place. (B) Critical angle between 1 and 2 is less than the critical angle between 1 and 3. (C) Critical angle between 1 and 2 is more than the critical angle between 1 and 3.
(D) Chances of TIR are more when ray of light travels from 1 to 3 as compare to the case when it travel from 1 to 2.
..Ji. . Select the correct alternative(s). (A) Minimum deviation from this prism can be 30° (B) Minimum deviation from this prism can be 45° (C) At angle of incidence 45°, deviation is minimum _(D) At angle of incidence 60°, deviation is minimum
37. A solid, transparent sphere has a small, opaque dot at its centre. When observed from outside, the apparent
position of the dot will be (A) independent of the refractive index of the sphere. (B) closer to the eye than its actual position. (C) farther away from the eye than its actual position. (D) the same as its actual position. 38. For a concave mirror (A) virtual image is always larger in size
(B) real image is always smaller in size (C) real image is always larger in size (D) real image may be smaller or larger in size
33. An equilateral prism has a refractive index
34. Parallel rays of light are falling on a convex spherical surface of radius of curvature R = 20 cm and refractive index µ = 1.5 as shown. After refraction from the spherical surface, the parallel rays
39. During refraction, ray of light passes undeviated, then (A) medium on both sides is same (B) angle of incidence is 90° (C) angle of incidence is 0° (D) medium on other side is rarer
40. A ray of light travelling in a transparent medium falls on a surface separating the medlUm from air at an angle of incidence 45° . The ray undergoes total internal reflection. If n is the refractive index of the medium
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Ray Optics with respect to air, select the possible value(s) of n from the following , · (A) 13 (B) 1.4 (C) 1.5 (D) 1.6 41. A convex lens made of glass ( µ,
=
¾) has focal length
f in air, The image of an object placed in front of it is real,
inverted
and
magnified.
Now
arrang~ment is immersed in water
the
(µ"' =¾)
whole without
changing the distance between object and lens, then (A) the,new focal length becomes 4/
(B) the new focal length becomes J_ .
4
(C) the new image formed will be virtual and magnified.
(D) the new image formed will be real and diminished. 42. . A thin, symmetric doubleconvex lens of power P is cut into ,three parts A , B and C as shown. The power of A
(D) If the_ entire arrangement is.immersed in water, the conditions will remain unaltered.
45. Check the wrong statement(s) (A) A concave mirror can give a virtual image. · (B) A concave mirror can give a diminished virtual image. (C) A convex mirror can give a real image. (D) A convex mirror can give a diminished virtual image. 46. ,When lights of different colours move through water, they must have different , (A) wavelengths (B) frequencies (C) velocities (D) amplitudes 47. A thin concavoconvex lens has two surfaces of radii of curvature R and 2R . The material of the lens has a refractive index µ . When kept in air, the focal length of the lens (A) will depend on the direction from which light is incident on it. (B) will be the same, irrespective of the direction from which light is incident on it 2 (C) will be equal to R .
µ1
(D) will be equal to B
(A) Ais·P
(C) B
lS 
_2
48. A convex; mirror is used to form .an image of a real object. The image , (A) always lies between the pole and the focus. (B) is diminished in size. (C) is erect. (D) is.real.
A is 2P
(D) B
. p
IS 
' 4
43. A watch glass having uniform thickness and having
average radius of curvature of its two surfaces much larger than its thickness is placed in the path of a beam of parallel light. The beam will (A) be completely unaffected. (B) converge slighUy. (C) · diverge slighUy. (D) converge or diverge slightly depending on' whether the beam ·is incident from the concave or the convex side.
44. A converging lens of focal length / 1 is placed in front of and coaxially with a convex mirror of focal length f, . Their separation is d . Aparallel beam of light incident. on · the lens returns as a parallel beam from the . arrangement. Select the correct'statement(s) .. (A) The beam diameters of tne incident and reflected beams must be the same.
(Bl
d =t, 211,I
(q
d :c1,
lt,1
µ1
C
(B)
. p
~.
·
.,.
49. Which of the following form(s) the virtual and erect image for all positions of object ? (A) concave _mirror (B) convex lens (C) convex mirror (D) concave lens 50. A ray of light is incident on a prism of refracting angle A . C is the critical angle for the material of ihe prism with respect to the surrounding material (say air/vacuum). . (A) An emergent ray will be there for all values of C . (B) An emergent ray will be there only for A < 2C . (C) Aray incident at an angle i can pass through the prism if sini> sin(AC) for C J.
4 ,
1 •
(A)
5..fi. i+ 5..fi_ I  ..fi. k
(B)
3/ +4]5k
(C)
5..fi.,  5..fi_ I+ ..fi. k
(D)
2i 3k
represent the focal length of paraxial and
marginal rays respectively, then correct relationship is
(D) 20,/6 units
The unit vector along refracted ray is
3 ,
19. If
(B) 90° (D) Cannot be determined
(A) 60° (C) 30°
3 ,
4 ,
1 •
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Ray Optics Comprehension 9 The lens governing the behaviour of the rays namely rectilinear propagation, laws of reflection and refraction can be summarised in one fundamental law known as Fermat's Principle. According to this principle a ray of light travels from one point to another $UCh that the time taken is at a stationary value (maximum or minimum). If c is the .velocity of light in a vacuum, the velocity in a medium of
refractive index n is
.£., n
hence time taken to travel a
Comprehension 10 Consider an equiconvex lens of radius R , made of a material of refractive index µ. Its focal length is / 1 when any one face is silvered. Now consider another planoconvex lens of radius R, made of same material having focal length, / 2 when no face is silvered, / 3 when plane face is silvered
and / 4 when curved surface is silvered. Based on above information, answer the following questions. 31.
distance I is nl . If the light passes through a number of C
media, the total time taken is ( ~)
L nl or ~ fndl
index varies continuously. Now,
Lnl
is the total optical
28. If refractive index of a slab varies as m = 1 + x 2 where x is measured from one end, then optical path length of a slab of thickness 1 m is
(B)
32.
(C) lm
(C) 33.
3
34.
2R (2µ1)
(D)
2R 2(2µ + 1)
equals R 2µ1 R (D) µ+1
(B)
R µ1
equals R (A) 2µ
(B)
2R µ
R 2(µ1)
(D) 2Rµ
f. equals · (A)
R µ
(B)
(C) 2Rµ
B
(B)
f,
(C)
 m
29. The optical path length followed by ray from point A to B, given that laws of reflection are obeyed as shown in figure is A
f,
(A) R
4 (D) None of these
3
R 2(2µ1)
(C)
path, so that Fermat's Principle states then the path of a ray
4
equals R (A) (2µ1)
if refractive
is such that the optical path in at a stationary value. This principle is obviously in agreement with the fact that the ray are straight lines in a homogeneous isotropic medium. It is found that it also agrees with the classical laws of reflection and refraction. Based on above information, answer the following questions.
(A)  m
f,
(D)
2R µ
R 2µ
Comprehension 11 SITUATIONI
(A) Maximum (C) Constant
Two identical pianoconvex lenses L1 and L2 having radii of curvature R = 20 cm and refractive indices µ, =1.4 and µ 2 =1.5 are placed as shown in the figure .
(B) Minimum (D) None of these
...... ...... ...... ...... ...... ...... ...... ......
30. The optical path length followed by ray from point A to B, given that laws of reflection are obeyed as shown in figure is A
SITUATIONII Now, the second pianoconvex lens is shifted vertically
p
(A) Maximum (C) Constant
+ 
B
(B) Minimum (D) None of these
downward by a small distance of 4.5 mm and the extended parts of L1 and L2 are blackened as shown in figure.
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~,,

~
+ + + +
Principal axis~ ,_ of lens Li
38. The nature of final image of the object when x = 2R is (A) Erect and.magnified (B) Inverted and magnified (q· Erect.and same size (D)· Inverted and same size
L,
Based on above information, answer the following questions.
39. Jt, is observed that for x = R , . a ray starting from 0 strikes the spherical surface .at grazing incidence. Th_e angle with the normal at which the ray emerges from
the plane S1.!I'face is
35. In SITUATIONI, the position of the image of the parallel beam· of light relative to the common principal axis is
100 cm 3
(A) 1~0 cm
(B)
(C)
200 (D)  c m 9
200 cm 3
36. In SITUATIONII, the new position of the image of the parallel beam is 2 0 ~ cm in front of the lens 2 mm below the (A)
(B)
Comprehension 13 The refractive indices of'the crown glass for blue and red lights are 1.51 and 1.49 respectively and those of the flint glass are 1.77 and ·1.73 respectively. An isosceles prism of angle 6~ is made of crown glass .. A beam of white light is incident at a small angle on this prism. The other flint glass isosceles prism is combined with the crOwn·glass prism such
that there is no deviation of the incident light. Based on the
principal axis of L, . l~O cm behind the lens 2 mm below the principal
40.
2 0 ~ qn
The refractive.index·of crown glass f9r yellow,colour is .
behind the lens 2.5 mm below the
2 0 ~ cm in front of the lens 2.5 mm below the
41. The refractive index of flint glass for yellow colour is (A) 1.70 (B) 1.72 (C) 1:73 (D) 1.75 42. The refracting angle of flint glase;prisni is (A) +2° (B) +4° . (C) 20 (D) 40
principal axis of L, . Comprehension 12 A small object O is placed in air at the principal axis at a distance x from the pole of the curved .surf~ce of a transparent hemisphere having refractive index 2 ,and radius R as shown. Based on above information, answer th~ following questions. .
n=2
0
43. The net dispersion produced by the combined system is (A) 0,02° (B) , 0.02° (C) +0.04° (D) 0.04° Comprehension 14 An equilateral prism AllC is placed in air with its base side BC lying horizontally along xaxis as shown in figure. A ray of light represented by equation ,J3z + x =10 is incident a:t a point P ·on the face AB of prism. Based on above information, answer the following questions. z • A
,_ X
+R+
(p, s) 7. A> (cj, r) 8. A> (p, q, s) 9. A> (p, s) 10. A> (p, q, r, s) B> (q) B> (r) B> (p, q) B> (p, q, r, s) B> (q) C ,· (p, q, s) C> (p, r, s) C> (r) C> (p, q, r, s) C> (p, q, r, s) r',D +(r). _______ D~(p, r) _________ ., D> (p, q, s~)_ _ _ _ _D_> (p, s~)_ _ _ _ _ _D_>~(p~,~q,~r~,s) 11. A>(P) 12. A>(p,s). 13. A>(S) 14c A>(S) 15. A>(q) ·] 1 ! B> (p) B> (p, q, r, s) B >.(p) B> (q) B> (r) ! C> (r, s) C> (p, q, r, s) C> (q) C> (p) C> (s) ~i'~D~>~(g,_p) D '>_(g,_s) D> (r) . .P >_(g)_ D ±(R).'
•' '"~IN'TEGER ANSW~R' TY~E,QUESTION$'· 0
. 24, 36 3. 3 4. 90 2. I 1. 30.' . . 8. 2 6. 7. 12 5. 36 9 11. 180 12. 15 10. 2 60 I9,  15.  5  16. 30 13. 5, 4 14. 8 19_._9_ _ _ _ _ _ _ _ _2_0._:..~? _______7__,· 18. 60 111. 30 · _____ , _____ _J
21.
! 25.
5
22.
100
10
26.
45
1.168
23.
12
24.
25

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1.
The angle between the incident ray and the reflected ray is 18020, so, we have
i+.,/3]
i.JJf
'
2
=>
:::::)
2
:'
X=4cm
So point of incidence of light from A should be at 4 cm from D on mirror.
'
''' '' e'e
3.
The image will be momentarily at rest when. the particle moves parallel to the mirror. Let at the time t the particle has a velocity v parallel to the mirror.
\180~28
'' cos(18O°20) =
( i+/31) 2
•
' ''. V
(i.13]) 2
1i+;13j11i;311 =>
(13) cos(20) =  4
1
=>
cos(20) = _ _! 2 1 cos(20) =
:::::) :::::)
20=60° 0=30°
=>
ucosa
vsin0=usinagt and
vcos0 =ucosa
:::::,.
V=
•.• (2)
COS0
From (1) and (2)
2
= usinagt ( ucosa)sin0 case . =>
2.
ucosa
... (1)
1
Drawing the ray diagram and using the Law of Reflection, we get
B
= ucosa(tanatanB)
g
4.
20cm
A .semi i
'
i=r :::::,. sini =sinr So we can say that ~DO and .6.OBC are similar
... (1)
From the figure, we observe 'that 30=180°
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Advanced JEE Physics =>
.." 3" ,,
0=60°
,, I
So, S, =180°2(30°)=120° (CCW) I
ands, =180°2(30°)=120° (CCW)
I
=>
S = 240° (CCW)
I
I
S = 180° + 8 = 240° (CCW) or 120° (CW)
I
\
1
\
I
I
I
\
2.' 1
I
I
\
"\
\\
,
0+0+0=180°
3 ,,
,•
\
I
'
\
\
,, J
I 1
\
'
•
5"
,,"1' ...... _..!_.. \.
I~ 6Q0 1\ \
/
\
I
\_.1'_...f ,
V
I',
I
I
\
\ \
\2' \ \ \
\
1) ,r•.,,::r,,c :, ,
r
I
\,: I \ J I r,5 1 , r3 1 I \. I' 5, / \ I' I 'f. ... _._.{ ). ... _. __ ( I 1
1
0=60°
\
\
\
1',, ',0.,
\J
~
\
'
Bi\
t
\
\
\,4" \
\';).
}..._
I
\
we observe that
I
\
~+7 ~~ I \ \ 1"
/___ Tr
Various angles made are as shown in figure. In triangle ABC,
=>
I
),,.    ... ../ ,, \ /'
I
Alternatively from the figure, we observe that
5.
2",f
I
I
So, total deviation 6 =61 + 62
\
I
,
,,
,,
1
4 ' .} _______ 4 ' ' \/ _______
1
So, the number of images formed by the combination is given by
N=61=5
Combination of Mirrors
6.
From the figure, we observe that . S, =180°2(50°)=100° (CCW)
Images Formed
AB & BC
1,2,3,4,5
AC & BC
1', 2', 3', 4', 5'
AB & AC
1", 2", 3", 4", 5"
These images along with the obJect must lie on a circle as shown in figure with an angular separation of
s, =180°2(20°)=140° (CW)
360° = 360° = 720 N 5
s, =180°2(10°)=160° (CW)
Similarly, the other two combination of mirrors also form 5 images each but we find from symmetry that 5 and 5' , 1 and 5" , 1 and 1" coincide. So the total number of images formed by three mirrors AB , BC and AC is
N'=(5)(3)3=12 9.
• M,
The ray diagram is shown in figure. We observe that
Hl=AB=d
So, total deviation S = 100°(CCW) + 140°(CW)+ 160°(CW) => S = 100° (CW) or 260° (CCW)
DS=CD=_cl_ 2
G 7.
The image is formed as far behind the mirror as the object is in front of it. Also in:iage formed by mirror 1 i.e., 11 , acts as object for mirror 2, so 1; is formed 50 cm behind the mirror 2 as
H
shown.
I',
I", 20
I, 60
10
10
I'
I,
0 30
30
20
'
I"
60
''
_________ .., __________ I
'
B
:E
'
F J
2 Taking all distances to be in cm and plotting them as shown (but not to scale), we get 011 = 20 cm 012 = 60 cm
01;
= BO cm
O1;'=100cm
=>
GH=2CD=2(%)=d
Similarly, IJ =d
01; = BO cm O1;=140cm
So, the respective distances are 20 cm , 60 cm , 80 cm , 100 cm and 140 cm 8.
Also, AH = 2AD
10.
=>
GJ=GH+HI+IJ=d+d+d=3d
(a)
For a one eyed man, the required size will be half the each dimension of the face i.e., 12 cm x 8 cm
(b)
For a two eyed man, the
Let us first consider the mirrors AB and BC , for which we have
360° = 6 60°
Smallest length of the mirror
=>
= Half the length of face
(Smallest l~ngth) = _!_ x 24 = 12 cm ' ofthe Mirror 2
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Ray Optics The smallest breadth of the mirror is calculated by using the fact that the rays from extreme part of face should reach one of the eyes after reflection from the mirror. The common overlapping portion is then the required breadth of the mirror. The ray
=>
13.
diagram is shown in figure.
v,=5(1+.JsH+s]
Let AB be the incident ray and angle of incidence at the mirror M1 be i, then
p  ..."""P' M ,...,,..........


:'
A
'' '' '
'' '''
'.,
I
 '.:JQ'
From figure, we get MM'= .!.pa __!_E,E 2 2 ' Smallest Breadth)= _!_( 16 _ 8) = 4 cm of the Mirror 2 So, the shortest size of mirror is 12 cm by 4 cm. 11.
=> =>
=>
the image of B2 in mirror ac and B4 is the image of 8 3 in point of intersection of ab with line AB 4
•
•
ac with 8 2
,
~
Point C is the
14.
E with 8 1 and F with 8 .
B4,,
:' ', ............ : .............
!
LDCB=202i LCDB=180°20 cr.=20
The angle between incident and emergent ray is 28 and it is independent of the angle of incidence i .
Let us now draw line
B 3 C from 8 3 and connect point D at which this line intersects
0
Using the Laws of Reflection, we get
Let us first find the image of point 8 in mirror bd (shown in figure). Let us then construct image B1 in mirror cd. Also, 8 3 is
mirror ab . Let us connect points A and 8 4
LCBO = go i LBCO = 180° 0 (goo il LBCO = go 0+i 0
(
Suppose that a plane mirror is kept horizontal as shown in figure. The reflected ray will make an angle of 30° with horizontal, or an angle of 60° with the vertical.
Incident ray
. . . . . . . . ... ..,.,,
I
I
83

,.,.,.
I
:
,,...
'
F
....... _,,"
_,,....
C
I

, __ 
·B ,.,.1 1
,,/'
1 l,'_,,                           
··
 
To make the reflected ray to go vertically upwards, the mirror is required to be rotated about O counterclockwise by 60°. To achieve this, therefore, the plane mirror is required to rotate about O by half the angle, i.e., by 30° , as shown in figure.
J 1
 ...
0
I
E',dl ..,
Reflected ray
~i/
. As b' lL"~~',C q_... I I
''' :
~ B2
It can be stated that the line ACDEFB is the sought path of the
Reflected ray
beam. Further, we observe that since, B3 CB4 is an isosceles triangle, CD is the reflection of beam AC . Similarly, we can show that DE is the reflection of CD and so on. This solution of the problem is not unique, as the beam will not necessarily always be sent initially to mirror ab .
Incident ray
12. Along xdirection i.e., perpendicular to the mirror, we have 30'
Relative Velocity of)= (Relative Velocity of) ( Image w.r.t. mirror Object w.r.t. mirror
=>
V1Vm=(vovm)
=>
v, (5cos30°)=(10cos60°(5cos30°))
=>
v, =5(1+./3) ms'
15.
In the direction parallel to the surface of mirror, i.e., along ydirection we have V1 =Vo
::::)
v,
Ray AB is incident on mirror OP at an angle e. The reflected ray BC is incident on second mirror OQ. Finally, the reflected ray CD is parallel to OP. Since CD and OP are parallel, and CO cuts them,
=> =>
LQCD = LCOP = 70° LDCN = gQ 0  LQCD = go 0 LNCB=LDCN=20°

70° = 20'
= 10sin(60°) = 5 ms1
So, velocity of the image
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16.
Reflection of a ray of light is just like an elastic collision of a ball with· a horizontal ground. The component of the incident ray
along the inside normal gets reversed while the component perpendicular to it remain!=i unchanged. So, the component of incident ray vector A= i +} k parallel to normal, i.e., i + J
A
gets reversed while perpendicular to it, i.e.,
k
remains
unchanged. So, the reflected ray is written as,
R=1]k A unit vector along the reflected ray will be,
>l.!::::3.:::==::::';;:== p B
Further, LOCB = 90°  LNCB = 90°  20° = 70° Now, in .6.COB , we ~ave LCBO = 180°(LCOB+LOCB) = 180'(70'+ 70°) = 40° => 0 = LNBC = 90'  LCBO = 90°  40'= 50°
• Fi
1]k
' =Fi =_Ja__,a_
=>
f= ~(i+j+i
.4=_=!Q_
10u u=12.5cm Please note that here, lul > !ti and we know that in case of a concave mirror, image is real when object lies beyond F . Case 2 (When image is virtual) : So, m=+4 1 Since m=1u
V
m=
=>
u
=> v=mu=(+4)x(2.5)=10 m Using the mirror formula, we get 1 1 1 +=10 2.5 f
f1 = 0.10.4 = 0.3 1 I= =_1_()_ m
=>
0.3 3 Since f is negative so, the mirror is concave. The radius of curvature of the mirror is given by
· ( 310)
=:>
3
According to the mirror formula, we have
5.
.!.+.!.=! V
U
f
M,
"'~"':~~4==::~~=='j,: 1::: ~r !oscm
optic;;;;~1;:,,~,\
~
!
M, i.
50 cm
Joi
4=_=!Q_
10u U=7.5cm
Please, note that here, lul => => =>
3=~ 12u 4=12u 12+U=4 u=8cm
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Here, m = 3
The following steps of construction for drawing the ray diagrams are used.
b)
3 = ~ 12u 4=12u
=>
=> c)
Here, m=
=> 6.
(i)
=>
From I or O drop a perpendicular on principal axis,
such that CJ= CD or OC =CD.
U:=c16 cm
1
(ii)
Draw a line joining D and O or D and I so that it. meets the principal axis at P . The point P will be the pole of the mirror as a ray reflected from the pole is always symmetrical about principal axis.
(iii)
From O draw a line parallel to principal axis towards the mirror so that it meets the mirror at M . Join M to I , so that it intersects the principal axis at F . F is the focus of the mirror as any ray parallel to principal axis after reflection from the mirror intersects the principal axis at the focus.
3 =>
36=12u
LI=4Bcm
Distance of image formed by the plane mirror is (ba) i.e., (b  5) cm and distance of object from mirror is (b+a) i.e., (b + 5) cm . Using mirror formula,
1
..!. + ..!. =! V
U
f
we get
9.
1
(b5)(b+S) = 20
Let the point A be at a distance x from the convex mirror as shown in mirror, then assuming the origin to be placed at the pole of convex mirror, we get
Solving this equation, we get
1
b=15cm
,r' :.:;,
The coincidence of the images can be established by observing the changes in the relative position of the images when the eye is moved away from the optical axis of the mirror.
When the images are at various distance from the eye the images will be displaced with respect to each other. When the images are at the same distance, they will coincide irrespective of the placement of the eye.
7.
(a)
[ :
,
'' '' ' 142R +1'
At any instant t, .we have
2
~ l :
t
:
I
,
• Incident Ray
Incident ., , Ray ,
A
'
"4X+: ,
2R~I
' .'
i+'1,_x_.......v,t,,t'
u = (2f + x) = (2f + f cos rot)
. hem1rror . formua, I +=,we 1 1 1 get Usmg,t V U f 1
For convex mirror, we have
1
1
2
v 2f+fcoscot ::::,
V=(2+coscot)t 1+coscot i.e., distance of image from mirror at time any instant t is
For concave mirror, we have 1 1
2 + cos oot)t ( 1+coscot Ball coincides with its image at centre of curvature, i.e., at
(b)
(2R x) ( 2R+ ~ )
2x+R Solving this equation, we get
X=O (c)
At t =
T
2
x=(1\F3)R and X=( v'3/ 1)R
, we have
Ignoring the negative value, as we have already used a negative sign with x, so the object should be placed at a distance
rot= re =>
x = fcos(x) = f
x = ( Js +
So, u = f i.e., ball is at focus. So, its image is formed at infinity, so m ), co
8.
(a)
Since the image is on the opposite .side of the principal axis, the mirror is concave. Because convex mirror·always forms a virtual and erect image. The ray diagrams for two different cases are shown in figure.
(b)
M,' A p,'
0
 ... 
0
2
10.
1
'M
A
:P
C'
D

,' I Case2
'' ' ''
1
1
1 1 f U
uf
==V
B
1
vu=1
'
' ''
1)R from the convex mirror.
Object is placed beyond C. Hence, the image will be real and it will lie between C and F. Further u , v and f all are negative, hence the mirror formula becomes
1
''t
Case 1
=
D, ,r
''
'' '
2 R
Uf
f V=f • 1u
B
Now since, uAs > uEo
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B' E' ...,....__ VMJ +I
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Ray Optics
/·: m=t) Therefore, shape of the image will be as shown in figure. Also note that v AB < uAB and .VEo < u~0 , 11.
lmABI < 1 and
=>
So, magnification, m
fmE 0 ] < 1
Length of Image Length of Object
_51_(21)
3'
Since the image is inverted, so the hlirror is concave.
=>
Now, u =30 cm
5f ~(21) 2 3 Case JI : When the other end lies beyond C
1
=>
v u f
=,
1
Since .:!.+.:!.=!,soweget V U f
1 1 1 +=v (i) f
Since image touches the rod, the rod ·must be placed with one
end at centre of curvature. However, two cases arise here. Case I : When the other end lies between C and F
=>
71
V=
4
For A, we have
U=(2t¾)=~ f=f
So, magnification, m = vA vc UA Uc
Since, .!.+.!=!,so we get 1
U
1
_
f =f
1
(15) + (30) f=10 cm
V
~2;__ _
U=(2t+¾)=~f
. 1 1· 1 Smee+==> =>
VA Ve
For A, we have
v=~=15 cm 2
1
m
UA  Uc
V
m=2=u
12.
51 2
V=
f
1
A'
F
p
=>
m
~(21) 4 _ 71 (21)
3
=4
3
+v (:') (0
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1.
· Magnification, m2 = ...:.'!...;,,. 4).55_ .. u .· J
d"=d= 100 =75 cm nm,attve 4/3
So, Liv;,, 0.55 cm
1 2.
The incident rays will pass undeviclted through_ the water surface and strike the mirror parallel to its principal axis. Therefore for the mirror, object is at oo . Its image A (in figure) will be formed
at focus which is ·20 cm from tl]e mirror. Now for·the interlace
m2 i::l1.1 m1
4•
between water and air, d =10 cm .
.
The refractive index of glass, V C µ·=~=..;_ vglass
:::) Air 10 crh
'' ' ''
...:~
R =40 cm d'
l
The colour remains yellow, as the colour depends on the frequency and not on the wavelength.
10 ·7 5 =(~:)=(4;3f . cm
l)sing equation, the total apparent shift is
AX =h,(1 ~,)+h,(1
Case I : When No Slab Is Inserted Accordingto mirror formula, we have
1 1 1 +=v u I
=>  Ax=2(1..:
~
6.
.. . V 1 Magm11cat1on, m1==,=0.5 u 2 Case II : When Slab Is Inserted
When we consider only two st8ps, then the ray of light starting from object O first gets refracted and then reflected. Distance of image 11 formed after refraction from the plane surface is given by
Again, applying the mirror formula, we get
1
v, (302)
=
1
10
O• X=nh+d=.±h+d
,1
Shitt=(1¾}=(1 5 )6=2 cm
~
4J3)+s(1 3J2)=1.5cm
10
V1 =15cm
1
µ:)
Thus, h=h,+h2 filC=2+31.5=3.5 cm
1 1 1 +=30
µ=v~T
A= i.., = 6000 =4000A µ 1.5
d
V1
c. A. 0
. S ince,
30cm
5.
3.
v=E.= 3x10a =2x10a nis1 µ 1~ .
Since "the frequency of light rerriains the same when it passes from one medium to another, so we have C=fAo and V=fA
water
4/3
v·
3
Therefore, distance of image 12 formed by plane mirror will be
.±h+d 3
v 2 =15.55cm
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Ray Optics 7
_
_ = sin(60°) · 18 s1nr => r =28.76"
=>
AB=(2)(1)(~)=
1
So, ·the distance between rays 1 and 2 is given by BE=ABsin(30°)= P
~
cm
N.
' I' ' ''' r' '
'
10.
Total deviation suffered by the ray is given by ()Total
=>
=Sp + C\:i
a=0r)+0r)
"
. 1r=2 Further, in ~OPQ , we have =>
0
Mirror
'
Since, MP= POtanr
... (1)
r+r+P=180°
=> MP= (6)tan(28.76°) => MP=3.3cm So, MN =2MP =6.6 cm 8.
cm
=>
r=90°.!l. 2
..• (2)
/
From Snell's Law, we have sin(45°) 3 sinr
i=
=>
sinr=¾sin(45°)
=>
r=32°
A
From equation (1),.we get
i=r+i=90°+(a;P)
.•. (3)
According to Snell's Law, we have sini µ=.smr.
=> F
=>
o+£>t
sin[90°+(";~)]
cos(P;")
sin( 90"%)
cos(%)
µ
cos(
p; a)= µcos~
Since, EF =ECtanr
=> EF=(3)tan32°=1.88 m Length of shadow at the bottom of the lake is £=DF=DE+EF=2.88 m
9.
· From Snell's Law, we have =:i
11.
. . r sm11= 2r
::::)
i1 =30°
sin(60°) smr
,l3 =·.
''
r=30°
E
r
l
60°160° 30° A• :r I
1
D
''' '' '' ' r'r C
Since; AB= 2(AD) =2(DCtanr)
''
' ''
2
I
1 cm
1
=2.
''
' ' \ ~,. ' ,
C Applying Snell's _Law at P , we get 3 sini1 2= sini2 => i2 =19.5° Now, Applying Sine Law (Lami's Theorem), on ACPR , we get 2r CR {·. LPCR = 60°} sin(180° 60° 19.5°) sin(19.5") => CR=0.7r
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Optics & Modern Physics
Advanced JEE Physics
12.
Let LAOM=0 Then by symmetry, LAIB= e
Since, sin i =~
=>
i = sin
1
V Eye
(~)
According to Snell's Law applied at A , we have
sini µ= sinr
1
__,.
B
. _,(sini) . '( L) r1 =sin =sm µ µR
Now from geometry of figure, we have
LA0!=2LIBA =>
LIBA = i:(90°  0) Lr= LIBA = 90' 9 2
Also Li=goo0+0=9O0+0 2 2
Deviation suffered by the ray is
S= ir =sin (ij)sin (µLR) 1
•.• (2)
Now, Snell's Law gives
1
1
This is also the angle r2 , so we have
. (90°+0)
sini
sin  2 
µ= sinr
s. , n  
. _,(L) . '( µRL) R ~sm
(90°0) 2
r2 =sin
0}' =µ'
cos+sine2 2 ( 0 . 0
From the knowledge of inverse trigonometry, we have
cossm
sin'(C)sin'(D) = sin,(cJ1D' 0J1C')
2
Ci']
Ct3 __!:__ Afµ2Fr µRff\2
r2 =sin,[!:_
... (1)
2
1+sin0
2
1sin0=µ '1
sin0=~ µ +1
Now, again applying Snell's Law at B , we get
sin0 smr2
µ=.
Since, sin
8 = sin1 (µsinr2 )
=>
a=~ 2a
1 d =2a(µ: ) µ +1
0=sin'(µLJ1 L' _!:_J1 L') R µ 2 R2 R R2 =>
13.
0 = sin
1
( ~2
2
2
2
~µ R L
 ~ 2
2
2
../R L
)
In the figure, let AB be the disc and O be the centre of the bowl.
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1.
(a)
Crltical angle between 2 and 3, is given by
1+R+1 l+X+I
sinC=~ 1.8 Now, applying Snell's Law, we get
1.6sinB = 1.8sinC = (1.8)(1..2) =1.3 1.8 13 8 =Sm . '(ffi ) .o::: 54 •34° (b)
2.
If 9 is decreased, the angle of incidence at the interface between 2 and· 3 gets decreased or i < C, so the light will refract into medium 3.
20cm
''
l
'''
µ=½
The path of ray is curved as shown in figure. As it travels successively into denser layers, it bends away from normal and TIR takes place at depth where angle of incidence approaches
So, the radius of shadow is R = ( 15 + B~O) cm
.::
~
R=
2
(b)
'' 6001'
,p
=>
max
=.!_(AB
J
a=. = 1sin(90°) 3 ~a!ax + 202
a
=>
16a!ax
Slab
=>
7a~~ = 9(20')
=>
a= = (
sin(i)
4.
(a)
= 9a!ax + 9 (202 )
~) ~ . Same is the case on left side of O .
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1".
The ray diagram for the situation discussed is shown in figure.
=>
A
2.
µ=~
r,+r2 =A Since i+e=A+c5 => 6=a+pA Further applying Snell's Law at incident surface and emergent surface, we get µ= sina. and sinr2 =! sinr1 sinf3 µ sin a= sinf3 sinr1 sinr2
(a)
From the figure, we observe that r1+r2 =r2 +r3 = 60° ~
r1 =ra
Applying Snell's Law at the faces AB and BC , we get sIn1 sinr1
sin(30°) sinr3
µ==
Since r1 = r3 ~
(b)
sina sin(Ar,)
sinp sinr2
sin(Ar,) sinr2
sina sinf3
sinAcosr.; sinr2
cosAsinr2 sinr2
. sina smAcotr2 =+cosA sinp
i=30°
Since. r1 + r2 =A= 60° ~
r1 =60°r2 :=60C
Further, µ
µ
sina· sinp
. sin~ +cotA sInf3smA
(·: r, = C} 5
sini sin(30°) sinr1 sin(60° C)
Sinceµ= !np =sinpcosecr2 smr2 µ =sin13:J1 + cot2 r~
0.5 · sin(60°)cosCcos(60°)sinC
. µ=sInp 1+
Since sin C = ! µ
cosc=J1
(
sma .  . +cotA s1nf3smA
)'
Since A=a+f3c5
1 µ'
1 µ(..J3)J1  2 _ _! =0.5 2 µ 2 ( ~)~µ'1 =1
. µ=smp 1+
3.
(
)'
sma . . (. ) +cot(a+p6) smpsm a+P,
µ=15
1 1=1 µ'
(sin45° = cos45°)
At minimum deviation, we,have
i = 58:8°
=> 6.
(a)
Applying Snell's law ai D , we·get
(1)sini =(¾)sin30° ..
..~ ~, i' =34.2~ => 6ro1a1=180°+2i'4r
=>
i =34:2°
A
OTotal =128.4°
Since, the condition for no emergence is A>2C 1
=>
A>2sin
(¾) 1
A> 2s1~1 ( ) 1.5
>83.62°
B~~~c ·
Therefor~, Ama:,,. ="83.62°, for escaping of the ray, thr~ugh the
5.
=>
2
. r. sini Since, v5  .. ( ) sin 22.5°
isini' = ¾sin(30°)
4.
cosC=~1 ;
r, =f2 =i=22,5°
Again applying Snell's ,Law for waterglass interface, we get
=>
=>
adjacent face.
Total deviation suffered bythe.ray·;s_ 6=60 +6E =260
T~e situation is shown in figure
=>
p
(b)
7.
6 = 2030°) = 8.4°
At near normal incidence, i 1:::: r1 = 0° Since r1 + r2 ~
=A
r2 =Cl
From Snell's Law apRlied at the face from where the refracted ray emerges, we get _ sine µ=.' s1nr2 ~
e = sin1 (µsina)
Now, deviation O= i+eA =sin1 (µsina)a
(a)
~
Since, e = 90° Also, r2 =C=sin
1
(¾)
8.
O=sin1 (µsina)a
{·: i = 0°}
For the ray to retrace its path, it must be incident normally to the face AC .:So, we have · "
r2 =Oo
Now, i=A=45° and r, =Ar2 =45°C • Applying Snell's Law at AB , we get
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Ray Optics A
sinA sinr1
=>
sinC
sinr1 = sinAsinC
... (3)
The ·ray does not ·emerge from the other face AC, when
r2 >C Since, r1 + r2
Since r1 +r2 =A
=> _r1 =A=30° From Snell's Law, we have µ= ~ini =../2 smr1 9.
From
11.
Ar, >C
=>
r1
sinr, < sin(AC)
=> => =>
sinAsinC < sinAcosCsinCcosA 1
1 _66 =
=>
e=63.1°
=:>
Sv=53.1°
For Red Light : According to Snell's Law, applied at the plane of incidence, we get
From Snell's Law applied at face AC , we get µ= ~ini =>/3
1.6 = sin(50°)
2
smr1 10.
sine sin(32.5°)
Since, 6v =Ci+e)A
{as e = O}
r1 =A=30°
=>
:::::)
Fron:i geometry, we observe that the angle of incidence at the face AB is A . Applying Snell's Law at face AB , we get
sinA µ=.smr1
f1
sinr1
=28.2°
=A
Since, r1 + r2
... (1)
r2 = Ar1 =31.8°
=:>
Applying Snell's Law at the plane of emergence, we get
A
1 _62 = :::::)
sine sin(31.8°)
8=58.6°
Since, SR
= (i + e)A =48.6°
So, angular dispersion is given by Sv 6R
B c __ _ _ _ __,C If C is the critical angle of th_e prism, then 1 µ=sine From (1) and (2), we get
12.
=4.5°
For minimum deviation, we have . (A+o.) S in
... (2)
µ=
2 sin(i)
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Optics & Modern Physics
:::::,
=> 1.3sin(45°)=sin(45°+B.) Solving, this we get 6m =22° For maximum deviation, we have the emergent ray to be grazing on the surface of emergence. So, e=90°, r2 =C and r, =90°r2 =90°C =>
137=~ · sin(45°) i = 75.6°
=> => 6red=i+eA=30.6° For violet ray, we have
6max =i+eA
We can find i by using µ
sini = .
s1nr1
. C 1 and s1r:i =
µ
1.42 = sin(75.6°)
. . . 8 ubstituting
s1nr1
the values, we get 6max ::= 550
13.
14.
Since, r, + r2
= 0°
r, =43°
=>
r2 =Ar1 =2°
f:
r1 +r2 =A}
1.4 2 = s!ne
smr2
17.
=A
=>
Again, applying Snell's Law at the emerging face, for violet rays, we get
Given that, i = 60° , A = 30° and B = 30° Since, 6=i+eA Substituting the values we get, e = 0° Now; e = 0°, means that the emergent ray is normal to the face through which it emerges.
Given that A= 30° and i = 0°, so r,
r, =A=45°
Now, according to Snell's Law, we have
=>
0=2.84°
=>
6v1o1ei=i+eA=33.4°
As the angles are small we can take,
sine;::: e
=> r2 =A=30° Further applying Snell's Law at the plane of emergence, we get
A
1 _5 =s!ne s1nr2 Substituting the values, we get e=49° =:>
15.
6=i+eA=19°
For no total internal reflection, when the ray leaves the prism, r2 =C But sinC =
1 .!µ = 1.6 
1 r, =C = sin•() = 38.7° 1.6
Further r1 + r2 =A= 45°
=>
r1 =45°38.7°=6.3°
Now, i = sin• (µsinr,) = sin' [1.6 x sin(6.3°)] = 10.1° 16.
C Applying Snell's Law for the two emerging rays at AC and AB , we get sin(A+1°) sin(4°) µ sinA sin(2A)
=>
A+ 1° 4° µ~~A=2A
=>
A=1° and µ=2
For reQ ray, we have e=0°=r2
Since, r, + r2
=A
= c:================================== 1.186
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1.
Let us see where do the parallel rays converge (or diverge) on the principal axis. Let us call it the focus and the corresponding
f • Using
length the focal length
J:.g_ = 11a V u
= µ2 
µ1
Now applying .&__,& = µ 2 µ, with the idea that 12 is formed
R
=>
appropriate values and signs, we get
4 3
1
3
4.
R
Applying, µ 2
=1.5 , we get
1
P = 2.5 dioptre= 2.5 D
1.5
v,
1
1.51 (2r) =  
i !E
.& = µ 2 R µ, , we get

V
U
1.6
11.6
PI  (3)
For first refraction at the unsilvered surface, we have
5cm
=>
PI =2.42 cm
=>
EI= (5 + 2.42) cm
=>
EI= 7.42 cm
(a)
Applying, µ 2
'•
&T
= ::S
0
V1 tco
i.e., rays become parallel to the principal axis. Hence the image formed by the curved mirror will lie at the focus of the mirror i.e., a distance
i
5.
=>
v,
(~r 1.5
µ,
= µ 2 R µ,
, we get
U
1.5
1
1.51
V
1
6
V=90 cm
So, the distance between object and its image is 80 cm
For second refraction at the unsilvered surface, we have
1

V
from pole of mirror. {from pole of the mirror}
=)
µ1
f=40cm=0.4m
1 1 P (in dioptre) =  (  ) =  f metre 0.4
:=:,.
1
(2Rµ+x +R)
x :::::,Q.75R
Hence,
2.
µ
(R+x)
Solving this equation for µ
Since, the rays are converging, its power should be positive.
=>
R
i_1
1~= +10 =:
u
V
at C, because light falls normally on the mirror.
with
(b)
11.5
Again applying µ 2
1.5
(r)

.& = µ 2 
=>
1
R
u
V
µ1
1.51
V 
at infinity.

V
AI ~ 2µR 1
12µ
1 µ 1µ BI, (AI, R) R Solving this equation, we get 2R(4µ1) BI, 3µ1 So, the distance between the final image and the object is d= 3R 2R(4µ1) (µ1)R 3µ1 (3µ1) Further,
+
5cm
*:t
5cm
8.
·p
We have to see the image of O from the other side
= c:::::==========::::;:====================== 1.188
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1.
When the images of both the sources are formed at the same point, then v will be same for both (in value). However for one case the image will be real and for the other case it will be virtual. So,
For 0 1 ,
·
1
1 1 we have+=V
X
1 1 1 For 0 2 , we have+=V 24X 9
... (2)
=
~
f=12 cm
Note that you would get the same answer by considering the othercase(m=3,for u=16cm)
... (1)_
9
1+3 24
=>
3.
The Lens Maker's Formula is given by
f=9cm
!=(µ1) f
Here, f =30 cm, R1 =10 cm, R2 =oo
o......______...oz
~= (µ1)
=>
30µ30=10
=>
µ~
30
~XS24xtt Adding equations (1) and (2), we get
x
24x
9
4.
= 6 cm
= µ1 10
4
3
1 1 2 +~SOiving we get x
(_!__.!) 10 ro
=>
o             < •
0,
(_!_~) RI R2
Applying Lens Maker's Formula,
!=(µ1)(_!_~)
Hence, the lens should be kept at a distance of 6 cm from
f
either of the object.
R1
R2
Here, µ=1.5, f=60 cm, R1 =+R, R2 =R
2.
Here the image formed can be virtual as well as real, so the
=>
6~=(1.51)(¾+¾)
~
R=60 cm
value of m should be +3 in one case (virtual image) and 3 in
the other (rea! image). Magnification of +3 can be obtained
Therefore,· the focal length of the spherical silvered surface, is given by
only when the object is placed within F {I.e., for smaller value of the object distance). The magnification of 3 is obtained when the object is kept between F and 2F (i.e., for greater value of object distance). So, m=+3,for u=Bcm Therefore, ·from the definition of magnification, we have
R 60 \,, ===+30 cm 2 2 (Positive, because it is a converging mirror)· The equivalent focal length of the lensmirror combination is then given by
V
m=
1 2 2 1 F=lt:"= 60  (30)
u
=> =>
+3=..!....
~ F=15cm The negative sign indicates that the combination behaves as a concave mirror.
8 V=3x(8)=24 cm
Using lens formula, we get
1 11
24
8
5.
According to the Lens Maker's Formula, we have
1 ( 1 ~ 1 ) ,where f=10cm =(µ1) f R1 R2
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Optics & Modem Plzysics
Advanced JEE Plzysics When placed in medium 1, then
f=(:,1)(~ ;J
..• (2)
=>
in the second case, let µ be the refractive index of the liquid,
When placed in medium 2; then
.!.t = (_!:_1)(_!__ _ _!__) µ R,, R
then
.•. (3)
2
2
From equations (1) and (2), we get 1= {µ1) 1.61 9
I
48
(:,1) G:~1) 10.
The system behaves like a mirror of focal length
given by, 1 _ 2(µ,/µ,)
FR,
1.61 102
(_!:_1)
(~1) 1.7
µ,
20
µ=1.37
From equations (1) and (3), we get 1= (µ1)
(5+ ~)
Solving, we get
=> \ =91=90 cm i.e., still it behaves as a converging lens
I
1 1 36 45=1 1=20 cm
.
2(µ,/µ,1)"
n,
R,
Substituting the values with appropriate signs, we
get
=> ~=10.2f=102cm i.e., it now behaves like a diverging lens. 6.
{·: R, >co)
Here R1 =+25 cm, R2 =25 cm, µ 1 =1 and µ 2
=>
=23
Image coincides with object, hence, u =v =x (say) + +ve
1 _ 2{3/2) 2(3/21) x  .  25 25 2 3 1 4 =+=x252525
=>
0
X=12.5cm
F =7.5 cm
So, the system behaves as a concave mirror· of focal length 7.5 cm. 11.
For a convex lens, the distance between an object and its real image is minimum when· u = 2~ and v = 2~ When concave tens is placed in contact, then we have
1 1
V
v' = 2\t
~2\
Given,\ =+5 cm and t =10 cm
Shilt of image Is
The combined focal length F is given by
11111 1 ++F I t 5 10 10 Since,
F=+10cm i.e., the combination behaves as a converging lens of focal
According to Lens Maker's Formula, we have
4\'
t
12.
_!__=(1.51)(_!__ __ 10 R, 10
=>
R1 =+10cm
.!. + 1
The path of the beam parallel to the optical axis of the
l_)
=>
Now, using,
R,
1
1
=> 9.
system and the image of object AB are· shown in figure. Image A'B' (direct and real) is obtainE!d to fu_ll scale with the object in any position. 2
2 (µ,/µ,)
2 (µ,/µ,  l)
vu R2 Substituting the values, we get
v 15 =
The following two cases are possible.
CASEI: The mirror is at a distance of d = f + R ,;, 2 m from the lens.
!f = (µ 11(_!__  _!__) R R 2
~
/J.V~
length 1o cm.
1
2 \t  2 \ = ~ t2\ t2\ » \ , so the shilt of image is
tJ.v=v'v=
=>
8.
fcomb
U
1+1=1_1_tl v'2\\~lt
Hence, the object should be placed at a distance 12.5 cm in front of the silvered lens.
7.
1
=
l+R+
l+1f.. .... ,, ______ ...... ., 
I
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Ray Optics CASEII:
f,=4a
The mirror is at a distance of d = f =R =1 m from the lens. The image of object A'B', also full scale, will be inverted
B
and virtual with the object in any position. A
Lens 1
Lens 2
1+ x,4a
1
V = V1 = ~ u x 4ax Similarly for Lens 2, we get
1 +    d   ..
m1 =
1 v2 1
For a convex lens, distance between an object and its real image is minimum when u =2~ . Hence,
V
2

v2 =x+12a
(1¾}=1
= i""=\
I
I
(b)
Using the lens formula,
~
(i.e., very
V
U
f
for the first lens
17.
_.!..._...!_=_.!...
15.
10
... (5)
v3 =(x+Sa)
::::,
m3 = ~
x+12a
...(6).
According to the lens formula, we have,
.!V _.!U = !f
Since u = mf , so we get
1
1
1
v+ mf = (fl
2(~:J. 2(~1J
R,
1
v2 4a = 4a
=>
v 2 =7.5 cm
Since, }
1 
same size but inverted.
For the second lens
=>
3
... (4)
the same place and m1.m 2m3 = 1 means final image is of the
v1 = 60 cm
30
x4a
Also, v3 = (x + Sa) means that the object and its image lie at
1 1 1 +=v, 30 20
V2
16 1a2' )
So, we observe from (2), (4) and (6) that m1m2 m3 =1
1 1 1 we get =, V
=>
m
1
If a concave lens of very large focal length
small power) is placed in contact with the convex lens, then its power and hence the focal length are almost unaltered. Therefore, there will be no shifting of the image.
14.
(
x+ a ,(16a2 ) x+12a x4a Similarly for lens 3, we get
=>
.1.=1.\_= 1\ µ
... (3)
16a2 =a 16a2
=>
I
... (2)
1 1 v1 4a a x4a 1
3\(1¾}=2\
µ
... (1)
4ax v1 = x4a =>
(a)
1
+=V1 X 4a
B
13.
1
Lens 3
_ __.,.__ 4a   . i
R,
=>
2µ _ 2(µ1)
V=·(m:Jf
So, lateral magnification is given by
r
V
1
ml =u= m+1
¾=( 4µr2)
Hence, the image will be (m + 1) times smaller than the object.
F=(4/2] i.e., the lens is equivalent to a concave mirror of focal length r
4µ2 16.
(a)
(b)
Let x be the distance between first lens and the object AB .
Applying the lens formula,
18.
Since a concave lens always forms an erect image whereas the given image I is on the other size·of the optic axis, so the lens is convex. Join O with I . Line 01 cuts the optic axis AB at pole (P) of the lens. The dotted line shows the position of lens. ·
.1.V  .!U = f1 thrice, we get for lens 1,
r:::::=================================1.191
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Optics & Modern Physics
Adva11ced JEE Physics
For first reflection, let us use the mirror formula, i.e.,
1 1 1 2 +==v u f R
I
:, +(µµ;1)=:
=>
2=(3µ1)
µR
V2
=>
v,=\=7cm
R
v: =(5~;1)
Similarly after nth reflections, we get
For the second lens, we have
1 1 1 v, (73)=6 4
µR
_!_;3µ1=2
Let parallel rays be incident on first'lens, then
V2
v1
For second reflection, similarly we get
From point O , draw a line parallel to AB which after refraction must pass through the focus F of the lens. 19.
=>
_!_ = [(2n + 1)µ1] v0
µR
Finally again u·sing the refraction formula,
6
~  ~1 = µ 2 ; µ 1
,
applied at the second curved surface, we get
y 2 =2.4 cm
_!__{(2n+1)µ1}= 1r, 20.
(a)
!f = (µ 1)(_!_  _!_) R R 1
R
V1
From Lens Maker's Formula, we get ~
R
R v,=~~
2(µn + µ1)
2
\ =µ,1= 1.71=7 =1.4 t µ,1 1.51 5 (b)
21.
In this liquid the first lens will be a diverging (as refractive index of liquid > 1.5) and the second a converging one (as refractive index of liquid < 1.7 ).
24.
1
\ 1
Since the incident beam is parallel, so we have
For the second lens; we have
1
1 5
So, the system behaves as a concave mirror of focal length
1 20
R() . The object will coincide with image when the object is
2 µ1
v2 =4cm For the third lens, we have :::::>
1 ;::;:>
V
3
1
2(µ1)
R \=2(µ1)
v, =\ =10 cm
V2
2µ
\=~R
placed at centre of curvature. So, we get R
1
x, =21\I=µ1
)00
In the second case,
... ,1)
2µ f1 =R
i.e., rays will become parallel to the optic axis. 22.
Since the focal length is equal to two times the radius of curvature, so f =2r
(~) =(µ, +a11J(±n
t=2µ
=>
x,= 2ltl=~ X
µ='x, x 2
2a. The rays will first get refracted, then n ·times reflected and finally again refracted. So, using
µ2 V

µ1
u
= µ2  µ1 R
for first
and R = ~
x, x 2
Now, according to Lens Maker's Formula, we have
! = (µ 1) (..!) = µ  1 = 2
refraction, we get .!:_ _ _!_=
v1 oo
... (2)
µ
Solving equations (1) and (2), we get
I= 32µo
23.
R
=>
f
µ1 R
~
R
R
{from equation (1)}
X1
f=X 1
(_.!:._)R µ1
v, =
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1.
[CJ
I,
4.
I,
f f(2/)
=>
m=
=>
m=3
1
[DJ Only one image will be fom,ed by this lens system, because the optic axis of both the parts coincide. Two images would have been formed if their optic axis would had been different.
i.
8 cm +1
5.
Distance of image from the plane surface is
4
X1 ==2.5cm 1.6,
µ {.. d =d"ci""} •
Maximum students are gripped by the misconception that when object lies at focus then image is formed at infinity. This is true but only for a Concave Mirror (Convex Lens) and is absolutely wrong when applied to Convex Mirror (Concave Lens) How To Proceed Then? If object placed at focus then just check out the MIRROR. If CONCAVE then image is formed at_ infinity and if Convex then
app
For the curved surface, we have 1.6 1 . 11.6 +=4 X2 8
=>
[BJ MISCONCEPTION
x2 :::::3 cm
1 1 1
The minus sign means the image is on the side where the object lies. So,
apply+=. V U f
1,1, =(82.53) cm=2.5 cm
~
=
~
v=10cm
V
2.
[CJ Area of object = 9 cm 2 Also, we know that
6.
=>
A
t= A,
[
Shift t.x=2520=(1t}
=>
1 5=(11.5
x9
)1
=>
=>
L
[CJ
10 ]' 25(10)
=rnJ
+20
Image will be formed at infinity, when the object is placed at focus of the lens i.e., at 20 cm from the lens. So, we have
. = mar = A,v'(f)' AIM rea agm.1. 1cat1on =2 = A0 u fu =>
20
W
~
The similar thing is extended and applied here too. Here the answer fabricated by the MISCONCEPTION is 1 (but we must know this is the answer only for a Concave Mirror (or Convex Lens). For Convex Mirror we have
f m=fu
[CJ Critical angle between glass and liquid interface is sinC = ~ = Zµ
3/2
3
Angle of incidence at face AC is 60° For TIR to take place, we have
i>C
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Advanced JEE Physics => => => 11.
2 sin(60') > ;
=>
µ
13.
m=~=3 u
=>
Since object always lies on negative side. So, u=20 cm
'
3 I => f(20) => 31+60=1 => f=30cm Negative sign indicates the mirror is concave. An Advice
J+30 cm30 cm~
I= 50 4 1=12.5 cm => Since R=2f => R=25 cm
=>
9.
Also Jul+lvl=D lul+3lul=B0 lul=20
~
0
1 +    50
cm+1
I would always advice you to write _lul+lvl =D wherever you are given the distance between object and image as no error will creep in these because the MOD signs prevent the errors.
[DJ Concave lens forms the virtual image of a real object. So, we
15.
have V 1 NOW,I"f m=u=4·
. 10 x=cm
i'=(2'+0)
3
U=cm
and
10 v= cm
143x+x
3
Substituting in
Since, 6to1a1 = 180°
0
40 3
=>
=>
0+1so 2i' =180'
=>
6=2i'
=>
2'=2(2+0)
4x
Here, negative sign implies that i gets decreased or i' =0 . i.e., light should fall normally on mirror.
U
16.
9
10. · [CJ · Jm.,,J=2Jm..,,,,J
[BJ The minimum length of the mirror required for the purpose ishalf the height of ttie person.
I= 40 f=4.4 cm
0
0= 2°
! =_!__.!., we get
f V 1 3 3 =+f 10 40
=>
[BJ Since, o=(µ1)A=(1.51)(4)=2'
=> 1=6=2° Let the mirror be rotated by an angle 0, then
u=4x,then v=x then 3x=10cm
=>
[AJ Since image formed is erect, hence it must be virtual. So,
..!.+.:!.=! V U f 1 1 1 +=10 50 f 51 f 50
[BJ Focal length of mirror is independent of the refractive index of medium in which it is placed.
U=50cm, V=+10cm
Since
[BJ 1 m=+2 1 20 =>  =  2 20u => 40 =20u => U=20 cm
B
=>
I 2 I 1(20) 1(15) 1+20=2f+30 l=10cm
17.
[BJ A divergent beam appears to converge behind the mirror thus giving a virtual image. So, a convergent beam will give a real
image.
~L Convergent Beam
 ~
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Ray Optics 19.
[CJ
10 _ 900 0 _ 180 cm 55 11 Negative sign with focal length implies that the mirror is a concave mirror.
.111 t Al ppymg vu=1,wege 111
=>
b+a=1
=>
25.
... (1)
f=~ a+b
Further in right triangle ACB , we have 2
AC + BC
2
=AB
S" 1 mce,
1 1 v+u=t
2
=>
(a'+c')+(b'+c')=(a+b)'
=>
a2 +b 2 +2c2 =a2 +b2 +2ab
=> =>
ab=C2 Substituting this in equation (1 ), we get :::>
c'
21.
1 10 20
1 20
V=Cffi
3
[A] Covering the lower half will just make the image less bright (not blurred) as less number of rays will be reflected as compared to the previous case.
27.
[BJ
[CJ If the mirror approaches the object or the object approaches the stationary mirror with speed v then image approaches object with speed 2v.
1 V
26.
f=a+b
20.
[BJ f=+20 cm u=10 cm
[CJ
I mrea1=n=fu
22.
20"..• ···
=>
nf+nu=f
=>
U=(n:1}
··20° 0
:
'b
g:ui
40"
[DJ The incident and the second reflected ray make the same angle 8 with vertical. Hence, they are parallel for any value of 8 .
28.
40"
[CJ
A'
23.
[D] 1
n=
B
I fu
fu=nf => u=(1n)f (n1)f According to the sign convention used u must always be negative.
height of wall.
=>
0
24.
29.
[DJ
0
=4.5
30.
f m=fu 4.5 =
=> =>
i=!:!3
360 n=>oo
[DJ mreal
B'
From symmetry we observe that length of mirror is one third of
[A] 3
:o =
f f(20)
n= 360 1=5 0
4.5190 =f 90 =5.51
!=~ 5.5
6 which is Even
31.
[BJ Ray's after reflections from two perpendicular mirrors are always parallel to incident ray irrespective of angle of incidence.
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Optics & Modern Physics
Advanced JEE Physics 32.
[CJ Power of concave lens must be Jess than that of the convex lens to form.a real image. So, net power will decrease or fqcal length will increase. Far real image v is positive, u is negative and f 1 1 1 is positive. Applying lens formula, = (substituting all V
U
38. [BJ Both concave and convex mirror give virtual image but a concave mirror gives a magnified virtual image (when object placed between F and P). Since the boy sees his image of diminished size, so the mirror must be convex.
39.
values withsign), we get 1 , ,
Let object be placed at a distance x from mirror.
u=X, v=(x+10), f=12cm
+=v u I
Since,
u is constant, f is iilcreasing. So v will also increase.
34.
[A]
1
1
v u I 1
[D] G C
==>
'D H ···j·······
t
:s i
d
!
B
So,
1
+=
1
1
+=
(x+10) x x =20 cm U=20cm, V=30 cm
12.
m=~:!.= (ao) =1.5 u 20
Negative sign with magnifi_catiori indicates image is real.
:E F
40.
,._ L.., J 142L+1
[BJ At face AB , the ray of light suffers no devicltio11, so applying
Snell's Law at face AC, we get 35.
[BJ
1
Since an elongated image is formed and it touches one end of the rod, so the rod must lie with one end at 2F and other end
µ = sinC = sin(45°)
=>
between 2F and F (shown in figure).
Forend A, U=
1
3
A
3
5
.=5151
1
µ,,,. = ,J2
51
. , , 1 S mce +=v u I 1 3 1 v51= I
2
.=51 51 V=2
·
1

/
f/3
A'
0
A
1+4~~~',~/
,
ROD
F
211+f~ fil +I 3
•
PA'= 51 2
B
41.
[CJ 1 1
C
1
+=v u I ==> v1 + u1 =r1 = constant
Take derivative w.r.t. time on both sides
0/>/=~21 2
!/v·')+!/u·')=O dv ( ) _, du (  1) V_, +  1 U = 0 dt dt
36.
:~ =
= 360 _ 1
when object moves towards mirror u decreases with passage of
90
time and hence du
n1 =3
dt
Now, when we consider the mirror on the ceiling then it will make a total of 4 images (one of the original object and th_ree images of the previous arrangement) so, total images formed equals 7. However, if the object were the observer himself then total number of images is 7 1 = 6
37.
=9 cms1
dv =(1)' du dt fu dt dv ( 24 )' du di= 24+60 di
[CJ Velocity of light is always normal to the wavefront.
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Ray Optics
dv =i(9) di 9 dv , =+4 ems
{·: 2sin(½)cos(½) =sini} =>
di
Positive value of ~: indicates that v increases with the passage
=>
of time i.e. image must be going away from mirror.
=> 42.
43.
[DJ A_:::: 60° for equilateral prism.
47.
[Bl The source cannot be seen if angle of incidence in the denser medium is greater than the critical angle.
Since, i+e=A+D => 2i=A+D
Since,
=>
2(¾A )=A+D
=>
D=~=30° 2
d h r==~µ'1 ~µ'1
h=(1)J(%)'1 4
h= cm 3
[BJ Since,
uab
> U00
48.
[A] sinC=..!. µ
Vab C i.e. 45° > C for Red Colour
Since, m="!... u
[m,,[
µ,
1.39 1.41
µ=V =>
sin CR > sin 45 Z= 3x10
=>
8
V
Green Colour
=>
V=1.5x108 ms1
=>
v=1.5x10 10 cms1
µG=1.44 1
=>
v,
45° 0
X

VlO
,
, ,,
, ,,
v,
. C ==Sin 1 1 . 45 Sin
[D] C
45.
1
i=e=~A 4
=>
44.
½) cos(½)=% i=2cos (%)
n =2cos(
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Advanced JEE Physics 51.
59.
[AJ
sine=
= vdllnsar
µrarer
µdenser
vrare,
=>
sin8 = vder,,;er
=>
sine=.!.
asa.
= vA
Vrarer
60.
Ve
53.
54.
[DJ
Applying Snell's Law µ = s!ni , we get s1nr
v,
=>
=> 52.
[BJ Since rays after passing through the glass slab just suffer lateral displacement hence we have angle between the emergent rays
sinr = sin I µ
[CJ Since, µv = constant
=>
µ 9V9 =~Vw
=>
~(2x10')=iv 2· 3 w
:::::>
vw =2.25x106 ms
s R 1
=>
[DJ After two reflections from two mirrors placed at right angles, the emergent ray will always be parallel to the incident ray for any value of i.
.
sInr=
sin60°
/3
=
1
2
=> r=30° Since, PC= QC => LCPQ = LPQC =Lr= 30°
[CJ
. Angle between reflected ray QR and refracted ray
Since light has to travel from denser to raser medium so, it must
other face Is 180°  r 60' = 90°
be made incident in the dens8r medium at an angle less than
as
at the
{·.· r = 30')
critical angle. 61. 55.
Since
==> 57.
[AJ According to Snell's Law
[DJ C oc 1. A.Red
> AVlolat
CRed
> CVlolet
sini sinr
[BJ According to the problem, we have, u = (11)
sinr=....!_sini
=>
sinr = ) sin60 43
I I
''
µ.
r = sin{
....: l
µ8
=>
V=O+1) f =+f
: : :
=µw
62.
............... ..
"'" ....
' , .._ ._
,.._
.......... ; .........
,.,_,_,___
;3)
3
[BJ LDBM=45' => DM = htan45° = h = 32 cm
A
sini smr
µ=.F
_,_
I+
.
smr =
=>
3 sin45° µsini = 4i'J = ,/2 4
tanr = sinr _ sinr cosr .J1sin2 r 3
f + 1+t
=>
. 1 1 1 =, we get Appymg I V u I 1 1 1 +=(1+1) 01) f
=>
=>
12 211=0
=>
=>
1=("2+1) cm
63.
4.J2 tanr= J1 9 = 32
3
m
CM=htanr=32x
_______ !!  
':i; ~~,_ili ~; Stone
= 3
=20 cm v23 CD =DMCM = 3220 =12 cm
(CJ sinC=_!_=~ µ. 4
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Ray Optics =>
µA =constant
C=49°
Apex angle of cone is 2C
= 98°
=>
.
''
A
: .;:ec,_c  •:•c'~ :'·~~: ____________'{9:  49;:~~==== ____________ =====~== __ ::::::, ::::::      F:        
75.
In 6OAN'
ON cosr=OA 1 OA=
••..•(ir)
••·••••.••.. N'
67.
second image is formed due to reflection at X2 Y2
i l ··
N
ti ;{::]''
...
cosr
[CJ For grazing incidence and emergence, we have
76.
[CJ In both A and B, the refracted ray is parallel to the base of prism.
79.
[AJ . (A+D.) sm 
r, =r2 =~=30°
s1nr, 68.
sin(~)
=2
4 sine=.!\:.= / 3 µ, 3/2
_
C
.
=Sin
1(8) Q
and light must go from denser to rarer medium.
69.
[BJ
71.
A.Red
1 (A+Dm) v1z=sIn  2 60+D. = 45 2 D. =15 2 Dm = 30°
=>
> A.Vlolel
80.
[CJ Critical angle C =sin
=>
2
Further we know that at minimum deviation i = A ~Dm
According to Snell's Law µA = constant Since
.J2sin30 = sin( A+ Dm)
"' "' "' "' "'
[AJ

2
µ
2 According to Snell's Law, we get µ ·= ~inl
and is
brightest all other images formed further are faint.
t
tsin(ir)
i=e=90°,
[BJ
''"!_.._.·~'.....··''
cosr
6x
µ,
This happens due to multiple refractions and reflections. The first image is formed due to reflection at X 1Y1 and is fainter. The
[DJ sin(ir)="" OA => 6x=OAsin(ir) In 6OAN
= µaA.a
•

64.
µaAa =·µgA.g
''
C = sin1
1
(¾)
i=45°
[CJ Normal Emergence implies e=O Since i+e=A+D For prism with small A,
D=(µ1)A
(i) =30°
=>
i=A+(µ1)A
"'
i~µA
When A > 2C , the ray does not emerge from the prism. So, maximum refracting angle can be 60° .
81.
[CJ Since, 1 µ 2 x 2 µ 3 x 3µ 4 x 4 µ 1 =1
73.
[AJ According to Snell's Law µv = constant
74.
4µ3 x3µ2 x2µ1 x1µ,1 =1 4
3
2
1 µ,
µ3x µ2x µ1=,
[AJ According to Snell's Law
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Advanced JEE Physics 82.
Optics & Modern Physics
[CJ Let thickness of slab be t and real depth from first side be x. Then µ
=
i
88.
Apparent Depth = !!. n
(when viewed from first side)
90.
µ = t~x (when viewed from second side)
(·.µ=¾) => => 83.
4
••. (1)
3X
From mirror formula, we get 1 1 1 +=2(6+x) 6+x f
[DJ
=>
sin(~)
Since, µ
[CJ From mirror formula, we get 1 1 1 +=3x X f
=>
X=9cm 3 t9 2 4 t=15 cm
=:>
[AJ
3 1 2(6+x) =1
... (2)
sin(~) cot(~) sin(~) 2 sin(~)
=>
6cm ,....__ X __..,.__ _ 3 X
•
142(6 +X)+i
=>
COS
Equations (1) and (2) give 4 3 3x = 2(6+x)
(2A) . (A+D ) =Stn
~
=>
=> =>
=>
Shift of screen = 3x  2 (6 + x)
9X =48+8X
x=48cm
Shift of screen = 3x482(6+ 48) = 36 cm
84.
85.
[AJ i+e=A+D => 60+e=30+30 =:> e=0° [BJ µ
µ=12
=>
[DJ
Apparent Depth =
92.
sin(45)=.!
86.
91.
..i.. + ..i.. 2µ, 2µ2
[A] In this case, one of the image will be real and the other will be virtual. Let us assume that image of S1 is real and that of S2 is
virtual. Then, applyin~ .!+.! = ! V U f
[BJ
=tr f =9cm
sinC1 =E:r_=.&_=..!._ µd µg µg sinC2
= µw µ,
S, I
> sinC1
....,__ X
I
I
S, I
y +I 24x~
87.
for 8 1 , we get
[BJ
. S mce r= =>
4 r~
r=4x3=3m 4 Diameter = 2r = 6 m
=>
1 1 1 +=y X 9
d
~
yµ2 1
.•• (1)
for S2 , we get 1 1 1 +=y 24x 9 Solving equations (1) and (2), we get x=6cm
... (2)
= ================================== 1.200
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a 4 =
CONCEPTUAL NOTE(S)
c 3
This questi~m may· have following ahsWer· 6 cm from S1 a~
18cm fr9m .52 and 18cm from.S; or6cm.fram S2 • 93.
98.
sin C
[CJ Optical path length = nt So time taken
µdenser
= nt
~
To a fish. the outer world is seen in a circle of radius

"µ21
=> ,d
99.
is the depth at which the fish swims.
:::::)
=> 95.
12 r=
[CJ
~~1
=>
µgVg =µtVt
36 r = ..ff cm
=>
( 1.5 )x ( 2 x10') = µ, (2.5x1 o')
=>
3 30 µ,=~==1.2 2.5 . 25
[AJ
1 sinC=_!_=·
(Rememberthatµwatar
~
(4/3)
µ.
=±3.)
100. [DJ
. C
sm
1 = F2
:::::
C = 49° with vertical
=>
C=45°
=>
0=9049°=41° with the horizon.
For i
=C
[CJ
= µw
·.
101. [CJ
µg
µdenser
=>
in denser medium angle of refraction {n rarer fJ.1edium
is 90°.
sinC = µmm,
97.
. C =5 sin 6
According to Snell's Law µv =con_stant
For the diver
96.
4/3 5/3
Real Depth µ
Apparent Depth
4 5
sinC==
=>
[BJ Applying Snell's Law at the interface sep8rating two media, we
get
=>
24 413 Apparent Depth = 18 cm Apparent Depth =
102. [CJ
Given that A
y
__________ J I 1
_i;.
I
I I
I).
8t+bj
' :
i=(A~~m )=60°
µ,=1.5=§. 2
b
' i
xz plane (i)'
103. [BJ
"' " j' r~·", : I
c
,, ••4L"
,I
I
r+i=90 i=90r For ray not to emerge frOm curved
•
surface
'f I
=·~
~      y ( o r j) From the figure, we get
(¾)(~ )= (/c'c+~') 2
a_7 + b}
~nd
ci + d}
= Om = 60°
At minimum deviation, we have
dl
Since
µA
=> · sinC= VA=_!_ V8 2.4
[CJ
.
= µrarer =,&_
Since, µ,._v,._ = µ 8 V8
C
94.
[BJ
i>C sini>sinC
=> => =>
sin(90r)>sinC cosr > sinC
=>
~ >1
=>
1 _ sin i >_!_
=>
1 >\( 1 +sin' i) n
=>
n2 > 1+sin2 i
n
... (1)
2
~
are unit vectors, so we get·
/a' +b' =ic'+d' =1 Substituting in equation (1 ), ~e get
,_
{·: sinC=~}
n2
n2
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n>./2 least value = ./2
104. [BJ
I
CONCEPTUAL NOTE(S) Critical angle incre~ses as the relative refractive index • is decreased.
110. [BJ
Dispersive Power
dµ dl.
= :~
2B
= )!
So, as B increases, dispersive power i~creas~s. 105. [CJ
For the near end of the rod applying
!f = ..:!.U + ..!.V
Since, here u and f are negative, so
lvl=~ uf The farther end of the rod is at infinity, so its image will be formed at focus. Hence, Length of the image is £ = lvl f
111. [BJ u=20cm, f=+20cm, V=?
Since.!+.!=! V u I 1 1 1 =>  + ~ = v 20 20 => V=10 cm Students generally give (A) as the answer to this problem because they have in their mind that if .an object is pl8.ced at centre of curvature C, then image is also formed at C, but this is true only for the case of a convex lens or a concave mirror. So correct answer is (B).
£=~f=__f_ uf uf · 106. [CJ For A
(1.5)t Total number of wa:ves = 1.(
... (1)
Total number)= ( Optical path length) of waves wavelength
113. [DJ ·
For Band C Total number of waves n,(½) (1.s>(¥) =+~~ l. 1. Equating (1) and (2) :::::) n8 =1.3
... (2)
Since no parallax exists between the images formed by two mirrors (convex and plane) hence the images for both coincide. But for a plane mirror an image is as far behind the mirror as the object is in front of it. Hence the image for plane mirror should be 30 cm behind it or 10 cm behind the convex mirror. So for convex mirror. U=50cm, v=+10cm Since.!+..!=! V U f
107. [CJ Focal length of curved mirrors is independent of the refractive index of the medium in which the mirror is.placed.
=> =>
108. [BJ C
n=v
1 1 1 +=10 50 I 1 51 i=so
=>
f = 50 4
=>
f=12.5cm
0
·
i+30 ,m_..:__30 cm~ 1450cm+1
Since R=2f => R=25 cm 114. [CJ
Since, 109. [DJ
6,~. =(µ1)A=(1.51)4°=2°
1 ·01 =1 an d sin ·02 = sin µg µ..,. Since, µg > µ .... , 01 < 02 The critical angle 8 between glass·water interface is given by
2'
sine=& µ,
2'
=1.202
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6,,.,, =(µ1)A+(1802i)
=>
6,,.,, =(1.51)4°+(1802x2°)
=>
()total
=2°+176°=178°
121. [CJ
sinC = 4000 6000
115. [CJ
==> 0 (I)
µ
(~)
123. [CJ
Ax=3[1(i)f
C
f).alr
3
vmad
fA.me.d
2
µ===
AX=1Cm
1 V Now, v=+Sm, m==3 u => U=24 m
Object distance = 1O+ 1 = 11 cm
116. [CJ
Since,
Apparent depth =
1
122. [BJ A virtual, erect image is obtained by using the mirror. The mirror can be both concave and convex. But a virtual image obtained by a concave mirror is always magnified and hence the mirror must be convex as we are getting diminished image.
i.3 cm+1 / >
sin i =nf sin r1
AIC
:::::, R=25cm According to Lens Maker's Formula
l',
r/ .·
s o.=90r,
~+ "2 = n2n1 u
n,
R
Consider origin to be at the pole P.
E _.0,\ .. 9 F
o· {:r
"·.·
n,
. 0
p"·. •.
2 => cos 2 i = n~ cos r3 Adding (1 ), (2), (3) & (4) => 1+n!=nf+n~
... (4)
156. [AJ
P=(µ1)(..!....!..J R, R 2
12 is the direct image of O by mirror M2
= c::::=================================== 1.206
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3 5 v"= 5151
r+C=45° r=45C
1 2 v"=51
51
Since sini = n sinr sini = n sinr
V=
2
PA'= 51
sini=n sin(45C) sini
2
=n(sin45cosC cos45sinC)
OA'=~21 2
sini= h"(cosCsinC)
DA'=!_ 2
f
DA' 2 3 m===OA .!. 2 3
162. [DJ For a plane refracting surface, the lateral magnification is 1. So, the image of the coin will be of the same size as the coin itself.
Further, sinC = ~
163. [DJ
n
:::::,.
sini=
~[~1~!~
6=~=
60
2
1 ]
:::::,.
. . = ../2 1 [ vn ~ srn1 n 1 n 1]
2
=30
The ray passes symmetrically through the prism parallel to its base.
Hence 0 = 0°
. .,[.fnCnfn,J .J2
164. [CJ
l=Sln
RAY2
159. [DJ
f, +f, =36 cm
... (1)
.l,_=5
... (2)
f, :::::,.
PA2 E
0:
PA1 'A ~ f2 ..l
RAY1
fo=30cm and fe=6cm.
~ d +! x+
f'=~(175) . 2 25 f' = 3.5R
=:172. [OJ
For refraction at plane Surface. 1 1.5 1.51 +=(mR) V oo =:v=1.5 mR For refraction at the curved surface u'=(1.5mR+R), v'>oo
Similarly we can calculate value of y . OBJECTIVE TRICK If t:. = 0 , y must be zero (satisfied by A, 8, C) · If d> \+~, then x>"' (satisfied by B, C, D)
=:,
If d> \, then x> \ (satisfied by C, D) So, we observe that OPTION (C) is .the only option satisfying both the ,x and y in special situations mentioned above.
Let d' be the diameter of refracted beam. Then d = PQcos(60°) and d' = PQcosr d' cosr :::::,. ==2cosr d cos60° :::::,. d' = 2dcosr
1.5 (1.5 mR+R)
"' =:,
3 (1.5 m+1)
=:,
3=1.5m+1 1.5m=2
=:,
m=
=:,
166. [CJ
1
11.5
R
1
4 3
174. [CJ
Deviation by a sphere is 20r) Here, deviation s = 60° = 20 r) :::::,. ir=30° => r=i30°=60°30°=30°
AccOrding to Snell's Law, we have µ
= s!ni
s1nr
sin60° ,::;3 µ==,~ sin30° 175. [AJ
Fa
. sini 2 1 Since, sm r = = _3_ = Fa
µ
2
Also,. cosr = .J1sin2 r
=> . cosr=J¾ cc:,·· d'
:::::,.
= (2)(2)JI
sinu sinr
,
=4JI
ct··~ 3.26 cm
=.!l n2 sina = .!lsinr
cm
n,
... (1)
For TIR at other end
sinC= " 2
167. [AJ
n,
1
1
1
X
f
I
~
I~
=+
Also C=90r
sinu =.!lsin(90C)
n,
170. [AJ
1 1.5 1.51 +=(x) x R 2.5 0.5
. =:,
{Put r=90C in(1))
sina.=~cosC
n,
x=R X=5R
1.208
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176. [DJ
181 .. [BJ
CD is parallel to AB, so both m$diamust .have equal refractive i · indices.
177. [CJ Since all the prisms P , Q arid R are made of same material, so the deviation suffered· by the combination remains the same as the combination forms a part of the bigger sphere. ·
·
__
Lens Maker's Formula, we get
1f =(1.51)(.!.R +.!.) R =>
R=f
For the water lens, again applying ttiB Lens Maker's Formula, We
2
I'
3f
Since, we know that for lenses placed in contact, we have
1
1
1
1
2
Ca~)
n=3O
'
A== 2(36')
.
f
(90)' (20)' 4(90) f=21.4 cm
184. [CJ .
4
F=T 3t = st
sin·i
Smeeµ·=.smr
=>
179. [BJ sini sinr
d
183. [CJ f= 02x2 4D
=>
1 1 1 1 =++F f f I'
2
=>
=>
·=++' F. I t t,.
1
2,/3
182. [CJ Amu. =2C
.
get
=>
I
= n==
i=45°
185. [CJ
n
Since
tanr= 2 h_=1 2h
=>
r=45° . • h srnr = h.JS
' ' 1 srnr = .JS 1
1 .JS ;; =1
.
./2
/
·1?
,!"
,'; 0 0 o.,g,;~==~~=
0 r 8=60°
186. [CJ
n=J%
180. [CJ
2=(µ1)·(1.._...:!...) f R1 R2 For no dispersion
d(D=O

0 C' C
1 0cm
AB'= Apparent Depth= x AB=1O cm
The face BC appear's to be shifted to B'C' . If x be the apparent
depth, then AB'=x and B'B=(1Ox).
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Advanced JEE Physics Also, according to Laws of Reflection an image is formed as far behind the mirror as the object is in front of it => 20+x=23.2+(10x) => 2x=13.2
=>
x=6.6cm
. Smce µ=
=>
Real Depth Apparent Depth
f
=3 f+u {Since we get a real image)
=>
10 µ==1.51 6.6
=> => =>
187. [AJ Since µ2t = µl
=>
. (  f )' m 9ar f+u
_f_=3 f40 3f+120=f 4f=120 f=30 cm
194. [DJ Axial Magnification is given by
v'
m = I along PA aK O along PA
dX=t't
188. [BJ
m., =¾=( f~J'
=>
i = b (  )' fu
1
195. [BJ From the figure it is clear that the angle between incident ray and the emergent ray is 90° . Emergent Incident ray ray
By Laws of Reflection i=r
Also i+r=90 => i=45 According to Snell's Law sini µ=.smr
=>
u'
Air Glass
1_5 =!!_=sin45 2
=>
sinr
189. [AJ Shift=i(1;) away
196. [BJ easel n = Real Depth Apparent Depth
=± 3
190. [BJ According to Newton's Formula, we have X1X2
=f
=>
(10)(40)=f
=>
f2 =400
=>
f=20cm
191. [BJ A=60° i=55°' 8=46° Since i+e=A+D => 55+46=60+D => D =41° So, Dm
=>
Now, PI,= µ(PI,) =
(µ1)
Since
=>
2
1 foc µRed
3
2
x20 = 30 cm ~
30 cm +1
< µ~"lolel
ted > (.,.,let ted > t1ue
Always keep in mind that whenever you are asked to compare {greater than or less than) u, v or f· you must not apply sign conventions for comparison.
10cm
198. [DJ Since, the refractive index is increasing linearly from top to the bottom, so the light cannot travel in a straight line in the liquid as
shown in options (A) and (B). Initially it will bend toward_s normal and after reflecting from the bottom it will bend away from the normal as shown below.
!
i Rarer
Rarer
! Denser
! i
!=(µ1)(1J...) f R R ... (1)
~(µ1)(¾)
=>
II Denser
On cutting
1=(µ1)(11) r R "' =>
f=2f
205. [CJ When a lens is cut parallel to principal axis its focal length remains the same. Hence each part will have a focal length 2f.
f m=f+u
=>
204. [CJ
j
199. [AJ U=(f+X)
=>
Hence, the rays will converge at a distance of 40 cm from the lens.
f m=
ffx f m=
x
Negative sign confirms that real magnification is negative. 200. [DJ For a convex mirror, both (A) and (B) are incorrect. 206. [BJ According to Lens Maker's Formula
201. [DJ
U={f+x,)
.1__.!_ =(µ1)(1._J...)
v=(f+x,)
v
=>
1 1 1 +=
=>
1(2f+x,+x,)=f+(x,+x,)f+x,x,
~
~
f
f+X 1
R1
R2
The lens is made of two materials, so for a single object distance, two different image distances are obtained i.e. two images are formed.
Since..:!._..:!_=~ V U f f+X2
u
f
208. [BJ Let ~ and
t
be the focal lengths of the lenses of refractive
indices µ 1 and µ 2 in water, then
=X1X2
·f=~X1X2
202. [CJ Two pianoconvex lenses of focal length f on combining give a
convex
lens of focal length
½.
... (1)
To obtain a real image of
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1 1 2(µ,µ,) += \ t µ.R => =>
::::::,
=>
...!.. = 2(µ, µ,) 30
µ.R
(µ1µ2 )
µ.R =so
ill
1 1.5 1.51 (50) + ~ 20
=> =>
R
v
...!..+12= 0.5
50 V 20 1.5 1 · v=4o5o 1.5
10
v = 15x200 =+300 cm 10 Positive side indicates that image is real.
=>
212. [CJ
r2 =Oo =>
r1 =A=30°
and
i1 =60°
1 1 1 ,=vu
1 1 1 1 1 vfu2040 => v=40cm The situation is shown in figure. From similar triangles PAI and IQB 5 40 h 20
=+=
=> 216. [CJ Since 0 2 = 1112
=>
4 = (0.5)!,
=>
l 2 =8cm
217. [CJ According to Newton's Formula
x1x2 =f
So, from Snell's Law, we get
µ = sini1 = sin60° = .Js sin~ sin30° .
. gives.
=>
f=.J16x25 =20 cm
= ==================================== 1.212
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222. [DJ For real image U=U1, V=2U1, f=20cm
(~)(~)
cos(1B0'20)
1i+~11i;13]1
!,
Substituting in ..:!. + .:!. = we get V U f
11
=2U U 20
'' '
1
''' ''' ' 0 ''0
=>
For virtual image u = u 2
''
=>
(13) cos20=41 1 cos20=2
=>
cos20=..!
=> =>
20 =60'
''
''
=>
'
=>
2
=> =>
9= 30°
220. [AJ Focal Length of Lens is f1 =20 cm . Focal length of combination is fc =20+5=25 cm
1
1
U2
20
u2 =1Dcm u1 u2 =30 cm10 cm=20 cm
f= 40x25 4025 f =66.67 cm 10 p f(in cm)
1.5 D
=>
P=10D 1 f=p=0.1 m
=>
f=10cm
225. [BJ
when the space between lens and mirror is filled with a liquid then
Object is placed at a distance of 2f from the lens of focal length f i.e., the image formed by the lens will be at a distance of 2f or 20 cm from the lens. So, if the concave mirror is placed in this position, the first image will be formed at its pole and it will reflect all the rays symmetrically to other side as shown below
MO••~,''°~., oo,_, ~, •• OOj)
1 1  =+f' I, f 1 1 1 =+25 f, 20
1 25
1 2U2
224. [AJ P=P,+P2
f=20 cm As object distance = 40 cm Image distance = 40 cm
1 f,
=2u2 , f =20 cm
223. (AJ 1 1 1 =+f 40 25
1 !f =(1.5 1)(..!. _.!.) + (1.21)(oo114 ) 14 oo
=>
v
So, the distance between two positions of the object is u1  u2
219. [BJ
=> => =>
,
. I . 1 1 1 Aga1nappy1ng +=,we get V U f
... ' 180'20.
=>
1
u, =30 cm
1 20
o
ft=100cm For this liquid lens
I
I)
14 20 cm +I+ 20 cm ....i
f,=(µ1)(~ ~)
.!.)
=>
__1_={µ1)(1__ 100 33 oo µ1=0.33
=>
µ=1.33=3
=>
4
226. [DJ Shift=d=t(1¾) =>
l=~
n1
227. [AJ Focal length of piano convex lens using Lens Maker's Formula is
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7=(¾ 1)C~ ~)
235. [BJ
:::::> f=20 cm If point object O is placed at a distance of 20 cm from the piano convex lens rays become parallel and final image is formed at second focus or 20 cm from concave lens which is independent
of y.
0
228. [DJ Focal length remains unchanged. I oc Area of Aperture
Since angles opposite to equal sides are equal. So, in MOB We have LABO =p . In .6.CAB , external angle equals the sum
I =kD'
of internal opposite angles. So,
D=(aP)+(ap)
I'= 31
=>
D=2(ap)
4
236. [CJ
229. [BJ
When magnification is m = +2 , then we have U=X
1'=~ f In moving from air to glass, f remains unchanged while· v
V=2X
f =+20
decreases. Hence, A should decrease.
Applying Mirror Formula,
230. [BJ f=xm1 m2
(See 'Displacement Method' in SYNOPSIS)
231. [BJ
02 x2
f=4D
=>
234. [BJ 1 2 1 =+F f, I,,
:::::>
1
1
237. [DJ When water is filled in the mirror, a piano convex lens is formed, so now combination contains a pianoconvex lens and a mirror. The effective focal length of combination is less than the focal length of above mirror, so image is shifted downwards.
239. [DJ
1 2 1 =.+ft
1
+=2y y 20 => y=30cm So, the object has lo be moved by yx=20cm
When 8 < C partial transmission and reflection will occur. When 8 > C , only ieflection takes place.
~
V=+2y f =+20
=>
232. [CJ
;:::}
1 1 1 +=2x X 20 ::::,. x=10cm To have a magnification of m' = 2, we have· U=y and
f= (0.9)' (0.2)' 4(0.9) f=0.214 m 1=21.4 cm
.!.V  .!.U =!1 , we get
The given lens is a convex lens. Let the magnification be m, then for real image, we have
00
1
F=!t_
1
1
+=mx x f and for virtual image, we have
2
Also
.!. =(µ1)(J_ _ _!_) t f 00
r
1 1 1 +=my y f From equation (1) and (2), we get
ft=
.. ,(2)
I= x+y 2
µ1
F=r2(µ1) Negative sign indicates that mirror is concave.
... (1)
=>
241. [BJ Let 11
(i) (ii)
(iii)
,
12 and 13 be the images formed due to
refraction from ABC reflection from DEF and again refraction from ABC
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D
I,, =180 cm
=>
1 2 2 1 =++F 60 180 10 1 13 => =cm F 90 90 cm => F= 13 So, required answer to get asked condition is 2F. Hence 180 u=cm 13 =>
0
B E
145 cm
C
•u
F
•1
2.5cm
Then BI,= (5)µ 9 = (5)(1.5) = 7.5 cm Now EI, =(7.5+2.5)=10 cm =>
EI2
=1O cm
246. [CJ For equiconvex lens, we have IR,l=IR,1=1=10 cm
behind the mirror
Now BI, =(10+2.5)=12.5 cm =>
Now P=2P, +P•
BI = 12.5 = 12.5 = 25 cm 3 µg 1.5 3
=>
_.! = 2(µ1)(J__...!_)~
=>
1 _.! = 2(1.51)(J_F 10 10
F
242. [CJ
02 x2 f=4D
(By Displacement Method)
=>
P=~
=>
P=~ 10.16
=>
p =__±__ 0.84 P=4.76D
=>
02 x2
1
=>
~J
=> =>
=>
2
1
1
1 1 ;; (40)
V 15 40 V=24 cm
Applying µ 2
_
V
and
2
1
=+F 60 10
=> F=7.frcm For image to be formed at the same place where object is situated we have u = 2F = 15 cm 245. [DJ 1 2 2 1 =++Ff, I,, i,, (Because here we have two refractions at the concave surface of tens and two refraction for water lens). Since
1
_& = µ 2
u

µ, , we get
R
1 1.5 1(1.5) ;; (u) = ~
20 t,==10cm
t=(t X6~ ±J .
1
=15
248. [DJ
(1 '1J
1 · :r,=(1. 5 l) 2060
2
F=i=15 cm
vu=F
Shift= 2 cm (downwards)
1
t, ~ co
Further 1
.!= (µ1)(J__...!..) t R1 Rz
=>
2 1 f, i,,
Since .
t =60 cm
J~
10 => F=2.5cm , Therefore, the system will behave like a concave mirror of focal length 2.5 cm.
F
244. [AJ 1 2 1 =+F f, i,,
=>
R2
247. [BJ
Shift=d(1¾J=1(1 1
=>
R2
=+
243. [AJ
=>
R1
obje~t
1 3 1 => +=v 2u 2R For v to be positive, we have 1 3 >2R 2u
=>
U>3R
249; [BJ ~inaUon ), OO
1
Since  =.:!.+.!~ fcomblna1ion ~ ~ ~~
=>
0=1+.! __1_0_ 10 ~ (10)~
==================================== = 1.215
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Optics & Modern Physics
Advanced JEE Physics 254. [DJ f . Smce m=f+u
0=1+.!.+.!. 10
=>
t t
1
2
=>
10=1
=>
t=20cm
=>
1=
,:u
==> ==>
fU=f U=2f When the lens is cut, the focal length of the remaining portion is
250. [BJ f f f+(12) f+(20) f12=(f20) f12=1+20 21=32 f=16cm
21. Applying Lens formula, we get 1 1 1
.(21) =21
==>
V",oo
255. [BJ 251. [CJ
1 ( R, 1 R, 1) 10=(1.51)
5=(o.5>(~ ~) ... (1)
=:
_!_ _ _!_=10
=>
R1
On immersing it in a liquid of refractive index 3
R2
1)( ~ :,)
1=('µ,1)10
~=( 'µ,
=>
(::1]=0.1
~=(\51)(¾)
=>
~=0.9 µ,
=>
f 10 => f=10 cm Negative sign indicates diverging nature of lens in the liquid.
=>
256. [CJ Since, we have, at time t d = y+ µ(hy) = µh(µ1jy
252. [DJ lul+lvl=D V
m=u => v=mu lul+mlul=D D => l u l = m+1
=>
f
U=(m~1)
r
Since image formed is real, so it must be on positive side. Hence lvl=+v= mD m+1 For a lens
1
=>
m+1 m+1 1 +=mD D f
=>
f=~ (m+1)'
253. [AJ 1 (11) =(1.51) f R eo
=>
If A is the area of the tank, then we have
at y =,;:_
{·: Ay = at}
1 1
vu=t
=>
f !
hy
_!_ = 0.5 .!_ 16 R R=8 cm
d,;µh(µ1)at A i.e., d t graph is a straight line with negative slope and positive intercept. But d becomes constant once y =H.
=>
257. [BJ On immersing in water f increases and hence P decreases. 258. [DJ 1 1 !=(1.51)(   ) f 0.5 0.5
=>
!f = o.s(..3..) 0.5
=>
P=2 D
= ================================== 1.216
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Ray Optics 259. [AJ
266. [BJ Rays from O must fall normally on the mirror, only then the lens forms a virtual image at C (the centre of curvature). C
0=$i; 0='14x16 =8 cm
f
260. [A] For a lens making real image I m=
36cm
l+u
=> =>
mfmu=f
=>
U={1+¾)
=>
lul={1+¾)
mu=l(1+m)
For lul to be MINIMUM m must be MAXIMUM i.e. m, oo
=>
lulmin = f
=> =>
1 +
1
(3612) x=15cm X
1 40
For lul to be MAXIMUM m must be MINIMUM i.e. m = 1
=>
267. [DJ 1
lul.,. = 21
1
1
F= (1/2) + (1/3) =5
261. [A] The ray diagram is as shown in figure A
=> F=0.2 m=20 cm Since
1 V
C
0
u
F
=>
1 1 ,(30) = 20
=>
V=60 cm
   21 _,...;.___ 21 ____,..
268. [BJ Pcomb =P1 +P2 xPf2
Since triangles CAI and NEI are similar, so we have
h I d/2 = 21
0=.!.+..!~ I ~ I~
2h
=>
ct= 2
=>
h=~ 4
=>
X=I+~
270. [DJ Apparent Separation = 2(Apparent Depth)
265. [AJ If mirror would have been absent then image is formed on the other side of lens ( at I'). So V
=>
20
=>
Apparent Separation
= 2h µ
271. [AJ For distant vision u = 00
15
v=60cm
=>
2:;=7
=> l=2 m => P=0.5 D For near vision u=D=25 cm
1
=> Since the mirror reflects the ray back, so O' serves as a virtual object and forms a real image I in front of mirror.
1 1 1 +=v 60 15 =>
v=+12cm
1
1
1 o.25=f
=> P=+3 D Hence he must use bifocal lenses with P = 0.5 D and additional+ 3.5 D (to give +3 D net)
273. [CJ In the first case. Let x be the distance of object from the mirror. Then
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Optics & Modern Physics
Advanced JEE Physics 277. [CJ
u=x, v=+2x and f=f
.
. 1 1 1 Usrng v+u=f,weget 1 2x
1
vv1025
m=m0 xm =9..x2.=x8 U0U82,55
=>
1
x=1
f 2 In the second case, let y be the distance of object from the mirror. Then
=>
X=
=>
u=Y, v=2y and f=f 1 1 1 2y
m=20
278. [BJ
1 (Shift),= 1(1~J =9(1) =3 cm µ, 3/2 (Shift),= 1(1 ~:) = 9(1%) = 1 cm So, distance between two images is given by
y=,
(Shift), (Shift), =2 cm
3 v=t 2
279. [AJ
So, object will have to be moved by a distance of y  x = f .
m = m0 x ffi 8 =m 0 (
1+{'J
274. [AJ
1=5
t
Separation =t + t 36 =4t =>
fa=9cm
=>
t=45cm
280. [CJ
m=1+~ f
275. [DJ For lenses placed in contact, we have 1 1 1 =+F \ t
J=(µ,1l(±+i)+(µ,
1)(~ ±)
.!_=µ,µ2 F R
281. [CJ f
m=...!!..=10
t,
=>
fo=50cm
Separation =
t +f
= 55 cm
8
F=_R_ µ, µ2
282. [AJ
m=1=5
276. [BJ V8 =D=25 cm,
1 1 1 ·==+
t
1 u,
Ve
t
t =6.25 cm.
Since Ue
1 1 =6.25 (25)
U8 =5Cm
=>
,
V0 =155=10 Cl!1
283. [A] Separation= t + t Separation= 0.3 + 0.05 = 0.35 m 285. [D] For an equilateral prism, we have A=60° Since, the ray inside the prism is parallel to its base, so we have the condition of minimum deviation. So, i = e = 60° and 6m = i+ eA = (60°+ 60°)60° = 60°
Also, t =2 cm and 1
1
1
t
Vo
Ua
2
10
U0
=
=>
sin(A+6m) 2
Since, µ =
sinrn) =>
µ =sin(60') sin(30')
=J3
u0 =2.5 cm
= ==================================== 1.218
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Ray Optics 286. [AJ .
f
.!_ = (µ,
\
must be maximum ::::,.
{·: R,=R,}
\too
Similarly,
289. [BJ 2 cm . .. . Lmear magn111cat1on m =   = 2 1 cm So, the image is real and inverted lvl=2lul Let lul = x then lvl = 2x Now lul+lvl=S0(40)=90 ::::,. x+2x=90· ::::,. x=30cm
1)(_!__...!.) =0 R, R,
t > oo
So, a hollDw, convex lens of any material will behave like a glass plate.
295. [Dj The ray diagram is as shown below
.Y
So, the distance of object from the lens is 30 cm and of object is 60 cm, i.e., the lens mt.ist be located at x = 10 cm as shown in figure.
0
T
x=+50cm
x=40cm
I
X=O
a . S ,nee, x =../2
•r 30cm

and
1+c+
3=sin·'(µsini)i
••. (1)
This is a nonlinear variation of 6 with i • Also,. we have 6 to be
i= 60° => (i is the Zaxi~}
angle ZO
=6,=~C 2
... (2)
FIJrther When i > C ,then TIA takes place and·thEI incident ray is reflected back in the denser medium as shown in the tiQure . •• i .......... .
:' , / 'L,' ~6=it2i· ,/
n"'k
µ But according to Snell's Law
,/2 sin60 =,/3 sinr =>
Sp, 6=rc·2i • i.e., 6 decreases linearly with i. So,
sinr = v'2 ,/3
,/3
.
2
6ma.; =02 ·=rc2C
1
s,nr = v'2 => 3.
62
r =45°
4.
[A, DJ Combl~ed Solution to 2 & 3
For i < C, ·no TIA will take place, so we have deviation (8) givenby 6=ri
sin C = µram, µdenier
=..&_ µd
i+r' =90° According to Snelrs Law µd sini = µr sin(r')
=> µ
=261
[A, BJ
=>
Now, according to Snell's Law, we have
•.. (3)
From.(2) arid (3), we get
=> .=> =>
~=..&.=sine sin(r') µ, sini sinC sin(90i) sin~= tani
Reflected Ray ·o(mser Riirer Refracted Ray
C = sin·' ( tani) C = sin·' ( tanr)
{Be~use i=r}
= t:::================================== 1.220
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Ray Optics 6.
[A, CJ According to Snell's Law, we have ~A,= µ2A2
=>
(1)(6000) = (1.5)).,
=>
i., = 4000
Also at m=O, v=a => O=fa => f=a
14. [A, B, C, DJ
A
All are the consequences of the
Since frequency does not change when light goes frOm one medium to another, so·
"DISPLACEMENT METHOD TO FIND FOCAL LENGTH OF A CONVEX LENS"
C
V=
15.
A, 3x10
5x1014 Hz
V = 6QQQx1Ql0
[A, B, C, DJ Objective and eye piece are separated by a distance (t+t)=16.02m {OPTION {A)}
8
Angular Magnification=
_1_ =  ~ = 800
{OPTION (B)}
9.
[B, C, DJ See Displacement Method.
f, 0,02 A telescope produces an image which is always inverted.
12.
[A, DJ
In a telescope an objective is larger than the eye piece.
(OPTION (C)} Since images are formed at the same place, So one image must
{OPTION (D)}
be real and other must be virtual.
t
[B, DJ 1 1 1 =+f 15 30 => f = 30 cm (Diverging in nature)
,i,'
Since red deviates the least and violet deviates the maximum.
16.
lxl+IYl=32 p
Q'
So a coloured pattern with red on the outerside is observed.
P' X+t+Y ~ i+
(32x) .i
17.
ForP
1 y x 15 1 1 1 +=y X 15
=>
y
on,
,,,(1)
1 (32x)
So total optical path length= I;11s,
32xx' =240 x' 32x +240 =0
=>
x2 20x12x +240 = 0
=> => => => 13.
x(x20)12(x20) = 0 (x12)(x20)=0 x=12cm,20cm
[B, CJ
{OPTION (C)}
_.,
~
1 15
1 1 1 =>  +   = y 32x 15 Adding (1) and (2), we get 1 1 2 +=x 32x 15 32 2 x(32x) =:is =>
[A, C, DJ Optical path length in passing from 1st medium is r1is1 • Optical path length in passing from 2nd medium is n2:52 and so
ForQ
1
{OPTION (D)}
.
1
m
Total time of flight t =  Ln;s1
{OPTION (A)}
C 1~1
For
inhomogeneous
media
optical
path
length
is
B
f
.,,(2)
OPL= n(s)ds and the ray must travel along a path in which A
time taken to go from A to B is minimum. Such paths are called stationary pathways and this is the statement of Fermat's Least
Action Principle or Fermat's Principle of Least Time. 18.
[A, DJ The final image is formed at infinity when the combined focal
length of the two lenses (in contact) is 30 cm i.e., 1 1 1 =+30 20 f
f=60cm So, when another concave lens of focal length 60 cm is kept in contact with the first lens. Similarly, if µ be the refractive index of a liquid in which focal length of the given convex tens becomes 30 cm. Then from Lens Maker's Formula, we have
fv
m=
f
m=1~ f
1 b
=>
Slope==
=>
f
!
="b
C
2~=(¾1)(~,;J
... (1)
3~=(3~21)(;,  ; J
... (2)
From equations (1) and (2), we get 9
µ=a
c::::=================================== = 1.221
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Optics & Modern Physics
Advanced JEE Physics 19.
[B, DJ
=>
When upper half of the lens is covered, image is formed by the rays coming from lower half of the lens i.e., the image will be formed by lesser number of rays. Therefore, intensity of image
=>
will decrease. However, complete image will be formed.
20.
When the Image is real, then u=y,so v=2y Again applying the mirror formula, we get
[A, CJ m = +2 , means image is virtual, erect and magnified. A virtual and magnified image can be formed only by a concave mirror
1
and that too when object lies between pole and focus.
21.
2y
[B, C, DJ
1
1
y=,
=>
When passing from vacuum to a medium, frequency remains
unchanged while speed and wavelength decreases µ times.
=> 22.
[A, B, CJ For convex mirror (having positive focal length) the image is always smaller in size. For concave mirror (having negative focal length) the image is smaller when the object Iles beyond 2f .
23.
[A, BJ
30.
31.
[A, CJ When the object moves from infinity to the pole of the mirror, the virtual image moves from focus to the pole.
[B, DJ Since, for refraction at a plane surface, we have
.&=µ2 U
... (1)
V
If x be the height of the bird above the water surface, then for the light travelling from the bird to the fish, we have µ1 =1, µ 2 =µ and u=x
:'
So, from (1 ), we get 1 µ
l+ f1    
:::::,
(x)=v V=µx
=>
lvl = µx
''' ''
    21,     
24.
Now speed of the bird is dx dt So, apparent speed of the bird is
[B, CJ Th~ tube length of an astronomical telescope, in normal adjustment, is (f0 + ~) and that of Galilean telescope, in normal adjustment is
(tt)
where
t
and
t
l~:I=µ ~~
: : :, 1~;1> ~;
are focal lengths of
objective and eye piece respectively. In this case, f0 = f, so difference in tube lengths is
32.
1,c, =(t +t)(tt,)=2t =21 27.
28.
[B, DJ A concave mirror can give both real and virtual magnified images. Since nothing is specified,,so m=±3 ±3 = ~
15u
+u=10cm u= 20 cm
29.
=
[C, DJ When the image is virtual, then U=X, SO V=+2X From the mirror formula, we get 1 1 1 +=
v u I
1.222
[A, C, DJ TIR takes place when ray of light traveis from denser to rarer medium.
[B, CJ A concave mirror and a convex lens_give virtual magnified.image for a particular object position (i.e. when object lies between F and P (or Cl)
=>
{·: µ>1}
Further, sinC12 Since,
µ2
µ,
>
=&.. µ1
and sinC13
:12 µ1
µ3
µ1
c,2 > C13 Smaller the value of critical angle, more are the chances of TIR. 33.
[A, CJ Since, µ
sin(~)
.A
sm2 For µ
=,/2 and om= 30°
A = 60' , we get
Further, at minimum deviation, we have
r, =r2 =
A
2
=30°
Applying Snell's Law. we get
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Ray Optics .,,, sini1 = µsinr1
34.
=>
sini, =(v'2)sin(30°)= ~
=>
i1 =45°
l+t=1.Q=~ 16 8 Now, from (2) we get 1(~1)=_1_ 8 16 16\'10\+1=0
=>
[B, CJ I
1.5
=>
1.51
1
16\'8\2\+1=0 8\(2\1)1(2\1) = 0 (8\1)(2\1)=0
=>
Using µ 2 &= µ 2 µ, we get V u R ' =V oo 20 ;::::, v=+60cm Since v is positive, the rays actually meet.
=>
1
\=.!. or \ =8 2 P, =8D or P1 =2D
=> =>
35.
[BJ
In air, !=(µ1)(_1_ _ _1_) R,
f
When
immersed
f=C~2=>
... (1)
R2
in
a
liquid
of
refractive
... (2)
f immersed
in
a
liquid
of
i=¾(~.:J
refractive
index
2µ ,
[A, DJ Real image is smaller in size if object lies beyond 2f and it is larger if object lies between f and 2f .
39.
[A, CJ A ray can pass undeviated when µ1 = µ 2 or the ray is incident normally i.e., angle of incidence is 0°.
..
(3)
'=2(µ1) •
40. [C, DJ For TIA, we have i>C => sini > sinC =:i, sin(45°) > sinC Since, sine=..:!.
f
n
1
[BJ When in contact, we have 1 1 1
I
n > F2
n>1.414
~
(·. P=i=10l
10=.!+.!.
I t
1+~=10\t
·1
~J
1)( RI1 R21)
From these two equations, we get
\yatar = 4falr = 4f
I~
6=100.25
It It = 16
(3/2
\.atar  4/3 
6=.!+.!. 0.25
1
Applying Lens Maker's Formula, we get
f~, =(¾1)(~,
1 1 1 X =+F' I t It ~
41. [A,C]
... (1)
When at a separation of 0.25 m, we have
I
1
v'2 >,;
=+F
R
38.
l=µ1
~=(µ1l(f)
36.
V
V=R
From (1) and (3), we get
=>
R where µ 1 =µ, µ 2 =1, u=R, R=A
2'
X~. :J
= µ 2 µ,
V
1
i=(:µ 1)(~, ~J =>
u
µ
index
¾=(~. :J
When
[A, DJ
Since, .&. + µ 2
From (1) and (2), we get 1 µ1 1=1=>
37.
... (2)
In air, the image was inverted, real and magnified i.e., the object must be lying between f and 2f . Now the focal length has changed to 4f . Therefore, the object now lies between pole and focus and so the new image formed will be virtual and magnified .
From (1), we get
==================================== = 1.223
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Advanced JEE Physics 43.
Optics & Modem Physics 45.
[CJ Let t be the thickness of the watch glass, R be the radi1:1s of
and P) but the image is always magnified and a convex mirror
curvature at inner surface, so (R + t) is radius of curvature of outer surface. If µ be the refractive index of this thin lens (watch
glass), from Lens Maker's Formula, we get
[B, CJ , A concave mirror can give a virtual image (object lies between F can never give a real image.
46. [A, B, CJ Since, the lights used are of different colours, so they have different frequencies and hence (8) is also correct.
R
47.
[B, CJ
!f = (µ 1)(1 ...!.) 2R R =>
=
' R1
.f= 2R
R2
µ1
The focal length of lens has nothing to do with the direction frOm which the light is incident" on it.
.!V = !f = (µ1)[...!.  _1_] R (R+t) l=(µ1)(1__ .!)= (µ1)t
1
1
1
=x V 20
.•. (1)
For concave lens : 1
1
(20v)
=>
(20x) 10 1 1 ' =20x 20v 10
Salving this equation, we get
x =>
45•
µ
4.
= sin45" =./2
3.
v=20(v'31) cm
[DJ
m = v, = v = :!_ ' LI, X X
[CJ
20(,13 1) 20( ,/3 1)/,/3
v'3
Magnification by concave tens is given by
STotal
= Sp + Sa + SR
SToS,
= (45° 30°) + (180°  2 X 30°) + (45° 30°) = 150°
v,
(20v)
20v
m,u,(20x) 20x
[BJ
m,
Image formed by concave lens is virtual for all positions of object i.e., image by concave lens lies between the two lenses. For both the images to coincide, image by convex lens should also lie in between the two lenses the two lens or image by convex
lens should also be virtual.
••• (2)
Magnification by convex lens is given by
sin30°
2.
20(,131) v'3 cm
4020,/3 20 20( ,/3 1)
( 2,13 _ 3 )
,J3 8.
[AJ Since refractive index decreases with increase of A and velocity decreases with increase of refractive index, so we have V red
9.
>
Vorang9
> Vyellow
[DJ Colour and frequency remains unchanged as it is property of source.
l+X+>420  X 111
10.
[BJ Dispersion depends on wavelength
====================================· = 1.227
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Advanced JEE Physics . 11.
Optics & Modern Physics
[DJ
21.
[AJ
Since light takes shortest path, so
60°
t = nxo C
12.
[CJ
''
Speed of light changes with refractive index.
'' '' '
/60°
~'' 6=180120=60° Laws of reflection is universally true 22.
[DJ
A=)(6/3)' +(a13)' +(10)'
Ao,
A= 20 units 23.
=> => => =>
1 1 1 +=V 20 15 1 1 1 =V 15 20 1 , =V 60
[CJ
. idii) (6/3i+a/3]1oi
r=45°
[DJ Let the vector representing the refracted ray be
A' = 6,/31 + a..Jaj + Ck
.
ii' .(1
( 6,/31 + 8/3] +Ck) ,(k) )(6/3)' +(a.fa)' +c'
[CJ
[AJ Since, e =·0° => i=r :::::,. K2 =1
=>
C cosr=;=~~~=e
=>
cos45°
,/108+192+C' C
,/aoo + c' ·;
1
C
F2  ,/aoo + c'
=> C =±10/3 Since the refracted ray travels downwards, => C=10/3 =>
[CJ
Since, f=R~seca
26.
R So,for 0>0, f
f'
f
=>
OPL= (1+x')dx
i4
_B_
2µ .·
0 .
=>
x' OPL= ( x~g
)I' =am 4
35.
[DJ Focal length of /eris
O
Li
and L2 is given by using the Lens
Maker's Formula, so
31.
[CJ peq
.! = (µ, 1)(..!.  ..!.) I R "'
=2P,ene + Pmirror
Simila~y _1_ = (µ 1)(..!.   1) ' f, ' "' (R) => f=40 cm
and F', =1=~ mrror f , R
So, equivalent focal length is given by 1 1 1
¾=2(µ1)(~(..!i))(~) => => =>
1 4(µ1)
\=50cm
=>
where P00,=i=(µ1)(~ ~,)
=+
t., I f,
2
=+
I R R .!= 4µ ±+~ I R R R 1
4µ
36.
2
t,=RR ¾=¾(2µ1)
=>
l2(2µ1)
=200 cm 9
[CJ Image formed by L; lies 50 cm behind it and on principal axis
of
=>
f ..
=>
Li . This will act as an object for
L2 . 86, for L2
,
R
......................... I,
PA of L 1
32.
[CJ
4.5mm
For a planoconv~x lens, we have _1_ = (µ
f,
.
PA of L,
11(..!.  ..!.) R "'
=>
33.
=>
we have u=+SO cm,
[BJ When plane surface is silvered, then
Applying Len's Formula, we get 1 1 1 V 50 40
1
2
1
I=,,~
{·. ~> «>}
=>
1 2(µ1)
I=R=>
34.
=>
[DJ
=+f, R H
=>
,:=R
1
PA of L, is
y = m, (4.5 mm) =>
4
y= (4.5mm)=2mm
9
Hence 12 is at a distance of (4.52) mm =2.5 mm from PA of
2(µ1). 2 2µ
9
its principal axis i.e., (PA of L2 ). So, distance ofimage 12 from
,.=,,r
=>
1
.±
When curved surface is silvered, then 1 2 1
1 (1 1) (R) 2 ,.=2(µ1) ~ R
1
=+v 50 40 200 V=cm
Magnification produced by L, is m, =~ = · u 9 However, for ,L2 , the image 11 is at a distance of 4.5 mm above
R
~ = 2(µ1)
=>
1
t =+40 cm
L,. 37.
[CJ Applying refraction at cuived surface formula, i.e.,
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Advanced JEE Physics
EL+h=µ 2 µ, ,we get u V R 1
2
1
+=x V1 R =>
2xR
= x R
{from pole of curved·surface}
O,
This image formed will be at a distance v'=(v,R) from the
I
V1
C
I
2
I I I I I I
plane surface. Now again applying the above formula, but at the
plane surface, we get
I
_!_3_=0 V
V'
RR
{·: forplanesurlace R>oo}
Since, refractive index of this is 2, so th~ critical ang!B C is given by
2 =
=>
v
V1 R
C =sin•(½)= 30°
2 · Substituting v 1 = xR , we get xR 2(xR) = R(x +R)
Since, i =C, so the ray grazes the plane surface.
v
40.
Since yellow is the mean colour, so µb +µ, 1.51+1.49 50 1. µ, 2 2
For the image to be virtual, we have
..!.
2(xR)
X
=>
 a, ) +(a;  a; )
=(µ.µ,)A+(µ;µ; )A'
=> m1=1 However for plane mirror, we have
sinr
.
[DJ For no deviation to take place, deviation (given by a=(µ ,1 )A
Since, m, =(~)(~)
39.
175
by one must be cancelling the deviation due to the other.· For this (a) the prisms must be arranged upside down
v,=4R
=>
1.77+1.73 2
µ=
2
Ja
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Ray Optics 45.
[BJ
Equating equations (1) and _(2), we get
For µ =
3
2
dy
the ray will be internally reflected, so, the ray is
= y3/4
dx
normal at face BC . Hence finally, the iefracted ray is parallel to zaxis.
::::::, :::::i
=>
y:314dy=dx
f.
f'
y314dy= dx
'
=X
4y1l4
' ... (3)
The required equation of trajectory is 4y114
'
=> 46.
[BJ If BC is silvered, the ray will retrace its path. Hence equation of ray coming out Ot the prism is .Jaz + x
47.
52.
= 10 .
[AJ The image formed by the objective must lie within the· focus of the eyepiece.
49.
50.
[AJ To obtain best magnification, the object must be placed just beyond the focus of the objective lens. In this case, the first image distance from the objective is very large.
X=(256)''=4m
So the coordinates are ( 4 m, 1 m)
53. 48.
[DJ . At the point of intersection on the upper surface, y=1m
=>
[CJ The objective lens r11ust form a ·real image for eyepiece to magnify it.
[AJ As nA siniA = np sinip and as nA = np Therefore, ip
54.
[CJ i+9=90°, 9=90°i,
... (1)
55.
1
2
F
f,
~)(!o)
=>
~=2(1.51)( 2~ 6
=>
+F 20 60 20 60
=>
F=
= 1 because iA =90° ( Grazing incidence) = ~Ky312 + 1 =~ys12 + 1 08 1
. .
Slnl
2
8
60 .=7.5cm 8
[AJ 7.5 22.5
3
57.
[CJ
58.
[BJ
59.
[BJ Combine solution of 57, 58, & 59 Each part will work as a separate lens and will form its own image. For any part, we have u = 0.3 m, f = +0.2 m .
V
=> ::::::,
+1
coti =~or y314
1
[BJ
'
1 ~y3/2
1
Therefore, from lens formul~,
(1)( 1) = ~( y' 1' + 1) sini
=>
1
I 7.5 m==1u 7.5+30
nA =1because y=O
because K=1.0(m)"
I,
Since F =0 , so combination behaves like a·concave mirror.
56.
siniA
i.e., the ray Will emerge ·parallel .to the
[CJ
[DJ but tan8 = dy dx dy = coli dx Applying Snell's Law at A and B nA siniA = n8sini8
= iA = 90°
=1
boundary at P i.e., at grazing emergence.
Slope of tangent = tan8 = tan ( 90°  i) = coti
51,
=x
... (2)
1.
U
f
1
;;o.3
1
= 0.2
V=0.6 m
So, each part forms a real image of the point object O at 0.6 m from the lens, as sho'Wn in figure.
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Optics & Modem Physics 64.
[DJ Using LensMaker's Formula for the li(luid concave lens,
_1_ =(µ1)(....!..._..!.) = _µ1 _ 50 · R oo R
=
_µ1 10
3
~=µ1 50 10 µ=1+0.6=1.6
=> =>
,>v=0.6m+1 Since the triangles OL,½ and OI,12 _?re similar. So, we have
65.
r1 = o Further, r, + r2 =A= 60°
1,1 2 _OB_u+v
1,L,  OA  u=> => 60
I,I, _ 0.3 + 0.6
1,L,
0.3
[BJ Since the ray is incident nonnally, so
0.9 = 3 0.3
==>
r2 =A=60°
A
111, =3(L,L,) =3(2x0.0005) =0.003 m
[AJ
f=(µ1)(~ ~,)
s~~
f=(¾1)(¾ ~)=ix¾=¾ => 61.
[CJ
X~
=>
~)=3~
f,=3R
=>
=>
. 4 . µ 1 sme= .Jaxstn60°
=>
sine=
.
F=3R
So,
P=er, =0
=>
e=r2 =60°
=>
i
63
R=10cm
=>
c:::l
=> 66.
p=O
2
../3
=../3 4
k, =../3
[CJ When µ 1
1
2=µ1 µ,
F
3 :=15cm
... (1)
x µ, = µ
sin(60°)=~ µ,
0
=>
../3
So, equation (1) becomes
2 Since, image coincides with the object so, clearly rays of light must have retraced their path after reflection. This is possible only when rays of light must have fallen normally on the plane mirror. For this, the object is at the focus of the lens system.
F
=
From the graph, we observe that when µ 1 =k2
_1_=.!.+_1_=_1_ __!_=~ 'Fif,R3R3R
~
4,/32,1312
..f3µ 1 x 2
Further, deviation is given by P=er2
[CJ
Now
= µ,sine
µsinr2
i =A
i=(¾ 1
62.
Applying Snell's Law at AC • we get
=>
sinr2 =sinC=~
=>
sin(60°)=~
=>
2= .,/3xµ,
=>
µ1=3
3
µ,
µ,
,/3
4
{·: r, =60°)
1
8
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1/.ay Optics Therefore, second image will be formed at a distance of m6. or
k =~ ' 3
67.
__ t_)(A) below its optic axis. ( t+\d
[BJ From the graph, we observe that
J3,
Therefore, ycoordinate of the focus of system is given by is the maximum deviation
Y=A(t+ttd)
and j32 is the minimum deviation
Y=(\d)A t +\d
=>
So, P2 = z60° = 30° and P, "
70.
[AJ Wavefronts are parallel in both media. Therefore, light which propagates perpendicular to wavefront travels as a parallel beam in each medium.
71.
[CJ All points on a wavefront are at the same phase
=60°
$,=$,and$,=$, $, $, = $, $, 72.
68.
[A] From the first lens parallel beam of light is focussed at its focus i.e., at a distance \ from it. This image 11 acts as virtual object
[BJ In medium2 wavefront bends away from the normal after refraction. Therefore, ray of light which is perpendicular to wavefront bends towards the normal in medium2 during refraction. So, medium2 is denser or its speed in medium1 is more.
medium1
for second lens L2 • Therefore, for L2
U=+(\d). f=+t 1 1 1 1 1 =+=+v
I u t t (\d)
\d
V=t +\d
y
•'
medium2 75.
[DJ For both the halves, position of object and image is same, however the only difference is of magnification. Magnification for one of the halves is given as 2(> 1). This can be for the first one, because for this, lvl > [ul . Therefore, magnification, 1ml =]ti> 1. So, for the first half, we have
l~l= L.i I +    d     · f,d +I      f,     Therefore, xcoordinate of its focal point will be
X=d+V=d+ t(\d) t+\d => 69.
X
It+ d(\d) \+td
[DJ Linear magnification for L2 , is given by
m
~ (~~\d~)(\~d)t+~d
2
=> . lvl = 2lul Let LI= x , then v = +2x
and lul+lvl=1.8 m :::::, 3X=1.8m :::::, X=0.6 m Hence, u=0.6m and V=+1.2m .1111 1 1 Using===! V U 1.2 0.6 0.4 :::::, f=0.4m For the second half, we have 1 1 1 ~+ f 1.2d (0.6+d) 1 1 1 =>  =   +
0.4
1.2d (0.6+d)
Solving this, we get d =0.6 m
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Advanced.JEE Physics Magnification for the second half will be V 0.6 1 m, =;;= (1.2) =2 12 and magnification for the first half is m1 = "!.... =·() =2 u  0.6
76.
79.
The distance of the mirror as observed by the eye is
[AJ For direct observation, the eye E appears to be farther that it actually is. So,
X=H+!:!+..!::!_ µ
=>
[BJ The distance of eye from mirror is
81.
µH+H
So, distance of eye as seen by the fish in the mirror is H
X=(µH+H)+ 2
78.
2µ
X=H(1+}µ)
[CJ 1sin8, =nsin02 . 9 sm 2
1 . = s1n0 1 n
3H
=:
X=2+µH
=>
X
9,
=H(µ+¾)
9,
[CJ For direct observation by the eye E , the fish appears to be closer than it actually is. So, we have H X=H+
.
·
H+ !:!µ . So,
the distance of the fish as seen by the eye in the mirror is
X=µH+i=H(µ+¾) 77.
[DJ
· 1.234
2µ
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1.
A> (r, s) B> (p, q, r, s, t) C> (q, r, s, t) D> (p, r, s) (A) m < O , means real image, possible for concave mirror and convex lens. So, (A) > (r, s). (B) m > O, means virtual image, possible for all i.e., plane mirror, convex mirror (always), concave mirror (when
(8)
object lies between focus and pole), concave lens (always) and convex lens (when object lies between optical centre and focus). So, (B)> (p, q, r, s, t). ·
(C)
Velocity of bird as seen from water = 6 x
(D)
Velocity of bird in water after reflection from mirror
{C)
(D)
4
6 6=12
1,
1)
A> (s, t) B> (p, t) C> (s, t) D> (q, t)
4
w.r.t..bird =6+6=0
Velocity of bird w.r.t. fish
vA. = si +2}
=>
"A'A =VA' VA
Similarly,
1)
5.
A> (p, q) B > (r) C> (s) D> (p, q)
6.
A> (p, s) B> (q) C> (p, q, s) D> (r)
7.
A> B> C> D>
=>
8.
A> (p, q, s)
=10i
v, =(1 +3]}
.±3 = 8 t
=8 + 8 =16 t = 8 J.
w.r.t. fish = 88= O
vA =1+at= f +{2f +j)
Velocity of fish w.r.t. bird = + t Velocity of image of fish after reflection from mirror in air
=8x,:l=6t
1ml < 1 , means diminished image, possible for concave mirror, convex lens, .convex mirror and concave lens. So, (C) > (q, r, s, t). . 1ml ~ means magnified (> and same sized (= image, possible for concave mirror/convex lens (both 1ml > 1) and plane mirror (1ml = 1) . So, (D) > (p, r, s).
2.
(A) Velocity of fish in air = 8, ,cl= 6 t
v,,=(i+aj)
(q, r) (r) (p, r, s) (p, r)
B; (p, q)
C> (r) D> (p, q, s) 3.
A> (s) B> (p,"q, r) C> (q, r,.s) D; (t)
9.
A> (p, B> (p, C> (p, D> (p,
4.
A> (r) B> (q) C> (p) D> (p)
10.
A; (p, q, r, s) B> (q) C> (p, q, r, s) D> (p, q, r, s)
s) q, r, s) q, r, s) s)
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Advanced JEE Physics 11.
14.
A>(p) B> (p) C> (r. s) D> (q. p)
A> (s) B> (q) C> (p) D> (q)
~x
· · 12.
~
1
1
V
f
~y
U
V=(u~f) 15.
~ 7
) 1
Again 1ml can be greater than or less than 1. For concave mirror : 1 1 1
(8)
~
+u
..'!.(x+y)= dx + dy =5+2=7 ms·•
(B)
dx =3+2=5 ms1
(C)
d(2x) = 2 dx = 10 ms·• dt dt
(D)
dx =5 ms_,
dt
dt
dt
dt
µ, = µ, (no change in path) Li0·
V=(f~u)
=>
So, v is always negative i.e., image is always real. .
dt
A> (q) B> (r) C> (s) D> (p) A µ 2 > µ 1 (towards normal) B
f
Further, !ml=[~=
B
µ 2 > µ 3 (away from normal)
+=v
C
(A)
Now, v may be positive and negative, depending on values .of u and f • Since, !ml=fu[=(u~ 1)=(
A
Ref.on ground
A> (p. s) B> (p, q, r, s) C+ (p, q, r, s) D> (q, s) (A) For convex mirror : 1 1 1 +=v u f
i
C
Lr·= o on the block
µ 1 > µ 2 (away from the normal)
µ 2 > µ3 (away from the normal)
+1
I
µ1 x
i.e., m is always less than 1 or image is always diminished.
13. A> (s) B> (p) C> (q) D> (r)
=>
1 . J2=µ 2 s1nr sin r =
}:;1
v2µ 2•
Since sinr < 1
=>
A+ when object lies between pole and focus image is virtual, magnified and erect. B+ when object lies between focus and centre of curvature,
D·
·image is real, inverted and magnified. C+ when object lies at centre of curvature, image is real, inverted and of equal size. p + when object lies beyond centre of curvature, image is real,
µ, < ../2µ2
ForTIR: 45°>C => sin45° > sinC
=> =>
_1_>&
1?. µ, µ, > !?.µ,
inverted and smaller in size.
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1.
Forward shift of point of incidence due to a single reflection
The position of the image of end b can be obtained using mirror formula, according to which we have 1 1 1 +=v u f where u = 40 cm , f = 30 cm
X = 0.2tan(30°) = ~ . Hence required number of reflections required is
N
Mirror Length Forward Shift
2.
2./3 _
30
1 1 1 +=v 40 30 => V=120cm Length of the image (rod) is given by e, =12060=60 cm
( ~)
=>
Since, for a concave mirror, we can have both virtual and real image, so for real image, m =5 and for Virtual image, m = +5
So.magnification rn=dv =(v2 v1 ) du U2U1
Since we know that m =_f_ fu
=>
where f = 30 cm and m = ±5 because the image can be real or virtual. . For real image, m = 5
=>
5=~ 30u => U=36cm For virtual image, m = +5
4.
lml=3
Applying mirror formula,
..!. + ..!. = ! V
f
U
for concave mirror, we get
1
v 60 = (40)
~30 5=30u => u=24 cm Hence, the object must be placed at 24 cm or 36 cm in front of the concave mirror
=>
3.
=> =>
m=(12060) 6040 m=3
=>
v=120cm
s
The image of the end a of the rod which lies at the centre of curvature C is formed at C .
>+ x +1+60 cm ...........i Now, for the rays to again converge at S , (after reflection from the plane mirror) b'
a C
b
F
''
I
,1
1+4Qcm~ 1+
Distance of S ) ( from Plane Mirror
p
1 +    60 crn.i 120 cm _ _ _ _ _ _.,.
'
= (Distance of Image formed) . by Concave Mirror
=> =>
X=120(X+60) x =·ao cm So, the desired distance is 90 cm From reversibility principle, it hardly matters whether the ray of light is first reflected from the concave mirror or plane mirror.·
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Optics & Modem Pliysics
Adva11ced JEE Physics 5.
Since the image formed is real, so for the first case, we have m=3
Bird(B)r
i
=> =>
' '
Since.!.+.!=! u
V
1 1 1 +=(3u,) (u,) I
=>
f = 3u, 4
Differentiating w.r.t. time, we get
(::)=M:;)+(:)
m=2  V2 =2
=>
16=~(v)+u,where u=4cms1
=>
v=9 cms1
u,
7.
Since..!.+.!.=! V U f 1 1 1 =>   +   = (2u,) (u,) I f = 2U2
should get internally reflected i.e., should sufferTIR. From the figure, it becomes Obvious that the ray with least angle of incidence is the one which is incident almost grazingly with the Inner wall.
... (2)
3
The necessary and sufficient condition for all the rays to pass around the arc is that the ray with least angle of incidence
From (1) and (2), we get 3U1
'
''
2u2
J.
••. (3)
=>
that sin a= R; d where d is the diameter of tube
u2 ~=6cm
=>
!!.u u =6
=>
u1 =48 cm
For TIA, n ;,.C
'
Since, v1 = 3u1
v1 =144cm
So, I= 3u, = ~(48) = 36 cm 4
Now u, =
4
sin a~ sinC
=>
,'2:::
=>
A~~
8.
Let at some instant bird is at a height of x · from the water
distance between fish and image of bird at this instance will be,
Since, 9 = 90°  i => lane= coli dy = coti • dx Ac_cording to Snell's Law at O _and P ~ we have µ 0 sin i0 = µP sin ip
... (1)
Since µ=~1+ay => At y=O, µ=1 =>
=
µ1
=>
surface and it is diving downwards with v cms"1 At this instant fish is at a depth y· below water surface. Then the
=>
1 µ
Hence, the least radius required is 12 cm.
(v,v,}=144108=36cm
S=µ>
Since, µ=J.5 and d=4cm,soweget
and v, = 2u, = 2(54) =108 cm =>
"a"
0 For this ray, let a be the angle of lncidenC:e, then we observe
But according to the problem, we are. given that the shift of the object is 6 cm , so we get
8 '
,,
Aidi /'R ', ~
7 =3
=>
t
Fish (F) ... (1)
Similarly, for the second case, we have
=>
y
u=4cms·'~,
I
=>
=>
X
'' ''
sln(so•) = (~1 + ay )sini
s=ix+y.
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Ray Optics
' '
y A
' ,B 'I
'
'
' P(x, y)
D
i': y _._.""'::'.'.:='':.le!L_;__ _ _ x
0
C
'}I.,.
0 I +    X+< . .
Slnl=
i=r+r=2r Since, we know that sini µ=.
1 ,1+ay
~
coli= .Jay= dy
s1nr
dx
µ= sin2r ~ 2r = 2 s1nr r
d • J'ovay l.=Jdx o
=!>
11. After reflecting twice from two plane mirrors at right angles, a ray
X=2l
of light gets deviated by 180° , irrespective the angle of incidence, so the emergent ray is hence angle of deviation is 180° .
Substituting y=2m and a=2x10a m1 ,weget xmax
9.
aniiparaliel to incident ray and
=2000 m = 2 km
The ray diagram for the situation,is shown in figure.
R
\.
4~
I.· .... ~~·~·"·
12.
From Archimedes Principle, we know that Yimmersed
= pbody
V101a1
Since, LPCQ=,2r, LPRO=s20r) From the property of a circle, we get 2(LPRQ) + LPCQ = 2s => 2s40r)+s2r=2•
P1;quid
"1mmarsed
r=2i!:
= Psphera = J:...
V
p~""'
V
2
2p
i.e., half the sphere is inside the liquid. For the image to coincide
2
with the object, light should fall normally on the sphere. Using,
According to Snell's Law, we have sini
µ2
s1n1
µ= sinr
_
.&. = µ 2  µ 1 twice, we have
V
sin ( 2i.!:)
R
u
2
../3 = =>
sini cos2i
sini 2sin2 i1
::::)
2;/3sin'isini,/3 =0
. ' 1±.,11+24
sm,
r,;
4v3
4
1±5
4,3
. 2
13.
i.= 60°
in~idence, then
3
Forthelens, U=2m, f=+1.5m
From the figure, we observe that
BO=OC So, i( LOBC = LBCO = r
4
Solving this equation, we get h=1·s cm
.. Fa Slnl=
10.
3
Further  3  _g_ = .L._g_ ' h10 8 2
r,;
Rejecting the negative value, we get
=>
v1 =12cm
::::)
(say)
and
be the angle of
1
1
1
V
2
1.5
V=6·m
!
Since, m = (
) = 3 2
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Optics & Modern Pl,ysics
Advanced JEE Pliysics I1P = I2P and I1P12 is .L to mirror.
2~ =(1.51>(¾
,,
==>
R=20cm
..!:l. _ _& = µ2  µ 1 twice with the condition that rays
Applying
0.3m
!i)
V u R must fall normally on the concave mirror, we get 1.5 1.2 1.51.2 v, 40 +20
0 1,
>+ d _ _ _ ' _..,
2 1.5 21.5 =d80 v, 20
>+6m~
... (1) ... (2)
Solving equations (1) and (2), we get d =30 cm and v1 =100 cm
Therefore, y coordinate of image formed by the lens is given by Y=m(0.1)=0.3 m
The ray diagram is as shown in figure.
In triangle PNI2 , we have tanB = 12N NP o.3 tan 0 = NP =0.3
=> => and
1,
{·. lane= 0.3}
NP=MP=1m d=61=5 m
l,
....,.___ 100 cm _______,.. 40 l+' ~c~m.... _ _ 30 cm_____,..
x coordinate of final image 12 is,
1+B0cm+1
X=d1=4m
14.
Applying lens formula
.JV _.!U =!f
17.
twice we get
f=6cm
Using the lens formula.:!._.!.=!, we get V U f
!=12cm
,/\ ,'
V
0
1+
.,...__ 15 cm _____..
12 cm >oi+ d  
1
1
1
v,
12
6
=
1 1 1 1 ,=1510=30
••. (1) =:>
1 1 1 oo v,d 12 Solving equations (1) and (2), we get v, =4 cm
and
'
••. (2) 18.
d=B cm
Using lens formula, _!_ _ ..:!. = ! , we get V U f 1 1 1 v,  (40) = 30 =:>
15.
Using lens formula, ..!_.!=!,we get V U f 1 1 1 +=v, 40 20
==>
1
1
V2
10
U
=:>
f
1 10
=
From Lens Maker's Formula, we get
1.240
=4 cm
Va= 180 cm
v2 =5 cm
so, the final image is formed at a distance of 5 cm from the mirror towards lens. 16.
8
Therefore, the screen has to be shifted away fromthe lens by a distance X=VaV1 =60cm
+===>
=(1 / )9
So. u' =(40Ax) =36 cm 1 1 1 => v,  (36) = 30
.:!.+.!=!, we get V
V1 =120cm
Shift due to the slab A>c =(1¾)d
V1 =40cm
Using mirror formula,
f=30cm
19.
Lateral magnification in first case is 3, so if U=X then V=+3X Since.!._..:!.=! V U f
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1 X
1 f
22.
+=4
1
3x
=1
02 x2 f=4D Substituting f =16 cm and x = 60 cm , we get
4f 3
X=
D' (60)'
.•. (1)
16
In the second case magnification is 2 , so now we have
U=(x+1,5), v=2(x+1.5) Since.!_.!=! V U f 1 1 1 => ~'~ +    =2(x+1.5) (x+1.5) f
3 =>
D'64D3600=0
=>
D' 100D+36D3600 = 0 D(D100)+36(D100) =0 (D100)(O+36) =0 D=100cm
f=3.(x+1.5) 3 Solving equations (1) and (2), we get f=9 cm
... (2)
23.
For the first flare spot, the lens acts as if its right face is silvered . The equivalent focal length is given by
Using Lens Maker's Formula for both the cases, we get
_1_ =(µ, 1)(..!..
ti,
and
R1
D'  3600 =64D
=> =>
f
=>
20.
..!..) R2
... (1)
2(~1)
2x1.5
2(1.51) +30
... (2)
R,
F=12 cm
Using mirror formula,
1
\..,., = (µ, 1) [~:
60
F ::::::,.
~.. =[~: 1)(~' :,)
.!= 2(~:) F R,
Dividing equation (1) by (2), we get
f,,.
4D
But D ~ 36 cm , because it happens to be less than 60 cm . So,
1
2(x+1.5)
=
Since we know that
1
..! + ..!. =..! with v u F
u t co , we get
1
+=v 00 12
1)
::::::,.
Substituting the values, we get
24.
v=12cm
Method 1 : Ray diagram is as shown
rn1)
14d+
21.
····..
3 ~ate,=4'8;r=4x10=40cm
··
!,
Using the lens formula, ..:!_ _ ..:!_ = we get V U f u=+15cm, f=+30cm 1 1 1.
=>
15 cm ,...____ 40cm ;~ d=25cm
=
15 30 => v =+10 cm Therefore, the focus of the rays will move 5 cm closer to the screen. The ray diagram is as shown in figure. V
Method 2: Since this combination just behaves as a plane glass plate, so the power of the combination is zero. Since
1
poomS
::::::,.
O=_.:!_+ 40
8 ....... 
1
1
X
=l.,m, =i:+t \t 1
X
(15)
(40)(15)
X
F
(40)(15)
 F
X
(40)(15)
15 40 4015 (40)(15)
X=25cm
25.
Applying Lens Maker's Formula, we get
1 ( 1+ 1) =(1.51) 40 120 R,
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26.
R1 =24 cm
o:
f
The ray diagram is as shown in figure for first two steps. If the rays reflected from the mirror are parallel after passing through the lens for the second time, then 12 must lie at first focus of lens. So, the desired distance is given by
"4>l
f/2
,6! 1
10cm
_x___., , _ _ 21+l Multiple Correct Choice Type,Questions ...............................................................2.66 :> Reasoning Based Questions (Assertion Reason Type) ........................................2.69 :,
Linked Comprehension Type Questions (Paragraph Type) •...•••..•••..•............•••..•••2.71
:,
Matrix Match Type Questions (Column Matching Type) ........................................2.TT
:,
Integer Answer Type Questions ............................................................................2.80
Answers to lri Chapter Exercise's.(ICE) & Practice Exercise Sets .....................................2.83 Solutions to In Chapter Exercises (ICE) .........................................................................., •• 2.86 Solutions to Practice Exercise Sets ...................................................................................2.92
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INTRODUCTION
he phenomenon of interference, diffraction and polarisation exhibited by light could not be explained on the basis of Newton's Corpuscular Theory. In 1678, Huygen suggested that light propagates in the form of waves. The first historic experiment in favour of wave theory was done by Focault, who in 1850 found experimentally that velocity of light in denser medium is less than that in the rarer medium which was contrary to Newton's Corpuscular Theory.
T
phase velocity or wave velocity. The energy travels outwards along straight lines emerging from the source, normally to the wavefront, that is, along the radii of the spherical wavefront. These lines are called the rays. For a point source in a homogeneous medium the wavefront is spherical.
NEWTON'S CORPUSCULAR THEORY
Newton proposed that light is made up of tiny, light and elastic particles called corpuscles which are emitted by a luminous body. These corpuscles travel with speed equal to the speed of light in all directions in straight lines and carry energy with them. When the corpuscles strike the retina of the eye, they produce the sensation of vision. The corpuscles of different colour are of different sizes (red corpuscles larger than blue corpuscles). The corpuscular theory explains that light carry energy and momentum, light travels in a straight line, Propagation of light in vacuum, Laws of reflection and refraction. However, it fails to explain the phenomenon of interference, diffraction and polarization. WAVE OPTICS
Wave optics is the study of the wave nature of light. Interference and diffraction are two main phenomena giving convincing evidence that light is a wave.
Spherical
wavefront
For a linear source of light the wavefront is cylindrical.

,"

s
.. 
Cylindrical wavefront
A small part of a spherical or cylindrical wavefront from a distant source will appear plane and is, therefore, called a plane wavefront.
WAVEFRONTS AND RAYS
The locus of all the points vibrating in same phase of oscillation is called a wavefront (WF) i.e. a wavefront is defined as a surface joining the points vibrating in the same phase. The direction of propagation of light (ray of light) is along the normal to the Wavefront. The speed with which the wavefront moves onwards from the source is called the
/ Plane wavefront
2.1
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 CONCEPTUAL ···NOTE(S) Different types of wavefront Type of wavefront
Intensity
Amplitude
Spherical
Point source
LAWS OF REFLECTION ON THE BASIS OF HUYGEN'S THEORY
Cylindrical Cylindrical
Let AB be the plane wave front incident on a plane mirror M 1M 2 at LBAA' = i, where 1, 2 are th_e cqrrespondin~ incident rays .perpendicular to AB .
Light ray
Cylindrical WF
2
Line :::: :· ...
source
~1:ne
Plane
WF
''.JJ'J:rJ ••'ft.'' J • ·..J··l 'a
I cc r
0
Aocr
0
~
rays ;+ Light
T
According to Huygen's principle every point on AB is a source of secondary wavelets, so BA'= ct, where c is speed of light The secondary wavelets from . A will travel the same distance ct in the same time. So, AB'=ct
~
Now, LAA'B=90i,sothat LA'AB=i, (0
I =0.1710
= 14.86 rad
Viewing Screen
1bis value is intermediate illumination closer to darkness.
Geometric construction fordescribingYoung's doubleslit experiment. Note that the path difference between the
two rays is r2  r1 =dsin8
THEORY OF INTERFERENCE : MAXIMA & MINIMA THEORY OF DIVISION OF WAVEFRONT: YOUNG'S DOUBLE SLIT EXPERIMENT
The phenomenon of interference of light waves arising from
Consider a point P on the viewing screen located a perpendicular distance D from the two identical slits S1 and S2 , which are separated by a distance d . Let us assume that the source is equidistant from the hvo slits and is 2.5
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monochromatic, that is, emitting light of a single wavelength A.. Under .these assumptions, the waves emerging from slits
S, and S2 have the same freque11cy and amplitude and are in phase. The light intensity on the screen at P is the resultant of light .coming from both slits. Note that a wave from the lower slit travels fartl!er_ than a wave from the upper slit by an amount equal to dsin0. This distance is called the path difference, x , where ·
fringe~ is.also equal to p = "~ . Since the quantities ;
and
d are both measurable, we see that the doubleslit interference pattern, together with equation (6), provides a direct determination of the wavelength ;\. . Youpg used this technique to make the first measurement of the wavelength oflight.
... (1)
x=r,r,=dsin9
The value of this path difference will ,determine whether or not the two :wavesare in phase when they arrive at .p. If the path difference is either zero or some integral multiple of the wavelength, the two waves are in phase at P and constructive interference results~ Therefore, the condition for bright fr4,ges, or constructive inte~eience, at P is given by
x=dsin9=n;\.
This result shows that the separation between adjacent dark
(n=O,±1,±2,±3,..)
· ... (2)
The index number n is called the order number of the fringe. The central bright fringe at 9 = O(n = 0) is called the
YDSE (QUANTITATIVE TREATMENT): METHOD 2
s;
Conside; that two coherent sources•of light and S2 are placed at" ; dist~ce ·d · ap:.rt and. a screen is placed· at a distance D from the plarie of th~ two sources._ ~t P be a point on the screen at a distance y frorri a point O exactly opppsite to the_ centre of the two sources S1 and S2 • If x .is path difference between the light waves reaching point P from the sourCes S1 and S2 , then
x;s,Ps,"r
zeroth order maximum. The first maximwn on either side, when n = ±1, is called the first order maximum, etc. 
Similarly, when the path difference is
pl
an odd multiple of ~,
the two waves arriving at P will. be opposite in phase and will give 'rise. to. destructive interference. Therefore, the
~
~ is,
condition for dark fringes, or destructive·interference, at Pis given by x = dsin9 = (2n + 1)2': 2
(n=O,±1,±2...)
s
It is useful to obtain expressions for the positio~ of. the bright and dark fringes measured vertically from O to P . We shall assume that D > d and consider only points .P that are close to O . In this case, 8 is small, and so we can use the approximation .sin9 = tan9. From the large triangle ·oPQ in Figure, We see ~t
~~  : l
d Q~ 
1n right angled t.S,BP,
B
we have
S,P' =52 B' +BP' =D' +(y+f)'
... (4)
Using this result .together with equation (2), we see that the positions of the bright fringes measured from O are given by
y
LJi...' D 
For Maxima, we know that path difference x must be an
$=(2t)(~)
even multiple of ~, so
x = (2n)
).
2
, where
n = 0 , 1, 2, .....
=>
~ =n').., where n =0, 1, 2, .....
=>
y.=n().~),where n=0,1,2, .....
Illustration 3
At y = 0 (i.e., for n = 0 ) we get a Central Bright Fringe. For Minima, we know that path difference x must be an odd multiple of ~, so 2
x = (2n + 1)
).
2
yd
D=(2n+1)
, where 11 = 0, 1, 2, .....
).
2
, where
In YDSE, the interference pattern is found to have an intensity ratio between the bright and dark fringes as 9. Find the ratio of (a) intensities. (b) amplitudes of the two interfering waves. Solution
In case of interference, we have n=O, 1, 2, .....
=>
l.D Y. =(2n+l) d, where n=O, 1, 2, .... . 2
=>
Y.=(n+½)).~ ,where n=0,1,2, .... .
I =11 +I, +2N, cos$ (a)
r_ =I, +I, +2N, =(JI: +F,)' and Imm = 11 + 12 2N, = (F,  F,)'
(F, +F,_)' (F, F,_)'
Since, 1max Illustration 2
[min
If the maximum intensity in YDSE is I 0 , find the
F, +.fl, F,F,_
intensity at a point on the screen where
(a)
the phase difference between the two interfering
).
(b) Since, I oc A2 ,
1S  .
4
Solution
=>
Imax
sources is
(~:J
=>
~=2 A,
is 10 i.e., intensity due to independent
~ . Therefore, at
~
I.i..=( I2 A2
=>
Since, l=lm,,cos'(t) where
3 1
I.i..=±=4 12 1
(b) the path difference between the two interfering beams
(a)
1
Solving, we get
beams is .'.::, 3
.
9
J
=4
Illustration 4
I =10 cos '(") =io 3
6
The intensity of the light coming from one of the slits in YDSE is double the intensity from the other slit. Find the ratio of the maximum intensity to the minimum intensity in the interference fringe pattern observed. 2.7
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Solution Since, we know that I/min
Position of second minima is obtained by putting n = 2 in the equation
=(K..p; .Jr, +.Jr,)'
11 = 210 and 12 = I0
Im~ =
1_
(,/2 ,/2 +11)'
34
=
Illustration 5 In a Young's doubleslit experiment the distance between the slits is 1 mm and the distance of the screen from the slits is 1 m. If light.of wavelength 6000 A is used, find the distance between the second dark fringe and the fourth bright fringe. Solution The position of the second dark fringe is given by
y 2 (dark)=(2nll"D =(4ll"D =~(,.D) 2d
2d
y 2 =(2.!.)1,.D =~P=~(0.2)=0.3mm 2 d 2 2
According to the problem, we have
=>
y, = (2n1) AD, so we get
2d
2
Illustration 7 Young's double slit experiment is canied out using microwaves of wavelength A = 3 cm . Distance between the slits is d = 5 cm and the distance between the plane of slits and the screen is D = 100 cm . Find the number of maximas and their positions on the screen. Solution The maximum path] [Distance between the] difference that = coherent sources [ can be produced i.e., 5 cm Thus, in this case we can have only three maximas, one central maxima and two on its either side (for a path difference of 1,. = 3 cm)
d
The position of the 4th bright fringe is given by . nJ,.D 41,.D y,(bnght)=a=d
f 81:
Therefore, the separation is given by
1s, ...r___:1
fly= y4 (bright)y, (dark)= (
=>
fl
10
5 6000 X 10y=zx 10"'
X
4¾Y~
For maximum intensity at P, we have S,PS1P=A
1 l.S X lQ°" = l.5 mm
(y;d)' +D'  (y;d)' +D' =A
Illustration 6
=>
In a YDSE , the separation between the coherent sources is 6 mm, the separation between coherent sources
Substituting d = 5 cm, D = 100 cm solving the equation, we get y=±75 cm
and the screen is 2 m . If light of wavelength 6000 A is used, then (a) find the fringe width. (b) find the position of the third maxima. (c) find the position of the second minima. Solution (a) Since fringe width is given by p= 1,.~ , so we have
and
1,. = 3 cm
and
Thus, the three maximas will be at
y=0 and y=±75cm FRINGE WIDTH & ANGULAR FRINGE WIDTH
The separation between two consecutive bright (or dark) fingers is called the fringe width (p), given by
10
(6000x10· )(2) =O.Z mm 6xl0'3 d (b) Position of third maxima is obtained by substituting P=i,.D
n=3 in the equation
2
y, =n(A~),soweget
31,.D
y, =d=3P=3(0.2)=0.6 mm D~ r:::::l 2.8
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.Wave Optics
:w
f3=Y11+1Yn·=7 Note that fringe width p is independent of n. That is, the interference fringes have same width throughout. The angular fringe width is given by
CONCEPTUAL NOTE(S) In VOSE alternate bright and dark bands obtained on the screen. These bands are called Fringes. ·
s LJ ·
. 'Hf ~u1·s,
~
·.
Screen 1 3_ Bright liiail!liil4 Dark . ht 1 3 Dark 2 Bng : 1 Bright lliil/illfiilll2 Dark
lillllliill
d)
e)
f)
14mm
1~~i~~;
=
21 (1000)( 400 X 10_.) 2x0.1 So, required separation is Lly=4214=28 mm
Yn
:::~:~:bri~~tn;;i~ge
2 Bright ~ 2 Dark .' 3Dark . ht_.. j, 3 Brig D . ~4.0ark
' at Central position Central fringe is always bright, because the path difference, x =.0 a,nd hence the phase ,illSo called the zeroth maxima. The nth ·minima_ comes before the'nth maxima'. The fringe pattern obtained due to a,slit is more bright than that due to a point. If the slit widths are unequal, the minima will not be complete dark. For very large width uniform illumination occurs. If one slit is illuminated with red· light and the other slit is illuminated with blue light, n9 interf~rence pattern is observed on the screen. If the two coherent sources consist of object ?nd ws· reflected im_age, the central fringe is dark instead of bright
.__ _:o::.n•:c·c____ _ "                ~
Illustration 8
In a YDSE bichromatic light of wavelengths 400 nm and 560 nm are used.' Th~ distance between the slits is 0.1 mm and the distance 'between the plane of the slits and the scree_n is 1 m . Find the minimum distance between two Successive regions·of complete darkness. Solution . . . Let 71i minima of 400 nm coincides with n,_ minima of 560 nm, then
(2n 1 1)400 = (2n, 1)560 2n 1 1 ='!...= 14 =21 2n,_1 5 10 15
42mm
Light from a source consiStS of two wavelength A1 = 6500 A and A1 = 5200 A . If the separation between the sources from each.' other is 6.5 mm and that from the screen is 2 m, find the minimum value of y(;t:O) where the maxima of both the wavelengths coincide. Solution Let n 1 maxima of A1 coincides with n2 maxima of A. 2 •
Then, Yn1 ~ Yn:z 71iXp d
difference, $ =0° and hence tjle Central Bright Fringe is
b) c)
7(1000)(400x10_.) · 2x0.1
Illustration 9
d = Distance between slits D = Distance between slits.and.screen A.'=' Wavelength of monochromatic llght emitted from source a)
Y,
Next , 11th minima of 400 nm will coincide .with 8th minima of 560 nm Location of this minima is
d~=I=~ D d
n
i.e., 4th minima of 400 nm coincides. with 3rd minima of 560nm The IoCation of this minima is.
= n,,.,D d
5.= ,., = 5200 =± n,_ '1 6500 5 Thus, fourth maxima of A 1 coincides with fifth maxima of
',. The minimum value of y(;, 0) is given by
41.,D y=d=
4(0.65x10.. )(2) 6.5x10..,
0.8mm
Illustration 10
Two coherent sources are ·o.3 mm apart. They are 0.9 m away from the screen. The second dark fringe is at a distance of 0.3 cm from the centre. Find the distance of fourth bright fringe from the centre. Also find the wavelength of light used. Solution Second dark fringe n = 2 will be obtained at = 3'D Y=(2n1)"D •'2d 2d
i..D
...
2
=>
=y d 3
=>
AD 2 P=a= (o.3)=0.2 m
3
Fourth bright fringe from the centre will be obtained at y, =4P=0.8 cm 2.9
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Advanced JEE Pl1ysics From equation (1), we get 1,_ 2yd 2 X 0,3 X 10,3 X 0.3 X 10' 3D 3x0.9
=>
=>
n+l n
=>
n=4
=>
n1,.D 4x6500xl010 xl.2 Y, =d= 2xl0_,,
=>
y, =l.56xl0_,, m=l.56 mm
7
1,.=6.67xl0 m
Illustration 11 In a Young's double slit experiment, two narrow slits 0.8 mm apart are illuminated by a source of light of
I
source of monochromatic light (,. = 5000 A) is placed in front of the lens at a distance·of a= 15 cm from it. Solution
10
P=""D d
5 4
Illustration 13 A convergent lens with a focal length of f = 10 cm is cut into two halves that are then moved apart to a distance of d = 0.5 mm (a double lens). Find the fringe width on screen at a distance of 60 cm behind the lens if a point
wavelength 4000 A. How far apart are the adjacent bright bands in the interference pattern observed on a screen 2 m away? Solution Since, d=0.BxlO_,, m, 1,.=4000 A=4000xl010 m, D=2m The distance between the adjacent bright bands or the fringe width is given by
6500 5200
Applying lens formula,
4000xl0 x2 m=lmm O.BxlO_,,
,! _,! = .! , we get V
f
U
1
1 1 15 10 => v=30cm V 30· , Smce, m===2 u 15
+=
Illustration 12 A beam of light consisting of two wavelengths, 6500 A and 5200 A, is used to obtain interference fringes in a Young's double slit experiment. The distance between the slits is 2 mm and the distance between the plane of the slits and the screen is 120 cm . (a) Find the distance of the third bright fringe on the screen from the central maximum for the wavelength 65ooA (b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide? Solution According to the problem, we have ", = 6500 A = 6500 x 1010 m
v
~ ~ s
________ _)'
~ 15 cm
Fringe width p "~
d=2 mm=2xl0_,, m
YJ
=
=>
(n + 1)1,.,D
d
d
"", = (n + 1)1,.,
D+1
60cm+1
(5xrn')(0.3) (l.5x10_,,)
In water (liquid), of refractive index µ the wavelength 1.17 x 10_,, m = 1.17 mm
the wavelength coincide
corresponds to that value of n for which
n1,.1D
0.5mm
INTERFERENCE EXPERIMENT IN WATER
(b) The least distance from the central maximum where the bright fringes due to both
30cm
0.25mm 0.25mm
P=O.lmm
D=l20 cm=l.2 m n1,.D Y, =d
=>
' ''
s2!'
0.5mm
Distance between two slits is d = 1.5 mm Distance between slits and screen is D = 30 cm
I
3x6500x1010 xl.2
''
1
1+
", = 5200 A =5200x1010 m
(a)
'
S,t'
'
"
to A.' =Therefore, if interference µ experiment is performed in water the fringe width decreases decreases from A.
from
p to P' , such that P=""D and P'=""'D =""D d d µd
=>
P'=~ µ
= c================================== 2.10
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Wave Optics Illustration 14
A Young's double slit arrangement produces interference fringes for sodium light (,. = 5890 A) that are 0.20° apart. What is the angular fringe separation if the entire arrangement is immersed in. water? Refractive index . 4 of water1s .
41,
'
;:,
·:'' '
3
Solution
The wavelength of light in water is Angular fringewidth in air,
So,
d
d
µd
µ
d
b)
B +y
±
passing through slits, then
11 _a~_ W 1 :7:l2
c)
Im,. Imaic
d)
2N, 11 + 12
=> e)
and if I_ =lmin, then V =0 f)
INTENSITY DISTRIBUTION
When two coherent light waves of intensity 11 and 12 with a constant phase difference $ superimpose, then the resultant intensity is given by
W2
(F.Jw,)' =(,µ,,µ,)' .jI;
(a,a,)' a1 +a2 = .{vi; +,Fi;
=
l1
best. Also if I_ =0 then V =1
a2
+.JI;
If point source is used to illuminate the two slits, the intensity emerging from the slit is ·proportional to area of exposed part of slit. In case of identical slits.
If [min = 0, V = 1 (maximum) i.e., fringe visibility will be the
= l2
a1 =a2
When white light is used to illuminate the slit, we obtain ,an interference pattern consisting of a central white fringe having on both sides symmetrically a few coloured fringes and then a uniform illumination. If qi
IR
I gl
=11 +12
h)
+2M cosq,
If x is the path difference, then
IR=I1
I 1 =I, =I0
!
+12 +2Mcos(~1txJ
In YDSE , if n1 fringes are visible in a field of view with light r of wavelength A1 , while n, witti light of wavelength A2 in \
!'
Since, IR= 11 + 12 + 2.jf;i;·cosq>, so we get (a)
Between_ nth bright and m 111 bright fringes (n > m)
I= 4!0 cos'(¼)
ilx =(nm)~
For maxima, . q> = 2mt
=>
In:un
=0
'
Separation (file) between fringes
I =2I0 (1+cos$)
For minima, $ = (2n + 1lit
I
is the phase difference .between two waves of /
I
In YDSE, usually the intensities 11 and 12 are equal, so
I= =410
I
intensities 11 and 12 , then
I =I, +I, +zN, cos$
=>
I
the intensity of light (with respective amplitudes a1 and a2 )
With the help of the concept of visibility, the knowledge about coherence, fringe contrast and interference pattern is obtained. Fringe visibility V is defined as
~
l
If w1 and w2 are the widths of the slits and 11 and 12 is
FRINGE VISIBILITY (V)
I= + Imm
D
i
3
v r_r_
B
PROBLEM SOLVING TRICK(S) a) lntt::rference occurs due to L_aw of Conservation of Energy. / Actually, redistribution of energy takes place. ,
ew = 'w
e  'w  ,.  0,  0.200  0,150 w
B
Intensity distribution on the screen as a function of yin YDSE Ima,= 410 for bright fringe and Im1n =0 for dark fringe.
µ
e, = !:
Angular fringewidth in water,
D
D
'w = !:
(b)
I i [
111
Between n bright and
(i)
mth dark fringe
If n>m then fil< =(nm+1)~ 2
(ii)
;
I
II n
Similarly, here once we calculated the path difference, then
FOR MAXIMA, Lu= (2n)
D,
Now, we observe that Lu= 201So, 20th order maxima is obtained at O .
where, y' is the distance of the source S above or below the
A,
At O, Lu= y,d = 102 mm= 0.01 mm
· O
i
~ 1+ £
(a)
=1.5 m+r
I=Im.,cos'(¾)
White light is used in a YDSE with separation between the sources to be 0.9 m and separation between the sources and the screen to be l_ m. Light reaching the screen at position y =1 mm is passed through a prism and its spectrum is obtained. Find the missing lines in the visible region of this spectrum. Solution
Path difference is given by
¾rm., =Icos'(¾)
,ix= yd= (9xl04 )(1x10') = 900 nm D
1 = 2'. 2
6
For minima Ill:= (2n + 1)~, where n = 0 , 1, 2, 3, ....
~=~=(2;)tµl)t
t=
A, 6(µ1)
6000 6(1.51)
A,=~= 1800 (2n+l) (2n+l)
2000 A
,_
_ 1800 1
missing
MISSING WAVELENGTH IN FRONT OF ONE SLIT IN VOSE
Suppose P is a point of observation in front of slit 51 as shown. The wavelengths missing are the ones obtained by using the condition of destructive interference, i.e., A, iix=(2n+l), where n=O, 1, 2, 3, .... .
2
1
1
1800 7
Of these, 600 nm and 360 nm lie in the visible range, so these will be missing lines in the visible spectrum. SUSTAlNEDINTERFERENCE
The interference pattern in which the positions of maxima and minima remain fixed is called a sustained interference.
a)
... (1)
iix=D[(1+~:)½1]
l
Sel+ ~, 2
So, (1) becomes d' A, iix =  = (2n + 1)2D 2
l J :.. .
1800 1800 3 5
CONDITIONS FOR SUSTAINED INTERFERENCE
Now iix=.JD'+d' D
Since, D»d, so (1+ ~:
1
'~ ,____ D _ _ _..
The fundamental condition for sustained interference is that the two sources should be coherent i.e., the initial phase difference between the two interfering waves must remain constant with time. b) The amplitudes of the two waves should be equal or nearly equal. This will give good contrast between bright and dark fringes. c) The two sources should be very closely spaced, otherwise the fringes will be too close for the eye to resolve. d) The sources should be monochromatic, otherwise there will be overlapping of interference patterns due to different wavelengths, which will reduce are contrast. e) The frequencies of the two interfering waves must equal. f) The sources should be nanow. Since hvo independent sources cannot be coherent, a sustained interference pattern can be obtained only if the two sources simultaneously and, therefore, the phase difference behveen them remains constant.
==================================== = 2.15
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Advanced JEE Physics ORDER OF FRINGES
If the slits are verticat as shown in figure, path difference is, M=dsiri0
n=8
If d =101. (say)
n=9 ++\
n = 1O
n= dcose ).
n=1Oat0=O'
s,~
CONCEPTUAL NOTE{S)
Screen
To calculate ,the .number of maximas or minimas that can be
obtained on the screen, we use the fact that value of
This path difference increases as 0 increases. The order of fringe H is given by·
sine(or case) can never be greater than 1'. For example in the first case when the slits are vertical.
dsin0 = HA. =>
sin9=~ d
,.
dsin0
n=
{for maximum intensity}
$ince, sine'/> 1.
The order of fringe increases as we move away from point 0 on the screen.
ni. ~ 1 d
=>
d i.
n~
Suppose in some ,question
n=2
t s, ~
=>
d ).
comes out say 3.6 then total
number of maxihlas on the screen wm be 7. Corresponding to, n = O, ±1, ±2, ±3.
n=1
d _ _,.,__ _ _ _ _ _ _ ___, n = o
So, highest order of interference maxima is
:t .s,~
nm~
9 n  dsin ).
=[P.] ).
where [ ] represerts the greatest integer function. So, total ~ number ofmaxlmas obtained are

n=Oate;,0°
N=2nmax +1
When the slits are horizontat as shown in figure, then the path difference is ax= dcose
.
nP
Similarly, highest order Qf interferenc~ minima is
nm,,
=H:+f]
where [ ] represent~ the greatest integer function. So, total number of·minirTlas obtained are N =.2n~;n
Illustration 20
~ c s,
II u
s,
Md+I
·
Two coherent narrow slits emitting light of wavelength A. in the same phase are placed par.Jlel to each other at a small separation of 2,. . The light is collected on a screen S which is placed at a distance D(» ,.) from the slit S1 as shown in figure. Find the finite distance x such
Screen
This path difference decreases as 0 increases The order of fringe H is given by
that the intensity at P is equal to intensity at O .
dcos0 = HA. :::::::>
ep f
,.
dcos0 n=
X
The order of fringe decreases as we move away from point
0.
~1.    ~ 2_ _ _ _ _ _ _ _ _ _ _ _ _ _ _
0
f+2).+l
>+ D     < " i i
=
s 2.16
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,, Solution
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pf
Path differen_ce for w9ves reaching.at O is
s,o  s,o = 21,.
y
e
(21,,)cos8=~
=>
n1.
=>
ny' 2D(D+n1.)
=>
y=
i+D+IIJ
PO =tane
s,o
~=tan60'=v'3 D
=>
x=v'3D
maxima. At point P , where path difference is l. (so, x = ../3D ) we get first order maxima. The next, i.e., zero order maxima will be obtained where'path difference, ~cos ff= O 9=90° X4CO
so; our answer, i.e., finite distance of x should be x=../3D., Corresponding to first order maxima.
Illustration 21
Two point sources are d =nA apart. A screen is held at right angles to the line joining the two sources at a.distance D from the nearest source. Calculate the distance of the point on the screen, where the first bright fringe (excluding the centre one) is obser_ved. Assume D » d. Solution
At point C path difference is n1,, . Therefore, nth bright fringe will be observed. Next bright fringe is observed where path difference is (n1)1,,, so S,PS,P=(n1)1,,
y' (nA.) 2D(D+n1,,)
n1.1,,
1
2D(D+n1.) n
OPTICAL PATH
CONCEPTUAL NOTE(S) At point O ·, pa,th difference is 2A. i.e. we obtain seco~d order
=> =>
(n1)1,, · ·
d
s
1 => cos0=2 => 8=60' Now in AS1 PO
=>
... (3)
It is defined as distance travelled by light in vacuum in the same time in which it travels a. given path length in a medium. If light travels a path length d in a medium. at d . speed v , the time taken by it will lie t =  . So, the optical V
path length is OPL=cl=c(;)=(;}=µd Since, for all media µ > 1, optical path length greater than geometrical path length. When two light waves arrive at a point by different distances in different media, the phase between the two is related by their optical path instead of simple path difference. So, 21t . Phase Difference = T( OPL).
I·µ=;) is always
travelling difference difference
DISPLACEMENT OR SHIFTING OF FRINGE PATTERN IN YDSE
When a transparent film of thiCkness t and refractive index µ is introduced in front of one of the slits, the fringe pattern shifts in the direction where the fiim is placed. How much is the fringe shift?, . Consider the YDSE arrangem~nt shown in the figure.
... (1) 2.17
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Advanced JEE Physics difference is (µ 1)t. d)
(µi)t . If shi~ is ·equivalent to n" fringes then n =  or ),,
t=~· (µ1)
e)
µ, t
~s,
Illustration 22 Interference fringes are produced by a double slit arrangement and a piece 1>f plane parallel glass of refractive index 1.5 is interposed in one of the interfering beam. If the fringes are displaced through 30 fringe widths for light of wavelength 6 x 10_. cm, find the thickness of the plate.
l+D    ~ . ,
A film of thickness t and refractive indexµ is placed in front of the lower slit.
·The optical path difference is given by x = [(S,Pt) + µt]S,P
=>
Solution Path difference due to the introduction of glass slab is
x=(S,PS,P)+t(µ1)
Since S2PS1P=dsin0
=>
ru:=(µl)t
x=dsin0+t(µ1)
Since sin 0 ~ tan 0 =
The Sh!ft, tlx is i_ndependent of the order of frihge n., i.e. Shift Of.Zero order maxima = Shift of nth order maxima: Shift is independent of wavelength.
Thirty fringes
are displaced due to the introduction of slab.
Y:D
So,
d' x=1&+t(µ1)
=>
(µl)t=3OA.
The maxima will be obtained when the path differenceis an
=>
t==
even multiple of ,?:. i.e., 2
=>
t=3.6x10__, cm
=>
Llr=3OA.
D
,. x=(2n)
2
=>
,. d ' (2n)=1&+t(µ1) 2 D
=>
y;= n~D (µl)t~
In the absence of film, the position of the nth maxima is given by equation nAD Y, =d
Therefore, the fringe shift (FS) is given by FS=y,y;c'~(µl)f=t(µl)t
\:~="'~)
Note that the shift is in the direction where the film is introduced.
a)
CONCEPTUAL NOTE(S) 'The "enlife pattern shifts t6wards the side where the Pl.ate
is:intrOduced and there is no other change in the pattern~ TO aj~asJ:Jrei this, shift .:,Vhit8 li9ht•·.must be used: bec1;1use , With, monochrom8.tic light,all the fringes will exactly~ be slinilar.and hehce no,Shift.Can"b8 observed. " · ' . (. . ,, c) .The·8ffective pat~ in air'is inCf~ased by an a,rnount'' (J..L:.1)t' .~due~tojntroduction oMhe ,PlateJe .. jhe :additionai' 'path b)'.
~
=
2.18
~
3OA. µ1
3Ox6x10·5 1.51
Illustration 23 In Young's double slit experiment using monochromatic light the fringe pattern shifts by a certain distance on the screen when a mica sheet of refractive index 1.6 and thickness 1.964 microns is introduced in the path of one of the·interfering waves. The mica sheet is then removed and the distance between the slits and the screen is·doubled. It is found that the distance between successive maxima (or minima) now is the same as the observed fringe shift upon the introduction of the mica sheet. Calculate the wavelength of the monochromatic light used in the experiment. Solution Shifting of fringes due to introduction of slab in the path of one of the slits is given by
,
uy
(µl)tD d
... (1)
Now, the distance between the screen and slits is doubled. Hence, the new fringe width will become
P' = ,_(2D) ·d
... (2)
Given, !:!ly = J3'
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(µl)tD
1.(2D)
d
d
(µ1)1 2
(1.61)(1.964x10
Illustration 2S
1,. = 0.5892 x 10< m = 5892 A
Illustration 24
In a YDSE, the two coherent sources are separated from each other by 6 mm and from the screen by 2 m . A light of wavelength 6000 A is used. A film of refractive index 1.5 is introduced in front of the lower slit such that the third maxima shifts to the origin. (a) Find the thickness of the film. (b) Find the positions of the fourth maxima. Solution (a) Since third minima shifts to the origin, therefore, the fringe shift (FS) is equal to three fringe widths i.e., 3P,
A Young double slit apparatus is immersed in a liquid of refractive index µ 1 • The slit plane touches the liquid surface, A parallel beam of monochromatic light of wavelength 1,. (in air) is incident normally on the slits. (a) Find the fringe width (b) If one of the slits (say S2 ) is covered by a transparent slab of refractive index µ2 and thickness t as shown, find the new position of central maxima. (c) Now the other slit S1 is also covered by a slab of same thickness and refractive index µ3 as shown in figure due to which the central maxillla recovers its position find the value of µ 3 •
sl
!=:~'i"  ()
*=: s. s
so we have
FS=y, =3(1,.~)
1 '1+,D
Since we know that the fringe shift (FS) is given by FS=(µ1/~
(d) Find the ratio of intensities at O in the three conditions (a), (b) and (c). Solution (a) Fringe ·width is given by
(µl)tD = 3 i..D d d
P=i..'D =i..D d µ1d
t=~
µ1
(b)
Position of central maximum is shifted upwards by a distance
Since, 1,. = 0.6 x 10< m , µ = 1.5
t (3)(0.6x10= "~=( ")(1,4) 1. · 1. _
=>
0
lsin0kl
=> 20:,(n40),;20 => 20,;n,;60 Heri.ce, number of maxima obtained is N=6020=40 (c)
To observe bright fringe at C, the mica slab should be placed in front of S2 • In that case, net path difference at
=>
µ, .. =1.2µ. =1.2(¾)=1.6
(2n1)1.D
y
2d
.
I
2
2(µ1)
(2n,1)1.;D (2n,1)1.;D 2d. 2d · .and t..;_ are wavelengths in water.
For blacklme,
(µl)t=!:'. =>
0.2
=1.2
For minimum intensity at C , we have
,..
0
(b) A black line is formed at the position where both the wavelengths interfere destructively. Distance of nth dark fringe from C is given by
Hence, maximum intensity will appear at C .

(2xl0·3 )sin(30°) 5x10'
where_
sooxio... = 500 ~2xo.s
·
t..;
,.,
A; · µw
A1
4000
=>·===,.; ~ ,., 5600
Illustration 28
In a Young's double slit experiment a parallel beam containing wavelengths 1.., = 4000 A and 1. 2 = 5_600 A
µw Substituting these values in equation (1), we get 2.21
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• Advanced JEE Physics
y 2 =asin(rot+$)
2n,:...1=z: 2n2 1 S
y, = asin( rot+ 2$)
.. For minimum value n, = "4 and n, = 3 •
,
r
y, =asin(rot+3$)
•
Henc!', d,istance of first blacJ< line is given by
y
(2x41)(4000x101~)40x102 x3 · 2x2x10' x4
=>
y=2.lx101 m
=>
y=0.21mm
y. = asin[rot+(N1)$] At angle 0, the path difference b~tween any two successive slits is tu = d sin 8 . So, the corresponding· phase difference $ is given by 2 2 $=( ; }~= :(dsin8)
MULTIPLE SLIT INTERFERENCE PATTERN Let us now look at the case where we have a general number N of equally spaced slits, instead of two equally spaced slits. AB an assumption, we have. shown in figure. a se.tup of six equally spaced slits.
RESULTANT WAVE AMPLITUDE (USING PHASORS) The· above set of equations can be represented by phasor · diagram shown in figure (for a set Of six sources generalised to N sources).
p .,/
r/
./
4>\
R

.,'
i
.. F
:~~~:::______ t , I
1 ' \ \
: Set up fur 6 Equally Spaced Slits
Similar to the N = 2 case discussed already, we will make the farfield ...... assumption that the distance of the sources from the screen is much larger than the total span of the slits, which is (N 1)d. We can then say, as we did in · the N = 2 case, that all the paths to a given point P on the screen· have approximately the same length in a ~ multiplicative (but not additive) sense, which implies that the amplitudes of the interfering waves are all essentially equal and we can also say that all the paths are essentially parallel (because of farfield assumption). A closeup version near the. slits is shown in figure. Also, each path length is _dsin8 longer than· the one just above it. So the lengths take the form of r. =r1 +(nl)dsin0.
· y1 =asin(rot)
=
2.22
r/
\
,''
'\
' 'A'
a
........
',
',
'' '
4'
.........
,\
.
...... D
'\
ii
'
If R be the amplitude of the resultant of N interfering waves, then from above phasor diagram, on extracting triangles DAG and· DAB, we get following figures to be used for evaluation of R . 0
G
,1\
~ I o,i'.&!
~
To find the total wave at a given point at·an angle 8 on the screen,. we need to add up the N individual waves. The Procedure is the same as in the N = 2 case, except that now we simply have more terms in the sum. If a_ be the amplitude due to an individual source, then the equations of waves interfering at the point P' are given by
\
y
B.2
L.)' .. 2
:t11' ' 2
r ''I
.
'
:
~I
'" ~
/i:1\
• .
A
.
2
/ 2 I2 \
/ I
'' ' A'
''
1
: '' I
' '''
\
''
\
'' ''
'B.
M
I+ .2. +t+ .2. +I
2
2
DAG and DAB are isosceles triangles so for triangle OAG , we have
¾=rsin(~$)
... (1)
Triangle OA~, we have ½=rsin(½)
 . . .. (2)
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Dividing (1) and (2), we get R
sm(~)
a
sm(½)
/ [ (1:~.,m:Hl ClC'lJ]1
~
y=a1~ e'·'
~
y=alm (e )
~
yalm e
So, the resultant amplitude R is given by ... (3)
R=[s:ntT]] If IR be the resultant mtensity, then IR =R2
~
[
,., [,{~',.m)J[,{',.m_,{½) ;!) _e{1¥)]]
_ [ {•••
.2 5,
N
2,
2
However, one exception to this is when
!2
0
2:_3,c7t5'1t31t
2
4
42
2"
'"
2
"
N=4
... (5)
$=Even Multiple of .':. N
·31t Sn n 3n n
2 4 42
!(a) l(O)
is also an integral
multiple of 1t, because the denominator in equation (4) is also zero.
So, f=m'rr.,where m'=0,1,2,3, .....
=>
$=(2m')1t
=>
$ = Even multiple of 1t
... (6)
So, from (5) and (6), we conclude that f, = 0, when $ = (2m); excluding $ = 0, 21t, 41t, 6,i, .....
·
Positions of Primary Maxima
i.e., $=(Even Multiple); , excluding Positions of Primary
a}
CONCEPTUAL NOTE(S) It is customary· 'not to deal With the resultant intensity
alone, buJ rather fo deal with the resultant inte!]sity relative to the maximuni intensity i.e.,· _!a_
Maxima (located at 0, 21t, 41t, 61t , ....)
•,

LOCATION OF SECONDARY MAXIMA(S)
derivative I, w.r.t. $(i.e., ~;) and then equating it to zero ..
(J,,__) =+.o lim (..!e...) =;
c)
?
~
has"a periodicity of 21t in
cl>
i.e., repeats itself for
"""
d)
This equation has to be solved numerically. However, for large N , the solutions of $ are generally very close to odd
e)
region. Just to make you understand, we are plotting
~
·
For N=2 (for two slits)~· ~e gBt,ZERO Secondary Maxima.. For N = 3 (for 3 slits), we get One Secondary Maxima. .
I, I,
The nUmber of .zeros between the ~fnain peaks· is, (N1) , where N is the number of slits used. The number of secondary maxima (little bumps) between. the main peaks is (N2), where N is the number of slits
used.
multiples of _2:_ excluding the values of q> = 21t±~, because N N these values will be lying well within the primary maxima Q
For N =4 (for 4 slits), we getTwo Secondary M~ima a~d so on. A point worth noting here is that the height of the se~ondary maxima (~ittle bui'nps).i.e., the bump sizes are symmetric around cj> = tt (o( in general. any, multiple of n:·).
_ _ Also,_ we know that since
=
... (1)
integratmultiples of 21t .
Ntan(½)=tan(~$)
(with$ for N=4 and N=8).
. (N$)]'
sin2
[
lim o.o Imax
d$ =>
I.
1, : = 1: = Nsin(½)
b)
So, dl, =0
·I
~
To find the locations of the secondary maxima (i.e., small bunips) we have to find the local maxima of I, by taking the
,lmax
2.24
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,, =,, [
s~~n~r]r
=1 t =1
d
Hence, the bump size is shortest at
4' =1t ,
because then
4, = re ,
due to which IR becomes the
least at 4' = n . Furthermore, the bump size grows as they get closer to the main peaks, as shown for various slits taken.
d

the denominator in the equation (1 ), will be having a maximum value at
t
I
i
D
(a) the distances from P where intensity reduces to zero. (b) the distances from P where next bright fringe are observed. (c) the ratio of intensities of bright fringes observed on the screen.
Single slit
./.__
Solution N=2
(a)
_ _._"~>
P=i,,o d 1'.=
2
cm
• C 3x10 8 7 Wavelength of the light 1'.==14 =5x10 m f 6xl0
pd= p(2a) D
(AS')(BS') BC
(190+5+5)(0.1) 10 P,P, =AP, AP1 =2 cm AP,
width p is given by
D
p can be determined, experimentally, by using a low power microscope. Knowing p, a and D value of 1,, can be calculated. CONCEPTUAL NOTE(S) Central spot, in case of Lloyd's single mirror is a dark one instead of being bright. This proves that there is a phase change of nradian when a transverse wave (light) is reflected from a denser medium.
Fringe width p = i,,~ Since, D=S'A =(190+5+5)=200 cm=2 cm, d=SS'=2 mm=2xl0" m p (5xl0')(2) 2x10" Number of fringes is
5xl04 m=.0.05 cm
N= P,P, =40
p
Illustration 30 The arrangement £or a mirror experiment is shown in the figure. S is a point source of frequency 6 x 10 14 Hz. D and C represent the two ends of a mirror placed horizontally and LOM represents the screen. Determine the position of the region where the
fringes will be visible and calculate the number of fringes.
Illustration 31
Two flat mirrors form an angle close to 180° . A source of light S is placed at equal distances b from the mirrors. Find the interval between adjacent interference bands on screen MN at a distance OA = a from the point of intersection of the mirror. The wavelength of the light wave is known and equal to 1,, • Shield C does not allow the light to pass directly from the source to the screen. N
5 cm 5 cm
cm+:
190
A
M
Solution
Fringes will be observed in the region between P, and P2 because the reflected rays lie only in this region. From similar triangles BDS' and S'P2 A
2.28
AP, = AS' '
BS'
BD
M
Solution
Fringe width is given by P=1'.D d where D=AB~a+b and d=S,S 2
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s,r
i
l
,  •
I
:
di 81 ____ 0' I
'
......
a.,..._ s
'!.t
\
1
..... .._
r;

A
_".,.
x=2µtcosr
82~
For maxima 2µtcosr =.n;>,
I+ D      . a
and for minima
In !1S15B, we have
2µtcosr=(2n1J.
.'!.=2b~ 2 2 d=2ba 2ba THEORY OF DIVISION OF AMPLITUDE REFLECTED LIGHT
If µ is the refractive index of material of film of thickness t, then path difference between the waves abc and abdef is 2µtcosr
Additional path difference due to reflection at denser medium ( at b) is .'.: 2
a
,. , 2
C
Illustration 32
A thick glass slab (µ = 1.5) is to be viewed in reflected white light. It is proposed to coat the slab with a thin layer of a material having refractive index 1.3 so that the wavelength 6000 A is suppressed. Find the minimum thickness of the coating required. Solution Optical path difference for the reflected light from coating and slab is l1x = 2µ!
For minimum intensity,
,. 2µ,t =2
m
n=l,2,3, ...
Obviously, the conditions of interference in reflected and transmitted lights are opposite to each other, therefore if the film appears dark in reflected light, it will appear bright in transmitted light and vice versa. With the use of white light, the colours visible in reflected light will be complementary to those visible in transmitted light, i.e., the colours absent in one system will be present in the other system; the sum of two constituting the white light.
p ;>,(a+b)
A.
TRANSMITTED LIGHT
In transmitted light system there is no phase difference or path difference due to reflection or transmission as all reflections take place from rarer medium. So, the effective path difference is
A/jBf Cl µ,~1.3 iDl
t=l:....= 6000 4µ 1 4xl.3
n
So, effective path difference is
,.
       . . ····3
.=>
x=2µtcosr+2
For maxima or constructive interference to take place, we have
,. =(2n) ,. 2 2 ,. 2µtcosr= (2n1) , n =1, 2, 3, ...

t=1154A
CONCEPTUAL NOTE(S) Both reflected rays (one from AB and the another from CD) get a phase change of 1t •

·
··
2µtcosr+ =>
Illustration 33 A parallel beam of white light falls on a thin film
2
4
For minima or destructive interference to take place 1,,
2µtcosr+ =>
2
,.
= (2n + 1)
2
2µtcosr=nt.., n=l,2,3, ...
whose refractive index is equal to  . The angle of 3
incidence i = 53° . What must be the minimum film thickness if the reflected light is to be coloured yellow ( ;>, of yellow = 0.6 µm) most intensively? Given tan53° =
i, 3
==================================== = 2.29
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Solution According to Snell's Law, we have sini
BASED ON INTERFERENCE (Solutions on page 2.86)
µ=.smr
=>
4 3
In a Yourig1s Double Slit Exp~riment.carried out in a liqµid Qf retr8.ctive index· µ = 1.3 , a thiri fi1in of air is formed in front
1.
sin(53°) sinr
37° •
4 5 sinr ·
5
of the lower slit as shown in the figiJre. If a maxima of third 0 order is f0rmed at the oriQin O ; find th_e 
4 0
(a) (b)
3 sinr =5 => r=37° From Figure (a) :
thickness of the air film. positions of the fourth·maxima. The wavel8ngth Of light i'n 'is
53'
=>
3
D
A. 0 _= 0:78 µrn ~ and·
d=1090.
.. sJ ~ Lt ·:__·. . _. ;~o
2
µ=1.3
I
,I
i s,U
I
D (a)
Path difference between 2 and 1 is
In YDSE , if light of wavelength thickness of ~•glass slab
ru; = 2(AD)
~
D 0
2.
~
~Airfilm_ ·
E,
I
=>
air
.
:sooo A is used, find the
(µ .=1,5)
.
~
;.,hlch should be piac~d . ' . , . ,..
befor~. the uppe;' the upper, sli( S1 so that the' c~r\tral
ru:1 = 2BDsecr = 2tsecr
maximum now lies at a point where ·sii bright fringe was 1 lying earl!er (bef_ore inserting the slab);:
Their optical path corresponding to ru:1 is 2µtsecr
From Figure (b) :
3.
A source S of wavelength A is kept directly behind the
~lit
S1 in a double slit apparatu$. Find.the phase.difference at a
E
point O which Is equidistant from · ~ 1 and S2
2
•
If D » d ,
what will be ~he phase difference ·at. P if a· liquid Of refractiv_e index; µ is filled (a) between the screen ar1d·the Slits? (b) , between the slits and th0 s9urce · s?
C
fi fils,j_·· . IP· OP=d/2
,~r ~io
1 D (b)
Path difference between 1 and 2 is given by
5
ru: 2 = ACsini = (Zttanr)sini =>
(ru:),,. =ru:1 ru:2 =2µtsecr2t(tanrHsini)
=>
.6x
=>
4 5 3 4 =2xxtx2xtxx,~ 3 4 45
L,\xnet
Since reflection takes place at the surface of denser medium, so phase difference between 1 and 2 is 1t • So, for constructive interference, we have 32t=~
15
=>
2
I= 151'. = 15x0.6 =O.l 4 µm 64 64
o
+1
'4.
_In Solar"cells, a smcdn Solar c;li'°{µ =:= 3.5) 18 cocite~·wiih .a
r
thin film "of silicon monoxide "SiO(µ =1~45) t9 ITliniITli_?:e
I
32 = 15 f
'1,__ _ _ D
I 5.
reflective lcisses from the s·urtace. 'Determine the mil1imum thickness .of· SiO that ProduC~s the le_ast reflectiOn ··a( ·a wavelength of 550' nm , near the .ceflt_re of the visible spectrum. • · · · A parallel beam of ·gr·een iighl' ·of waVelerlgth · 546 'nm passes 'through a slit of width 0.4 mm ., The tran?l"!litted: llgh_t is .. collected on a screen 40 cm .away;· Find the· distance between the two first order· ITlinima. Calculate the ·minimum. thickness Ct a soap bubble film (µ =1.33) that results in construt;;tive interference in the
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Wave Optics reflected light if the filiTI is illuminated With light whoSe ~ wavelength in free space is A= 600 nm.
7.
(b) (c) {d)
Monochromatic light of wavelength 5000 A is used in YDSE , with slit separation 1 mm , distance between screen and slits 1 m . If intensity at the two slits are, 11 =41 0 , 12 =1 0 , find ~
(a)
fringe width
(b)
distance of 5 minima from the central maxima on the
.
(c)
white light (e)
14.
th
screen.
intensity at y =
i
mm.
distance of the 1000 maxima.
(e)
distance of the 5000th maxima. 15.
8.
A. 1 = 400 nm and A= 700 nm . Find minimum order of A, which overlaps with A2
=6 µm , respectively.
Determine the phase difference (in radian) at the
16.
between x =0 and x = 1.5 µm . What then is the phase difference (in radian) at the origin between the radiation from S, and the radiation from S 2 ?
thickness 8000 A is placed in front of the lower slit, it is : observed that the intensity at a point P, 0.15 mm above the 1 1 central maxima does not change. Find the value of '1 ·
•
In a Young's double slit experiment using monochromatic light the fringe pattern shifts by a certain distance on the screen when a mica sheet of refractive index 1.6 and thickness 1.964 micron is introduced in the path of one of the interfering waves. The mica sheet is then removed and the distance between the slits and the screen is doubled. It is found that the distance between successive maxima (or minima) now is the same as the observed fringe shift upon the introduction of the mica sheet. Calculate the wavelength of the monochromatic light used in the experiment.
17.
(b)
18.
In a double slit pattern
(1' = 6000 A), the first order and
a particular reference point. If A is changed to 5500 A , find the position of zero order and tenth order fringes, other arrangements remaining the same. In YDSE, the two slits are separated by 0.1 mm and they are 0.5 m from the screen. The wavelength of light used is 5000 A . Find the distance between 7~ maxima and 11"' minima on the screen. 13.
•
Find the number of bright fringes between P and the central fringe.
In a Young's double slit experiment, two wavelengths of 500 nm and 700 nm were used. What is the minimum distance from the central maximum where their maximas ' coincide again? Take
~ =103 •
Symbols have their usual
meanings.
ray falls onto the plate perpendicularly. The refractive index
tenth order maxima fall at 12.50 mm and 14.75 mm from
Compare the intensity at a point P distant 10 mm from the central fringe where the intensity is 10
In Young's experiment a thin glass plate is placed in the path of one of the interfering rays. This causes the central , light band to shift into a position which was initially occupied
of the plate is 1.5. The wavelength is 6 x 107 m. What is the thickness of the plate?
In a Young's double slit experiment set up, the wavelength of light used is 546 nm . The distance of screen from slits is 1 metre. The slit separation is 0.3 mm . (a)
by the fifth bright band (not counting the central one). The
12.
In Young's double slit experiment setup with light of :
Suppose a slab of transparent material with thickness 1.5 µm and index of refraction µ = 1.5 is placed
origin between the radiation from s, and the radiation
(b)
•
wavelength A.= 6000 A, distance between the two slits is 2 mm and distance between the plane of slits and the screen is 2 m. The slits are of equal intensity. When a , sheet of glass of refractive index 1.5 (which permits only a , fraction T\ of the incident light to pass through) and
from 8 2
11.
Bichromatic light is used in YDSE having wavelengths ,
radiating waves in phase with each other of wavelength 400 nm . The sources are located on xaxis at x = 6.5 µm and x
10.
In a Young's double slit experiment, the slits are 2 mm apart and are illuminated with a mixture of two wavelengths A, = 750 nm and A. 2 = 900 nm . At what minimum distance :
S1 and S2 are two point sources of radiation that are
(a)
9.
The whole experiment is carried out in a medium of ' refractive index µ
from the common central bright fringe on a screen 2 m from the slits Will a bright fringe from one interterence , pattern coincide with a bright fringe from the other?
111
(d)
The (monochromaticf so\Jrce is· replaced by another (monochromatic) source of shorter wavelength The separation between the two slits is increased The monochromatic source is replaced by source of
19.
When a thin sheet of a transparent material of thickness 7.2 x 104 cm is introduced in the path of one of the interfering beams, the central fringe shift to a position occupied by the sixth bright fringe. If A= 6 x 10s cm, find the refractive index of the sheet.
DIFFRACTION : INTRODUCTION & CLASSIFICATION
When light waves pass through a small aperture, an interference pattern is observed rather than a sharp spot of light cast by the aperture. This shows that light spreads in various directions beyond the aperture into regions where a shadow would be expected if light travelled in straight lines.
What is the effect on the interference fringes in a VOSE due to each of the following operations? (a)
The screen is. moved away from the plane of the slits
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rIIJr0
+11).~1   d
  
!______   
t +    D +1
Uniform intensity distribution
Light passing through two slits does not produce two distinct bright areas on a screen. Instead, an interference pattern is observed on the screen which shows that the light has deviated from a straightline path and has entered the otherwise shadowed region. Other waves, such as sound waves and water waves, also have this property of being able to bend around corners. This deviation of light from a straightline path is called diffraction. Diffraction ·results from the interference of light from many coherent sources. In principle, the intensity of a diffraction pattern at a given point in space can be computed using Huygens' principle, where each point on the wavefront at the source of the pattern is taken to be a point source. The phenomenon of bending of light around the comers of an obstacle/ aperture of the size of the wave.length of light is called diffraction. The phenomenon resulting from the superposition of secondary wavelets originating from different parts of the same wave front is define as diffraction of light. Diffraction is the characteristic of all types of waves. Greater the wave length of wave higher will be it's degree of diffraction.
phase of the light at each point in the aperture is the same. This can be achieved experimentally either by placing. the observing screen at a large distance from the aperture or by using a converging lens to focus parallel rays on the screen, as in Figure. Note that a bright fringe is observed along the axis at 8 = 0, with aitemating bright and dark fringes on either side of the central bright fringe. · Common examples : Diffraction at single slit, double slit and diffraction grating.
10
11:.' Incoming wave
B.
Screen
FRESNEL'S DIFFRACTION
When the observing screen is placed at a finite distance from the slit and no lens is used to focus parallel rays, the observed pattern is called a Fresnel Diffraction Pattern. Fresnel diffraction is rather complex to treat quantitatively. Common examples : Diffraction at a straight edge, narrow wire or small opaque disc etc.
?,
I
o~  
!j I+
D +1
Slit
Screen
A Fresnel diffraction pattern of a single slit is observed when the incident rays are not parallel and the observing screen is at a finite distance
CONCEPTUAL NOTE($) Diffraction, can be regarded as a consequence of interference
from many, coherent wave sources. In other words, t_he phenomena of diffraction and interference are basically ~jv_alc.e_nt_._ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
from the slit.
1
FRAUNHOFER DIFFRACTION AT A SINGLE SLIT
Diffraction phenomena are usually classified as being one of two types, which are named after the men who first explained them. The first type is called Fraunhofer Diffraction and the second is called Fresnel's Diffraction. A.
FRAUNHOFER DIFFRACTION
This occurs when the rays reaching a point are approximately parallel i.e. when both the source and screen are effectively at infinite distance from the diffracting device. In this case, the incident light is a plane wave so that the
Consider that a monochromatic source of light S , emitting light waves of wavelength 1,., is placed at the principal focus of the convex lens L,. A parallel beam of light i.e., a plane wavefront gets incident on a narrow slit AB of width d as shown in figure .. The diffraction pattern is obtained on a screen lying at a distance D from the slit and at the focal plane of the convex lens L2 •
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Wave Optics SCREEN
SLIT
amount equal to the path difference
L,
J'l=,..iPt 
y
====::::c=~ ot l+D+< According to rectilinear propagation of light, a bright image of the slit is expected at the centre O of the screen. But in practice, we get a diffraction pattern i.e., a central maximum at the centre O flanked by a number of dark and bright,; fringes called secondary maxima and minima on either sid~ of the point O . The diffraction pattern is obtained on the screen, which lies at the focal plane of the convex lens L2 • It is found that (i)
the width of the central maximum is twice as that of a secondary maximum and (ii) the intensity of the secondary maxima goes on decreasing with the order of maxima. These observations are explained on the basis of the phenomenon of diffraction using the following mathematical treatment.
Consider Frawlhofer diffraction by a single slit as shown in Figure. Important features of this problem can be deduced by examining waves coming from various portions of the slit. According to Huygens' principle, each portion of the slit acts as a source of waves. Hence, light from one portion of the. slit can interfere with light from another portion, and the resultant intensity on the screen will depend on the direction 0 .
}ine, where d is
the width of the slit. Similarly, the path difference between waves 2 and 4 is also equal to
(f}ine. If this path
difference is exactly one half of a wavelength (corresponding to a phase difference of 180° ), the two waves canCel ·each other and destructive interference results. This is tr'ue, in fact, for any two waves that originate at points separated by half the slit width, since the phase difference between two such points is 180° . Therefore, waves from the upper half of the slit interfere destructively with waves from the lower half of the slit. when
d . ,_ sin 8 =2
=>
2
. 8
Slll
,_ =
d Similarly, destructive interference (minima) occurs when the 3
5
path difference (r!.)sin8 equals 1,., ', 21,., ', etc. These 2 2 2 points occur at progressively larger values of 8. Therefore, the general condition for destructive interference is
·e =m,_ sm d where
EXPLANATION & MATHEMATICAL TREATMENT
(f
(m=±l,±2,±3, ... )
... (1)
1ml,; '!_,_
Equation (1) gives the values of 8 for which the diffraction pattern has zero intensity. However, it tells us nothing about the variation in intensity along the screen. The general features of the intensity distribution along the screen are shown in Figure. A broad central bright fringe is observed, flanked by much weaker alternating maxima. The central bright fringe corresponds to those points opposite the slit for which the path difference is zero, or 8 = 0 . All waves originating from the slit reach this region in phase, hence constructive interference results. The various dark fringes (points of zero intensity) occur at the values of 8 that satisfy equation (1). The positions of the weaker maxima lie approximately halfway between the dark fringes. Note that the central bright fringe is twice as wide as the weaker maxima. Angular width of central maxima is
2 ' and. width of central d
maxima is Z~D , where D is the distance of the screen from Diffraction of light by a narrow slit of width d. Each portion of the slit acts as a point source of waves. The path difference between rays 1 and 3 or between rays 2 and 4 is equal to (d/2) sin0
To analyze the resultant diffraction pattern, it is convenient to divide the slit in two halves as in Figure. All the waves that origi:rate from the slit are in phase. Consider waves 1 and 3, which originate from the bottom and center of the slit, respectively. Wave 1 travels farther than wave 3 by an
the slit. The intensity distribution of the diffraction pattern is quite different from the interference pattern produced due to superposition of light from two coherent sources. The point 0 on the central ax:s is the brightest. The angular position (8) of n"' diffraction minima is given by dsin8 = n/,. n=l,2,3,4, ...
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Y,
  ·
I,
Y, ""Rine  'i.ld
sine  o ...,_ I, +:d'lil ntralaxis ___ frol~~~_"_""' _""'_':~:,>ln't+ensity (I) ce .t lnte nsity
y.
sin0=}Jd
Y2
sine= 2A/d
Central
Secondary maxima
sine= 2/Jd
Sec nd
First __!_,__
121
2l. er
),_
d
).
0
d
D+
0=30° (b) At C, 0=0°,soweget
and ;\.=500nm=5x104 mm So, fringe width is given by
l!vc = dsin~(µl)t
4
P= ;\.D = (5x10 )(10') mm=! mm d (1.5) 3 Now, as the point A is at the third maxima
=>
=>
Problem 3 Light of wavelength ;\. = 500 nm falls on two narrow slits placed a distance d = 50 x 104 cm apart, at an angle ~ = 30° relative to the slits shown in figure. ON the lower slit a transparent slab of thickness 0.1 run and refractive
~
is placed. The interference pattern is observed on 2 a screen at a distance D = 2 m from the slits.
l!vc=(50x10'i(½)(¾1)co.1)
=> l!vc.= 0.0250.05 = 0.025 mm Substituting, l!vc = n;\., we get n l!vc 0.025 50 A. 500x10~
OA=3P=3(½)=1 mm
(ii) If the gap between I, and L, is redu,~ed, d will decrease. Hence, the fringe width p will increase or the distance. OA will increase.
i~dex
Hence, at C there will be maxima. Therefore the· order
(c)
of minima closest to the C are 49 . Number of fringes shifted upwards is N=(µ1)1 ;\.
( i2 1)co,1J ~~.,....,..... 100 500x10~
Problem 4 In a modified Young's double slit experiment, a monochromatic uniform and parallel beam of light of
wavelength 6000
A ~d intensity
(1;) Wm_, is incident
normally on two apertures A and B 0.002 m respectively. A perfectly thickness 2000 A and refractive wavelength of 6000 A is placed in (shown in figure). (a) Locate the position of the central maxima. (b) Find the order of minima closest to centre C of (c)
screen. How many fringes will pass over C, if we remove the transparent slab from the lower slit?
Solution (a) Path difference is given by l!vc = dsin~+ dsin0(µl)t
~ 2.42
C p

of radii 0.001 m and transparent film of index 1.5 for the front of aperture A
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Wave Optics Calculate the power (in W) received at the focal spot F of the lens. The lens is symmetrically placed with respect to the apertures. Assume that 10% of the power received by each aperture goes in the original direction and is brought to the focal spot.
Solution
Applying the lens formula _!:. _ _!:. = l_, we get V
2
PA= I( rrr,;} =: (it)(0.001) = 10_, W
{: I=:}
10
~
v=lOcm
~
V lQ m===2 u 5
i.e. two virtual sources are formed with distance between them 1
Power received by aperture_ B is given by
P, = I( itr;) =
2
(it)(0.002) = 4x 10C5 W
d=0.Smm
"
Lens
Only 10% of PA· and P, goes to the original direction, so
s,
'' ''
Portion of PA going to _original direction is P1 = 10' W Portion of P, going to original direction is P, = 4 x 10' W = (1.51)(2000) = 1000 A
F. "dth"p nngew1
Now, resultant power at the focal point is given by
!
10 cm       D. 
1'.(D+lO) d
Fringes are observed between the region P and Q (waves interfere in this region only), where L d =D 10
P = P1 + P, + 2JP1 P, cos~
~
a,:
~
i+
~=(:")1'x= 6~; x1000=i 0
•L
0
5
Corresponding phase difference is given by
P,i"
1i
s
'' s'' '
Path difference created by slab is given by 1;x =.(µl)t
f
1 1 1 +=v 5 10
Solution Power received by aperture A is given by
U
P = 10< + 4 xlO' + 2J(10., = 4000 A is observed to be 600 times the fringe width in
the screen of figure (b) using the wavelength :>., = 6000 A . If D (as shown in figure) is 1 m then find the separation between the coherent sources S1 and S2 • Given that
d>31c,_
~o)
2
/=20cm
Since source lies in focal plane of lens. So, all the emergent rays will be parallel. So, 1 . d/2 d tana.===:;::sma 20 40 400
SI'
S1 S2 =d
sI
S,S,=d
}o
s,I
Screen
Initial path difference (fil)ini.., = dsinu
s,I
I
I "" ~
(a)
(b)
0 ;,• en
"
Solution
Had the screen been perpendicular to S2 P, then P and Q' would have been the positions of first and second minima Qast two). .,
'''
s)
'' ' '' ~o ::,·"=
d=6x104m=0.6mm
D'
For maximum intensity, we have path difference to be an even multiple of
!:: , so 2
l!.x = (2n)!: = nJ.. 2
Substituting the values, we get
Problem 8
In a Young's double slit experiment setup source S of wavelength 5000
A
illuminates two slits S1 and S2 ,
which act as two coherent sources. The source S oscillates about its shown position according to the equation y=O.Ssin(nt), where y is in millimetres and t in seconds. y
Is, Lx ~        p
0.5sin(1tt)+ 0.25 = 0.5n
=>
. ( rt I) sm
0.5n  0.25
=>
sin(1tl)=0.5
=>
nt =
=>
1 I=  = 0.167 s
0.5 For minimum value of t, we have n =1 1t 6
6
s Problem 9
1 mm
i s i+1mm•1• (a)
2
2mm,~.,,
Locate the position of the central maxima as a function of time.
Two parallel beams of light P an\i Q (separation d ) containing radiations of wavelengths 4000 A and 5000 A (which are mutually coherent in each wavelength separately) are incident normally on a prism as shown in figure. The refractive index of the prism as a function of 2.45
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,.
wavelength is given by the relation, µ(:\.) = 1.20 +..!', where
(b) For, 4000 A condition of TIR is just satisfied_ Hence, it will emerge from AC, just grazingly_
A is in A and b is posiHve constant. The value of b is such that the condition for total reflection at the face AC is just satisfied for one wavelength and is not satisfied for the other.
A
,, /
c ,/
A 8
p
r
_ sin8=0.B
 S;;,90°C
For4000 A
B
d
~
,
C
So, deviation for 4000 A is given by
go• B
~
C
6.,.,A =90i=90sin1 (0.8)~37°
For 5000 A , we have
(a) Find the value of b (b) Find .the deviation of the beams transmitted through the face AC (c) A convergent lens is used to bring these transmitted beams into focus. If the intensities of the upper and
8xl05 b µ=1.2+,=1.2+, =1.232 '(5000) A
the lower beams immediately after transmission from
the face AC, are 4I and I respectively, find the resultant intensity at the focus.
Solution
(a)
Total internal reflection (TIR) will take place first for those wavelength for which critical angle is small or µ is large.
Applying, µ
From the given expression of µ, it is more for the wavelength for which value of :\. is less_
,,
,
,,
,,
i=C for 4000A.
~
9=C sin9=sinC
~
(c)
C
Since sin 9 = 08 and sinC =_!_,so we get µ 0.8 = _!_ µ
~
0.8
The intensity of the upper beam (4000 A) and lower beam (5000 A) after transmission are 4I and I respectively, then I, =I1 +I, +2.p;i; cos~
Since no phase change takes place for the waves refracted from the lens, so qi = 0° .
~
r.=4I+I+2.J(4Ilicos(0°)
~
I• =9I
(for 4000 A.) Problem 10
1
b 1.20+(4000)2
Solving this equation, we get b = 8_0 x 10' (A)'
2.46
0.8
65000 A =i.,,,im
a}
b)
c)
sin(60°) = ,J3 sinr1
=>
sinr1
=>
'1 = 300
=>
2x3.6
A
1
=2 =A
r2 =Ar1 =30°30°=0°
Therefore, ray of light falls normally on the face AC and angle of emergence e = i2 = 0° .
648 7.2
CONCEPTUAL NOTE(S) For a wave (whether it is sou_nd or electromagnetic), a medium is denser or rarer is decided from the speed of wave in that medium. In denser medium speed of wave is less. For example, water is rarer for sound, while denser for light compared to air because speed of sound in Water is more than in air, while speed of light is less. In transmission/refraction, no·phase change takes place. In reflection, there is a change of phase of n when it is reflected by a denser medium and phase change is zero if it is reflected by a rarer medium. If two waves in_ phase interfere having a path difference of ax; then condition of maximum intensity would be
Ax=nA., n=_0,1,2, ... But if two waves, which are already out of phase (a phase difference of n ) interfere with path difference ax, thf;ln Condition of maximum intensity will be Ax·= n=1, 2, .... ~ 
=>
Since, r1 + r2
=A
t. =nm
=
(a) the angle of emergence. (b) the minimum value of thickness t of the coated film  so that the intensity of the emergent ray is maximum. Solution (a) Applying Snell's Law, we get sini1 = µsinr1
A 1 2 B
µ, = 2.2
(nf}·,
 ·
(b) Multiple reflection occurs between surfaces of film: Intensity will be maximum if interference takes place in the transmitted wave. For maximum thickness, we have
1'x=2µt=A where t is the thickness of coated film
=>
t =__?:_ =~ =125 nm 2µ
2x2.2
Problem 12 A vessel ABCD of 10 cm width has two small slits S1
and S2 sealed with identical glass plates of equal tnickness. The distance between the slits is 0.8 mm. POQ is the line perpendicular to_ the plane AB and passing through O , the middle point of S1 and S2 • A monochromatic light source is kept at S, 40 cm below: P and 2 m from the vessel, to illuminate _the slits as shown in the figure alongside.
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Optics & Modem Physics
Calculate the position of the central bright fringe on the other wall CD with respect to the line OQ. •
•
L
•
(b) Now, a liquid is poured into the vessel and filled upto OQ • The central bright fringe is found to ~e at Q . Calculate the refractive index of the liquid.
"1·
0
Llx,  Llx, = 0
0
Lll'.2
0
,,,
dsin0 = 0.16 (0.8)sin0 = 0.16
D
0
sin8= 0.16 =l. 0.8 5
t fo  a
0
1 tan8= .fiA
0
. 0 ::::1· sm
s,
40cm
S1
I.
5 . . y2 1 tan8R:"sin8==D, 5
2 m +t
=>
~=(~)n
_!. Ignore dispersion] 3
Now, I(~)=Icos'(t)
Solution
Given A=600 nm=6x10·' m,
=>
l=Imaxcos '(13n)
=>
I=~I
d=0.45 mm=0.45x10°" m and D=l.5 m
SI1 (c)
s,8
 
For maximum intensity at O , we have
M=nA.,where n=1,2,3, ..... .
Thickness of glass sheet, I= 10.4 µm = 10.4 x 10 m
=¾
And refractive index of glass sheet, µg
(a)
=>
6x 6x 6x , , ...... and soon A.=l,
=>
6x=(~1)(10.4x10'm) 4/3
=>
6x =(~1)(10.4xl0 nm)=1300 nm
=1.5
Let central maximum is obtained at a distance y below point 0.
=>
4=
At O ,path difference is 6x=6x2 =(~:1}
1
Refractive index of the medium, µm
6
6x1 =S1 PS2 P= ~
Path difference due to glass sheet is given by 6x,=(~:1}
2 3
3
4/3
So, maximum intensity will be corresponding to
=>
A=l 300 nm, 1300 nm, 1300 nm, 1300 nm, ... 2 3 4 , = 1300 nm , 650 mm, 433.33 nm, 325 nm, ....
The wavelength in the range 400 nm to 700 nm are 650 nm and 433.33 nm .
Net path difference will be zero, when we have Lll'.1
= M2 2.49 r1
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Advanced JEE Physics Problem 14 In Young's experiment, the source is red light of wavelength 7 x 10' m. When a thin glass plate of refractive index 1.5 at this wavelength is put in the path of one of the interfering beams, the central bright fringe shifts by 10" m. to the position previously occupied by the 51h bright fringe. Find the thickness of the plate. When the source is now changed to green light of wavelength 5 x 10' m , the central fringe shifts to a position initially occupied by the 6th bright fringe due to red light. Find the refractive index of glass for green light. Also estimate the change in fringe width due to the change in wavelength. Solution Path difference due to the glass slab,
t.x = (µ1)1 = (1.5l)t = 0.51
Problem 15 A double slit apparatus is immersed in a liquid of refractive index 1.33. It has slit separation of 1 mm and distance between the plane of slits and screen is 1.33 m. The slits are illuminated by a parallel beam of light whose wavelength in air is 6300 A. . (i) Calculate the fringe width. (ii) One of the slits of the apparatus is covered by a thin glass sheet of refractive index 1.53. Find the smallest thickness of the sheet to bring the adjacent minimum as the axis. • Solution Given
(i)
Due to this slab, 5 red fringes have been shifted upwards. So, we have l!u=5Ared
0.51=(5)(7x10'm)
~
I = thickness of glass slab = 7 x 10' m
Let µ' be the refractive index for green light, then t.x'=(µ'1)1 Now the shifting is of 6 fringes of red light. So, we have
d=lnun,
D=l.33 m
and
i,,=6300 A Wavelength of light in the given liquid is 6300 1,,' = 2: = A "' 4737 A = 4737 x 1010 m µ 1.33 ~
~
µ=1.33,
'dth . Frmgewi ,
1,,'D P=d
~
p (4737x1010 m)(1.33 m) (lxlO'm)
~
Pc"6,3x10 4 m=0.63 mm
(ii) Let t be the thickness of the glass slab.
t.x' = 61,,red ~
(µ'1)1=61,,..,

( 'l)= (6)(7x10') µ 7x10'
~
µ'=1.6
t+o 06 ·
Since the shifting of 5 bright fringes was equal to 10" m ~
sp.., = 10°" m, where p is the Fringe width
~
10" Pre, = m=0.2x10" m 5
A P'green
A.red
=A
i,,,,..,. =(0.2x10")(5xl0') 7 X 107
1'red }..red
~
P,.... =0.143x10"m
~
t.p = P""" p.., =(0.1430.2)x10" m
~
t,p =~5.71 X 10' ID
=
2.50
't,.x
= 0.151
Now, for the intensity to be minimum at 0, this path 1,,' difference should be equal to 2
P,,..,. = ,,,,..,. • Pred
~
153 t.x = ( µglas, 1)1 = ( 1)1 µliq•ld 1.33 ~
Nowsince P=i,,o d ~ poci,, ~
Path difference due to glass slab at centre O is given by
1,,'
~
b.x=2
~
o.1st =
~
t = 15790 A = 1.579 µm
4737 2
A
Problem 16 A point source S emitting light of wavelength 600 nm is placed at a very small height h ;;_bove a flat reflecting surface AB (shown in figure). The intensity of the reflected light is 36% of the incident intensity. Interference
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fringes are observed on a screen placed parallel to the reflecting surface at a very large distance D from it.
n~:
''
l' h A
(a)
L~
•S •S
,:::!
D
•S'
'
ts'
•S'
Initial
Final
Since the ray is reflected from the surface of a denser
'
What is the shape of the interference fringes on the
medium, so it suffers an additional path change of .?:_ or 2
screen?
a phase change of
(b) Calculate the ratio of the minimum to the maximum
(c)
!;. .
p
For maximum at P, path difference equals nA .
intensities in the interference fringes formed near the point P (shown in the figure).
If
H the intensity at point P corresponds to a maximum,
path difference of
calculate the minimum distance through which the reflecting surface AB should be shifted so that the intensity at P again becomes maximum.
Solution (a) Since there is symmetry about the line SP, so the shape
it .
AB is shifted by x, then this will cause an additional
2(x~) (for object and its image
taken as coherent sources). Since reflection takes place at surface of denser medium, so this will produce an ·
additional phase change of
it
,.
or a path change of  . 2
So, we get
of the interference fringes will be circular.
2(x1)=ni..
(b) Intensity of light reaching on the screen directly from the source 11 =I, (say) and intensity of light reaching on the screen after reflecting from the mirror is I,= 36% of I,= 0.3610 •
!i_=_I_,=112
0.3610
0.36
=>
x=(n+1)
,. where n=0, 1, 2, 3,....
2
Now, to get minimum value of x, n must be minimum i.e., n = 0
1 JF,=0 6
(.2..1)' 0.6 1 )' ( +1 0.6 . (c)
=>
2xi..=ni.. 2x =(n + l)i..
1 16
,.
=>
X=
=>
X=
2
600 2
=300 nm
Initially path difference at P between two waves reaching from S and S' is as shown.
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This section conti.'ms Single Correct Choice Type Questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is c~rrect. 1.
In ail interference pattern produced by two identical slits, the intensity at the site of the central maximum is I . Th'7 intensity at the same spot when either of two
slits is closed is I (A) 2 (C)
I
2,/2
(B) (D)
(B)
wavefront at a later or an earlier instant
(C) (D)
3.
6.
Longitudinal waves do not exhibit (A) refraction (B). reflection (C) _diffraction · (D) polarization
7.
The idea of the quantum nature of light has emerged in an attempt to explain (A) interference (B) diffraction (C) polarization _ (D) radiation spectrum of a black body
8.
In the spectrum of light of a luminous heavenly body the wavelength of a spectral line is measured to be 4747 A while actual wavelength of th_e line is 4700 A. The relative velocity of the heavenly body with respect to earth will be (velocity of light is 3x10 8 ms·') (A) 3 x 105 ms·1 mov;,.;g towards the earth (B) 3x10 5 ms·1 moving away from the earth (C) 3x10 6 ms·' moving towards the earth (D) . 3 x 10 6 ms·' moving away from the earth
9.
A grating has 5000 lines cm·'. The maximum order
I
,/2
In YDSE bichromatic light of wavelengths 400 run and 560 run are used. The distance between the slits is 0.1 mm arid the distance between the plane of the slits and the screen is 1 m . The minimum tjistance bet_ween two·successive regions of complete darkness is (A) 4 mm (B) 5.6 mm (C) 14 mm (D) 28 mm_ · In the ideal doubleslit experiment, when a glassplate (refractive index 1.5) of thickness I is introduced in the path of one of the interfering beams (wavelength 1'. ), the intensity at the position where the central maximum ocqrrred previously remains unchanged. The minimum thickness of the glassplate is
(A) (C) 4.
21'.
,.
(B)
21'. 3
(D) ,.
3
Young's double slit experiment is carried out by using green, red and blue light, one color at a time. The fringe
widths recorded are (A) Pa >P, >P, (C) P,>P,>Pa 5.
Pa, P,
and P, respectively. Then, (B) P, >P, >P,
(D)
is used to'determine the velocity of light is used to explain polarization of light
I 4
I
2.
is a geometrical method to find" the position of a
P,>Pa>P,
Huygens' conception of seco11dary Waves
visible with·wavelength 6000 A (A) 2 (B) 3 (C) 4 (D) 0 10. A beam of monochromatic light enters from vacuum into a medium of refractive index n . The ratio of the wavelengths of the incident and refracted waves is (A) n:1 (B) 1:n 2 (C)· n :1 (D) 1:n'
(A)
helps us to find the focal length of a thick lens
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Wave Optics 11. In Young's double slit experiment, 62fringes are seen in visible region for sodium light of wavelength 5893 A. If violet light of wavelength 4358 A is used in place of sodium light, then number of fringes seen will be (A) 54 (B) 64 (C) 74 (D) 84 12. Monochromatic
light
of
wavelength
0.01 mm
20.
21.
(D) 1.0mm
13. Air has refractive index 1.0003. The thickness of an air
column, which will have one more wavelength of yellow light ( 6000 A) than in the same thickness of vacuum is (A) 2mm (C) 2m
Ray optics is valid when characteristic dimensions are (A) of the same order as the wavelength of light (B) much smaller than the wavelength of light (C) much larger than the wavelength of light (D) of the order of 1 mm
of the oil film should be of the order of (A) 1 cm (B) 10 A
(C)
24. The deflection of light in a gravitational field was
predicted first by (A) Einstein (C) Max Planck
16. Though quantum theory of light can explain a number of phenomena observed with light, it is necessary to
retain the wave nature of light to explain the phenomenon of
(A)
(B) (C) (D)
25.
photoelectric effect diffraction compton effect black body radiation
17. Which of the following cannot be polarized?
(A) (C)
Radio wave Infrared radiation
(D) 10000 A
white light is observed through a red filter. The pattern seen is (A) a red cross on a black background (B) a blue cross on a red background (C) a red cross on a blue background (D) a black cross on a blue background
How many.colours are possible, then (A) 3 (B) 1
(D) None of these
5000 A
23. The blue cross on a white background illuminated with
15. There is a wavelength corresponding to each colour.
7
In Young's double slit experiment, carried out with light
22. In order that a thin film of oil floating on the surface of water shows colours due to interference, the thickness
(B) 2 cm (D) 2km
14. Two coherent monochromatic light beams of intensities I and 41 are superposed. The maximum and minimum possible intensities in the resulting beam are (A) 41 and I (B) SI and 31 (C) 91 and I (D) 91 and 3I
(C)
scattering is reduced at noon
of wavelength 1,. = 5000 A, the distance between the slits is 0.2 mm and the screen is at 200 cm from the slits. The central maximum is at x = 0. The third maximum (taking the central maximum as zeroth maximum) will be at x equal to (A) 1.67 cm (B) 1.5 cm (C) 0.5 cm (D) 5.0 cm
5000 A
illuminates a pair of slits 1 mm apart. The separation of bright fringes in the inteiference pattern formed on a screen 2 m away is (A) 0.25 mm (B) 0.1 mm
(C)
(D)
(B) Xrays (D) Sound waves in air
18. A blue object on a white background when seen through a blue filter will appear (A) blue on a white background (B) black on a blue background (C) blue on red background (D) invisible 19. illumination of the sun at noon is maximum because (A) the sun is nearer to the earth at noon (B) rays are incident almost normally (C) refraction of light is minimum at noon
(B) Newton (D) Maxwell
Both the particle and wave aspects of the wave aspects of light appear to be used in (A) photoelectric effect (B) gamma emission (C) interference (D) classical mechanics
26. At sunset, the sun seems to be (A) higher than it really is (B) lower than it really is (C) exactly where it really is (D) lower than it would be at sunrise 27.
In Huygens' wave theory, the locus of all the points in the same state of vibration .is called a (A) half period zone (B) vibrator (C) wavefront (D) ray
28. In Young's experiment, monochromatic light is used to illuminate the two slits A and B. Interference fringes are observed on a screen placed in front of the slits. 2.53
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Now if a thin glass plate is placed normally in the path of the beam coming from the slit
33. Two coherent point sources s1 and s2 vibrating in
phase emit light of wavelength ). . The separation between the sources is 2A . The smallest distance from s2 on a line passing through s2 and perpendicular to s1s2 , where a minimum of intensity occurs is (A)
(CJ (A) The fringes will disappear (BJ The fringe width will increase (CJ The fringe width will increase (D) There will be no change in the fringe width but· the pattern shifts ·
(CJ
.!1 2
(DJ
0
¼I,
30. In the diagram, CP represent a wavefront and AO and BP, the corresponding two rays. Find the condition on 0 for constructive interference at P between the ray BP and reflected ray OP
0
(BJ
12
4
).
(DJ
,2
(A)
(CJ
1.75 1.25
(BJ (D)
(A) B
(CJ
sec8cos8=a
(BJ ).
).
cos8=4d
(CJ
41. (D) sec9cos8=d
31. In Young's doubleslit experiment the separation between the slits is doubled and the distance between the slit _and the screen is halved. The fringewidth becomes (A) onefourth (B) half (CJ double (DJ quadruple 32. In Young's double slit experiment, the separation
between the slits is halved and the distance between the slits arid the screen is doubled. The fringe width is (A) unchanged (B) halved (CJ doubled (D) quadrupled
1.50 1.00
s, d
3).
(¾}, The
35. Consider a usual setup of Young's double slit experiment with slits of equal intensity as shown in the figure. Take O as origin and the Y axis as indicated. If . . . .b 1.D d 1.D verage mtens1ty etween y1 =  an y 2 = + 4d 4d equals n times the intensity of maxima, then n equals (take average over phase difference)
t
cos8=2d
4
refractive index of second plate is
f
(A)
3).
of same thickness, the shift of fringes is
R
A
151.
34. In a Young's double slit experiment, the fringes are displaced by a distance x when: a glass plate of refractive index 1.5 is introduced in· the path of one of the beaII1S. When this plate is replaced by another plate
29. When an unpolarized light of intensity ! 0 is incident on
a polarizing sheet, the intensity of the light which does • not get transmitted is (A) ZERO (B) ! 0
n
½(1+;) (1+;)
y
s,
0
D
(B) (DJ
2(1+;) ½(1;)
36. A plate of thickness ! made of a material of refractive index µ is placed in front of one of the slits in a double slit experiment. What should be the minimum thickness ! which will make the intensity at the centre of the fringe pattern zero? (A)
(CJ
37.
(µ1)~
(B)
(µ1)1.
(DJ
(µ1)
).
2(µ1) Two polaroids are placed in the path of unpolarized beam of intensity ! 0 such that no light is emitted from the second polaroid. If a third polaroid whose polarization axis makes an angle 0 with the polarization axis of first polaroid, is placed between
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Wave Optics these polaroids then the intensity of light emerging from the last polaroid will be (A)
(~}in' 20
(B)
(~}in' 20
(C)
(; )cos' e
(D) I, cos' e
(¾) of the maximum intensity. Angular position of this point is
sin
(C)
sin'G'a)
(~)
sin (:a)
(D)
sm
(A) (B) (C)
. 1(1,.) 4d
(A)
(2n + 1)?:
(B)
(C)
(2n +1).1:
(D) (2n + 1)_!':_
2
8
(2n + 1)?:
16
If two slightly different wavelengths are present in the the sharpness of fringes will be more than the case when only one wavelength is present
the sharpness of fringes will decrease as we move away from the central fringe the central fringe will be white the central fringe will be dark
41. Two identical coherent sources placed on a diameter of a circle of radius R at separation x( « R) symmetrically about the centre of the circle. The sources
emit identical wavelength A each. The number of points on the circle with maximum intensity is (x =SA) (A) 20 (B) 22
(C)
24
(D)
(D) 26
Laser is (A) intense, coherent and monochromatic (B) only intense and coherent (C) only coherent and monochromatic (D) only intense and monochromatic 43. Imagine a hypothetical convex lens material which can transmit all the following radiation. This lens will have minimum focal length for (A) ultraviolet rays (B) infrared rays (C) radio waves (D) Xrays 42.
44. A star emitting yellow light starts accelerating towards earth, its colour as seen from the earth will
many wavelengths uncoordinated wavelengths coordinated waves of exactly wavelength divergentbeams
the
same
47. The wavelength. of light observed on the earth, from a moving star is found to decrease by 0.05%. Relative to the earth the star is
(A) (B) (C) (D)
4
light used in Young's doubleslit experiment, then
(C) (D)
refraction oflight polarization oflight dispersion oflight
consists of
1
(B)
39. In the Young's double slit experiment using a monochromatic light of wavelength 1,. , the path difference (in terms of an integer n ) corresponding to any point having half the peak intensity is
(B)
(B) (C) (D)
46. Laser light is considered to be coherent, because it 1
(A)
(A)
tum gradually red tum suddenly red remains·same tum gradually blue
45. The transverse nature of light is shown by (A) interference of light
38. In Young's double slit experiment intensity at a point is
40.
(A) (B) (C) (D)
Moving away with a velocity of 1.5x105 ms1 Coming closer with a velocity of 1.5 x 10 5 ms1 Moving away with a velocity of 1.5 x 10 4 ms1 Coming closer with a velocity of 1.5 x 10 4 ms1
48. A beam of electron is used in an YDSE experiment. The slit width is d . When the velocity of electron is increased, then (A) No interference is observed (B) Fringe width increases (C) Fringe width decreases . (D) Fringe width remains same 49. The ratio of the intensity at the centre of a bright fringe to the intensity at a point onequarter of the distance between two fringe from the centre is 1 (A) 2 (B) 2
(C)
(Q) 16
4
50. The ratio of intensities of consecutive maxima in the diffraction pattern due to a single slit is
(A) 1 : 4: 9
(C)
1~4_ . 91t2 . 251t2
(B) 1 : 2: 3 1 9 (D) 1 : ' : ' n
n
51. A beam of natural light falls on a system of 6 polaroids, which are arranged in succession such that each polaroid is turned through 30° with respect to the preceding one. The percentage of incident intensity that passes through the system will be (A) 100% (B) 50% (C) 30% (D) 12%
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52. The maximum number of possible interference maxima for slitseparation equal to twice the wavelength in Young's doubleslit experiment is · (A) Infinite (B) Five , (C) Three (D) Zero 53. A rocket is going towards moon with a speed v , The astronaut in the rocket sends signals of frequency v towards the moon and receives them back. on reflection from the moon. What will be the frequency of the signal received by the astronaut (Take v « c)
(A) (C)
C
C
(B)   v
v
cv 2v
58. In the ideal' doubleslit experiment, when a glassplate (refractive index 1.5) of thickness t is introduced in the
path of one of the interfering beams (wavelength 1,, ), the intensity ai the position where the central maximum ,occurred previously remains unchanged. The minimum thickness of the glassplate is 21,,
(A)
c2v
2c
(D) ~v
v
intensity of light at the midpoint of the screen in the first case to that in the second case is (A) 1:2 (B) 2:1 (C) 4 : 1 (D) 1 : 1
(C)
V
C
54. In Young's double slit experiment the.ycoordihates of central maxima and lQth maxima aie 2 · cm and 5 cm respectively. When the YDSE,apparatus is immersed in a liquid of refractive index ,1.5 the corresponding ycoordihates will be · · · '· (B) 3 cm, 6 cm (A), 2 cm, 7.5 cm 4 10 (C) 2cm,4cm. (D) cm, cm
3
.
3
,55. Jn·Yoting1s double slit exp°eriment how many maxim.as
can be obtained on a screen (including the central maximum) on, both • sides of" the ·central fringe ·if 1..=ioooA',,;,d d=7oooA 
(B)
1,,
3.
(D) 1,,
3
59. In a Young's doubleslit experiment, the intensity ratio of maxima and minima is infinite. The ratio of the
amplitudes of two sources (B) is unity (A) is infinity (C) is two , (D) cannot be predicted 60. Figure represents a glass plate placed vertically on a horizontal table with a beam of unpolarised light falling on its surface at: the polarising angle of 57° with the normal. The electric vector in the reflected light on , Screen .. S . vib:ate Vfith respect. t~ the plan:e .of incidenc~. in a.
rill
(B) 7 (D) 4
(A) 12 (C) 18
56. A monochromatic beam ,of· light falls on YDSE apparatus at some angle (say 8) as shown in figure. A thin .sheet of glass is inserted in front ofthe. lower slit S2 , The central bright fringe (path _difference = O_) will be obtained (A)
Vertical plane
(B) Horizontal plane , (C) Plane makihg an angle of 45° with the vertical (D) Plane making an angle of 57° with the horizontal (A)
61. A clear sheet of pol!'}'oid is placed on the top of similar
At 0
(B) Above 0 (C) Below 0 (D) Anywhere depending on angle 8, thickness of plat_e 't and refractive index of glass_ µ ' 57. In Young's double slit experiment, the two slits act as coherent sources 'of equal amplitude A and wavelength A . In another experiment with the same set up the two slits are of equal amplitude A .and wavelength i,.. but incoherent. The ratio of the
are
1
sheet so that their axes make,.an ru:ig~e ~~~
(¾) with
each other. The ratio of intensity of the emergent light · to that of unpolarised incident light is ·(A) 16:25 (B)· 9:25 (C) 4:5 (D) 8:25 62. Optically active substances are those substances which
(A)
(B) (C)
produce polarized light rotate the plane of polarization of polarized light produce double refraction ·
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convert a plane · polarized light into circularly polarized light
63. If I, is the intensity of the principal maximum in the single slit diffraction pattern, then what will be its intensity when the slit width is doubled
(A) . I, (C)
2I,
(B)
~
(D) 4I,
64. The critical angle of a certain medium is sin1
(¾). The
polarising angle of the medium is (A)
sin'(¼)
(B)
(C)
tan''(¾)
(D) tan'(½)
tan'(¾)
65. The ray of light is incident on glass of refractive index 1.5 at polarising angle. The angle of deviation of the incident ray in glass is (A) 57° (B) 33° (C) 24° (D) 114° 66. Double refraction of light is shown by (A) quartz and calcite only (B) calcite only (C) calcite and ice only (D) calcite, ice and quartz
1.3 µm
(D) 2.6 x 104 cm
68. The resolution of the human eye is 1'. The resolving power of the human eye is nearly
(A) (C)
360 36000
(B) 3600 (D) 360000
69. Colours of thin films are due to (A) dispersion of light (B) interference of light (C) absorption oflight (D) scattering of light 70. A person standing at a distance of 3.6 km can just resolve two poles. The distance behveen the poles is
(A) (C)
0.1 m 1m
(B) 100 m (D) 10m
·
71. A heavenly body is receding from earth such that the fractional change in A. is 1, then its velocity is
(A)
C
(B)
3c 5
(D)
5
2c 5
72. A star is moving towards the earth with a speed of 4.5 x 106 ms' . If the true wavelength of a certain line in the spectrum received from the star is 5890 A, its apparent wavelength will be about (c = 3 x 10 8 ms') (A) 5890 A (BJ 5978 A (CJ 5802 A (DJ 5896 A
73. Lights of wavelength 1,,1 = 4500 A, 1,,2 = 6000 A are sent through a doubleslit arrangement simultaneously. Then (A) no interference pattern will be formed (B) the third bright fringe of A. 1 will coincide with the fourth bright fringe of 1,, 2 the third bright fringe of fourth bright fringe of 1,,1
(D)
1,, 2
will coincide with
the fringes of wavelength 1,, 1 will be wider than the fringes of wavelength 1,,2
74. Two slits separated by a distance of 1 mm are illuminated with light of wavelength 6 x 10' m. The interference fringes are observed on a screen placed 1 m
from the slits. The distance between the second dark fringe and the fourth bright fringe is equal to (B) 1.0 mm (A) 0.5 mm (C) 1.5 mm (D) 2.0 mm
67. A slit of width a is illuminated by red light of wavelength 6500 A . The first minimum will fall at 9=30° if a is (A) 3250 A (B) 6.5 X 104 mm
(C)
C
(C)
75. Interference fringes were produced in Young's doubleslit experiment using light of wavelength 5000 A .
When a film of thickness 2.5 x 10" cm was placed in front of one of the slits, the fringe pattern shifted by a distance equal to 20 fringewidth. The. refractive index of the material of the film is (A) 1.25 (B) 1.35 (C) 1.4 (D) 1.5 76. In Young's doubleslit experiment, an interference pattern is obtained on a screen by a light of wavelength 6000 A, coming from the coherent sources S1 and S2 •
At certain point P on the screen third dark fringe is formed. Then the path difference S1 P  S2P in microns is (A) 0.75 (B) 1.5 (C) 3.0 (D) 4.5 77. Young's double slit experiment is made in a liquid. The 10th bright fringe in liquid lies where 6th dark fringe lies in vacuum. The refractive index of the liquid is approximately (A) 1.8 (B) 1.54 (C) 1.67 (D) 1.2
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Advanced JEE Physics 78.
A point source emits light equally in all directions. Two points P and Q are at distances 9 m and 25 m respectively from the source.. The ratio of the amplitudes of the waves P and Q is (Al 9:25 (Bl 25:9 (C). ' 9' : 252 (Dl 252 : 9'
79. In Young's double slit experiment the ycoordinates of
central maxima and 10th maxima are 2 cm and 5 cm respectively. When the YDSE apparatus is inunersed in a liquid of refractive index 1.5 the corresponding · ycoordinates will be (Bl 3 cm, 6 cr_n (A) 2 cm, 7.5 cm 4 10 (C) 2cm,4cm (Dl cm, ~cm
3
observation be taken from :point PB  PA = !:. . Then the phase 4 waves from A and B reaching (Al 156° (B) (C) 136° (Dl
P , such . that
difference between the . P is 140° 126°
86. Two coherent sources S1 and S2 are separated by a
distance four.times the wavelength A of the source.'The sources lie along yaxis whereas a detector moves along +xaxis. Leaving the origin and far off points the number Of points where ni.axima are observed is (Al 2 (B) 3 (C) 4 (D) 5 87. The first muumum due to a Fraunhofer diffraction
80. In Young's double ~slit experiment
~ = 10""
(d =
distance between slits, D·= distance of,screen from the slits). At a point P on the screen resulting intensity is equal to the intensity due to individual ~lit I0 • Then the distance. of point P from the central maximum is (A =6000f) (A) 2mm (Bl 1 mm (C)_ .0.5mm (Dl 4mm. 81. Th~·yci~g's doubleslit experiment is carried out with
light of wavelength 5000 A . The distance b~~een the slits is 0.2 mm and the screen is at 200 cm from the slits. The central maximum is at x = 0 . The third maximum will be at .. x equal to (Bl 1.5 cm (Al 1.67cm (C) 0.5 cm (Dl 5.0 cm 82.
using light 'of ·wavelength 500 nm and a slit of )Vidth 0.5 mm will be formed at an angle (in minutes) (Al 2.42 (B) 3.43 (C) 4.84 (Dl 1.7188. Aperture of the human eye is 2 min. Assuming the
mean wavelength of light to be 5000 A , the ..,;gular resolution limit of the eye is nearly
(A)
(C)
89. In the Young's double slit experiment apparatus shown in figure, the ratio of maximum to minimum intensity on the screen is 9. The wavelength of light used is A,
then the v~lue of y is Screen
83. The blue colour of the sky is explained by
(Al
(C)
refraction polarisation
(B) reflection (D) scatteriog
84. A Young's double slit experiment uses a monochromatic source. The shape of the interference fringes formed on a screen is_
(A)
(C)
Straight line Hyperbola
(B) Parabola (D) Circle
ho d/2
in
a Young's . double slit experiment,' 12 fringes are observed to be formed in a certain segment of the screen when light of wavelength· 600·nm is used. If the wavelength of light is changed to 400 nm , number of fringes obseived in the same segment of the screen is given by (A) 12 (Bl 18 (C)' 24 (D) 30 .
(Bl 1 min'!}te . (D) 1.5 ~ute
2 minute 0.5 minute
y
d/2
(A)
(C)
AD d AD 3d
.) 
2d.
132. A diffraction pattern is obtained using a beam of re'd light. If the red light is replaced by blue light, then (A) the diffraction pattern remains unchanged (B) diffraction bands become narrower and crowded together (C) bands become broader and farther apart (D) bands disappear 133. At sunrise or at sunset the sun appears to be reddish while at midday the sun looks white. The reason is that (A) the sun is less hot at sunrise or at sunset than at noon (B) diffraction sends red rays to the earth at these time (C) refraction is responsible for this effect (D) scattering due to dust particles and air molecules
(D)
129. A ray of light is incident on the surface of a glass plate at an angle of incidence equal to Brewster's angle $. If µ represents the refractive index of glass with respect to air, then the angle between reflected and refracted rays is
=
. 1(n:>.)
(A)
134. Light of wavelength 6328 A is incident normally on a slit having a width of 0.2 mm . The distance of the screen from the slit is 0.9 m. The angular width of the central maximum is (A) 0.09 degree (B) 0.72 degree (C) 0.18 degree (D) 0.36 degree
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Wave Optics 135. Two point sources X and Y emit waves of same
frequency and speed but Y lags in phase behind X by 2nl radian. If there is a maximum in direction D the distance XO using ri as an integer is given by
X~D
\~
143. An optically active substance
y
(A)
A. (n1) 2
(B)
!:en+ l)
(D) )..(n0
2
(A) .produces polarized light (B) rotates the plane of polarization of polarized light (C) converts a plane polarized light into circularly polarized light (D) converts a circularly polarized light into plane polarized light
,.(n+I)
136. In a Young's double slit experiment, .the slits are 2 mm
apart and are illuminated wii:h a mixture of two wavelength A.0 = 750 nm and A. = 900 nm . The minimum distance from the common central bright fringe on a screen 2m from the slits where a bright fringe from one interference pattern coincides with a bright fringe from the other is (A) 1.5 mm (B) 3 mm (C) 4.5 mm (D) 6mm 137. If sound waves can be assumed fo be diffracted, which of the following objects will diffract sound waves in air from a 384 Hz tuning fork (A) A sphere of radius 1 cm (B) A sphere of radius 1 mm (C) A sphere of radius 1 m (D) A sphere of radius 10 m
138. A beam of light of wavelength 600 nm from a distant source falls on a single slit 1 mm wide and the resulting diffraction pattern is observed on a screen 2 m away. The distance between the first dark fringes on either side of the central bright fringe is (A) 1.2 cm (B) 1.2 mm (C) 2.4 cm (D) 2.4 mm 139. A nicol prism is based on the action of (A) refraction (B) . double refraction (C) dichroism (D) both (B) and (C) 140. A thin sheet of glass (refractive index 1.5) of thickness 6 micron,_ introduced in the path of one of the interfering beams in a doubleslit experiment, shifts the central fringe to a position earlier occupied by the fifth bright fringe. The wavelength of light used is (A) 3000 A (B) 6000 A
cq
4500
A
(DJ 7000 A
141. Which of the following cannot be polarised?
(A) (C)
Radio waves Infrared rays
(B) p rays (D) y rays
142. When light is incident on a transparent surface at the polarizing angle, which of the following is completely polarized? (A) Reflected light (B) Refracted light (C) Both reflected as well as refracted light (D) Neither reflected nor refracted light
144. Diffraction pattern of a single slit consists of a central
bright band which is (A) wide, and is flanked by alternate dark and bright bands of decreasing intensity (B) narrow, and is flanked by alternate dark and bright bands of equal intensity (C) wide, and is flanked by alternate dark and bright bands of equal intensity (D) narrow, and is flanked by alternate dark and bright bands of decreasing intensity 145. In Young's experiment with one source and two slits, one of the slits is covered with black paper. Then (A) the fringes will be darker
(B) (C) (D)
the fringes will be narrower the fringes will be broader no fringes will be obtained and the screen will have uniform illumination
146. The distance between two coherent sources is 0.1 mm. The fringewidth on a screen 1.2 m away from the sources is 6.0 mm. The wavelength of light .used is (A) 4000 A (B) 5000 A
(CJ
6000 A
(D)
noo A
147. If three slits are used in Young's experiment instead of two,weget
(A) (B) · (C)
(D)
no fringe pattern the same fringe pattern as that wit!) two slits a pattern with fringe width reduced to half of that in the two slit pattern alternate bright and dim fringes
148. When a transparent parallel plate of uniform thickness t and refractive index n is interposed normally in the path of a beam of light, the optical path is (A) increased by nt (B) decreased by nt (C) decreased by(n1)1 (D) increased by (n1)1
================================:::: = 2.63
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Optics & Modern Pl,ysics
Advanced JEE Pl1ysics 149. In Young's experiment, monochromatic light is used to
illuminate the two slits and interference fringes are observed on a screen placed in front of the slits. Now if a thin glass plate is placed normally in the path of the
(A)
(C)
I 2
I
(B)
(D)
I
..fi. I 4
beam coming from one of the slits, then
(A) (B)
(q (D)
the fringes will disappear the fringewidth will decrease the fringewidth will increase there will be no change in the fringewidth
150. If the Young's double slitexperiment is performed with white light, then the colour which will have maximum fringe width is · (A) Blue (B) Green (C) Yellow (D) Violet , 151. In the interference pattern, energy is
(A) (B) (C) (D)
created at the positions of maxima destroyed at the positions of minima conserved but is redistributed not conserved
152. Fluorescent tubes give more light than a filament bulb of same power because
(A) (B)
the tube contains gas at low temperature ultraviolet light is converted into visible light by fluorescence
(C) (D)
light is diffused through the walls of the tube it produces more heat than bulb
153. Energies of photons of four different electromagnetic radiations are given below. The energy value 1 eV 5 eV
(B) 2 eV
(D) 1000 eV
(C) (D)
intensity in the two situations are related as
(A)
I =I0
(q
I =3I0
15B. The phase difference between two wave trains giving rise to a dark fringe in Young's doubleslit experiment is ( where n is an integer ) ZERO
(B)
(C)
2nn + 1t
(D) 21tn +.!:
2
4
159. A Young1s doubleslit setup for interference is shifted from air to within water. Then the
(A)
fringe pattern disappears
(B) fringewidth decreases (C) fringewidth increases (D) fringewidth remains unchanged 160. Two interfering beams have intensities in the ratio of 9 : 4 . Then the ratio of maximum to minimum intensity
162. The fringe pattern observed in Young's doubleslit experiment is (A) a diffraction pattern (B) an interference pattern (C) a combination of diffraction and interference patterns
(D) 156. A beam of unpolarized light of intensity I is passed first through a tourmaline crystal A and then through another tourmaline crystal B oriented so that its principal plane is parallel to that of A. If A is now rotated by 45° in a plane perpendicular to the direction of the incident ray, the intensity of the emergent light will be 2.64
2,m+2:
(A)
fringe width is proportional (A) to wavelength (B) to inverse wavelength (C) to square of wavelength (D) to inverse square of wavelength
pure emission line spectrum emission band spectrum absorption line spectrum absorption band spectrum
155. If a torch is used in place of monochromatic light in Young's experiment (A) fringes will appear as for monochromatic light (B) fringes will appear for a moment and then they will disappear (C) no fringes will appear (D) only bright fringes will appear
=
(B) I= 2I0 (D) I =4I0
161. In interference with two coherent beams of light, the
154. Atomic spectrum should be
(A) (B)
intensity of the central bright fringe is I . When one of the sources is blocked, the intensity become I0 • The
iri the interference pattern is (A) 25:1 (B) 13:5 (C) 5:1 (D) 3:2
corresponds to a visible photon is equal to
(A) (C)
157. Interference pattern is obtained ol1 a screen due to two identical coherent sources of monochromatic light. The
neither a diffraction nor an interference pattern
163. In Young 1s interference experiment with one source and two slits, on~ slit is covered with a cellophane sheet
which absorbs half the intensity. Then (A) no fringes are obtained (B) bright fringes will be brighter and dark fringes will be darker (C) all fringes will be dark
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Wave Optics (D)
bright fringes will be less bright and dark fringes will be less dark
164. The distance between sources in a biprism _of angle a and refractive index µ , if the source is placed at a distance a from it is
(A)
2(µl)cx
{B)
(C)
(µl)cx
(D) (µl)cxa
170. In Young's doubleslit experiment, we get 60 fringes in
2(µl)aa
165. To obtain a sustained interference pattern, we require two sources which emit radiation of (A) the same frequency.
(B) (C) (D)
~69. In the doubleslit experiment, the distance of the second dark fringe from the central line is 3 mm. The distance of the fourth bright fringe from the central line is (A) 6mm (B) 8mm (C) 12mm · (D) 16mm
nearly the same frequency. the same frequency having a definite phase relationship. different wavelengths.
166. A thin mica sheet of thickness 2 x 10< m and refractive index (µ =1.5) is introduced in the path of the first
wave. The wavelength of the wave used is 5000 A . The central bright maximum will shift (A) 2 fringes upward (B) 2 fringes downward (C) 10 fringes upward (D) None of these 167. In Young's double slit experiment, the slits are 0.5 mm apart and interference pattern is observed on a
screen placed at a distance of 1 m from the plane containing the slits. If wavelength of the incident light is 6000 A ,.then the separation between the third bright fringe and the central maxima is (A) 4 mm (B) 3.5 mm (C) 3 mm (D) 2.5 mm 168. Interference fringes are obtained due to the interference
of waves from hvo coherent sources of light with amplitudes a1 and a,_ ( a,. = 2a,). The ratio of the maximum and minimum intensities of light in the interference pattern is (A) 2 (B) 4 (C) 9 (D) 00
the field of view if we use light of wavelength 4000 A . The number of fringes we will get in the same field of view if we use light of wavelength 6000 A is (A) 40 (B) 90 (C) 60 (D) 50 171. With a monochromatic light, the fringewidth obtained in a doubleslit experiment is 1.33 mm. If the whole setup is immersed in water of refractive index 1.33, the new fringewidth will be {A) 1.33 mm (B) 1mm 1.33 . (C) 1.33x1.33 mm (D) mm 2 172. Two waves having amplitudes in the ratio 5: 1 produce interference. The ratio of the maximum to the minimum
intensity is {A) 25:1 (C) 9: 4
(B) 6:4 (D) 3:2
173. If the intensities of the two interfering beams in Young's doubleslit experiment are I1 and 12 , then the contrast between the maximum and minimum intensities is good when (A) [I,  I, [ is large
(B) (C) (D)
[11 I,[ is small either J1 or 12 is zero I,=I,
2.65
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This section contains Multiple Correct Choice Type Questions. Each question has four choices (A), (B), (C) and (D), out of which ONE OR MORE is/ are correct. 1.
< i,_,
(B) If ).. d < at least on~ more maximum (besides the central maximum) will be observed on the
In an interference arrangement similar to Young's doubleslit experiment, the slits S1 and S2 are illuminated with coherent microwave sources, each of frequency 106 Hz . The sources are synchronized to have zero phase difference. The slits are separated by a distance d = 150.0 m. The intensity I(8) is measured as a function of 8, where 0 is defined as shown. If ! 0 is
screen
(C) If the intensity of light falling on slit 1 is reduced so that it becomes equal to that of slit 2, the intensities of the observed dark and bright fringes will increase (D) If the intensity of light falling on slit 2 is increased so that it becomes equal to that of slit 1, the intensities of the observed dark and bright fringes
the maximum intensity, then I(8) for O,; 8,; 90° is given by ·1
f d/2
f
will.increase ;
Os,
3.
In a modified YDSE ·. experiment if point source of monochromatic light O is placed in such a manner that QS1 OS, = ~, where ).._ is the wavelength of light and
 8 
S1 , S2 are the slits separated by .a distance 2).. . Then
d/2
value(s) of 8 for which a maxima is obtained is/ are
* (A) I(8)= I, for 8=30°
 (!_ _____
2
(B)
I(8) =.!._. for 8 = 90°
(C)
I(8)=I0 for 8=0°
s,
4
(D) I(8) is constant for all values of 8 2.
In a Young's double slit experiment, the separation between the two slits is d and the wavelength of the light is )._ . The intensity of light falling on slit 1 is four times the intensity of light falling on slit 2. Choose the correct choice(s). (A) If d =).., the screen will contain only one
4.
. 1(1) 8 . 1(5) sm 6
. 1(1) 4 . 1 ( 7)8
(A) sm
(B)
sm
(C)
(D)
SID

Two coherent waves represented by 2
rot+i) and . (2nx, y, = Asm T  rot +6") = 2.66 ================================::i maximum
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y1 =Asin( ';:'
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Wave Optics are superposed. The two waves will produce
(D) the amplitude ratio is 2
. _inter . ference a t x  x = M ll:\. (A) constr.uctive 1 2
9.
. mter1.erence . ' (B) constructive at x1  x2 = 23,. 24
(ii) y 2 =a,sin(rot+~) (iii) y, = a1 sin(2rot)
(q destructive interference at x1  x 2 = ~: . mter . £erence a t x  x =~ ll:\. (D) d estructive 1 2
5.
6.
To obse~e a stationary interference pattern formed by two light waves, it is not necessary that they must have (A) the same frequency (B) the same amplitude (q a constant phase difference (D) the same intensity Two point monochromatic and coherent sources of light
of wavelength A. are placed on the dotted line in front of a large screen. The source emit waVes in phase with each other. The distance between S1 and S2 is d while
their distance from the screen is much larger. Then for
S1
Four coherent light waves are represented by (i) y, = a, sin(rot)
(iv) y, =a2 sin(2rot+~) Interference fringes may be observed superposition of (A) (i) and (ii) (B) (i) and (iii) (C) (ii) and (iv) (D) (iii) and (iv)
due
to
10. If one of the slits of a standard Young's double slit experiment is covered by a thin parallel slit glass so that it transmits only one half the light intensity of the other, then (A) The fringe pattern will get shifted towards the covered slit (B) The fringe pattern will get shifted away from the covered slit (C) The bright fringes will become less bright and the dark ones will become more bright (D) The fringe width will remain unchanged 11. If the first minima in a Young's slit experiment occurs directly in front of one of_ the slits, (distance between slits and screen d = 5 cm ) then the wavelength of the radiation used can be (A) 2cm (B) 4cm
S2
•• 0 1+d+t
Screen
2
4 3
(q cm 3 2 d = 3,. , there will be a total of 6 minima on· screen d = A. , there will be one maxima on the screen d = 2A , there will be two maxima on the screen
(D)  cm
3
(A) d = ,. , 0 will be a minima (B)
(q (D)
12. Two coherent sources A and B emitting light of wavelength A. are placed at positions (D, 0) and
(D, 3A.) respectively D»1,, y
7.
8.
If white light is used in a Young's doubleslit experime~t, then (A) bright white fringe is formed at the centre of the screen (B) fringes of different colours are observed clearly only in the first order (q the firstorder violet fringes are closer to the centre of the screen than the first order red fringes (D) the firstorder red fringes are closer to the centre of the screen than the first order violet fringes
In the Young's double slit experiment, the interference \ pattern is found to have an intensity ratio between bright and dark fringes as 9. This implies that (A) the intensities at the screen due to the two slits are 5 units and 4 units respectively (B) the intensities at the screen due to the two slits can be 4 units and 1 unit respectively (C) the amplitude ratio is 3
o~•x (A) number of minima on yaxis is 6 (B) number of minima is more than nurnbef of maxima on yaxis (C) number offfiaxima on xaxis is 3 (D) number of maxima on xaxis is more than number of minima on xaxis 13. In the figure A , B and C are three slits each of them individually producing the same intensity 10 at P
when they are illuminated by parallel beam of light of
2.67
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Optics & Modern Physics
Advanced JEE Physics wavelength 1,. • it is given that BP '
AP =!:.2 . Also given
that d « D, then wavelength 1,. and resultant intensity I at P will be
14. White light is used to illuminate the two slits in a Young's doubleslit experiment. The separation between the slits is b and the screen is at a distance d(» b) ·from the slits. At a point on the screen directly in front of one of the slits, certain wavelengths are missing.. Some of these missing wavelengths are b' 2b' (A) 1..=~ (B) 1..=
(C)
d b' 1..=3d
d 2b'
(D) 1..=3d
'
 p t +    D ~>d
10. In a YDSE (young double slit experiment) screen is
placed 1 m from the slits wavelength of light used is 6000 A . The fringes formed on the screen are observed by a student sittingdose to the slits. The student's eye can distinguish two rieighboring fringes, if they subtend an angle more than 1 minute of the arc. Calculate the maximum distance between the slits, in nun,. so that fringes are clearly visible. Give your answer to the nearest integer.
6.
7.
8.
Two slits are separated by 0.32 mm. A beam of 500 nm light strikes the slits producing an interference pattern. Determine the number of maxima observed in the angular range 30° < 8 < 30° . Interference fringes were produced by Young's double slit method, the wavelength of light used being 6000 A . The separation between the two slits is 2 mm . The distance between the slits and screen is 10 cm . When a transparent plate of thickness 0.5 mm is placed over one of the slits, the fringe pattern is displaced by 5 mm . If µ be the refractive index of the material of the plate, then find 5µ .
11. A parallel beam of white light falls from air on a thin film in air whose refractive index is ,J3 .The angle of incidence is i = 60° . Find the minimum film thickness (in nanometer), if the reflected light is most intense for i..=6oooA. 12. In a modified YDSE the region between screen and slits is immersed in a liquid whose refractive ind~x 5 · w1'th' vanes time as µ1 =  t , un til"1t reaches a stead y
2 4
state value
¾.A glass plate of thickness T = 36 µm
and
refractive index µ = ~ is introduced in front of one of 2 the slits. If the separation between the sources and the screen is 1 m and the separation between the sources is
2 mm, then calculate the speed ·of the central maxima, in mms1 , when it is at O,
In Young's double slit experiment mixture of two light wave having wav~engths 1,. 1 = 500 nm and i.., = 700 nm are being. used. Find the position next to
y
central maximal where maximas due to both waves
coincides. ( Given 9.
~ = 1000)
0
Consider the interference at P between waves emanating from three coherent sources in same phase located at 51 , S2 and S3 • If intensity due to each source is ! 0 =12Wm' at P and ~=!::, then 2D 3 resultant intensity at P , in wm2 •
find the
X
µ,
13. In a Young's double slit experiment, 12 fringes are observed to be formed in a certain segment of the screen when light of wavelength 600 nm is used. If the
wavelength of light is now changed to 400 nm , what will be the new number of fringes observed in the same segment of screen.
'
==================================== = 2.81
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Optics & Modern Physics
Advanced JEE Physics 14. A glass wedge of angle 0.01 radian is illuminated by monochromatic light of wavelength 6000 A falling normally on it. Find the distance from the edge of the wedge, in mm where the lQth fringe will be observed due to the reflected light.
15. In Young's double slit experiment the two slits act as coherent sources of eq"i'1 amplitude A and wavelength :>.. • In another experiment with the same set up the two slits are source of equal amplitude A and wavelength :>.. but are incoherent. Find the ratio of intensity of light at the midpoint of the screen in the first case to that in second cas.e.
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•
ICE! 1.
(a)
7.8 µm
(b)
0.6mm
8.
2.
5 µm
3.
(a)
2"[(,/d' + !' d) + µd']
(b)
2 "(µ(Jd'
A
A
2D
+I' 1)+~) 2D
5n
(a)
2
(b)
25n 4
9.
5890
A
10.
1.2 µm
11.
Zero order maxima will remain unchanged. Tenth order will now be at 4.55 mm.
4.
94.8 nm
5.
0.546 mm
12.
8.75 mm
6.
112.78 nm
14.
4.5mm
7.
(a)
0.5 mm
15.
7
(b)
2.25 mm
16.
0.21
(c)
31,
17.
(a)
3x104l 0
(b)
5.49
(d)
,/3 meter
(e)
•
1
n = 5000 is not possible
18.
3.5
19.
1.5
mm
ICE II
.BAS_l=D ON Dli=FRACTJ.0N &!f'()~_filS~IOl;fj
1.
11.8mm
9,
±45° I ±135°
2.
9x104 m
10.
(a)
0.75
3.
0.2mm
(b)
0.25
4.
40 m
11.
37.5%
5.
12.5 cm
12.
!..
6.
1.3µm 0.2mm
13.
1:3
7.
16.
30°' 45°
8.
3.534 x 10.., rad
4
2.83
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1.
B
Optics & Modern Physics
2.
D
6. . D
7.
D
8.
D
L11. 16.
0
12.
D
13.
A
B
17.
D
18.
D
19.
21.
C
22.
C
23.
D
24.
26.
A
27.
C
28.
D
29.
33.
A
34.
38.
C
39.
31.
A
32.
O·
36.
C
37.
A
41.
A
46.
C
51.
3.
   
D
5.
B
9.
B
10.
A
14.
C·
15.
D
B
20.
B
B
25.
D
C
30.
B
A
35.
A
B
40.
B
4.
43.
0
44.
D
45.
C
47.
B
48.
B
49.
A
so.
C
D
52.
B
53.
B
54.
C
55.
B
56.
D
57.
B
58.
A
59.
B
60.
A
61.
D
62.
B
63.
D
64.
B
65.
C
66.
rn.
D
67.
C
68.
B
69.
B
70.
C
A
72.
C
73.
C
74.
C
75.
C
76.
B
77.
A
78.
B
79. 'C
80.
A
81.
B
82.
B
83.
D
84.
C
85.
A
86.
B
87.
B
88.
B
89.
C
90.
A
91.
C
92.
B
93.
D
94.
C
95.
B
96.
D
97.
D
98.
D
99.
A
100.
A
A
102.
0
103.
D
104.
D
105.
C
106.
D
107.
C
108.
B
109.
A
110.
C
111.
D
114.
B
115.
C
116.
A
117.
D
118.
A
119.
C
120.
B
121.
A
126.
B
[
~
[ !
42. A 
A
112. A 113. C ··122. A 
123. . D
124•
A
125.
C
127.
A
128.
B
129.
C
130.
A
D
134.
D
135.
A
L 131.
C
132.
B
133.
136.
C
137. • C
139.
D
140.
B
B
142.
143. 
D
141.
B
144.
A
145.
D
A
138.
146.
B
147.
D
148.
D
149.
D
150.
C
151.
C
152.
B
153.
B
154.
A
155.
C
156.
D
157.
D
158.
C
159.
B
160.
A
161.
A
162.
163.
D
164.
B
165.
C
166.
A
167.
B
168.
C
169.
B
170.
A
171.
B
172.
C
173.
D
, __1:.:..__:_A:.:C:..... ~
6.
ABC
     2.
7.
·C
I
AB
3. AD 4. BD 5. BO ~=__:c=....J
ABC
8.
BD
9.
AB
10.
AGO
L__1c...1:.:.·__:_A:.:C:.__ _ _ ____::12~~c_______1_3_._B_D_ _ _ _ _ _ 14_._A_c_ _ _ _ _ _1_5_._·,_A_B_ _ _.  . J
= c:=================================== 2.84
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Wave Optics
                    
I
L _ ~_ _ _ _ _ _ _ _2.:___D_____ _ _ _ ~'.~ _________4:_ _B____ ~ D [
I
6.
A
11.
A
16.
C
21.
A
7.
D
17.
D
8.D 18.
______
9.D
D
19.
10.A
A
20.
A
_ ________________________________________________

1.
J
~                

j
1
11.
A _ _ _ _ _ _ _ 2_._D_________ 3·~  4.  A   5.__ A ___________ I _ A 7. C 8. A ____ 9. A 10. B D_______ _ C       1  2 . D _________ 13. __ C ________ 14. __ C _ _ _ ______ 15. __
16.
D
~
6.
I
17.
A
22.
C

01._~_ 26. C 31.

____
18.
B
23.
D
19. B 20. B .       . __________24. ___~ ____________ 25:_!________ )
 
~B ___
~A
aD
me
:n:D~__________~.:__~ ____  ____ 34._ C _____________ 35. C________; C
36.
B
OT
B
46.
C
51.
D
~
~
A
D
__ 42_. _ ~ _____ :13_._ __[?__~ 47. D 48. D
=__=:: __4_4.__c_______~= ~s=_c~===~~J ~
C

A
49.
A
50.
B
___: ~~~=__ ·=:==:::=]
IANSWERr
1.
I
I
6.
A; (p, s) B> (q, r) C> (p, q, r) D> (q, r, s) A; (r) B> (t) C> (r) D; (p)
A> (p, q) B; (s) C> (p, q) D> (r)
2.
A> (p) B> (s) C> (r) D> (q)
7.
3.
A> (q) B> (p) C> (s) D; (r)
4.
·A> (p, q, r, s) 8.
 5.: =: l~!·1
A> (r) B> (s) C> (q) D> (p)
C> (r) ____ D>(q,~

B> (p, q, r, s) C> (p, q, r, s) D> (p, q, r)
·   ·  , _ _ _ _ _ _ _ _ _3_. (a)45,(b)59 _ 4._(a)B0,(b)4,(c}60 _ _ j 2. _32 I_ 1. _3_ _ _ _ _ _ _ _ _ 5.
361
6.
739
7.
6
8.
3
~~                       '== ____ 13.
18
10.
2
11.
100
14.
3
15.
2
12.
3
...J
2.85
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Therefore, net path difference t:.x. = (..Jd2 + t 2
t) +
i;:;
So, phase difference Is given by
$= 2•[(../d' +l' d)+ µd'] (b)
,_
Upwards (b)
yd (1.!)1= 4,_ D µ µ y=4.2mm
t,x
=> Downwards
=>
. 1(1.!)+yd='4,. µ
2.
(µ1)tD =>
d
4.
$=~(µ(../d'+t't)+i~)
The reflected light will be minimum when rays 1 and 2 (shown in figure) meet the conditi6n of destructive interference.
Rays 1 and 2 'both suffer an ad~itional phase change of 1t (or 180°) after being reflected at surface of a denser medium. The net change In phase due to 1 reflection is therefore zero and the 2
SW d
condition for a reflection minimum
s, 2sooo A =5 µm l===50,000 µ1
=µ(t../d' + I' t)+ ~)
µ
Since.we are given that Shift= 5 (fringe width)
=>
3.
D
y=0.6 mm
=>
requires a path difference of
1.51
Hence,
=>
=>
=>
5.
,550 I= 4µ = 4 (1 _45 ) = 94.B nm
The desired distance is
=>
546x109 x0.4 0.4x103
y = 0.546 x 10"' m = 0.546 mm
·First order minima
f
y =0
y'
First order minima
i
1
=~
2D
Their optical path difference
AD =d
, ~,I sil 1 ~
Geometric path difference between S2P and Sf is
D
Air
2
,.
y= I'nngew1'dh t
YJ~)d
,.
2µ1=2
f>x=SS,0SS,O=..Jd'+l'l
(a)
2D
When liquid is filled between slits and source S , then
= ,:;;
6.
For constructive interference In case of soap film,
= ================================== 2.86
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Wave Optics 1' = (µ1)t = (1.61)(1.964 x 10"') 2 2
n=1,2,3, ....
2µt=(ni}
For minimum thickness t, n =1 1' => 2µ1=2
10.
1=~=~=112.78 nm 4µ 4x1.33
=>
A= 0.589x10e m
=>
1'=5890A
Shift = 5 (fringe width)
=> 7. · )a) . P= 1'D = (5ooox10·")(1) d
(b)
1x10:i
0.5 mm
t
t=1.2µm
y=(2n1)w
d
At y= =>
1
3
=> d A
11,
P= (14.7512.50) mm=0.25 mm 101 , ),.' 5500 p = i:P = x 0.25 mm~ 0.23 mm 6000 Zero order maxima will remain unchanged. Earl/er it was at 12.5 mm12.25 mm. Tenth order will now be at 12.25 mm+ 10P' = 14.55 mm.
12.
Given,
mm, y«D
,Ix= yd D
Now resultant intensity, is given by 1=11 +I2 +2Mcos.6.4, ·
(d)
I= 41 0 + 10 + 2 g cos.6.4' =510 +4l 0 cos( 10"' 0.5x10v
~1t) =31
t.y=(Y,,).,. (y,),,..,
0
l:i.y= 7AD 2d
2000
13.
=> 0=30° Since, y =Dtane
As d increases, fringe width
(d)
The interference pattern due to different component colours of white light overlap (incoherently). The central bright fringes of different colours are at the same position. Therefore, the central fringe is white. Since blue colour has the lower A, the fringe closest on either side of the central white fringe is blue; the farthest Is red.
(e)
Since in a medium the wavelength of light is 1'' =
4
9.
Since, shift =2(Fringe Width)
=>
(µ1)1D 1'(20) d =d
(Pa::{)
fringe
width
decreases.
is
given
!:µ , by
Thus, fringe width decreases by µ .
14.
For bright fringes to coincide, we have
n1"P n2A2D d=d
ITT
2 $'=( " )(1.25x10"') 400x10_,
$' = 25n
the
of the fringes remains constant.
P=1''D=W. d µd
$= 5n 2 t.X'=(µ1)1=(1.51)(1.5)=0.75 µm
=1.25 µITT= 1.25 X 1Q6
=~)
8.75x103 m=S.75 mm
(c)
therefore
=>
(.6.)c)ne! =_rue+ flx'
7x5x10~xO.S 2x10
?AD d
(b)
Highest order maxima
.1>c =0.5 µm =5 x 101 ·m So, phase difference is given by
(2x111)1'D 2d
But the linear separation or fringe width increases in proportion to the distance (D) from the screen. As 1' decreases, fringe width (P oc 1') decreases.
$ = (~)(,Ix)= ( 400~·10_, )
Ym1n=4.5x103m
=>
Ymrn =4.5 mm
2 " $'=.!:+ (1.51)(8000+10") 2 6QQQ X 1olO
=>
15.
Let n1 bright fringe of A1 overlaps with n2 bright fringe of A. 2
Then,
I= 11 +1 2 +2~ cos~
n1A1D = n2 A2 D =>
d d n1A1 =n2 A2
=>
.!1='2 = 700 =!... n,
l.,
=>
400
$' =11.
6 The intensity at P is now given by
•
1 I'=l0 +11I0 +2~cos( ;1t)=2l0
{·: I'= 21,}
Solving this equation, we get, ~
4
4
7
17.
(a)
= 0.21
Since, file
~ (o. 3 xl0,;(10xl0_,) _3x10'
m
6 3 2
5
A,
1=989'
l.,
2
Since, Ip = I0 cos
=>
implies that 7th ' bright fringe of A. 1 will overlap
(b)
2
(!)
1p=3x10"'I0
Fringe width
p
l.D
d
(546x10,)(1) (0. x _,) 3 10
1.82x103 m
with 4" bright fringe of A. 2 • Similarly 140, of A1 will overlap with 811'1
=>
of A2 and so on.
Therefore, number of fringes between point P and the central fringe are given by
So the minimum order of A.1 which overlaps with A. 2 is 7. 16.
$=(~")(file)=(546~"10• )(3x10
N=~=5.49 1.82
Without inserting the slab, path difference at P is given by
.6.x
=>
yd D
0.15x10a x2x10a 2
P=1.82mm
18.
Let n1 bright fringe co"!"esponding to wavelength i..1 = 500 nm coincides with n2 bright fringe corresponding to wavelength
~X=1.5'x107 m
i..2 =700 nm. => =>n1A.27 n2 =i;=s This implies that 7"' maxima of i..1 coincides with 5th maxima of
i..2 • Similarly 14~ maxima of i..1 wilf coincides with 10~ maxima of A.2 and so.on. So, minimum distance is
Corresponding phase difference at P is
$=(
2:
Ym1n
)
=> 19.
=>
Yrrin =3.5x103 m=3.5 mm
Here i..=6x10s cm, t=7.2x104 cm
Displacement of the central bright fringe, file = sp 0
Since, intensity at P is given by I= 410 cos
=>
= n1~p = 7x5x107 x10 3
1=4l COS (¾)=21
2
(%)
2
0
0
Since file= f(µ 1)t l.
=>
f(µ1)t=6P l.
=>
l=6l. 6x6x10_. =0, 5 µ t 7.2x104 µ=0.5+1=1.5
Phase difference after placing the glass sheet is
$'=$+~(µ1)t
=>
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1.
For first dark band, sine = ~
For 10 bright fringes, we get, 0, 0 =
As the diffraction pattern is obtained in the focal plane of lens, therefore
The angular width of the central maximum in the diffraction pattern due to slit of width a is
X
A
f
d
=>
10!:
a::;~ 5
For small e, tane = sine
x=(~} Since, l.=5900A=59x10' m, f=50cm=0.50m d=0.025 m=2.5x10s m X
59 x10axO.SO 2.5x105
a=..!. mm=0.2 mm 5
11.8x10a m=11.8 mm
4. Given, A, = 5890 x 1010 m , J..2 = 5896 x 1010 m ,
.
Smee, DF
Distance of first secondary maximum from the centre of the screen is
=)
2 d
For the two wavelengths, we have
5.
X1=~DA, and X2 =~DJ.. 2
40
m
1
40 2
km=20,000 m
Since size of Fresnel's zone, DF
=>
2x2x106 The linear separation between n bright interference pattern on the screen is given by
Here size of Fresnel zone DF at the middle hill must be less that
D=
3x2x(58965890)x10' SxlO~ m
fringes
in
= .Jw
..fw « 50 AD«2500 2500 2500 = =0.125 m=12.5 cm i.« D 20,000
an
Thus wavelengths longer than 12.5 cm will undergo serious diffraction effects.
nW
Xn=d Since, xn « D , the angular separation between n bright fringes
8 =x"=nA " D d
4x1QJ
Distan\:e of either of the two hills from the middle hill is
3D t.x =X, x, = 2d(l., J.,)
should be
_(4x10"')'
50 m.
2 d
Spacing betweeQ the first two maximum of sodium lines is given by
I=~
4
I
; =loCOS 2 8
Thus the intensity becomes one~fourth of the maximum transmitted intensity.
1 C0S8=±,n_
=>
a= ±45°,
13.
±135°
The same effect occurs no matter which sheet is rotated or in which direction it is rotated.
10.
e
According to Mal us Law, I= 10 cos 2 8
(a)
Here e =30°
=>
(b)
l=I0
cos (30°)=1 x( 2
=>
__r__=0.75 I,
Here
e =60°
0
:J
(
=¾l 0
Intensity of emerging)= (Intensity of emerging) beam A beam B
=>
IA cos 2 (30°) = le cos 2 (60°)
1_ cos 2 (60°)
2 l=l0 COS (60°)=1 0 ( 1)'
z'
le
=t I
__r__=0.25. 1, 11.
The planes of polarisation of light beams A and B are mutually at right angles. Initially, the beam B shows zero intensity. Therefore, 8 = 90° for beam B and 8 = 0° for beam A . When the polaroid is rotated through 30° , we have 0 = 60° for beam B and = 30° for beam A In this position, we have
14.
Let 10 be intensity of incident unpolarised light. Then the
cos 2 (30°)
G)'
(~J
=.!=,,3 3
sini P = µ From Snell's Law, ___ s1nrP sini From Brewster Law, taniP =~ = µ
intensity of light emerging from the first polaroid will be
COSIP
I _1_ '  2
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Wave Optics ::::>
sinrP = cosiP
=>
sinr,=sin(90°~)
::::>
rP =90°iP
::::>
iP+rp=90°
The intensity of light emerging from P3 is given by 2
13 = 12 cos (90° 0)
. '( 2e)BI,_8x3_3 10 32 4
Sin
Hence the reflected and transmitted rays are perpendicular to each other.
15.
Let P1 , P2
,
=¾1 0 cos2 0sin2 0 = ¾1 0 sin2 (20)
P3 be the three polarisers and 0 be the angle
between the transmission axes of P, and P2 • As P1 and P3 are
sin(20) = ../3 2 20 = 60° 0 =30°
=> => =>
crossed, the angle between P2 and P3 is (90° 0) .
So, 13
Let 10 be the intensity of the unpolarised light falling on P1 •
13
Then the intensity of light emerging from P1 will be
= ~sin2 (20)
will be maximum when
sin 2 (20) = 1 (maximum), i.e.,
sin(28) = 1 = sin(90°) ::::> 0=45°
I, l1=2 By Malus Law, the intensity of light emerging from P2 is given by
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1.
[BJ
lmax=I=(a+a) :::::>
2
1=4a2 =4l 0
When either of the two slits is covered then
I'=(a+o)' =•' =¼ 2.
~
Let nlh m1mma 560 nm, then
Ot
~
400 nm coincides with mth
minima of
5 0 4 0 (2n 1i( ~ ) =(2m1)( : )
10.
of
Location of this minima is,
Required distance
3.
=Y2 
42mm
Y1 =28 mm
11.
,Ix= ni. . (µ1)t=n1.
=>
t=~
[BJ The relation among angle of diffraction 8 , order n and number of lines per cm of the grating N is sine= NnA.. The maximum
4.
maximum
value for
n = N~ .
µ1)..1
=>
(1)1., =ni.,
=>
~=n
i.,
[DJ
=n2A2
=)
62 x5893 = n2 x4358
=)
n2 =84
[DJ
p = ).D
µ1
d
t = 2:._ =_i._ = 2).
=>
P= (5ooox10")(2) (10"')
=)
P=103 m=1 mm
1.51
[DJ As AR>AG>'e :::::>
Hence
=µ2)..2
=>
For minimum value of t, n =1 µ1
sine= 1.
n=1,2,3•..._.. 12.
=>
of
[AJ Since µ).. =·constant
n,A1
[AJ Path difference due to slab should be integral multiple of A or
=>
v& =3x106 ms~1 away from earth
order of spectrum is an integer. Thus we cannot see the fourth order, but can see the third order.
560 nm y _(2x111)(1000)(400x10_.) 2 2x0.1
=>
1.=6000x10_,, cm, N=5000 lines/cm, we get n=3.33. The
==
Next 11 th minima of 400 nm will coincide with B" minima
47x3x10 9 4700
vs =i.=
value
2n1 7 14 2m1 5 10 i.e., 41h minima of 400 nm coincides with 3rd minima of 560 nm. Location of this minima is, y (2x41)(1000)(400x10_.) 14 mm 1 2x0.4 =>
lU·C
=>
PR >J3o >Pa as j3ocA
13.
[AJ Let the required thickness be t A. So, number of wavelengths,in vacuum is _t_.
6000
= =============:;:===================== 2.92
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Wave Optics Number of wavelengths in air is 1.0031 6000 According to the problem
=:,
Einstein's photoelectric equation appears to be interlinking of the two aspects.
26.
_t_= 1.0031 +1
6000 =:,
14.
lm; 0
17.
6000
t=2mm
[CJ
I,
15.
N
=(A +F,)' =91 =(F,F,)' =[
[DJ Theoretically infinite colours wavelengths.
are
possible,
hence
....
Sun
infinite
[DJ
N' 28.
[DJ In the presence of thin glass plate, the fringe pattern shifts, but no change in fringe width.
29.
[CJ If an unpolarised light is converted into plane polarised light by passing through a polaroid, it's intensity becomes half.
30.
[BJ
[DJ The blue filter will allow only blue light to pass through. So light
from the object passes through filter. Similarly, the whi_te background will also look blue through the filter. Thus we have a
blue object under a blue background, which makes it indistinguishable.
19.
,
Image
1./
Because sound waves are longitudinal in nature. 18.
[AJ This is the effect of refraction. Rays from the sun are passing from vacuum to air when they enter·into the earth's atmosphere. They bend towards the normal making the sun to appear at a higher altitude as shown in figure.
[BJ Intensity of illumination at a point is decided by three factors; (i) power of the source P (ii) distance from the source r and (iii) angle of incidence of rays cos
e
=:,
0
I= Pcos 2
r At noon the sun's rays are normally incident making
e= 0 .
Hence I is maXiJ'l'.IUm at noon.
A
20.
[BJ Distance of third maxima from central maxima is 3l.D 3 x 5000 x 10" x (200 x 10') 1.5 cm X d 0.2x10s
22.
[CJ For observation of colours, the thickness of the film should be of the order of the wavelength of visible light.
23.
[DJ The red filter will pass only red light. Hence a blue cross will appear as black. The white background will appear as red when seen through a red filter. Thus the observer will see a black cross under a red background.
24.
B PR=d PO~dsecB and CO= POcot20 = dsec0cos20 path difference between the two rays is, 6x = CO+ PO= (dsec0+ dsec0cos20) phase difference between the two rays is Li~= 1t (one is reflected, while another is direct) Therefore, condition for constructive interference should be
i'.x!: 3l.  2' 2
[BJ Newton's concept of light is that it is made of corpuscles or particles. All particles are deflected by earth's gravitational field. Hence they are also deflected by gravitational field. In the general" theory of relativity. Einstein predicts deflection of light by the gravitational field. The angle of deflection predicted by both are however not the same. Einstein's prediction agrees with experimental results.
31.
25.
[DJ In photoelectric effect, the photon is treated as a particle having a quantum of energy hv. It emits one electron. Here the particle and wave aspect of light appear in two sides of equation.
l.
=>
dsec0(1+ cos20) =
=>
B) =!: ( _d_)(2cos' case 2
=>
case=_!:_ 4d
2
[AJ
l.D
~ =  and
d
P' ~ l.{D/2) =.!!_ 2d
4
2.93
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Optics & Modern Physics
Advanced JEE Physics 32.
36.
[DJ
~='°
[CJ
Intensity at the centre will be zero if path difference
d d is halved and D is doubled
So, Fringe width 33.
=~
(µ1)t=!:
p will become four times.
2
1='
[AJ Path difference at s 2 is 2/,,. . Therefore for minimum intensity at
p
2(µ1)
37.
[AJ No light,.is emitted from the second polaroid, so P, and P2 are
perpendicular to each other
P, 0
90°0 P,e':.L='="P,
2l.
Let the initial intensity of light is 10
Let x be the minimum distance from s2 • Then
31. 2
l. 2
~
X
transmission from first polaroid
11 = 1cos2 0
2
2 Intensity of light transmitted from last polaroid i.e., from
Solving this equation we get 7l. X=12
P2 = 11 cos2 (90° 0) =~cos2 0sin 2 0 l.
2
in equation (1) we get
2
P2 =1..(2sin0cos0) =1..sin 2 20
8
x =15A which is greater than TA . 4 12 34.
38.
[CJ 0
X=
= (µ  1) ID
~= 2n 3
d (1.51)tD d
and !l.x_
2
(µ 1)t0 d
..• (1)
.6.xx(2,.,7')= 231t =,i
••. (2)
sin0=
Dividing equation (1) by (2)
3d
39.
[BJ
I= 1_ cos'(½) ¾=cos'(½)
[AJ Phase difference corresponding to
y,
= " 2
=>
COS$=0
and that for
=+2"
=> =>
Average intensity between y1 and y2
I,. =.!n_~1' I= cos' (±)d~ =I= (n2n+ 2) 2 2
So, the required ratio is ,½(1+;)
41.
[AJ Path difference at P is .6.X
=
~
sine=..?:.
2 0.5 3 µ1 2µ2=1.5 2µ =3.5 µ = 1.75
Y2
8
I= 41 cos'(½)
[AJ Shift
35.
2
Intensity of light emitted from P3
= 3l.
NOTE: If we substitute s,Ps2P=
So Intensity of light after
=1.
... (1)
s 1Ps 2P='F
•
2.94
= 2(icos8) = xcos0
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Wave Optics For intensity to be maximum . ax= nl. => XCOS9=nA
=>
{n=0,1,2,3, ...) p
cos a= nA X
Since, case 'fo 1
=>
nl.t,1
.,/
X
=>
,/
,/
Using (1) and (2),
, , ''
a·
n;t,~
50.
x
l=lo[si:a.r ,where a.=! 1 2 For nth secondary maxima dsin8 = ( n + )A 2
nt,5 n=1,2,3,4,5, .....
=>
[AJ Laser is intense, coherent (all photons in phase) monochromatic (only single wavelength).
43.
and
2
44.
47.
51.
]'
I,
[DJ If I is the final intensity and 10 is the initial intensity then 1=1(cos2 30°)5 2
=> 52.
_I_=.!.x(./3)" =0.12 10
2
2
[BJ For maxima 11 = d sin 8 = nA ::::) 2Asin8=n1 ::::)
sine=%
since value of sine cannot be greater 1. n=0,1,2 Therefore only five maximas can be obtained on both side of the screen.
V C
53.
::::) V=1.5x10 5 (Since wavelength is decreasing, so star coming closer)
48.
2
[ (2n/1)n
[DJ As the star accelerates towards the earth, its velocity will increase. By Doppler effect in light, a source approaching an observer shows violet shift. So the colour of the star will shift toward~ the shorter wavelength region, that is blue.
[BJ 6). l.
=.P.2 =.'.:.[dsinS] =(2n+l)• l. 2 Sin  • . (2n+1)
.where A is wavelength. In the given
set of radiations. Xrays have the minimum wavelength. Hence focal length is minimum for Xrays. ·
a
I =I 0
[DJ The focal length increases with increases of wavelength. ioc (n1) nearly cc :
2Io
[CJ
Therefore, in all four quadrants there can be 20 maximas. There are more maximas at 8 =0° and e = 180° . But n =s· corresponds to e=90" and 8 =270° which are coming only twice .while we have multiplied it follr times. Therefore, total number of maximas are still 20 i.e., n =1 to 4 in four quadrants (total 16) plus four more at a= 0°, 90°, 180° and 270°.
42.
.!t = 410 = 2 l2
''' ' :
s,o~.. s,
l. Substituting x = SA , we get
=>
••. (2)
[BJ In this case, we can assume as if both the source and the observer are moving towards each other with speed v. Hence , cu c(v) c+v v =  0 v=v=v cus cv cv
, (c+v)(cv) c'v' v ~,~~~v=~~~v 2 2 2
[BJ Momentum of the electron will increase. So the wavelength
(cv)
(A=~) of electrons will decrease· and frinQe width decreases
Since v
0
+v 2vc
c' C « c , therefore v' =  ,   =  v c 2vc
c2v
as ~a:l..
49.
54.
[CJ Fringe width p oc A. • Therefore, A and hence p decreases 1.5 times when immersed in liquid. The distance between central 111 maxima and 10 maxima is 3 cm in vacuum. When immersed in liquid it will reduce to 2 cm . Position of central maxima will not 111 change while 10 maxima will be obtained at y = 4 cm .
55.
[BJ For maximum intensity on the screen dsin8 =nA
[AJ
1=41,cos'(!) At central position 11 = 410
.:.(1)
Since the phase difference between two successive fringes is 2x , the phase difference between two points separated by a distance equal to one quarter of the distance between the two, successive fringes is equal to 6 = (2n>(¾) =% radian
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56.
sine= n1. = n(2000) =_I!_ d 7000 3.5 Since maximum value of sin 8 is 1 So n = O , 1, 2, 3 only. Thus only seven maximas can be obtained on both sides of the screen.
the central maximum in the diffraction pattern you add N phasers, all in the same direction and each with the same
[DJ If d sin 8 = (µ 1)t , central fringe is obtained at 0
proportional to (2N>2" and is, therefore, four times the Intensity for the narrow slit.
If d sin 8 > (µ 1)t , central fringe is obtained above O and
amplitude. The intensity is therefore N2 • If you double the slit width, you need 2N phasers, if they are each to have the amplitude of the each to have the amplitude of the phasers you used for the narrow slit. The intensity at the central maximum is
64.
If d sin 8 < (µ 1) t , central fringe is obtained below 0 57.
sin C =
[BJ
Resultant intensity I
¾. Hence
i=tan1 (n)=tan1
= 11 + 12 + 2Jrj; cos qi
At central position with coherent source and 11 = 12
[BJ
= 10

... (1) Icon = 410 In case of incoherent at a given point, lj, varies randomly with
65.
58.
= l 1 + 12 =210
Hence 1coh
= _g_
llncoh
1
%. If i is the polarising angle
tan i = n .
(%).
[CJ Reflected ray
Incident ray
time so (cos$)~=0 I,nco11
n=
... (2)
[AJ
According to given condition (µ1)t = nA. for minimum t, n =1 So, (µ~1)tm;, =1.
Refracted
1. 1. t, ===21. ma µ1 1.51
ray For glass of refractive index 1.5, polarising angle is 57°, tan i = 1.5 . At this stage the angle between the refracted ray and the reflected ray is 90° . The required angle from figure is
8=5733=24°.
66. 60.
[AJ When unpolarised light is made incident at polarising angle, the reflected light is plane polarised in a direction perpendicular to
67.
Therefore E in reflected light will vibrate in vertical plane with respect to plane of incidence.
=>
6.5 x 101
[DJ
=> =>
a =13x101 m a=1.3µm
If I is the intensity of the incident unpolarised light, the intensity transmitted by the firs~ is
½. This is the intensity of incident light
68.
2
[BJ
subtended
by
two
objects
dB
just
is the smallest
polaroid is ( ½)cos 8, where 8 is the angle between the axes.
angle
Here sine=¾, case is therefore¾.
d8 = 1' = ~0 .,Converting this into radian and taking 1 radian as 6
distinguishable,
nearly 60° (Actually 57° 18') we get the resolving power as 3600.
2 2
..!.cas e=..!.x(i) =~I
2 2 5 25 8 is the required ratio. 25
=
= a sin 30
The resolving power of the human eye is
on the second polaroid. Intensity transmitted by the second
63.
[CJ For first minimum A.=asine
the plane of incidence.
61.
[DJ The property of double refraction is shown by quartz, calcite and ice also.
70.
[CJ
[DJ If you divide the original slit into N strips and represents the light from each strip, when it reaches the screen, by a phaser, then at
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Wave Optics If x is the distance between poles and D is the distance of the eye from the poles as shown in figure we have d0 = tand9 = ~. D
Here D = 3.6 km= 3600 m
and
d0 =
~ 36 0
=> 75.
radian. Hence
72.
[AJ M
V
).
C
=>
1=~
=>
V=C
C
[CJ ).' = 1.(1f) = 5890(1
45 ·
xio')
3x10 8
[CJ Shift, ilx=[(µ1)t]~
X=1m.
71.
= 5802 A
=>
tD 20~=(µ1)d
=>
1 20().~)=(µ1) ~
=>
201.=(µ1)1
=>
µ= 20). +1 t
=>
µ
=> =>
73.
20 X 5QQQ X 1 2.5x105
o
1 0
+1
4 µ=+1 10 µ=1.4
[CJ
i., _ 4500 _ 3 i.,  6000 4 ::::)
4A.1
76.
[BJ For dark fringe at P
= 3A.2
i.e. the fourth bright fringe of A.1 coincides with the third bright fringe of A. 2 •
74.
llx:=1.5 mm
[CJ From here, we must take a note that central maxima ( n = O) is
Here n = 3 and A =6000
So, 11= 77.
flanked by first minima (n = 1) and then first maxima (n = 1).
[AJ
liquid, A and hence w is reduced µ (refractive index) times.
1D x,., =(2n+1) d 2
10oo'=(5.5)oo
So, position of second dark fringe is
=>
101.·(%) =(5.5l~
=>
'i:= 5.5 =µ
=>
µ=1.8
x, =[2(1)+1]~~
=>
""< ____ ___,
._
~v,·
Second Maxima {n = 2) Second Minima (n First Minima (n
=~~, ._ _,
_____
x0 =n(i.~) =>
79.
[CJ Fringe width ~ oc A. Therefore, A and hence w will decrease 1.5 times when immersed in liquid. The distance between central th maxima and 10 maxima is 3 cm in vacuum. When immersed in liquid it will reduce to 2 cm. Position of central maxima will not th change while 10 maxima will be obtained at y = 4 C1"!1 .
First Maxima (n = 1)
Second Minima (n = 1) Second Maxima (n = 2)
Similarly, pbsition of nth bright fringe is
10
[BJ The intensity of light received is inversely proportional to the square of distance. Hence the amplltude of the light wave received is inversely proportional to the distance.
= 1)
=0)
1.
78.
First Maxima (n = 1)
l"'""''1 Central Maxima (n = O) '===~1 a@ftFirst Minima (n = 0) i=="1
"
/.Violet
and all other colours have wavelength value lying in this region.
Further, we have X cc A , so X(Blue) < X(Green) 96.
 , / / ,,
[DJ
Hence corresponding phase difference
I 91 From figure 11 =  and 12 = 4 64
Resultant intensity at P
=>
_!g_=~
I= I~ cos
16
1,
2
Screen
$ = 21t x!: = ~ ,
4
2
½= $ cos ¾J =~ 2
0
(
101. [A)
If i is the polarising angle, n = tan67°
n = tani
(Brewester's law).
is greater than 2. We also know
1
  = n. sine
sine=.!, where e is the critical angle, sine is less than 0.5.
n
Hence C is less than 30° . The only possible answer is 22° .
I By using "" =
lmin
l
1
VI, Vtti+ rx;_+,l = [ J¼ /9 Jf' ~1 [ ~ 1 16
11
97.
[DJ Since, according to Malus Law I= l 0 cos 2 ~
98.
[DJ
102. [DJ Only transverse waves can be polarised, while the other properties are common for all wave motion.
49 =1
103. [DJ In the arrangement shown, the unpolarised light is incident at polarising angle of 90° 33° = 57° . The reflected light is thus plane polarised light. When plane polarised light is passed through Nicol prism (a polariser or analyser), the intensity gradually reduces to zero and finally increases.
16
Since P is ahead of Q by 90° and path difference between P
and Q is
¼. Therefore at
A , phase difference is zero, so
intensity is 41 . At C it is zero and at B , the phase difference is
90° , so intensity is 21 .
99.
104. [DJ If shift is equivalent to n fringes then (µ1)t
,
n=
=>
noct
=>
_!g__= n2 11 n1
[AJ P is the position of 111h bright fringe from Q. From central position O, P will be the position of 10th bright fringe. Path difference between the waves reaching at P = S 18 = 10A =10x 6000x1010 = 6x10o m
100. [AJ Suppose P is a point infront of one slit at which intensity is to be d calculated from figure it is clear that x = . Path difference
t2 =
20 30
x4.8=3.2mm
105. [CJ Path difference between the waves reaching at P, A = .6.1 + .6.2
where .6. 1 = Initial path difference
2
between the waves reaching at P
,\=xdJ~)ct =_i_=S,=_1: D10d20204 Screen
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Advanced JEE Physics ~=Path difference between the waves after emerging from'slits.
6, =SS, SS, =~D2 +d' D
113. [CJ
and 6 2 =S,0S,O=~D'+d' D
Interference is a physical effect.of superposition of wave motion. This can happen for any kind of wave motion, i.e., light waves,
2
=>
6=2((o'+d )Lol=2{(0'+:~)o}
d'
sound waves, matter waves, etc.
{From Binomial expansion}
6=D
For obtaining dark at 0, 6 must be equals to (2n 1)~ i.e.,
114. [BJ The extra path difference produced by the glass plate of thickness t and refractive index µ is ( µ 1) t . Due to this if n is
the
d' ). =(2n1)D 2
=>
number
of
fringes
which
t=O.O1x1O...:a m, A.=6OOOx10
10
m,
shift,
(µ1)t=n1..
µ=1.5. This gives a
value for n = 8 .
d=/2n21)1.D
For minimum distance n =1 so d =
117. [DJ
~
I= a~ + a! + 2a 1a 2 cos4> Put a~ +a! =A and a 1a2 =B ~
106. [DJ
l=A+Bcos4>
For destructive interterence, path difference must be an odd multiple of
118. [AJ
~.
l!J..=A.'!.. and v=rro C
107. [CJ
These concentric bright and dark fringes are called Newton's
V=7x1O 6 x
Rings.
=> 108. [BJ I= 10 + 10 + 2l 0 cos(21t)
=>
2 rc , C=3x1O8 ms1 25x24x3600
61. =o.o4 A
119. [CJ Shift =.!!.(µ1)t
1=410
).
A. 21t 1t For X=. ~=X=
4 ). I'=l0 +10 +0
2
=>
7p =.!!.(µ1)t ).
=.!.2
1'=210
120. [BJ
109. [AJ
. P=w =>
d J3ocD
131
D,
= D2 P, P2 = D, D2
132
132
6P
D2
P,
ao=o2
1., =d2
3x105 =A.2 = Sx _2 x103 =6x107 m=6000A 10
6x=dcose
[(n1)t]~= 1.(2D) d d 1.= (n,lt =5892A 2
=> 112. [AJ
).L
X=
... (1)
For maxima at P
6x=~ From equation (1) and (2) nA. = dcos0
~
110. [CJ
=
Here path difference at a point P on the circle is given by
·~
1
Scos ( n;) = cos1( ~)
121. [AJ The film appears bright when the path difference
(2µtcosr) is equal to odd multiple of!: 2 (2n1)). i.e., 2µtcosr ~~ where n=1, 2, 3 ..... 2
d
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A
4µtcosr (2n1)
4x1.4x10,000x10"'10 xcoso
A,
56000 (2n1)
(2n1) .
i. = 56000
I=~
=>
A,
18666
A,
0000
A,
5 X 48QQ X 1olO
(µ, µ,)
(1.71.4)
8x10.. m=8 µm
A
6222
.
126. [BJ
A,
5091
A,
4308 A, 3733 A. The wavelength which are not within specified range are to be refracted.
At the polarising angle, the reflected ray is fully polarised while the transmitted ray is partially polarised. In fact a method to produce plane polarised light is by reflection at the polarising angle.
128. [BJ 122. (AJ M
).
a=B
V
6500 x1010
C
=>
(401.8393.3) 393.3
=>
v
=6.48 x 106
V
ms1 =6480 kms1
560 nm then (2n 1)400 = (2m 1)560
a=1.24x10~ m a=1.24µm
130. [AJ As here polariser is rotating i.e., all the values of 8 are possible. 12:rr 1 2. Iav =Jid8== J10 cos 2 8d8
i.e., 4th minima of 400. nm coincides with 3rd minima of 560nm The location of this minima is 7(1000)(400x10.. )  14 mm 2x0.1 Next, 11th minima of 400 nm will coincide with 8th minima of 560nm Location of this minima is 21(1000)(400 X 42 mm 2x0.1 Required distance =28 mm
10. )
t
:::, =>
(i)
Using Mal us Law, I= 10 cos 2 8
=>
Shift rue=
a
3x1Q8
123. [DJ Let nth minima of 400 nm coincides with mth minima of
124. [AJ
=>
21to
21t
On integration we get Iav
where I ~
I
0
av
Energy Area x Time
o
=~ p A
10"'
3x104
10watt
3rfi2
=.!x.!Q=E, Watt
2
3
3
. 2n: 2x3.14 1 andTimepenod T=== sec ro 31.4 5 So, energy of light passing through the polariser per revolution
=I xAreaxT=~x3x10"""x..:!.=104J ~ 3 5 131. [CJ
(µ1)1
p
w B'e
'
Path difference = 2dsin8 For constructive interference 2dsinB = ni.
11+ o, ___.,. Screen Shift due to one plate rue, =
t
=>
(µ;1)
'(n:>.)
. B =Sin2d
133. [DJ
Shift due to another path rue, =f(µ, 1)t Nets~ift rue=rue,rue,=.!1.(µ,µ,)~ ).
... (1) .
During sunset and sunrise we see the unscattered part of the sunlight. Sun's rays have to travel long distance and hence are incident obliquely. All shorter wavelengths are scattered by
Also it is given that rue = 5P
... (2)
Rayleigh's law for scattered intensity I a: h.~ . Only the red
.
Hence 5P =
t(µ,
µ,)t
orange rays reach us. During noon the sun's rays are normally incident on the earth. Hence a portion of both the shorter
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wavelengths and longer wavelengths reach us. The sun looks
white.
The action of a nicol prism is based on double refraction and dichroism. When a ray of light enters into a calcite crystal, it is split into two rays O and E. Both are plane polarised in mutually perpendicular planes. One of them suffers total reflection and absorption whil_e the other comes out as plane polarised. Selective absorption is called dichroism.
134. [DJ
P='° d So, angular width, .20 is
140. [BJ
29=2p=21 D
9
d
Shift, ,1x =[(µ1)t]%
6328x1010  0.2x10""
Here ,1x = sp
0 = 3164x10a radian
=>
e = 3164x10e x
=> =>
9 = 0.18 degree
180
•
degree
29 = 0.36 degree
For maxima 2,n = ~"(X0)27tl
=>
2 ;cxo)=2a(n+I)
=>
(XO)=1"(n+l)
s(1"~)=[(µ1)t]%
=>
51"=(µ1)t
=>
1"=(µ1)½ 1"= (0.5)(6x10"') 5
~
A.=6x107 m
=>
1"=6000 A
141. [BJ Only transverse waves can be polarised. (A) and (C) and (D) are electromagnetic radiations, which are transverse waves while
136. [CJ
(P,) (P,)
(B), that is, p rays are electrons.
From the given data, note that the fringe width ·
for
i, = 900 nm
for
=: 750
is
greater
than
fringe
width
nm . This means that .at though the central maxima of
= 900
nm will be
=750 nm
and so on.
the two coincide, but first maximum for l further away from the first maxima for A.2
1
A stage may come when this mismatch equals (32
,
=750
146. [BJ
P='° d =>
then again
maxima of A1 = ~00 nm , will coincide with a maxima of A. 2
=>
=>
135. [AJ
A. 2
139. [DJ
=>
s
M1.2)
1000 =( 0.1) 1000 1" = 5000
A
111
nm , let this correspond to n order fringe for A.1 • Then it
will correspond to (n + 1t order fringe far A. 2 •
150. [CJ Using the equation p = ~). , p ~A. Of the given colours, yellow
Therefore n~D  (n + ~J.. 2D
has the maximum wavelength and hence the maximum fringe
width.
=>
n x 900 x 10 = (n + 1)750 x 10·'
::::)
n=S
152. [BJ The tubelight has coating of fluorescent material which converts ultraviolet light into visible light. Hence the light is more intense.
Minimum distance from
C
t I
.
nA.p
en ra maxima = d =
9
5x900x10 x2 x aa 2 1
153. [BJ
= 45 x 104m = 4.5mm
We take one wavelength in the visible region and calculate the energy just to know its order. Let us take 6000 A (red) for
137. [CJ One of the essential condition to observe diffraction is that the dimension of the object should be of the order of wavelength of the wave. The velocity of sound in air as nearly 350 ms1 • Hence the wavelength of sound waves produced by a tuning fork of frequency 384 Hz is A=
f
= 1 m near1y. Hence the diameter
of the sphere should be of the same order.
138. [DJ
convenience of a rough calculation. Energy E =
~.
Taking
h=6.6x1034, C=3x108 , A=6000x1010 m we get :c in joules. Dividing this by 1.6x1019 to convert to eV, we get
E = 2 eV. For energy 1 eV, wavelength will be 12000 A. Energy 5 eV means wavelength
f
of 6000, that is 2400
A.
Both these ar8 invisible, recalling that the visible range is
Distance between the first dark .fringes is
4000 to 1000 A .
21D
Ax=d=2.4mm
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Wave Optics 154. [A].
167. [BJ
The energy levels of atoms are well defined. Hence transition
between them should give a well defined frequency, which means a single frequency, and hence a single line.
Separation n111 bright fringe and central maxima is xn
3 X 6000 X 1olO X 1 3.5mm. 0.5x103
155. [CJ The torch produces light waves which are noncoherent in nature.
168. [CJ lmax 1nm,
156. [DJ For unpolarised light
I'= .!P..cos 2
I ' =2COS I
2
~
2=9 =(81+82) a a 1
2
169. [BJ For second dark fringe, n = 1
'(n) 4
Since, x = (2n+1)~~
I'=.!.
=>
=n~D
3 = 3W
=>
4
... (1)
2d
For fourth bright fringe, n =4. Since
158. [CJ The path difference of two waves producing destructive interference or darkness is (2n +
1)( ½)
1, 2, 3, etc. A path difference of A corresponds to a phase
1t.
::,
½corresponds to
=>
Hence the corresponding phase
=>
difference of 2v. rad and a path difference of
a phase difference of
X=n(l.~)
where n is an integer 0,
difference is (2n + 1)x .
.•. (2)
X=4(l.~) X
4
3=(3/2) X=B mm
170. [A]
159. [BJ
'.
For same field of view
n1P1 = n2P2
P'=l
160. [A] 11
a~
9
½=a~=4
n,(7)=n,(l.~D)
=:>
n1A1 = n2A2
=>
60(4000) =n,(6000)
=:>
n2 = 40
=> 171. [BJ I= S.ince = (a,+a,)'   = (3+2)' 
, Im;n
a1 az
P'=£=1mm
32
µ
=>
172. [CJ
5=~ a,
161. (A]
Using the equation ~ = ~l. ; p « l.
=:>
164. [BJ
166. [A]
Shift = i(µ1) t ( p ")(1.5)x2x10 .. =2P 5000x10l. i.e., 2 fringes upwards.
=(a,+a2)' 81  82
=>
In a biprism, the distance between sources is due to deviation produced by two small angled prisms. The deviation produced by each prism is (µ1)o:. The total deviation is 2(µ1)o:. Separation of sources 2(µ1)aa.
lmax lmin
173. [DJ I=
=(Jf; +,/I;)'
I..,
=(Jf;,/1;)'
For contrast to be good 11 = 12 and hence
Imax
=411
Im1n =0
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1.
[A, C]
For, n=0, sin8=¾
The intensity of light is !(9) = 10 cos'(%)
n=±1,·sin8=!,+! where ii=~"(,i,c)=(~)cdsin9) (i)
n=±2_,
For 9 = 30' C
3x1Q 8
(ii)
ii
2
(~)= ~
{option (A)}
8 8
.... th en
01s a·m1mma ··
The path difference on screen is > 0 For d = 3A , we have 0;5 l.
i=.': and 1(0)=0
=>
s,
[A, B, Cl Path difference at O is d if d = A, 2~, 3A, .... ., then O is a maxima. 3" 5 '" 1,fd ="2' 2' 2'
For 9 = 90' ii= ( :• )
s1n8=a· 8
0=(~)(150)(1)=" 300 2 2
2=4
8 •3
ri=±3, sin8=~.?...
l.=;,=10'=300m and d=150m,weget
=>
.
81
2 2
1.5 l.
For 9=0', ii=O or i=o 2 !(9) =10
2.5 l. 0     +        t , i ) ( = 3l.
{option (C)}
2.5 l. 3.
[A, DJ The phase difference corresponding to 0S1 0S2
=
Hence, the net phase difference is given by
=>
[B,
DJ
I= (,µ, +,fi;)'
g·
I,,., =(,µ,,[I;)'
if,=2:41tsin8 2
2:4nsin8=n1t 2
=>
,ff; +F, =3(,Jf; F,) 2,µ, =4,fi;
=>
1 n 2sln0=
11 =41 2
Since,
For maxima, we have
=>
0.5 l.
8.
2
~=" "(2l.sin9) 2 l.
=>
1.51.
¼iS i
4' = n1t , where n = o, ± 1, ± 2 ...
=>
4
I =(A)'
....1.
1.
12
A2
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Wave Optics ,/o'+4d' O=~ 2
10.
[A, C, DJ I1 =1 2 =I
11.
=(,if,
=> 12.
A=D C
+F,)' =(.JI +J½J I"'=(./JJ½J >0
1_
4d 2
. gives,
14.
[A, CJ x = Path Difference = ..Jb2 + d2

d
l.=~
2n1
A=2cm,
2
3
2
cm,
5
=>
X=d(n1)
=>
X=d(1+£1) 2d'
cm ........ .
[A, B, CJ
{ b'
00
'' y ''
Since d2 « 1, Hence
~
3 l.
(D, 3 l.)
b' 2d
X=
!s, \"
•++x ''
(0, 0)
....··
d
P
!s, _;.:,...···~
0.5 l.
d
1.5l. 2.5 l. 3 l.
.rue at co on yaxis is
2
For wavelengths to be missing we must find positions of minima.
2.5 l. 1.5 l. 0.5 l.
''''
d v1rt1 +d2 =1+ 2b'}
J
At minima path difference is an odd multiple of
'' co'' 3). and 3A. there will be 6 minima and
=>
b' l. X= d =(2n+1) ; n =0, 1, 2, 3, ..... . 2 2
=>
l.
Smaxima
y
½
b' (2n+1)d;n=0,1,2, ............ .
b2
b2
b2
l.=d. 3d' 5d'"""""·"·
(0, 3 l.) .......... 15.
Ax= 3l. (0, 0)
o+
~, 2.5l. 1.5l. 0.5l.
Ax=0
No. of maxima = 3
13.
[B,
[A, BJ
P= l.D
d For increasing J3 , d must be decreased and A must be
increased (i.e. frequency must be decreased).
DJ
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2.
[DI Fringe width
p=
1 ~
shall remain the Same as the waves travel
in air only, after passing through the thin transparent sheet. Due to .introduction of thin sheet, only. path difference of the wave is changed due to .which there is shift of position of fringes only, 1 which is given as .rue D(n; )t , where n is refractive index of
[DI Ax=d=nA, for n=1, d=A and here we will have three maxima.
9.
[DI Statement1 is false but Statement2 is tn.ie.
14.
thin sheet arid t is its thickness. 3.
5.
(DI Statement1 is false, Statement2 is true.
[DI
15.
If both the slits are illuminated by two bulbs of same power, no interference pattern will be observed on the screen. This is because waves reaching at any point on the screen do not have a constant phaSe difference. as phase difference from two non coherent sources changes randomly. Therefore, maxima and minima would also change their positions randomly and in quick succession. This will result in gen·eral illumination of .the screen.
[A] Statement1 Is true & Statement2 Is true and Statement2 .is correct explanation. of the Statement1.
1
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1.
2.
[AJ It will be shifted upward
[DJ
.
Cannot be predicted without knowing the thickness of thin plate. .
3.
10.
[BJ Power received at 8,
[BJ
P8 =.!_CJ_n(0.002)' =4x10" W n
(µ, µ,)t =rue= o.3t 4.
11.
[AJ
l. (µ,µ,)l=rue=2
[CJ Power transmitted through A, p~ =~xrn'W=1o«,i,v
100
t=l.=J:... 2(0.3) 0.6
=>
5.
12.
[DJ Power transmitted through B,
[A]
P~ =_:!_Q_x4x10o W=4x106 W 100
3 ,~ ,/3 I0 =I0 COS  = 4 2 2
13.
~="'
·
6.
A=(µ1)t=(1.51)x2000x10·" m rue="~,.=~ 2n 6 ;. ;_ 5000 A (µ, ~ µ,)t = 6 t = (6l(o_3 ) = :i:a = 2777.7 A
.6.=107 m
14.
2n1tmln
a.
tmln
l.
=2
550 = =99.64 nm 4_x1.38
{·: n, < n,)
.6.
2n: X 107
6000x1010
qi=~ radian 3
[DJ We know that, P cc a2 or P =ka 2
,
where k is any arbitrary
positive constant.
Now, aA =~1~e 'aa =~4x~oe
~ra'""!
2
::::)
15.
t
2
Resultant aniplitude, 8., = +~'""+_2a_A_a._c_O_S_~
[AJ
2n,t = 3l.
9.
=>
[CJ For destructive interference, we have
=>
[CJ Phase difference, qi=
[A] . Optical Path Difference (OPD) =2n,t
7.
[CJ Path difference,
3
=>
·
t = 3tm1n =298.9 nm
[AJ Power received at A,
Substituting values and simplifying, 7x10..a 2 ar=kResultantpower, P, =k~ =7x.106 W =7 µW
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Advanced JEE Physics 16.
[DJ Anywhere on screen because there is no relation between 8 ,
=>
d2 nyd +=nA.
... (1)
Also,
~ + ny:; 1 =(n+1)1.
... (2)
µ. 17.
19.
(2)(1) we get !P=1
For central bright fringe to be obtained at O , we have dX=O
=>
(µ1)t=dsine
24.
[BJ &~=(2n1h,where n=5 [BJ
[BJ increase and hence b.x increases, so that the central maxima will shift upward.
=>
25.
[BJ Net path difference is given by dsin91 + dsin82
I
=½
Is,
I
D>>d
26.
Is,
22.
=>
d=n
Since, =>
rR
=3nR _ nR = 2nR = nA 2 2 2 =11 +12 +2$j; cos~
In=(,µ,+.µ,)' when ~=2n•, n=0,1,2,3, ....
[CJ
27.
~+dL=o D 2D => Y=2d So, the POS:ition of first bright fringe is given by d 3 y=2d+2=2d
=>
So, the first brightfringe is located
=
[CJ
Since the wave generated is divided equally in two parts, so we have
• For central maxima, we have dsin01 +dsin82 =0
23.
[AJ In transmitted light system, for destructive interference, we have
AX
drn)+d(!) =½
[BJ In reflected light system, for constructive interterence, we have
=>
Since, filC=dsin9(µ1)t, so if 9 increases then dsin9 will
21.
2D
P=1d
2µt = (2n  1).1: 2 For minimum thickness, we have n =1 ;,_ => 2µ1=2
Fringe width P=~ =4x105 nm 20.
2D
[AJ Total path difference is given by dX = (µ1)1dsine
=> 18.
D
[DJ dsln81 + dsin92
2.108
=nA.
¾d below O 28.
[AJ Since &x = xR For maxima, we have &$=(2n)n, where n=O, 1, 2,3, ....
=>
(~)
,._
=>
Amax= nR
max
=~=
nlT&ll
nR 1
[BJ Similarly, for minima, we have &$=(2n+1)n,where n=0,1,2,3, ....
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WaveOpHcs
2; )
(
=>
). = 2nR
=>
Arna~= 2nR
yd= 10). D
34.
29.
Y=10().~)=%x10 4 m
=>
2n+1
max
[CJ At P, we have
[DJ
~=
Since, d., =2a(µ1)a
and
dwater
=2a(:w 1)a
=>
l:__1
dalr
+(µ, 1)tµ.dsin0
d>C= 102x3x104 + 1 0A= 10).+ 3A 8x1 4
,I~= 2•(10i.+ 3l.) = 20•+ 3, l. 4 2
dwator=~
=>
i
µ.1
1t) Since cas(201t+ ~n)=cos(3;)=o So, Ip
=>
=>
=>
= 12 + 12 + 212 cos ( 201t + 3
2
I,
=.!21
Further I™ =½+½+2~(½)(½)cos(oo) 30.
[CJ Since, and
Pair
2a(µ1)a
lmax =1+1=21
=>
1, I™
DA"
Pwaier
2a(µµ.)a
35.
~1 = ~ = _2_ µµ. 3 4
=>
Pwatar
=>
Pwaier=3x1=3mm
P.,
=>
./21
1
=21= ./2
[CJ Path difference is given by ~
= dsin~+dsin0(µ1)t
2 3
31.
 LI~
[CJ
//1" 
S"mce, p=d " w
32.
1 xA. 0.5x103
=>
1x10a 
=>
).=5x10·' m=5000A
=> =>
= (µ, 1)tµ.dsin0
ffi(
4 1 = Q.5 X Q.41 X 1 Q3 3X 3 X 104 X2
=>
sin8=(µ 1)t_sin$ d
(~1)
sin8 ;Ox10a
=>
0=30°
~
A , so we observe that
=10).
So, 1011\ bright fringe, will be formed at 0. [DJ For central maxima, we have
~=0 where, ~=i+(µ,1)tµ.dsin0
36.
2
[BJ
At C, 8=0°,soweget ~=dsin~(µ1)1 =>
~=(50x10"")(½)(¾1)
~
= 0.025  0.05 = 0.025 mm
Substituting, AX= n). , we get
n = LIX A
=>
sin(30°) = .!
.1X=5x10..f.l m
Since, A =5000
33.
~
For central maxima, ax= o
[DJ At O , optical path difference is ~
e 
i+(µ,1)tµ.dsin0=0
SubstitutinQ values, we get
0.025
_50
500x106 Hence, at C there will be maxima. Therefore the order of minima closest to the C are 49.
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Optics & Modern Physics
Advanced TEE Physics 37.
=> =>
[A] Number of fringes shifted upwards is
N= (µ1)1 ).
38.
(¾1)(0.1) 500x10
ID
ax at R will be zero if Ax:1 =dX:2
=>
dsina=dsin8
=> =>
a=0
Substituting the values, we get
=(~ 1)10.4x10.. (1.5) Y 4/3 0.45x10"
tan a= tan0
11. = _lg_ D1
(.&_  1)1~d µ_
y=
D2
=>
y, = D, ·Y, =(_!Q_)(40) cm D,, 200
=>
y2 =2 cm
=>
y=4.33x10am
=>
y=4.33 mm=
13
3
mm
44. , [CJ
At O,· dl
a
Now, I(~)=I=cos'(f) :::::,.
s => => => => 43.
$=(~)n
(µ1)t=dl 45.
[CJ At O , path difference is Ax= Ax 2 = (:: 1}
[DJ Given A=600 nm=6x107 m,
For maximum intensity at O , we have ~x =nA, where n=1, 2, 3, ......
d=0.45 mm=0.45x10"m and 0=1.5 m
=>
A=
dX dX dX , , , ...... and so on 1 2 3
dX
=(!i:1)t10.4x10 .. m)
dX
=(!i:1)(10.4x10' nm)=1300 nm
sl1 =>
s,s
So, maximum intensity will be corresponding to
1
=
And refractive index of glass sheet, µg
1300  nm, 1300  nm, ...  nm, 1300 4 3 2 i. = 1300 nm , 650 nm, 433.33 nm, 325 nm, .... A=1300 nm
Thickness of glass sheet, t =10.4 µm =10.4 x 1a.fl m Refractive index of the medium,. µm
(1!1t)
i
=>

The wavelength in the range 400 nm to 700 nm are 650 nm
=1.5
and 433.33 nm(=
Let central maximum is obtained at a distance y below point 0. :::::,
yd
Ax 1 =8,PS 2 P=[)
Path difference due to glass sheet is given by
Ax,=(~: 6.x1
=.1.X2
yd
=(!11)1
D
nm)
[CJ The optical path difference between the two waves arriving at P
is y,d y,d (1)(10) (5)(10) dl
46.
:o
13
D1
D2
10
2x10
:::::,. Ax=3.5x10 2 mm=0.035 mm To calculate the order of interference, we shall calculate dX
n=1
µ.
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Advanced JEE Physics =>
n ~~00~5xx1~~:
Optics & Modern Physics
~
Since, m = _f_
f+u
=> n=70 => 6x =70l. So, 70111 order maxima is obtained at P .
47.
=>
X
=>
X=20 cm So,weget U=20, v=+80
[DJ
Since.!_.!=!
At O, 6x = y,d =10_,, mm =0.01 mm
V
D,
=>
Now, we observe that fix= 20A. So, 20111 order maxima is obtained at O . 48.
50.
f
l.D _ 6000x10" x1
'"'
51.
U
f=16 cm
[BJ 0
[DJ (µ1)1 = 0.01 mm I =o.o,   = 00 . 2 mm=20 µm 1.51 Since the pattern has to be shifted upwards, therefore, the film must be placed lnfront of S, . ·
49.
100x = 4
d
0
3x10a
2 · mm
[DJ
P= l.D d
When D is increased, then
p is increased.
[AJ Using the concept of Displacement Method, we get 0=$i; =3mm
=2.112
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1.
A> (p, s) B> (q, r) C > (p, q, r) D> (q, r, s)
2.
A> (p, q) B> (s) C> (p, q) D> (r)
3.
6.
s,
width remains same. 7.
A> (p) B> (s) C> (r) D> (q)
8.
A4' (p, q, r, s) B> (p, q, r, s) C > (p, q, r, s) D > (p, q, r)
A> (q) B> (p) C> (s) D> (r)
4.
A> (r) B> (s) C> (q) D> (p) .
5.
A> (p)
A> (r) B> (!) C > (r) D > (p) this effective path If glass slab is introduced across increases so central maxima will be shifted downward but fringe
B> (s) C> (r) D> (q, r)
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1.
Shift dY =I(µ, 1)1, (µ,
Substituting d = 201. , we get
1)t,I~
sinB=sin(45°)+
=> => => 2.
dy=i(1.21)x15x10 .. (1.61)x10x10_.1(
l.S __,) 1.5x10
=>
',
3
,.1.y=3x1ff" m
dY =3 mm
4.
(a)
Since, 1 =0.25 m and d =2 m = 81. At A ·and C, AX=d=81Since this is an even multiple of A, so maximum intensity is obtained. At Band D, AX=0 Since.this is an even multiple of A, so again maximum intensity will be obtained. · B
'' ' '''
9::::59°
1~ =(¾1)(¾+¾) => A=10cm Now applying µ 2 V
~
u
= µ2 
R
µ, twice, we get
3/2 __1_=3/21 and v, 20 10
•.• (1)
4/3 _3/2 4/3 3/2 V2
V1
.•. (2)
10
Adding equation (1) and (2), we get
v2 =80cm (b)
' C ••A S1 1 Sa
..,._~... '
''
Ray 1 has a longer path than that of ray 2 by a distance dsin(45°), before reaching the slits. Afterwards ray 2 has a path longer than ray 1 by a distance dsin8. The net path difference is therefore, dsin9 dsin45° (a) Central maximum is obtained where, net path difference Is zero, so we have dsin8dsin(45°)= 0 => 8=45° (b) Third order maxima is obtained where net path difference is 3" , so we have dsin8dsin(45°) = 31. 3 => sine"= sin(45°) + di.
.
The Image formed by (lens + water) system will act as an object for the mirror. This is below the axis of m1 and at the same distance as the centre is, therefore, its image will be fanned vertically above at 1 mm from AB . Similarly m2 will fonn an image of 11 , 1 mm below CD .
D Further, between A and B seven maximas corresponding to AX = 7A , 61. , 51. , 41. , 31. , 21. and i. are obtained. Simila~y between B and C, C and D , and D and A . So, total number of maximas is N = 4 x 7 + 4 =32
3.
::i.
+
c1 ·D i+20
=> (c)
d=1112 =4 mm, .
Since, 1c
=>
cm......
1112 =1+1+1+1=4 mm D =80 cm
V
=f
i."( 3x10' x3)  µ.f 7.5x10 14 4
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Wave Optics P=).D =~( 3x10'14 J(80x10""J m= 6 x 10_. m d 4 7.5x10 4x103 P=60 µm
=> => 5.
=> 3d] 3d2 S 3PS,P=dsine, = d [ ~ =w=l.
The intensities of the rays due to successive reflections and
refractions are shown in figure. So, 11 =0.11 0 , 12 = 0.08110
=>
,i~
= 2,
s,
pl
s,
2
3d

I ._ 
0.09!,
u 0.11,
I
 e, 
s, If IR be the resultant intensity of the wave at the point P , then IR=l1 +1 2 +2Mcoscji
=>
=>
6.
!
1
m~
I.,,.
=>
r, = 1, + 41 0 + 2J(1,)( 41,) cos(120')
=>
l,=31,=3(12Wm"")=36Wm""
=[Jf +ll' =(19)' =361
{I_1
1O.
vi;
Angular fringe width e = !:, __ ._ d180x60
Therefore_, number of fringe widths in a distance y are given by
r = __E.__ P ,/al,
=~
According to the given condition
Fringe width p = AD and y = ';, d v3 n=
!
0.32 X 10_, (,/3)(500x10_,)
d < 6x107 x180x60
=>
369.5
dmax 11.
=2.06 X 1"Q3
m =2.06 mm
According to Snell's Law, we have
1sin(30') = F3 sinr :) r=30° 2 air
~__c,.""'JL
film Therefore, total number of maxima obtained in the angular range 30° < 0 < 30° (including the central one) is
N=2x369+1=739 7.
air The Optical path difference is given by llX. = 2µtsecr2ttanrsini
Here, d=2mm=2x104 m, 0=10 cm=0.10 m, t= 0.5 mm =0.5x10a m, llX =5 mm=5x10a m,
A=6x107 m
AX =2,/a(tsec30')2ttan(30°)sin(6D')
=>
AX= 4t2{
Since, AX= ~(µ1)1 µ 1 = .1Xd Dt
=>
5x103x2x10: =0. 2
0.10x0.5x10 µ=1+0.2=1.2 5µ=6
lJ( ~)
=31
Since the Ray 1 is reflected at the surface of the denser medium so it suffers an additional phase change of re or a path change of
~ . So, for constructive interterence, we have 3t=!: 2
9.
=>
t=!:=1000 A=100 nm 6
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AdvancedJEE Physics
Since the wavelength is decreaf?ing from 600 nm to 400 nm ,
12.
so fringe width ~Isa decreases by a factor of
number of fringes will Increase by a factor of
~ = ~ . So
the
¾. Hence the new
number of fringes formed ls N=_:3_(12)=18 2 The path difference at O Is
14.
0 = 0.01 radian n =10
AX=(S,PT).,., + T (S,P).,.,
J..=6000x10~ AX=yd+(µT _T) µt 1quld D
Since, 8 =.!.
At the central maxima, we have path difference AX=O
=>
cm
AX=2t=n). X
Y=_DT(!1=11,)=TD(4:J) d µ, d 10t
Cen~ maxima will be at O when y = O
=> =>
l=0X 2Bx =
=>
1 x=" =3x101 m=3mm
n,
0 l+X
20
~
t=4s Now, the speed of the central maxima is given by lcl)tJ 6DT V=[dt]= (10t)'d At t=4s,wehave v=
=>
13.
6DT OT d =6d 36
1 3 v ( )( exrn) (6)(~x10_,)
Fringe width,
3x10'° ms' =3 mms'
15.
When coherent, then IDC at centre is zero, so, we have the resultant intensity to be
1, =41 When·incpherent, then the sources will not interfere and it wi11 be a general illumination at the point, so we have the resultant
intensity to be, I, =21
=>
.!i.=2
p = '~ «).
=2.116
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I
Dual Nature of Radiation & Matter
IContents
     _______ _,
DUAL NATURE OF RADIATION & MATTER···············································............ 3. 1 Solved Practice Problems ..................................................................................................3.15 Practice Exercise Sets
:> Single Correct Choice Type Questions ..................................................................3.21 :> Multiple Correct Choice Type Questions ...............................................................3.33 :> Reasoning Based Questions (Assertion Reason Type) ........................................3.36 :> Linked Comprehension Type Questions (Paragraph Type) ...................................3.38 :> Malrix Match Type Questions (Column Matching Type) ........................................3.40 :i
Integer Answer Type Questions ......................••..•••...•••..••...•.................................3.43
Answers to In Chapter Exercises (ICE) & Practice Exercise Sets .....................................3.45 Solutions to In Chapter Exercises (ICE) .............................................................................3.48 Solutions to Practice Exercise Sets ...................................................................................3.52
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MATTER WAVES
Light possesses dual nature i.e. it behaves both as a wave and as a particl~. In some phe~omena e.g., interference, diffraction and polarisation, it behaves as waves because they are explained on the basis of Wave theory while in some other phenomena e.g. photoelectric effect, Compton effect, it behaves as particles (photons). Since nature demands symmetry, therefore deBroglie thought that matter must have dual nature. The particle nature of matter is well known and hence deBroglie thought that material particles must possess wavenature.
According to deBroglie a material particle in motion must have a wave like character and the wavelength associated with it is given by . p
For charged particles accelerated through a potential difference of V volts, Kinetic energy i.e. EK = qV ,.
p = momentum of the particle deBroglie assumed this expression in analogy with photon because momentum of photon is
,.
h p=
H m is the mass of particle and v the velocity, then momentum of particle is p =mv .
h ,/2tneV
m=9lxl0°"1 kg,
Substituting
h = 6 •62 X lO""'Js ,
e=l6x10"19 C,weget
J1~Q i..=J1~o A=1'% A A. =
=>
... (5)
X 1010
This is expression for deBroglie wavelength associated with electron accelerated through a potentjal difference of V . For neutral particles (like neutrons, atoms) at temperature T , kinetic energy of most of particles, EK= k8 T, where k8 is the Boltzmann's constant.
p
p=~2mEK.
... (4)
For electrons accelerated through a potential difference of V volts
1,,
A.=!!.
So, deBroglie wavelength ).. =.!!.._ mv If E1: is kinetic energy of particle, then
h
~2mqV
... (1)
where, h is the Planck's constant whose value is given by h =6.63x10°"' Js and
=>
... (3)
deBroglie wavelength )..
deBroglie's POSTULATE
,. =!!.
1..=h~2mEK
h ~2mk,T
... (6)
The wave nature is possessed by all particles neutral of charged. The wave nature was first verified by Davisson and Germer for slow electrons.
.•. (2)
{·:EK=;:}
Illustration 1 Obtain the deBroglie wavelength associated with thermal neutrons at room temperature (27 °c) . Hence explain why a fast neutron beam needs to be thermalised
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Advanced JEE Physics
Optics & Modern Physics
with the environment before it can be used for neutron diffraction experiments.
deBROGLIE WAVELENGTH ASSOCIATED WITH THE CHARGED PARTICLES
Solution
The energy of a charged particle accelerated through
Average kinetic energy of a neutron at absolute temperature T is
potential difference V is E =.!.mv' = qV
.!.mv' =!l_k T 2 2 ' =a, =a,
2
Hence deBroglie wavelength
L =!3.k T 2m 2
'l.=!!.= h p .J2mE
{·: p=mv)
'
Using the above formula, we get
p=~3mk,T
,.
= 12.27
.Jv
Eectnm
deBroglie wavelength is given by '/,.
h p
h )2mqV
,.
h ~3mk8T
= 0.286 Proton
.Jv
A
A
Given m, = 1.675x10" kg, k, = 1.38x1023 J mol' K1 T=27+273=300K, h=6.63x10" Js '/,.
6.63x10""' .J3 X 1.675 X 1027 X 1.38 X 1023 X 300
=a,
'/,.
deBROGLIE WAVELENGTH UNCHARGED PARTICLES
6.63x1020 m=1.45x10" m 4.56
'l.=1.4sA As this wavelength is comparable to interatomic spacing =a
WITH
For Neutron deBroglie wavelength is given as 10
,.
0.286 x 10 m ~E(in eV)
N~1
41t
=a>
h m!;,xilv ?. 41t
3.2
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Dual Nahtre of Radiation & Matter
This principle is universal and holds for all microscopic and
6.63x10" 3.32 X 1027 X 1911
macroscopic particles. The principle is also unaffected by experimental techniques.
).
If ~ = 0 , then 11p > oo and if 11p = 0 then ~ > oo ,i.e. if we are able to measure the exact position of the particle (say an electron) then the uncertainty in the measurement of the linear momentum of the particle is infinite. Similarly, if we are able to measure the exact linear momentum of the particle i.e., tlp = 0 , then we cannot measure the exact position of the particle at that time.
). = 1.04; 10rn m = 1.04
tJ
Viewertj lncide~
·
\
photo~: \
a
Original momentum of electron
ti /
~
Reflected
photon
~a,. Final momentum of electron
An electron cannot be observed without changing it's momentum This principle is also applicable to energy and time, angular momentum and angular displacement. Hence,
!'.EM;,;!!_
ill
A
QUANTUM NATURE OF LIGHT
Some phenomena like photoelectric effect, Compton effect, Raman effect could not be explained by Wave theory of light. Therefore quantum theory of light was proposed by Einstein who extended the Planck's hypothesis to explain Black Body radiation. According to quantum theory of light or radiation, the energy of an electromagnetic wave is not continuously distributes over the wave front Gust like the energy possessed by water waves). Instead Plank proposed that an electromagnetic wave travels in the form of discrete packets or bundles of energy called Quanta. So, according to Plank, "light is propagated in bundles of small energy, each bundle being called a photon and possessing en~rgy", given by he E=hv=,_
where, v is frequency, 1,. is wavelength of light, h is Planck's constant whose value is 663xl0_"Js and c=3x10 8 ms1
4,r PROPERTIES OF PHOTONS
M,/10 ;,; .!!_
1.
4,r
PROBLEM SOLVING TRICK(S)
a)
For numerical problems, we shall use
b)
Ax.l.p ~ h t>EAt z h &t.8 ~ h If the radius of the nucleus is r then the probability of finding, .the. electron inside the nucleus is L\x =2r and uncertainty in momentum is t\p =..!!.... .
2
3.
4.r
In the interaction of radiatio~ with matter, the radiation behaves as if it is made up of particles called Photons. This fundamental is also called as Corpuscular Theory of Light. So, light behaves both as a particle and a wave. All photons emitted by any source travel through free space with a speed equal to the. speed of light i.e. c=3x108 ms1 • Each photon has a definite energy depending upon the frequency v of the radiation and this energy is independent of the intensity. So,
,.
E=hv=hc (injoule)
Illustration 2
Find the de Broglie wavelength corresponding to the rootmeansquare velocity of hydrogen molecules at room temperature (20 °C) • Solution . Smce,
=
v
Vrms=
5.
M
~3RT
3x8.3lx293 2x 103
nns
4.
=1911 ms'
6.
h h Now, A==p mvrms
7.
Mass of one hydrogen molecule is given by
8.
m
2
6.02x1026
kg = 3.32 X 1027 kg
12375 u,.·IS m · A , then E =,_eV .
If the intensity of the light of given wavelength is increased, then there is an incr~ase in the number of photons incident per second per unit area on a surface. However, energy of the photon remains the same as long as the frequency or the wavelength is unchanged. The speed of the photon changes as it travels through different media due to the change in its wavelength. The frequency of the photon does not change when it goes from one medium to the other. In the situations wizen a photon collides witlz a material particle, the total energiJ and momentum remains conserved. However, the number of photons may not be conserved in a collision because during the 3.3
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Advanced JEE Physics collision photon(s) may be absorbed or new photon(s) may be created. 9. A photon is an electrically neutral particle which is not deflected by electric and magnetic field. 10. Rest mass of photon is zero. Since the mass of a particle m moving with a speed v is given by
14. Intensity of light (1) Energy crossing per unit area normally per second is called intensity or energy flux ,i.e. I=
m,
p
'
g
=>
I oc ]:_ r' 15. Number of photons falling per second (n) If P is the power of radiation and E is the energy of a p photon then n =  .
2
Now, for a photon, v = c, hence m0
E
=0
So from here we can conclude tl:iat the rest mass of a photon is always zero, i.e. we cannot have a frame of reference where the photon is at rest. 11. Dynamic or kinetic mass of a photon, is determined by using the Einstien's MassEnergy Equivalence, i.e., E=hv=mc2
=>
MOMENTUM.AND RADIATION PRESSURE
An electromagnetic wave consists of photons capable of transporting linear momentum. The linear momentum P possessed by an electromagnetic wave is related to the energy E it transports according to the relation E P=
hv
h m==c2 cJ...
hv
W
C
12 The Linear Momentum of a photon is found by using the deBroglie relation according to which, we have h C :\.=,where :l.=p V
=>
h
p==c ,.
However, this resuit is also obtained by using the fact that the total energy of a subatomic particle of rest mass m0 , moving with a velocity v , having momentum p is
given by E2 =p2c2 + ni;c4
If the wave is incident in the direction perpendicular to a surface and gets completely absorbed by the surface, then equation (1), tells us the linear momentum imparted to the surface. If surface is perfectly reflecting, the momentum change of the wave is doubled. Consequently the momentum imparted to the surface is also doubled. According to Newton's Second Law, the force exerted by an electromagnetic wave on a surface is given by the equation, F=t.p
M
l(t.E)
. t.p=  Fromequation(l), 
M c M
Now, for a photon, m0 =0, so we have from above
=>
expression that
F=!(t.E) M
Intensity (I) of a wave is the energy transported per unit area per unit time.
Since, E=hv=:
(2) A M
he
,.
=>
I=
=>
pc=
hv h p =~=i
=>
=>
t.E =IA M
13. The number of photons N , each of energy E , emitted from a source of monochromatic radiation of w~velength ;\. and energy W and power P
E
... (2)
C
E=pc
N=w
= P = radiation power)
I=41tr2
is the rest mass of the particle.
m,=m
(f
=:
At a distance r from a point source of power P intensity is given by
mgm,
where
!t
Substituting in equation (2), F = IA C
=>
=w =Pt hv
hv
t,.E
=>
I
F
 =pressure =A
C
I prad
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D11al Na hire of Radiation & Matter
!_ is also equal to the energy density (energy per unit C
(b)
p I(4itR') n=E
volume) u. Hence, P,., = u
... (3)
The radiation pressure is thus equal to the energy density (Nm2 = Jm"). At a perfectly reflecting surface the pressure on the surface is doubled. Thus, we can write,
I
{wave totally absorbed}
prad==U C
21 and P.., ==2u
{wave totally reflected}
C
=>
n
E
(1400)(4,r)(l.49 X 1011 ) 3.3xl019
2
l.18xl0 45
Illustration 5
A small plate of a metal is placed at a distance of 2 m from a monochromatic light source of wavelength 4.8 x 107 m and power 1 watt. The light falls normally on the plate. Find the number of photons striking the metal plate per square meter per second.
Solution Illustration 3
Find the energy, the mass and the momentum of a photon of ultraviolet radiation of 280 nm wavelength. Solution
(6.6 X 1034 )(3 X 10 8 ) (4.8x10')
.1 X _ J 4 25 10 19
Number of photons striking the metal plate per square meter per second is
Given, ,. = 280 x 10" m n=(f)(4:,,)
E =l!c . Smce, ,_
=>
E (4.316x1015 eV sec)(3x108 ms1 ) (280x10" m)
=> 4.6eV
Mass of photon is m = ~
Illustration 6
C
=>
4.6 X 1.6 X 1019
m
(3x10')
Find the number of photons entering the pupil of our eye per second corresponding to the. minimum intensity of white light that we humans can perceive (10" wm2 ) • Take the area of the pupil to be about 0.4 cm' and the average frequency of white light to be about 6 x 1014 Hz.
8.2 X 10J6 kg
2
Momentum of a photon is E P=
Solution
C
=>
p
1 1 n=( =4.82xl016 m2s1 19 ) . 4.125xl0 (4it)(2)2
4.6 X 1.6 X 1019 3xl0 8
Minimum intensity, I =1010 wm2
2.45 X 1027 kg ffiSl
Area of pupil, A=0.4 cm' =0.4x10< m Average frequency, v = 6 x 1014 Hz
Energy of one photon
Illustration 4
The intensity of sunlight on the surface of ear\h is 1400 wm2 • Assuming the mean wavelength of sunlight to be 6000 A, calculate
E=ltv=6.63xl034 x6xl014 J=4x1019 J Let n be the nllmber of photons crossing per square metre area per second.
the photon flux arriving at 1 m' area on earth perpendicular to light radiations, and (b} the number of photons emitted from the sun per second assuming the average radius of Earth's orbit is 1.49 X 1011 ffi ,
Now Intensity second
Solution
=>
(a)
(a)
Energy of a photon 12400 E = ltc = = 2.06 eV =3.3xo" J ,_ 6000 Photonflux
IA E
(l400)(l), 4.22x10 21 3.3x1019
=> =>
=>
= Energy incident per square metre area per
I= Total Energy of
11
photons
I = n x Energy of one photon Intensity 11 Energy of one photon 1010 wm2
I =:,',,,CC.2.5xl08 m2 s1 19 4xl0 J
So, the number of photons entering the pupil of our eye per second is
N = n (Area of the pupil)
=>
N
= 2.5 x 108 x 0.4 x 101 ., =104 s' 3.5
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Advanced JEE Physics Illustration 7 Find the number of photons emitted per second by a 25 W source of monochromatic light of wavelength
6600 A . What is the photoelectric current assuming 3% efficiency for photoelectric effect? Given h = 6.6 x 10"' Js . Solution
Energy of each photon is
Eh:
10 66 · ::~:;;_: '
19
3x10 J
Illustration 9
A cylindrical rod of some laser material 5 x 102 m long and 102 m in diameter contains 2 x 1025 ions per m 3 • ff on excitation all the ions are in the upper energy level and deexcite simultaneously emitting photons in the · same direction, calculate the maximum energy contained in a pulse of radiation of wavelength 6.6 x 10' m. If the pulse lasts for 107 second, calculate the average power of the laser during the pulse. Solution
Total energy emitted per second by 25 W source is
Total number of ions in the rod is
E=25J Number of photons emitted per second is 25 n= 8.33 X 1019 3 x1019
N
per unit volume
N=7.85x10 190 As all the ions deexcite simultaneously, the number of photons emitted in the same·direction is also 7.85x10 19 • So, the energy contained in a pulse of radiation of wavelength 6.6 x 107 m is
3 % of photons ) x (charge on) emitted per second electron
3 25 _,, I=x xl. 6 xl 0 =0.4A 100 3x1019 Illustration 8 A source emits monochromatic light of frequency 5.5 x 1014 Hz at a rate of 0.1 W . Of the photons given out, ·0.15% fall on the cathode of a photocell which gives a current of 6 µA in an external circuit.
(a)
Find the energy of a photon.
(b) Find the number of photons leaving the source per (c)
second. Find the percentage of the photons falling on the cathode which produce photoelectrons.
Solution
(a)
20
E = he x7.85x1019 A,
E
E=2.27eV (b) Number of photons leaving the source per second is
n, = O.l 5 x2.75x1017 =4.125x1014 100 Number of photoelectrons emitting per second .is 6x10_. n2   ~ ~ 3.75x1013 l.6x1019
So,
%age of Photons ) = n2 x 100 ( falling on Cathode n,
"
...
BASED ON PHOTON PROPERTIES / (Solutions on page 3.48)
, 1.
!
The intensity of direct sunlight before it passes through the ; earth's atmosphere is 1.4 kWm 4 • If it is completely : absorbed find the corresponding radiation pressure. \ I
! 2.
\ 3.
I
According to the maxwell theory of electrodynamics an electron going in a circle should emit radiations of frequency ] equal to its frequency of revolution. What would be the wavelength of the radiation emitted by a hydrogen atom in ground state if this rule is followed? An electron is accelerated by a potential difference of 25 volt. Find the deBroglie wavelength associated with it.
I
. 4. 13
= % %age of Photons ) = 3.75 x 10 x ( falling on Cathode 4.125 x 1014 100 9
23.55 J
,A: .... ,, _, .,., .. ,.,,,,,,. . 1~~~1e·1 ~·~v~:: :'~~:.:~·,,,t1H:,'" ~,,>~~;:a.lSEt~i5m4flN 1.
n=!'._=
O.l 2.75x1017 E 36.3x1020 Number of photons falling on cathode per second is
6.6x1034 x3x10 8 6.6 x107
Energy Average power P=.Tune 2355 P= J = 23.55 x 107 W = 235.5 MW 10' s
Since, E=hv=(6.6x10"')(5.sx10")=36.3x10 J
=>
(c)
the rod
N=(2x10 25 m3)x(3.14x(0.005)2 x5x102 m 3 )
Photoelectric current (I) is
I=(
=(Number of ions)x(Volume of)
15.
;
Find the number of photons emitted per second by a MW ! transmitter of 1O kW power emitting radiowaves ofj wavelength 500 m . lf 5% of the energy supplied to an incan9escent light bulb Is radiated as visible light, how many vi.Sible _light photons are .. .!31J1itted_by 1O_O_'!V~tt b_ulb~b~sume. waY.E!l~i:i91Ji. of_~IJ _visible
,.____... 3.6
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Dual Nature of Radiation & Matter photons to be 5600 A . Given h = 6.625 x 10" Js . 6.
Calculate the number of photons in 6.62 J of radiation
The minimum energy required for the electron emission from the metal surface can be supplied to the free electrons by any one of the following physical processes.
energy of frequency 1012 Hz. Given h = 6.62 x 1034 Js.
THERMIONIC EMISSION 7.
Monochromatic light of frequency 6 x 1014 Hz is produced by a laser. The power emitted is 2 x 103 W .
8.
(a)
What is the energy of each photon in the light?
(b)
How many photons per second, on the average, are emitted by the source?
Show that a free electron at rest cannot absorb a photon and thereby acquire kinetic energy equal to the energy of the photon. Would the conclusion change if the free electron was moving with a constant velocity?
9.
An electron microscope uses electrons accelerated by a voltage of 50 kV . Determine the deBroglie wavelength associated with the electrons. If other factors (such as numerical aperture etc.) are taken to be roughly the same, how does the resolving power of an electron inicroscope compare with that of an optical microscope which uses
yellow light? 10.
An electron and proton are possessing the same amount of kinetic energy. Which of the two have greater wavelength?
11.
An electron and a photon have same de Broglie wavelength (say 1 A ). Which one possesses more kinetic energy?
12.
An electron and a proton have same wavelength. Which one p_o~sesses more energy?
EMISSION OF ELECTRONS
As we are aware of the fact that metals have free electrons (negatively charged particles) which are responsible for their conductivity. However, the free electrons cannot normally escape out of the metal surface. If an electron attempts to come out of the metal, the metal surface acquires a positive charge and pulls the electron back to the metal. The free electron is thus held inside the metal surface by the attractive forces of the ions. Consequently, the electron can come out of the metal surface only if it is supplied some minimum energy to overcome the attractive pull of the metal. This minimum energy required by an electron to escape from the metal surface is called the work function of the metal. It is generally denoted by W or $0 and is measured in eV (electron volt). One electron volt is the energy gained by an electron when it has been accelerated by a potential difference of 1 volt, so 1 eV=1.6x1019
J
This unit of energy is commonly used in atomic and nuclear physics. The work function, W depends on the properties of the metal and the nahrre of its surface. The work function for platinum is the highest WP, = 5.65 eV, whereas it is the lowest for caesium i.e., W0 = 2.14 eV.
By suitably heating a metal, sufficient thermal energy can be imparted to the free electrons to enable them to come out of the metal. The free electrons so emitted are called as Therm ions. FIELD EMISSION
When the metal surface is subjected to very strong electric field of the order ranging from 10 3 vm1 to 108 vm1 , the electrons (beyond a certain limit) start coming out of the metal surface. This method of emission is dangerous and less efficient. This method of emission is also called the Cold Cathode Emission. PHOTOELECTRIC EMISSION
When light of certain minimum energy (or m1rumum frequency or maximum wavelength) illuminates or falls on a_ metal surface, electrons are emitted from the metal surface. TI1e emitted electrons are called pllotoelectrons. In case of Photoelectric emission, the rate of emission of photoelectrons is very low. SECONDARY EMISSION
When fast moving electrons strike a metal surface, then some of their energy is transferred to the free electrons of the metal. Due to this, when free electrons gain energy more than the work function, then they are emitted from the metal surface. These emitted electrons are called the secondary electrons. PHOTOELECTRIC EFFECT
The phenomenon of emission of electrons from a metallic surface by the use of light (or radiant) energy of certain minimum frequency (or maximum wavelength) is called photoelectric effect. The emitted electrons are called as photoelectrons. The phenomenon was discovered by Hallwach in 1888. For photoelectric emission the metal used must have low work function e.g., alkali metals. Cesium is assumed to be the best metal for photoelectric effect. To escape from the surface, the electron must absorb enough energy from the incident radiation to overcome the attraction of nucleus of the atom of the metal surface. The explanation to the photoelectric effect given by Einstein is based on the Law of Conservation of Energy. Before discussing the effect further, we must understand the following terms. A.
WORK FUNCTION (OR THRESHOLD ENERGY) (W,)
The minimum energy of incident radiation, required to eject the electrons from metallic surface is defined as work 3.7 C=:J
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Advanced JEE Physics function of that surface. It is the characteristic of a metal surface.
a)
W0 =hv0 = he (in joule), where 'I., v O = Threshold frequency and '/.
0
= Threshold wavelength
Work function in electron volt is given by W.{eV)=~ e'l.0 B.
'/.
b)
1237~ (in A)
0
THRESHOLD FREQUENCY
c)
(v0 )
The minimum frequency of incident radiations required to eject the electron from metal surface is defined as threshold frequency. If incident frequency v < v0 => No photoelectron emission.
d)
For most metals the threshold frequency is in the ultraviolet (corresponding to wavelengths between 200 and 300 nm), but for potassium and cesium oxides it is in the visible spectrum ( 'I. between 400 and 700 nm)
e)
C.
THRESHOLD WAVELENGTH
('1.0)
The maximum wavelength of incident radiations required to eject the electrons from a metallic surface is defined as threshold wavelength. If incident wavelength 'I.> 'l.0 , then No photoelectron emission will take place. Illustration 10 In an experiment on photoelectric effect light of wavelength 400 nm is incident on a caesium plate at the rate of 5 W . The potential of the collector plate is made sufficiently positive with respect to emitter so that the current reaches the saturation value. Assuming that on the average one out of every 106 photons is able to eject a photoelectron, find the photocurrent in the circuit. Solution
E = 12375 = 3.1 eV 4000 Number of photoelectrons emitted per second 5 (!.)( 10 3.lxl.6x10. q Ne (N) t=;=f e=ne
n=
. smce, =:,
19 )
= 1 x 1013 per second
= 1
i=(ne)=lxl013 xl.6xl019 =1.6x10' A=l.6 µA
LAWS OF PHOTOELECTRIC EMISSION
We thus have the following laws of photoelectric emission, derived from the experimental observations.
=
f)
For each emitting metal, there is _a certain minimum frequency v 0 (or maximum wavelength 'l.0 ), called the threshold frequency of the incident radiation, below (above which) which no emission of photoelectron takes place, no matter how great is the intensity. The value of v0 ( or 1..0 ) is different for different emitting surfaces. The process of emission of photoelectrons is an instantaneous process. There is no time lag ( < 10..,, s) between the incidence of radiation and the emission of photoelectrons. Photoelectric effect is a one photonone electron phenomenon i.e. even if photon has an energy sufficient to strike off 3 electrons (say) it can only strike off one electron with the excess energy being imparted to the struck off electron as kinetic energy. The number of photoelectrons emitted per second, that is, photoelectric current is directly proportional to the intensity of the incident radiation but is independent of_ the frequency (or wavelength) of light. The velocities (or the energies) of the emitted photoelectrons vary between zero and a definite maximum (vmax). The proportion of photoelectrons having a particular velocity is independent of the light intensity. The maximum velocity, vmax, and hence the maximum kinetic energy is independent of the intensity of the incident light, but depends on its frequency, increasing linearly with the increase of the frequency of the incident light.
EINSTEIN's EFFECT
EXPLANATION
OF
PHOTOELECTRIC
The wave theory of light could not explain the observed characteristics of photoelectric effect. Einstein extended Planck's quantum idea for light to explain photoelectric effect. According to his idea, the energy of electromagnetic radiation is not continuously distributed over the wave front like the energy of water waves but remains concentrated in packets of energy content hv, where v is frequency of is universal Planck's constant radiations and h (=6625x10""" Js). Each packet·of energy moves with the speed of light. The assumptions of Einstein's theory are a) The photoelectric effect is the result of collision of two particles, one of which is a photon of incident light and the other is an electron of photometal. b) The electron of photometal is bound with the nucleus by Coulomb attractive forces. The minimum energy required to free an electron from its bondage is called work function, W = $0 = hv 0 • c)
The incident photon interacts with a single electron and loses its energy in two parts
3.8
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Dual Nature of Radiation & Matter. (i)
Firstly, in getting the electron released from the bondage of the nucleus. (ti) Secondly, to impart kinetic energy to emitted electron.
d)
The efficiency of photoelectric effect is less than 1 %, i.e. less than 1 % of photons are capable of ejecting photoelectrons.
EXPERIMENTAL SETUP FOR PHOTOELECTRIC EFFECT
It consists of two conducting·electrodes, the anode (Q) and
cathode (P)· which are enclosed in an evacuated glass tube ,,.~
as shown in figure.
0,0~o