24.wave optics

T

Chapter Overview

,. Introduction • Newton's Corpuscular Theory ,. Huygen's Wave Theory • Coherent Sources • Interference of Light

~> Introduction

" Diffraction of Light "' Difference between Interference

and Diffraction g Polarisation of Light " Doppler's Effect in Light

The nature of light has been the source of one of the longest debates in the history of science for centuries. Different scientist's theories like Newton's corpuscular theory, Huygen's wave theory, Planck's quantum theory, Maxwell's electromagnetic theory, Einstein's photoelectric effect etc presented light as waves and particles. Finally the quantum electromagnetic dynamics completely explained the light and electromagnetic interactions. According to this theory, light shows wave like behaviour in certain situations eg, interference , and diffraction and it behaves like particles (photon) in other circumstances (eg, photoelectric effect).

In this chapter we deals about wave nature of light interference and diffraction.

R> Newton's Corpuscular Theory In 16 7({> , Newton propounded the corpuscular theory of light to define

its nature . He assumed that

/

936 , I Chapter 24 • Wave Optics

(i) Light is composed of nearly massless tiny particles called corpuscles.

(ii) The corpuscles travel in a straight line with a tremendous speed in a homogeneous medium.

(iii) The speed of corpuscles changes when they travel from one medium to the other.

(iv) The corpuscles oflight of different colours are of different sizes.

(v) When the corpuscles strike on retina, we get the sensation of vision.

~> Huygen's Wave Theory In1678, a Dutch scientist, Christian Huygen propounded the

wave theory of light. According to him (i) Light travels in the form of waves.

(ii) These waves travel in all the directions with the velocity of light.

(iii) The waves of light of different colours have different wavelengths.

(iv) Initially the light waves were assumed to be longitudinal. But later on while explaining the phenomenon of polarization the light waves were considered to be transverse.

(v) The whole universe with all matter and space is filled with a luminiferous medium called ether of very low density and very high elasticity.

(vi) Huygen's theory could explain reflection, refraction, diffraction, polarization but could not explain photoelectric effect and Compton's effect.

(vii) Wave theory introduced the concept of wavefront.

Wavefront A surface drawn

at any instant in the medium affected by

p

the waves originated Rays from a source, on which S -E----+---r----'-- Q

each· particle vibrates in same phase is called the wavefront.

The wavefront may also be defined as the hypothetical surface on which the light waves are in same phase.

As the wave travels forward .

R

S =source of light, AB =wavefront and SP, SO

and SR are rays of light.

forward, the wavefront also moves

The perpendicular to the surface of a wavefront gives the direction of light ray. The disturbance from the source propagates in all directions, with the same speed c. Hence, after a time t, all the points on the hypothetical sphere of radius ct, with the point source Sat the centre, are in the same phase of vibration. Hence, this hypothetical sphere acts as a spherical wavefront of light. The energy transfers along the direction of rays.

Wavefronts in Different Cases (a) Light emerging from a point source Wavefronts are

spherical and eccentric with point source at their centre

as shown in figure by circles 1, 2, 3 and 4. Rays are radial as shown by arrows.

In spherical wavefront

Amplitude oc _!. r

I . 1

ntens1ty oc 2 r

Direction of propagation

(b) Light coming out from a line source Wavefronts 1, 2 and 3 are cylindrical and co-axial with the straight source as their common axis.

(c) When light sources is emitting parallel rays, or when the light is coming from a very far off source Wavefronts will be planes as shown in figure.

Plane wavefronts

Huygen's Principle of Secondary Wavelets In 1678, in order to explain the propagation of wave in a

medium, Huygen propounded a principle known as Huygen's principle of secondary wavelets by which at any instant we can geometrically obtain the position of wavefront. He made following three assumptions

(i) Every point on a given wavefront (called primary wavefront) can be regarded as fresh source of new disturbance and sends out its own spherical wavelets called secondary wavelets.

(ii) The secondary wavelets spread in all directions with the velocity of wave (ie, velocity of light).

(iii) A surface touching these secondary wavelets, tangentially in the forward direction at any instant gives the position and shape of the new wavefront at that instant. This is called secondary wavefront.

A'

B'

(a)

> Coherent Sources

(b)

'TWo sources are said to be coherent if they have the same frequency and the phase relationship remains independent of time. In this case, the total intensity I is not just the sum of individual intensities ! 1 and ! 2 due to two sources but also includes an interference term whose magnitude depends on the phase difference at a given point.

I= ! 1 + ! 2 + 2JT;I; cos<j>

where ~is the phase difference.

The 2M cos <1> averaged over a cycle is zero if

(a) the source have different frequencies. (b) the source have the same frequencies but not constant

phase difference. (c) for such incoherent sources I =I 1 +I 2 , where ~ does

not change with time, we get an intensity pattern and the sources are said to be coherent.

In practice coherent sources are produced either by dividing the wavefront or by dividing the amplitude (as in the case of thin films, Newton rings etc) of the incoming waves.

·> Interference of Light When two light waves of exactly equal frequency having

phase difference which is constant with respect to time travelled in same direction and overlap each other, then intensity is not uniform in space.

At some points the intensity of light is maximum (more than the sum of individual intensities of those waves), while at some points the intensity of light is minimum (less than the difference of individual intensities of those waves).

Thus, formation of maximum intensity at some points and minimum intensity at some oilier points by the two identical light waves travelling in same direction is called tlle interference of light.

The interference at the points where the two waves meet in same phase, ie, the intensity of light is maximum, is called the constructive interference while at the points where the two waves meet in opposite phase, ie, the intensity of light is minimum is called the destructive interference.

Chapter 24 • Wave Optics 937

Resultant wave Second wave

(a) Constructive interference

First wave Resultant wave

(b) Destructive interference

Conditions of Maxima and Minima Lety1 =A1 sincotandy2 =A2 sin (cot+~) be two sin1ple harn10nic

waves of same frequency travelling in tlle same direction, A 1 and A2

are tlleir amplitudes, ~ is tlle initial phase difference between tllem and co/2n is the common frequency of the two waves.

By the principle of superposition the resultant displacement is y = y 1 + Y2 = A 1 sin cot+ A2 sin (cot+ ~)

From this expression, the resultant amplitude is given by A 2 =A~+ A~+ 2A1A2 cos<j>

The resultant intensity I is given by I =A2

So, I= A~+ A~+ 2A1A2 cos<j>

I == I 1 + I 2 + 2M cos <I>

Thus, the resultant amplitude (or the resultant intensity) at a point depends on the phase difference at that point between the two superposing waves.

(i) For constructive interference (maximum intensity) The intensity I is maximum, when cos ~ = + 1, ie, when phase difference ~ = 2nn; n = 0,1, 2, ... etc,

n = 0 stands for zero order maxima n = 1 stands for Ist order maxima n = 2 stands for lind order maxima

Path difference = n'A

So, I max= A~+ A~+ 2A1A2 = (A1 + A2i (ii) For destructive interference (minimum intensity)

The intensity I is minimum when cos 4> = - 1 ie, when phase difference

~ = (2n -1)n; n = 1, 2, .. . etc. n = 1 for first order minima, n = 2 stands for second order

minima 'A path difference = (2n -1)-

2 So, I min= A~+ A~- 2A1A2 = (A1 - A2 )

2

Figure shows the variation of intensity I with the phase difference !]> due to the superposition of two waves of equal amplitudes. The graph is called the intensity distribution curve. , .' . '. . ,

2 .

when!]> = 0, ±21t,±41t, ... , !max= 4A

and when <1> = ± 1t, ±3, ±51t , ... ,!min= 0

Instance 1 Light waves from two coherent sources of having i11tensity ratio 81 : 1 produce interferenr;e. Then the ratio of maxima and minima in the interference pattern will be

( ) 18 (b) 16 a

23 25

25 (c)

16

Interpret

or

(d) 23 18

. I A2 81 GIVen _l = - 1 = -

I 2 \ A~ 1

~=~ A2 1

A1 = 9A2

I max - (Al + Az)2

I min - (Al- Az)2

From Eq. (i), we have

I max (9Az + Az)2

Imin = (9Az-Az)2

c1oi 25 C8i = 16

... (i)

Instance 2 Light waves from two coherent sources having intensities I and 2I cross each other at a point with a phase difference of 60°. The intensity at the point will be

(a) 4.414 I (b) 5.455 I (c) 4 I (d) 6.441 I

Interpret Here, I1 =I, 12 = 21 and !]> = 60°

we know that the amplitude A of resultant wave is

A= ~a2 + b2 + 2 abcos<jl

A2 =a2 +b2 +2abcos<jl

We also know that I oc A2

:. Resultant Intensity, I= 11 + 12 + 2ji'J"; cos<jl

= I + 21 + z.J I x 21 cos 60°

= 4.414I

Young's Double Slit Experiment In the given figure, Sis a slit illuminated by a monochromatic

light. Light waves from the slit S are incident on slits 51 and 52•

Slits 51 and 52 are situated such that the distance SS1 is equal to the distance 552• Hence, the waves emanating from the slit S reach 51 and 52 simultaneously, ie, the waves at 51 and 52 are in same phase. Infront of the slits 51 and 52, there is a vertical

. screen P. We are to find the positions of bright and dark fringes on the basis ofYoung's experiment.

Suppose the distance of the point P from centre of the fringe pattern be x then in b.S2PQ.

(S2P)2 = (S2Qi + (PQi = D2 +( x +%Y

In b.S1PM,

(SlP)2

= (SlM)2 + (PM)

2 = D

2 + (X- %Y So, (S2P) 2 - (S1P) 2 = 2dx

or (SzP + S1P) (SzP- S1P) = 2dx Now since, P is very close to 0, we can have to a first

approximation assuming that SzP + S1P = D + D = 2D

dx .. 52P-51P=D

So, path difference between the interfering waves is dx . D

(i) Position of bright fringes or maxima on the screen Constructive interference occurs when the path difference is an even multiple of A/2.

xd A ie, -=2n-=nA

D 2

or nAD

X=--d

(where n = 0, 1, 2, 3 , .. )

When n = 0 gives central fringe, hence , x = 0, ie, x0 = 0

AD n = 1, x1 =-

d 2AD

n = 2, x 2 =-d-

nAD n=n,xn =d If

· (1st bright fringe)

(2nd bright fringe)

(n'h bright fringe)

(ii) Position of dark fringes or minima on the screen Destructive interference occurs when the path difference is an odd multiple of A/2.

ie,

or

xd A -=(2n-l)D 2

(2n-l)AD X=

2d (where n = 1, 2, 3 ... )

Let the positions of the minima be represented by x~ instead of xn (so that there is confusion as xn represents the 'position of nth maxima).

When n = 1, xi= AD 2d

n = 2 x' =~AD , 2 2 d

, (2n-l) AD n = n, xn =--2-d

(1st ~nima)

(2nd minima)

(n'h minima)

Fringe width The distance between two consecutive bright or dark fringes is called fringe width.

Let the distance of nth and (n + l)th bright or dark fringes from the central fringe are xn and xn +l' then

and

nW X=-

n d

(n + 1)W xn+l = d

:. Distance between (n + 1)th and nth 'fringes

(n + 1)W nW ')..]) x - x = - --=- [n + 1- n]

n+l n d d d

')..]) or xn+l -xn =d

:. Width of bright or dark fringe, ~ = ')..]) d

Important Features 1. Consider two coherent sources 51 and 52• Suppose two

waves emanating from these two sources superimpose at point P. The phase difference between them at P is ljl (which is constant) . If the amplitudes due to two individual sources at P are A1 and A 2, then resultant amplitude at P, will be

s1

p

A= ~Af + Ai + 2A1A2 cos<!J

Similarly, the resultant intensity at P is given by

I= / 1 +12 +2M cos<!J

Here, 11 and 12 are the intensities due to independent sources. If the sources are incoherent then resultant intensity at P is given by

I = Jl + lz 2. For two coherent sources the resultant intensity is given

by

Putting,

We have,

l=l1 +12 +2Mcos<!J

ll=lz=lo

I= !0 + !0 + 2~10 x /0 cos<!J

Simplifying the above expression, we get

I= 4J0 cos2~ 2 3. Resultant intensity due to two coherent sources is given by

I =!1 +!2 +2M cosljl

!max= cJf; + ji;)2 where cos <I>· = + 1 , and /min = cJf;- Fz )2 where cos ljl = - 1


If both the slits are of equal width, 11 = !2 = 10 (say) I max = 4!0 and Jmin = 0

Thus, I = I cos2 ! max 2 4. Optical path length We can show that a thickness t in

a medium of refractive index J..l is equivalent to a length J..lt in vacuum (or air). This is called optical path length. Thus, optical path length= J..tt.

5. If the whole YDSE set up is taken in another medium then A. change, so~ changes 1

eg, in water A = 2::9_ w f..lw

I =? ~w =fu

f..lw

6 . Displacement of fringes If a thin transparent plate of thickness t and refractive index J..l is placed in the path of one of interfering waves (say in path S1P) , then effective path in air is increased by an amount (J..l- l)t due to introduction of plate. Effective path difference in air

For maxima

= S.J>- [S1P + (J..l- 1)t]

dx =(S2P-S1P) -(J..t-1)t=-n -(J..t-1)t

D .

dx _ n -(J..t-1)t=nA D

D xn =d[nA+(J..l-1)t]

Fringe width ~ = xn+l - xn = DA d

It means fringe width remains unchanged. Position of maxima in absence of plate.

nDA (xn)t=O = -d-

Displacement of fringes

L'.x = xn - Cxn )t=O '

D nDA =-[nA+(J..t-1)t]--

d d

= E.cJ..t-1)t = ~CJ..t-1)t d A

7. In the problems of YDSE our first task is to find path difference. Let us take a typical case. In the figure shown, path of ray 1 is more deviated than path of ray 2 by a distance

940 Chapter 24 • Wave Optics

L1x1 = d sin r:i and L1x2 = (J..l1 -1) t1

and path of ray 2 is greater than path of ray 1 by a distance

L\x3 = d sin~ and L1x4 = J!-!2 -1) t2

Therefore, net path difference is / LlX = (LlX1 + LlX2)-(LlX3 + LlX4)

Interference in Thin Films Interference effects are commonly observed in thin films

such as thin layers of oil on water or the thin surface of a soap bubble.

c

A

Consider a film of uniform thickness t and index of refraction J..l as shown in figure. Let us assume that a monochromatic ray of light AB incident at angle i, on the upper surface PQ of the film. The ray of light will suffer reflection and refraction. The rays BC and FG are said to interfere in reflected system and the rays DE

and HI are said to interfere in transmitted system. It can be shown that (i) In the reflected system

For constructive interference 2J.!t cos r = (2n - 1) A/2

For destructive interference or 2!-lt cos r = n'A.

(ii) In the transmitted system For constructive interference

2!-lt cos r = nA. For destructive interference

2J.!t cos r = (2n- 1)A/2 Obviously, conditions for thin film to appear bright/dark in

reflected system are just the reverse of those in the transmitted system.

Instance 3 In a YDSE, if D = 2 m, d = 6 mm, /.. = 6000 A, then the fringe width and the position of the 3'd maxima are

(a) 0._2 mm, 0.6 mm (b) 0.8 mm, 0 .1 mm (c) 1.2 mm, 0.2 mm (d) None of these

Interpret We know that . . AD 6000 X 10-10

X 2 Fnnge Wldth ~ =-=

3 = 0.2mm

d 6x 10-Position of 3rd maxima

3AD Y3 = d = 3~ = 3 x 0.2 = 0.6 mm

Instance 4 White light is used to illuminate the two slits in a Young's double slit experiment. The separation between the slits is b and the screen is at a distanceD > > b from the slit. At a point on the screen directly infront of one of the slits. Find the missing wavelengths.

b2 b2 b2 b2 b2 b2 (a) (b) D' 3D' SD ... 2D' 4D' SD .. .

b2 b2 b2 (c)

3D' SD' 6D ... (d) None of these

Interpret According to theory of interference, position y of a point on the screen is given by

D y =- LlX

d

For missing wavelengrhs intensity will be minimum ( = 0) , if L1 x = (2n- 1) A/2

D(2n -1)A So, y=--- - -

2d Given, d =band y= b/2

b2 A = where n = 1, 2, 3, ..

(2n -1)D So wavelength when n = 1

b2 A----1- (2-1)D

b2 n = 2, A2 (4-1)D

n = 3, b2 b2

- --=- etc will be (6-1)D SD

absent or missing at point P.

Instance 5 The intensity of the light coming from one of the slits in a Young's double slit experiment 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.

(a) (.J3 + 1 )2

.J3 -1

(c) (-}2 -1 )2 -J2 + 1

(b) (-12 + 1 )2 -J2 -1

(d) None of these

Interpret !max __ (.jf; + Fz)2

!min .jf; -.[!;


Since, Interpret Here, d = 0.1 mm = 10-4 m

Instance 6 In YDSE, the two slits are separated by0.1 mm and they are 0.5 mfrom the screen. The wavelength of light used is 5000 A. What is the distance between 7th maxima and ll th minima on the screen ?

(a) 5.65 mm (b) 6.75 mm (c) 8.75 mm (d) 7.8 mm

D = 0 .5 m, f..= 5000 A= 5.0 x 10-7 m

· · Llx =(Xu )dark- (X7 )bright

(2xll-1)AD 7/..D

2d d

ilx= ?f..D = 7x5x1o-7 x0.5 8_75 x10

_3 m 2d 2x10-4 ~

= 8 .75mm

lntext Questions 24.~.1

(i) What is the geomenical shape of wavefront of light emerging out ofa convex lens; when point source is placed at

(ii) Can twoi,J:tdependent sources of light be coherent ? What happens to the interference pattern, if the phase diffel1enc:j'! blemreeJ!l. sources connnuously varies?

(iii) Why does an excessively thin film appear black in reflected light?

(iv) What is the effecjpn the i.J:!,terference pattern in Young's double slit experiment due to each of Jhe following ot)eratio:ils ? (a) The width of the slits are increased equally. (b) If the source is not exactly on the centre line between the slits.

> Diffraction of light Diffraction of light is the phenomenon of bending of light

around corners of an obstacle or aperture in the path of light. This bending light penetrates into the geometrical shadow of the obstacle. The light thus deviates from its linear path. This deviation is more effective when the dimensions of the aperture or the obstacle are comparable to the wavelength of light.

The phenomenon of diffraction is divided mainly in the following two classes

(a) Fresnel diffraction (b) Fraunhofer diffraction

1. The source is at a finite distance.

2. No optical.s are required.

3. Fringes are not sharp and well defined.

Fraunhofer Diffraction

The source is at infinite distance. Optical.s are in the foi:m of collimating lens and focusing lens are required.

I Fringes are sharp and well 1 defined.

1. Fraunhofer's arrangement for studying diffraction at a single narrow slit (width = a) is shown in adjoining figure. Lenses, L1 and L2 are used to render incident light beam parallel and to focus parallel light beam

Slit Screen ,

2. Angular position of nth secondary minima is given by

. e A. sm =9= n-a

and linear distance

X = De = nDA. = nf2A. n a a

where f2 is focal length of lens L2 and D = f2. Similarly, for nth maxima, we have

sine= e = (2n + 1)/... 2a

and x = (2n + 1)DA = (2n + 1) f 2 A. n 2a 2a

3. The angular width of central maxima is 29 = 2

" or linear a

width of central maxima

= 2Df.. = 2f2 /..

a a The angular width of central maxima is double as compared to angular width of secondary diffraction maxima.

4. Condition of diffraction minima is given by a sinS= nf..

where n = 1, 2, 3, 4, ... But the condition of secondary diffraction maxima is

a sinS= (2n + 1)2:. 2

where n = 1, 2, 3, 4, ... 5. Diffraction at plane grating When polychromatic

or monochromatic light of wavelength A. is incident normally on a plane transmission grating, the principal maxima are

where (e +d) sinS= nf..

n = order of maximum 8 = angle of diffraction

e + d = grating element d = opaque part


6. Diffraction of circular aperture When monochromatic light of wavelength A is incident on a circular aperture of diameter d, then angular width of dark fringe

dsin8=1.22A.or sin 8 = l.Z21c d

Angular radius of central maximum

. 8

1.221c Sill =--

d

If 8 is small, sin 8 = 8 = 1.22"-d

Diffraction pattern is as shown below

Principal maxima

Secondary

3/Je

• Difference between lnterterence and Diffraction 1. Interference is the superposition of waves from two

different wavefronts, whereas diffraction is due to superposition of wavelets from different points of the same wavefront.

2. Interference all maxima are equally bright and minima equally dark (perfectly black). In case of diffraction maxima are of decreasing intensity, minima are not perfectly black.

3. In case of interference, there are large number of maxima and minima, but in case of diffraction bands are few.

Interference

,

Instance 7 In a single slit diffraction experiment first minima for A1 = 660 m coincides with fi rs t maxima for wavelength A2, then A2 will be equal to ,

(a) 240 nm (b) 345 nm (c) 440 nm (d) 330 nm

Interpret Given, A = 660 nm Position of minima in diffraction pattern is given by

a sin 6 = n'A For first minima of A1, we get

asin81 = 1/....1

or . e A.! sm 1 = -a

For the first maxima approximately of wavelength A2

. 3, asm82 = -~~.2 2

. e 3/....2 sm z = 2a

The two will be considered if,

81 = 82 or sin81 = sin82

A. 3/.... 2 2 . _l = --2 or /.... 2 = - /....1 = - x 660 nm a 2a 3 3

/.... 2 = 440 nm

::~ Polarisation of light Light is an electromagnetic wave in which electric and

magnetic field vectors vary sinusoidally, perpendicular to each other as well as perpendicular to the direction of propagation of wave of light.

The phenomenon of restricting the vibrations of light (electric vector ) in a particular direction, perpendicular to the direction of wave motion is called polarisation of light. The tourmaline crystal acts as a polariser.

Thus, electromagnetic waves are said to be polarised when their electric field ve~tor are all in a single plane, called the plane of oscillation/vibration. Light waves from common sources are oscillation/vibration. Light waves from common sources are unpolarised or randomly polarised.

The plane ABCD in which the vibrations of polarised light are confined is called the plane of vibration. The plane EFGH which is perpendicular to the plane of vibration is defined as the plane of polarisation.

Brewster's Law

A ---------,o E : -----r-----7F

I /

I /

-----t- --G I I a ---------c

According to this law, when unpolarised light is incident at an angle called polarising angle, iP on an interface separating air from a medium of refractive index !-! , then the reflected light is fully polarised (perpendicular to the plane of incidence) , provided

This relation represents Brewster's law. Note that the parallel components of incident light do not disappear, but refract into the medium, with the perpendicular components .

Note If ~l =tan iP then iP + r =goo

siniP From Brewster s law,~= taniP =--.

COSIP

siniP and from Snell s law,~=-.-

smr cosip=sinr

sin (goo iP) =sin r Hence, or

or

law of Malus

r=goo iP i + r =goo p

1 tani = ~=--

P sinic cot iP = sinic (i, =critical angle)

When a beam of completely plane polarised light is incident on an analyser, the resultant intensity of light (I) transmitted from the analyser varies directly as the square of cosine of angle(8) between plane of transmission of analyser and polariser.

ie, .I = cos2 8

If intensity of plane polarised light incidenting on analyser is !0, then intensity of emerging light from analyser is

I 0 cos2 8

~ \ ·n No::_te:;___,_ ___ _ We can prove that when unpolari sed light 'of intensity /0 gets polarised on

f. 1

passing through a polaroid, its intensity becomes hal 1e, I= ~/0 2

Polaroids Polaroids are thin and large sheets of crystalline polarising

material (made artificially) capable of producing plane polarised beams of large cross-section.

A polaroid sheet can also be obtained from a sheet of polyvinyl alcohol. When such a sheet is subjected to a large strain, the molecules get oriented in the direction of the applied strain. When the sheet is impregnated with iodine, the material becomes dichroic. The polaroid sheet so obtained is called H-polariod.

If the stretched sheet of polyvinyl alcohol is heated in the presence of a dehydrating agent such as HCl, it becomes strongly dichroic as well as very stable. Such a polaroid sheet is called K-polaroid.

A polaroid has a characteristic plane called transm1ss1on plane . When unpolarised light falls on a polaroid, only the vibrations paralled to the transmission plane get transmitted

Polariser

,


Uses- - ~' -t t 1 ..... ' • ~

' L In sun glasses Polaroids are ·use-d· in sun glasses. They protect the eyes from the glare.

2. In window panes The polaroids are used in window P'lnes of a train and especially of an aeroplane . They help to contr;ol the light entering through the windows.

3. In three dimensional motion pictures The pictures taken by a stereoscopic camera, when seen with the help of polaroid spectacles, create three dimensional effect.

4. In wind shield in automobiles The wind shield of an automobile is made of polaroid. Such a wind shield protects the eyes of the driver of the automobile from the dazzling .light of the approaching vehicles.

In addition to the above uses, polaroids are used as filters in photography to produce and detect the plane polarised light in the laboratory and to study the optical properties of the metals.

Instance 8 Light reflected from the surface of glass plate of refractive index 1.5 7 is linearly polarised. What is the angle of reji·action of glass ?

(a) 32.SO

(c) 30.2° (b) 42S

(d) 44S

Interpret Given that ~t = 1.57 According to Brewster's Law

~l =tan iP = 1.57, iP = tan-1 (1.57) = 57.5° As r = 90° ~ iP orr= 90° ~ 57.5° = 32.5°

> Doppler's Effect in light The phenomenon of apparent change in frequency

(or wavelength) of the light due to the relative motion between the source of light and the observer is called Doppler's effect in light.

When the source of light and the observer approach each other, the frequency of the light appears to be more than the actual frequency of the light emitted from the source (or the wavelength of the light appears to be lesser than the actuial wavelength). On the other hand, when the source of light and the observer recede from each other, the apparent frequency of the ljght decreases or the apparent wavelength increases. The change in frequency (or the wavelength) of the light for the given relative velocity of the source and the observer is independent of the fact, whether the source is in motion .or the observer is in motion.

Consider that a source of light emits light waves of frequency v and wavelength A. If the light travels with velocity c, then

c = , . A. A stationary observer will receive v waves per second.

In case , the observer moves towards the source of light with velocity v along the direction of propagation of light, then the number of waves received per second ie, apparent frequency of light will be

v' = v + number of waves contained in distance equal to v

v v Y+-=Y+ --

A C/Y

v' = v (1 + ~ )

or v' = v (1+~ ) •


It follows that v' > v ie, when the source and the observer approach each other, apparent frequency of the light increases.

When the source of light and the observer move away from each other, then the apparent frequency of light can be obtained from above equation by changing v to -v. Therefore, when the source of light and the observer move away from each other, the apparent frequency of the light is given by

v' = v(l-~) It follows that v' < v ie, the apparent frequency of light

aecreases, when the source of light and the observer move away from each other.

For apparent frequency of light may be written in the combined form as

v'=v(l± ~) Doppler shift. The apparent change in the frequency (or

the wavelength) of the light due to relative motion between the source and the observer is called Doppler shift.

The apparent change in frequency is given by

v'-v = ±~v c

The apparent change in frequency v'- vis denoted by ~v and is called Doppler shift. Therefore,

v ~v = -v

c The positive sign is to be considered, when the source and

the observer approach each other and the negative sign, when the two move away from each o,ther.

,.,

by Doppler shift in terms of the wavelength of the light is given

~A.= ±~A. c

In equation, the negative sign corresponds to the case, when the source of light and the observer approach each other. The negative value of M indicates that the wavelength of the light decreases, when the source of light and the observer move towards each other. It is called blue shift.

On the other hand, in equation the positive sign corresponds to the case, when the source of light and the observer move away from each other. The positive value of D.A. indicates that the wavelength of the light increases, when the source of light and the observer move away from each other. It is called red shift.

Instance 9 The spectral line for a given element in light received from a distant star is shifted towards the longer wavelength by 0.32%. Deduce the velocity of star in the line of sight.

Interpret Here, M = 0·032 (postion, as the .shift is towards

A 100 longer wavelength) Now, Doppler shift is given by

~A= -~A c

V = - !:iA C =- 0.032 X 3 X 108

A 100

=- 9.6 x 104 ms-1

The negative sign indicates that the star is receding.

uestions 24.2

,

Chapter 24 "' Wave Optics 945

Chapter Compendium

1. Ray of light is drawn perpendicular to the wavefront.

2. Huygens' Piinciple (i) Each pomt on given Or primary wavefront acts as a.

source of secon~ary:vavelets, sending out disturbance~ all directions in a similar manner as the origil1al sogrce oflight. ·

(ii). The new positi<m of the wavefront at any instant is the em,.~lope of the secondary wavelets at that instant.

3. If two coherent waves with intensities 11 and lz are superimposed with a phase difference off, the resulting wave intensity is

I= 11 + 12 + 2~I1 12 cos$

For maxima, x"' .n'A (A.= optical path difference)

Forminima,x=(rt-~)A. or (n+~)A. 2 2

Phase difference, q, = 27t (x) 'A

Ratio of maximum to mininmm intensity

lmax _ Cal + az)2

[min - (al- azP f1 P1 a~

Also, Iz = Pz =a~ 4. Young's double slit experiment

(i) .. For maxima ,

or y = (ii) For rninirna,

DA- 3DA Y =+- +-·- so on - 2d ,_ 2d ?·

(iii) Fringe width, p = AD d

(iv) When a fil:nl of thickness t and refractive index f.t is introqw:ed in the path?f one ofth~source's light;then fringe shift occurs as the optical patl;ldifference changes. Optical path. difference at P

,

L'lx = S2P"-(S1P·qtt t ]

= S2P- 5 1 P- (!+ -l)U y . (d/D)--; (Jl - l)t

The fringe shift is given by, 4)>.= D(J.t-l)t d

(v) Intensity variation on screen

IfJ0 ~. the intensity of light beam coming from each slit, then the resultant intensity at a. pqi.):lt where they have a phase difference pf ¢ is given by

I== 410

cos2 %, where q, = 27t(d~ine)

5 . . Diffraction is the phenomenon of bending ()flight round the sharp corners and spreading into the regions of .geometrical shadow is called diffraction. .

6. Diffraction from a slit (i) Angular position of the n th secondary minimmn

nA. a

(ii) Distance of the nth secondary nDA.

centre of the screen xn = -;-

maximum from the

(iii) Angular positions of the nth

9,_ (2n + l)A.

secondary m imum

n··-2a

(iv) Distance ofttte n'h secondary maximum from the c~ntrt=

of the screen x' = (Zn + l)DA. 11 2a

( v) Width of a secondary maximWP- or mininlurn p =

(vi) Width of the centraLmaximum p0 = 2,Df:: a

a

phenomenon due to which vibrations light restricted in a parti<;ular plane is called polarisation.

8. Brewster's Law states that when light is incident at polarising angle, the refl:ected and refracteq rays are perpendicular to each other. Mathematically J.l=tan iP

9. Law o~Malus states that when a completely plane po~ris~d light beam is incident on an analyser tlie intensity of tht= emt;rgent: light varies as the square pf the cosh1e of the angle between the plane of trans:rni5~ion of ·the analyser and the polariser Mathematicap.y e

10. Doppler shift.

(a) Llv=±~v c

(b) Ll'A= c

Example 1 Ifamplitude rqtio of two sources producing interference is 3 : 5, th,e ratio pfinten,sities bfmaxima cirid'minima is

(a) 5 : 3 (b) 16 : 1 (c) 25 : 16 (d) 25 : 9

Solution We know in ~terlere~ce, ,

1max = c.ji; + JI;Y while 1min =.c.Ji; ~ J[;y I • J , I . 1 I l • • ~

' ' ' ' I;,~ _ C.ji; + .ji;y _ (A1 + A2 ) 2

!min- (.ji;-;~12 ) 2 -:-' (Al -A2)2

According to problem . ·. ·. ·. ·

'1 ~·, rll r ~~ ·~ lr{,ax ~ (3+5)2

; 16, -1 ' .

.·.: .... ·'·' flz' ·s 'dmin (3 '- 5)2 1 • r ~ t , ' '~.: ·· ' t • ' 1 •• - ... r, I:.'

·'

'.' ''.

Examp'le 2 . ' Monochromatic green light; of wavelength 5 x 10-7 m illuminates a pair of slits 1 mm apart. The separation of bright lines on the interference pattern formed •on a screen 2m away is

· (a) Ldmm1 • • • (b) o.b1 mm (c) 0.1 mm (d) 0.25 mm

Solution Separation ofprig~t line.$ o.r.fl;i.nge width

D'A Sxl0-7 x2 _3 •,· .. ·-.~=~-=. · ··· ·:3·· m=lO m=l.O·mm .. ·,·.· ! J -~~~-r 1 1~ ~ fll ,' lQ,- •JI 11,·;·1 \ .~,

Example 3 Young's double slit eX:p~riment is'· ccirried out wing microwaves of wavelength A "''-·3 em. Distance between the slits is d = 5 em and the distance between the plane of slits and the screen is D = 100 em. Then ·what is the number 6pin'ax:imas 'and their positions on the screen? ' -

·• ·(a),)' 3/0 1Qnd ±'7s ~J1i' ~ r - I

(c) 3, 1 and± 55 em Solution

·(b} 4;1 and-± 6'0 d;z (d) None of these

PT s1 Y

s2 ~----- ----------- Jl I I

The maximwn path difference that can be produced = distance between the sources ie, 5 em. Hence, in this condition we can have only 3 maximas one . central maxima and two on its either side for a path difference of/.. or3 em. Now, for maximwn intensity at P

S2P-S1P ='A ,

or ~(y + d/2)2 + D2 - ~(y- d/2)2 + D2 = i Putting, d = 5 em, D = 100 em and A = 3 em, we get

y = ± 75 em Thus, the three maximas will be at

y = 0 andy = ± 75 em

Example 4 In Young's double slit experiment, an interference pattern is obtained for A = 6000 A, .coming from. two coherent sources 51 'and 52. At certain point P on the screen third dark fringe is formed. Then the path difference S 1P- S2 P in microns is

' (a) 0 .75 (b)o 1.5 '<' (c) 3.0 ·fd) 4.5

· •' ' · • (2n-1)/.. Solution For dark fringe at-P, ,S1P- S2P = --~ 2

5/.. S1P-S2P=-

2

= 5 x 6000

= 15000- = 1.5 micron 2

Exam pi~ 5 · Tw.o slits. are separated .by a distance of 0.5 .mm and illuminated with light of A = 6000 A. If, the screen is placed 2.5 m from the slits . .The ~tance .of the third bright image froTI! the centre will be

(a) 1.5 mm ( b) 3 mm

. (c) 6 mm (d) ,9 mrn.

Solution Distance of n'h bright fringe from the centre

nD'A Yn=d

Y3 = 3 X 6000 X 10-lO X 2.5

0 .5 X 10-3

------~9 x 10-3 m = 9 nun.

0 I • ~

Example 6 In a double slit arrangement fringes are produced using light of wavelength 4800 A. One slit is covered by a thin plate of glass of refractive index 1.4 and the other with another glass plate of same thickness but of refractive index 1. 7. By doing so the central bright shifts to original fifth bright fringe from centre. Thickness of glass plate is

(a) 8 ~tm (b) 6j..i.m (c) 4!lm (d) 10j..i.m

Solution When one slit is covered by a·thin plate of glass, then shift

when another slit is covered by a thin plate of glass of different refractive index, then shift

~ ~Yz = i(llz -1)t

Net shift ~y = ~y2 - ~y1

~ = i(llz -111) t

Here given,

t..y=5~

5 ~ = t (Jlz -Ill) t

Example 7 What is the minimum thickness of a soap film needed for constructive interference in reflected light, if the light incident of the film is 750 nm? Assume that the index for the film is f.l = 1.33

(a) 282 nm (b) 70.5 nm (c) 141 nm (d) 387 nm

Solution Here, 2 f!t = A/2 A .

tmin =-=141 nm 411

Example 8 A beam of light of wavelength 600 nmfrom a distant soilr'Ge faits on a single slit 1.0 mm wide and the resulting diffraCtion ' pa~tern is observed on a screen 2 m away. What is the distance between the first dark fringe on either side of the central bright fringe ?

(a) 1.5 mm (b) 1 mm (c) 1.2 mm (d) 2.4 mm

Solution For the di.ffiaction at a single slit, the position of mininla is given by

a sin 8 = n/.. For small value of 8

,

sin 8 == 8 = y/D D

a (y!D) =/..or y=-A a

... l


Substituting the values, we have

2x6x1o-7

y = = 1.2 x 10-3 m= 1.2 mm 1 x l0-3

. . Distance between first minima on either side of central maxima = 2y = 2.4 mm

Example 9 A parallel beam ofmenochromatic light ofwaveiength 450 nm passes through ~~siit of ~i~tlf. ~r.2 mrri. ·Find th~ q'ngular divergence in which most of the light is' diffracted.

(a) 4.5 x 10-3 rad (b) 5.5 x 10-3 rad (c) 3.2 x 10-2 rad (d) 3.5 x 10-3 rad

Solution Most of the light is diffracted between the two first order mininla. These minima occur at angle given by,

. · .· A A5ox·lo-9,',',. asm9=±nA,sm9=±-=± ,

·' a 0.2x10~3 '

' =·± 2.~5x10-3 rad

:. angular divergance = 4.5 x'10-'-3 'riid

Example 10 · Two poldr~ids are ~rosseil to each 'other. Now, one of them is rotated through 60°. What percentage of incil}.ent unpolarised light will pass through the system ?

(a) 37.5 % (b) 45.6% (c) 52.3% (d) 66.7%

Solution Let I0 be intensity of the incident unpolarised light on the system. As the in<;ident .light unpo~aris~q, the intensity of light on being transmitted from the first polaroid will become

· I - I cos;~,Ei '--'- Io (:.' eos2 8=_!_) ' - a -2 . . 2

If 8 is the angle between the trans~sion planes pf the_ two polaroids, then intensity of light on passing through the s~ond polaroid is given by

·f=Jc6s29 =·· 10 cos2 e ) .· .. ~ .... 2

Since, the ··!Wo polaroids were initially cros~ed; -the angle between the transmission planes of the two' polaroids on r~tating thr~mgh 60° will become , , .. , . , , , , ,

,8 = · 90° -:- 60° = 306 · ' · • . . ( ~)2 ' . '• ·I 0 -·-· 2••• I v3 ·---

., J'c= .-XCOS 30"=_Q_ X - =0.37510 , · · ' '' 2 '' I, • 2 ' 2 •' '.'

·.

Hence, per.centage of tncident light that pass~s through :the ' system, · ' · ~ ' I_ ' ~ ~

I' -x100 = 0.375 X 100 = 37.5% Io

'I '' 1' ... , • • •• ~ •

~ ' ' ' ' .\ . . ~ .. . I I ,~ l J l

. r ·:I!

'1.;,..;, f.( r. t''

,· ,o ~ I

'·

Chapter Practice

Exercise I

Theories of Light and Interference 1. Light propagates 2 em distance in glass of refractive index

1.5 in time t0 . In the same time t0 , light propagates a distance of 2.25 em in a medim:n. The refractive index of the medium is (a) 4/3 (c) 8/3

(b) 3/2 (d) None of these

2 . The wave front due to a source situated at infinity is (a) spherical (b) cylindrical (c) planar (d) None of these

3. Which of the following is not an essential condition for interference? (a) The two interfering waves must be propagated in

almost the same direction or the two interfering waves must intersect at '\'ery small angle

(b) The wave must have the same period and wavelength

(c) The amplitude of the two waves must be equal (d) The two interfering beams of light rriust originate

from the same source

4. In Young's double slit experiment with monochromatic light of waveiength 600 nm, the distance between slits is 10-3 m. For changing fringe width by 3 x 10-5 m . (a) the screen is moved away from the slits by 5 em (b) the screen is moved by 5 em towards the slits (c) the screen is moved by 3 em towards the slits (d) Both (a) and (b) are correct

5 . In Young's double slit experiment, the distance between slits is 0.0344 mm. The wavelength ~f light used is 600 nm. What is the angular width of a fringe formed on a distant screen? (a,) 10 (c) 3°

6. A Young's double slit experiment uses a monochromatic source. The shape of the interference fringes formed on a screen is (a) hyperbola (c) straight line

(b) circle (d) parabola

7. If Young's double slit experiment, is performed in water (a) the fringe width will decrease (b) the fringe width will increase (c) the fringe width will remain uncha'nged (d) there will be no fringe

,

8. In Young's double slit experiment, the spacing between the slits is d and wavelength of light used is 6000 A. If the angular width of a fringe formed on a distance screen is 1°, then value of d is (a) 1 mm (c) 0.03 mm

(b) 0.05 mm (d) 0.01 mm

9. In Young's double slit experiment, when violet light of wavelength 4358 A is used, the 84 fringes are seen in the field of view, but when sodium light of certain wavelength is used, then 62 fringes are seen _in the field of view, the wavelength of sodium light is (a) 6893 A (b) 5904 A Cc) 5523 A Cd) 6429 A

10. In a double slit experiment, 5th dark fringe is formed opposite to one the slits. The wavelength of light is

d2 d2 (a) - (b) -

6D 5D

11. Two non-coherent sources emit light beams of intensities I and 4I. The maximum and minimum intensities in the resulting beam are (a) 9I and I (b) 9I and 3I (c) 5Iandi (d) 5Iand3I

12. The maximum intensity in the case of n identical incoherent waves, each of intensity 2 wm-2 is 32 wm-2 •

The value of n is (a) 4 (c) 32

(b) 16 (d) 64

13. In an interference pattern the position of zeroth order maxima is 4.8 mm from a certain point P on the screen. The fringe width is 0.2 mm. The position of second maxima from point P is (a) 5.1 mm (b) 5 mm (c) 40 mm (d) 5.2 mm

14. 51 and 52 are two coherent sources. The intensity of both sources are same·. If the intensity at the point of maxima is 4 wm-2

, the intensity of each source is (a) 1 wm-2 (b) 2 wm-2

(c) 3 wm-2 (d) 4 wm-2

15. n incoherent sources of intensity I 0 are superimposed at a point, the intensity of the point is

16.

(a) n !0 (b) !.9_

n (c) n2 I 0 (d) Noneofthese

In the given figure , C is middle point of line S1S2• A monochromatic light of wavelength A is incident on slits . The ratio intensity of S3 and S4 is

(a) zero (c) 4; 1

~c~: ~~ .}J._o---

(b) 00

(d) 1 : 4

17. In the given arrangement, S1 and S2 are coherent sources (shown in figure). The point Pis a point of

ft ·IP Sz

(a) bright fringe (b) dark fringe (c) either dark or bright (d) None of the above

18. When two coherent monochromatic light beams of intensities I and 4 I are superimposed, what are the maximum and minimum possible intensities in the resulting beams? (a) SI and I (c) 91 and I

(b) SI and 31 (d) 91 and 31

19. A parallel beam of light of intensity 10 is incident on a glass plate, 25% of light is reflected by upper surface and 50% of light is reflected from lower surface. The ratio of maximum to minimum intensity in interference region of reflected rays is

(a) [~+If]z 2-H

5 (c) -

8

(b)

(d) ~ 5

20. In Young's double slit experiment, the length ofband is 1 mm. The ring width is 1.021 mm. The number of fringes is (a) 45 (b) 46 (c) 47 (d) 48

21. In a Young's experiment, one of the slits is covered with a transparent sheet of thickness 3 .6 x 10-3 em due to which position of central fringe shifts to a position originally occupied by 30th fringe . The refractive index of the sheet, if A = 6000 A, is (a) 1.5 (b) 1.2 (c) 1.3 (d) 1.7

22. Light waves travel in vacm1m along the .Y-axis. Which of the fi !lowing may represent the wavefront?


(a) y = constant (c) z = constant

(b) x = constant (d) x + y + z = constant

23. The correct curve between fringe width ~ and distance between the slits (d) in figure is

(a)~~ ~)~~~ -d -d

~1 / 1 ~ (c) j L__ (d) ~ L_

- d -d

24. The correct formula for fringe visibility is

(a) V = [max- [min (b) V = I max+ !min

[max +[min [max -[min

(c) V =I max (d) V =I min

I min [max

25. Two coherent waves are represented by y 1 = a1 cos rut and y 2 == a 2 sin rut, superimposed on each other. The resultant intensity is (a) (a1 + a2) (b) (a1 - a2)

26. Light appears to travel in straight lines since (a) it is not absorbed by the atmosphere (b) it is reflected by the atmosphere (c) its wavelength is very small (d) its velocity is very large

27. The equations of two interfering waves are y 1 = b cos rut

and Yz = b cos (rut + <!>). For destructive interference the path difference is

- (a) 0° (b) 360° (c) 1806 (d) 720°

28. The Young's double slit experiment is performed with blue and with green light of wavelengths 4360 A and 5460 A

'·· respectively. If x is the distance of 4th maxima from the · central one, then'

' (a) x (blue) = x (green) (b) x (blue) > x (green) (c) x (blue) < x (green) (d) x (blue)/x (green) = 5400/4360

29. In a double slit interference experiment, the distance between the slits is 0 .05 em and screen is 2 m away from the slits. The wavelength of light is 8.0 x 10-5 em. The distance between successive fringes is (a) 0.24 em (b) 3.2 em (c) 1.28 em (d) 0.32 em

30. Two light rays having the same wavelength A in vacuum are in phase initially. Then the first ray travels a path L1 through a medium of refractive index n1, while the second ray travels a path of length L2 through a medium of refractive index n2• The two waves are then combined


to observe interference. The phase difference between the two waves is

(c) 2rr (n2LI - ni L2) (d) 2rr (LI - L2 ) A . A ni n2

31. Through quantum theory of light we 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) Photoelectric effect (c) Compton effect

(~) Diffraction (d) Black body radiation

32.- If the intensities of the two interfering beams in Young's double slit experiment be II·. and I 2, then the contrast between the maximum and minimum intensity is good

·,·.when (a) II is much greater than I 2

(b) I I is much smalle'r than I 2

(c) II = I 2

(d) Either II = 0 or I 2 = 0

' 33. The fringe width a:t' a distance of 50 em from the slits in Young's experiment · for light of wavelength 6000 A is 0.048 em. The fringe width at the same distance for A = sooo A will bE~ ,

35.

36.

(a) 0.04 em (c) 0.14 em

(b) 0.4 em (d) QAS em

Two waves originating from sources SI and 52 having zero phase difference and common wavelength A will show complete destructive interference at a point P, if (S1 P-S2 P) = r.

- (a) 5 A (b) 3A 4 ,

(d) llA 2 '.·

Two coherent sources of intensities I I and I 2 produce an interference pattern. The maximum' intensity in the

·interference pattern will be '. (a) I1 .+ ·I2 (b)' I~ +I~ ''

(c) CII +I2i (d) ·cji; +fi;f In Young's double slit experiment, the s;v€mth maximum with wavelength A1 is at a distance di and the same maximum with wavelength A2' is at a distance d2. Then dl/d2 =

) AI (a -· '·A2

r c) Ai , A~

)._2 (d) A.~

I

37. When monochrbm:atic light is replaced by white light in Fresnel's qiprism arrangement, the central fringe is (a~ col.0ured 1 · • ' • (b) white (c) dark (d) None· ofthese

38. In i:he ~ei: up shown in figure, tlie tw~ slits, 51 and 52 are not equidistant from: the slit S. The central fringe at 0 is, then , ·

I sLf----------- o ~s2

I (a) always bright (b) always dark (c) Either dark or bright depending on the position of S (d) Neither dark nor bright

39. In a double slit interference experiment, the distance between the slits is 0.05 em and screen is 2 m away from the slits. The wavelength of light is 6000 A. The distance between the fringes is (a) 0.24 em (c) 1.24 em

(b) 0.12 em (d) 2.28 em

40. The separation between successive fringes in a double slit arrangement is x. If the whole arrangement is dipped under water, what will be the new fringe separation? [The wavelength of light being used in soooA]

41.

42.

(a) 1.5 X (b) X

(c) 0.75 x (d) 2x

In a Young's experiment, two coherent sources are placed 0.90 mm apart and the fringes are observed one meter away. If it produces the second dark fringe at a distance of 1 mm from the central fringe, the wavelength of monochromatic light used will be (a) 60 x 10- 4 em (b) 10 x 10- 4 em (c) 10 x 10-5 em (d) 6 x 10-5 ern

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 two slits are 5 units

and 4 units respectively (b) the intensities at the screen due to two slits are 4 units

and 1 unit respectively (c) the amplitude ratio is 3 (d) the amplitude ratio is 2

43. Interference fringes are being produced on screen XY by the slits 51 and 52• In figure, the correct fringe locus is

S1 ! w1 _,---W3 , /

s \

' Q

Sz ! ' ----· w4 Wz

(a) PQ (b) W1W2 (c) W3W4 (d) XY

44. Microwaves from a transmitter are directed normally towards a plane reflector. A detector moves along the normal to the reflector. Between positions of 14 successive maxima, the detector travels a distance of 0.14 m. The frequency of transmitter is (a) 1.5 x 1010 H (b) 1010 H (c) 3 x 1010 H (d) 6 x 1010 H

I I.

I I

45. Two waves having the intensities in the ratio of 9 : 1 produce interference. The ratio of maximum to minimum intensity is equal to (a) 10 : 8 (c) 4 : 1

(b) 9 : 1 (d) 2 : 1

46. In Young's double slit experiment, phase difference between light waves reaching 3rd bright fringe from the central fringe when A = 5000 A is

47.

48.

49.

50.

51.

52.

53.

(a) 6 n (b) 2 1t

(c) 4n (d) zero In an interference pattern produced by two identical slits, the intensity at the slit of the. central maximum is I . The intensity at the same spot when either of the slits is closed is I 0 • Therefore (a) I = 10

(b) I= 2 I 0

(c) I = 4 I 0

(d) I and I 0 are not related to each other In a two slits experiment with monochromatic light, fringes are obtained on a screen placed at some distance

·from the slits. If the · screen is moved by 5 x 10-2 m towards the slits, the change in fringe width is 3 x 10-5 m. If separation between the slits is 10-3m, the wavelength of light used is (a) 4500 A (c) 5000 A

Cb) 3ooo A (d) 6ooo A

In Young's double slit experiment, let 51 and 52 be the two slifs, and C be the centre of the screen. If LS1CS2 = 8 and A is the wavelength, the fringe width will be

A • (a)

e (c) 2 A

e

(b) A8

(d) _!:____ 28

In Young's double slit .experiment, tl).e intensity on screen at a point where path difference .is A-is K. What will be intensity at the point where path difference is A/4? (_a) K/4 (b) K/2 (c) K (d) zero .

In the Young's double slit ex'Perimefif, a · mica slip of thickness t and refractive index 11 is introeluced in the ray from first source 51'. By how much distance fringes pattern will be displaced.

d (a) -C~-t-1)t

D d

(c) (f..L-1)D

D -(b) -C~-t-1)t

d .. -D

(d) d(f..L-1)

Oil floating on water looks coloured due to interference of light. What should be the order of magnitude of thickness of oil layer in order that this effect may be observed? (a) 10,000 A (b) 1 em (c) 10 A Cd) 100 A

"' !')'l

Two waves y 1 = A1 sin(wt- ~1 ) and .h .= A2 sin(wt- ~2 ) superimpose to form a resultant wav~· wnose amplitude is

(a) ~~ + 2A1 A 2 cos(l31 - -132 )

(b) ~A~+A~+2A1 A2 sin(l31 -13 2 ) (c) A 1 + A 2

(d) 1IA1 + A2 1


54. In Young's double slit experiment, 12 fringes are. obtained 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 observed in the same segment of the screen is given by (a) 12 (b) 18 (c) 24 (d) 30

55. In Young's double slit experiment, distance between two sources is 0.1 mm. The distance of screen from the sources is 20 em. Wavelength of light used is 5460 A. Then-angular position of first dark fringe is (a) 0.08° ~b) 0.16° (c) 0.20° (d) 0.32°

56. Two beams of light having intensities I and 4I interfere to produce a fringe ,pattern on a screen. The phase difference between the beams. is n/2 .at point A and 1t at point B. Then the difference between the resultant intensities at A andB is (a) 2I (b~.4I (c) 5 I (d) 7 I

57. In a Young's double slit experiment usilig red and blue lights of wavelengths 600 .nm and 480 urn respectively,

. · .. the value of n from which the nth red fringe coincides with (n + 1) the blue fringes is -(a) 5 (b) .• 4 . , , "~"-' (c) 3 (d) 2

58. In Young's double sfit experiment, distance between sources is 1 mm and distance betw~en the screen and source -is 1m. -If the frin.ge width on the screen is 0.06 em, •then A is (a) 6ooo A Cb) 4ooo A (c) 1200 A Cd) 24oo A

59. In Young's double s.lit experiment, the central bright fringe can be identified (a) as it has greater: intensity than the other bright fringes (b) as it is wider than the other bright fringes (c) a,s ,it is 11arrower than the other bright :(ringes

,_: , (d) by using white J,ight inst~iid of ffi9nochromatic light

60. Two slits, 4 mm apart are illuminated by-light of wavelength 600 A. Wnat Will be the fringe width on ·a screen placed 2 m from the slits ? (a) 0.12 mm (c) 3.0 mm

I i ' , ~ ' '

(b) 0.3 mm' · (d) 4 .. 0.11llT]- .

. Diffraction of Light , . ' 61. A parallel beam of light of wavelength 3141.59 A is

incident on a small aperture . After passing through the aperture; the beam is no longer parallel but diverges at 1° to the incident direction. What is the diameter of the aperture? (a) 180m (b) 18 ~-tm (c) 1.8 m (d) 0.18 :U

62. To observe diffraction,• the size of. an fiperture (a) should be ofthe same order as wavelength should be

much larger than the wavelength (b) should be much larger than the wavelength (c) have no relation t9 wavele~gth ' ' -- . (d) should be exactly A/2


63. Air has refractive index 1.003. The thickness of air column, which will have one more wave length of yellow light (6000 A) than in the same thickness of vacuum is (a) 2 mrn (b) 2 em (c) 2m (d) 2 km

64. The distance between the first and the sixth minima in the diffraction pattern of a single slit is 0.5 mm. The screen is 0.5 m away from the slit. If the wavelength of light used is 5000 A, then the slit width will be (a) 5 mm (b) 2.5 mm (c) 1.25 mm (d) 1.0 mm

65. Plane microwaves are incident on a long slit having a width of 5 em. The wavelength of the microwaves if the first minimum is formed at 30° is (a) 2 .5 em (b) 2 em (c) 25 em (d) 2 mm

66. A plane wave of wavelength 6250 A is incident normally on a slit of width 2 x 10-2 em. The width of the principal maximum on a screen distant 50 em will be (a) 312.5 X 10-3cm (b) 312.5 X 10-4 em (c) 312 em (d) 312.5 x 10-5 em

67. The main difference between the phenomena of interference and diffraction is that (a) diffraction is caused by reflected waves from a source

whereas interference is caused due to refraction of waves from a source

(b) diffraction is due to interaction of waves derived from the same source, whereas interference is that bending of light from the same wavefront

(c) diffraction is due to interaction of light from wavefront, whereas the interference is the interaction of two waves derived from the same source

(d) diffraction is due to interaction of light from the same wavefront whereas interference is the interaction of waves from two isolated sources

68. Light of wavelength 6000 A is incident on a single slit. The first minimum of the diffraction pattern is obtained at 4 mm from the centre. The screen is at a distance of 2 m from the slit. The slit width will be (a) 0.3 mm (b) 0.2 mn1 (c) 0.15 mm (d) 0.1 mm

69. The Fraunhofer 'diffraction' pattew of a single slit is formed in the focal plane of a lens of focal length 1 m. The width of slit is 0.3 mm. If third minimum is formed at a distance of 5 mm from central maximum, then wavelength of light will be (a) 5000 A (b) 2500 .\ (c) 7soo k CC.:) s::;oo A

70. What should be refractive index of a transparent medium to be invisible in vacuum ? (a) 1 (b) < 1

(c) >1 (d) Noneofthese

71. A slit 5 em wide is irradiated normally with microwaves of wavelength 1.0 em. Then the angular spread of the central maximum on either side of incident light is nearly

,

(a) 1/5 rad (b) 4 rad (c) 5 rad (d) 6 rad

72. Which of the following phenomena is not common to sound and light waves? (a) Interference (b) Diffraction (c) Coherence (d) Polarisation

73. A beam of ordinary unpolarised light passes through a tourmaline crystal C1 and then it passes through another tourmaline crystal C2, which is oriented such that its principal plane is parallel to that of C

2• The intensity of

emergent light is 10 • Now C2 is rotated by 60° about the ray. The emergent ray will have an intensity (a) 2 10 (b) lof2 (c) lof4 (d) Iof..fi

74. A ray of light strikes a glass plate at an angle of 60°. If the reflected an refracted rays are perpendicular to each other, the index of refraction of glass is

(a) 1 2

(c) 3 2

(b) #. (d) 1.732

75. An unpolarised beam of intensity 2 a2 passes through a thin polaroid. Assuming zero absorption in the polaroid, the intensity of emergent plane polarised light is (a) 2 a2 (b) a2

2 (c) J2a 2 (d) 9:_

2 76. 80 g of impure sugar when dissolved in a litre of water

gives an optical rotation of9.9°, when placed in a tube of length 20 em. If the specific rotation of sugar is 66°, then concentration of sugar solution will be (a) 80 gL-1 (b) 75 gL-1

(c) 65 gL-1 (d) 50 gL-1

77. If for a calcite c1ystal, f.!o and 1-le are the refractive indices of the crystal for 0-ray and E-ray respectively, then along the optic axis of the crystal (a) 1-lo = 1-le (b) f.le = f-lo (c) 1-le = 1-lo (d) None of these

78. 'A.a and 'A.m are the wavelength of a beam of light in air and medium respectively. If 8 is the polarising angle, the correct relation between 'A.a, 'A.m and El is

(a) 'A.a == 'A.m tan2 e (b) 'A.m == 'A.a tan2 e (c) 'A.a == 'A.m cote

(d) 'A.m == 'A.a cote

79. Ordinary light incident on a glass slab at the polarising angle, suffers a deviation of 22°. The value of the angle of refraction in glass in tllis case is (a) 56° (b) 68° (c) 34° (d) 22°

80. At what angle should an unpolarised beam be incident on a crystal of f.! == J3, so that reflected beam is polarised? (a) 45° (b) 60° (c) 90° (d) 0°


Exercise II

Only One Correct Option 1. A beam of circularly polarised light is completely absorbed

by an object on which it falls. If U represents absorbed energy and m reperesents angular frequency, then angular momentum transferred to the object is given by

u (b) u (a)

0/ 2ro

(c) U (J)

(d) 2U (J)

2. In an interference pattern by two identical slits, the intensity of central maxima is I . What will be the intensity of the same spot, if one of the slits is closed? (a) I l 4 (b) I12 (c) I (d) 2 I

3. In Young's double slit experiment we get 60 fringes in the field of view of monochromatic light of wavelength 4000 A. If we use monochromatic light of wavelength 6000 A, then the number of fringes obtained in the same field of view are (a) 60 (b) 90 (c) 40 (d) 1.5

4. The phenomenon which does not take place in sound waves is (a) scattering (b) diffraction (c) interference (d) polarisation

5. Two Nicol prisms are first crossed and then one of them is rotated through 60°. The percentage of incident light transmitted is (a) 1.25 (b) 25.0 (c) 37.5 (d) 50

6. Find the thickness of a plate which will produce a change in optical path equal to half the wavelength A of the light passing through it normally. The refractive index of the plate ~is

A (a) 4(~ - 1)

')..

(c) (~-1)

2A (b) 4(~-1)

A (d)

2(j.!-1)

7. A beam of light of wavelength 600 nm from a distant source falls on a single slit 1.00 mm wide and the resulting diffraction pattern is obsetved 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 em (c) 2.4 em

(b) 1.2 mm (d) 2.4 mm

8. Light of wavelength 2 x 10-3 m falls on a slit of width 4 x 10-3 m. The angular dispersion of the central maximum will be (a) 30° (b) 60° (c) 90° (d) 180°

9. nth bright fringe if red light (/..1 = 7500 A) coincid;s with (n + l)'h bright fringe of green light (A2 = 6000A) . The value ofn =? (a) 4 (c) 3

, (b) 5 . (d) 2

10. In Young's t"v'IO slit experiment the distance between the two coherent sources is 2 mm and the screen is at a distance of 1 m. If the fringe width is found to be 0.03 em, then the wavelength of the light used is (a) 4ooo A Cb) 5ooo A c c) 5890 A c d) 6ooo A

11. In Young's experiment the wavelength of red light is 7.8 x 10-5 em and that of blue light 5 .2 x 10--2 em. The value of n for which (n + 1)th blue bright band coincides with nth red band is (a) 4 (b) 3 (c) 2 (d) 1

12. Two slits separated by a distance of 1 mm are illuminated with red light of wavelength 6.5 x 10-7 m. The interference fringes are observed on a screen place 1 m from the slits. The distance between the third dark fringe and the fifth bright fringe is equal to (a) 0.65 mm (b) 1.63 nun (c) 3.25 mm (d) 4.88 mm

13. In Young's double slit experiment, the 7rh maximum wavelength A1 is at a distance d1 and that with wavelength A2 is at a distance d2 • Then (d11d2) is (a) (/..1 I A2 ) (b) (/..2 I /..1 )

(c) C"-i I A~) (d) (/..~I A.i) 14. If white light is used in a biprism experiment then

(a) fringe pattern disappers (b) all fringes will be coloured (c) central fringe will be white while ot.hers will be coloured (d) central fringe will be dark

15. Two waves of same frequency and same amplitude from two monochromatic source are allowed to superpose at a certain point. If in one case the phase difference is 0° and in other case is rr I 2, the ratio of the intensities in the two cases will be (a) 1 : 1 (b) 2 : 1 (c) 4 : 1 (d) None of these

16. 1\vo polaroids are kept crossed to each other. Now one of them is rotated through an angle of 45°. The percentage of incident light now transmitted through the system is (a) 15% (b) 25% (c) 50% (d) 60%

17. In Young's double slit experiment, the two slits act as coherent sources of equal amplitude A and wavelength A. In another experiment vlith the same setup, the two slits are sources of equal amplitude A and wavelength A but are incoherent. The ratio of the intensity of light at the mid-point of the screen in the first case to that in the second case is (a) 2 : 1 (b) 1 : 2 (c) 3 : 4 (d) 4: 3

18. If the two waves represented by y 1 4 sinmt and Y?. =3sin(ffit+n:l3) interfere at a point, the amplitude of the resulting wave will be about (a) 7 (b) 5 (c) 6 (d) 3.5

19. In a biprism experiment, by using light of wavelength 5000 A, 5 mm wide fringes are obtained on a screen

954 Chapter 24 e Wave Optics

1.0 m away from the coherent sources. The separation between the two coherent sources is (a) 1.0 mm (b) 0.1 mm (c) 0.05 mm (d) 0.01 mm

20. How will the diffraction pattern of single slit .change when yellow light is replaced by blue light? The fringe will be (a) wider (b) narrower (c) brighter (d) fainter

21. White light is used to illuminate the two slits in a Young's doubel slit experiment. The separation between slits is b and the screen is at a distance d (>>b) from, the slits. At a point on the screen dir~ctly in fTont of one of the slits, certain wavel~ngths are missing, figure. S<;>me of these missing wavelengths are

. '

(a) b2 2b2

A=-- (b) 0. b2

. 3b: ' 'A.ot •-,~-·

d '3d 2d 2d .'

2b2

(d) '- 3b2

(c) A=- 'A=-3d 4d '.'

, 22. A beam of unpolarized light having t1U:X io-3 W falls normally on a polarizer of cross sectional area 3 X 10-4 m2

•

The polarizer rotateS\,Yith an angular freqilt3rtcy of"31.4 rads-1. The energy of light passing through the polarizer per revolution will be (a) 10-4 J (b) 10-3 ,J

(c) 10~ 2 J (d) 10-1 J

23. In a biprism experiment, 5'h dark fringe is ·obtained.' at a point. If a thin transparent film 'is placed in' the path of one of waves, then 7th bright' fringes is obtained at the same point. The thickness of the film in terms of wavelength l

·and refractive index ~ will be 1.5A

(a) (Jl-1)

(c) 2.5(Jl-1)A

(b) j :5c~-1)~ -~:

(d) J.5A (Jl-1)

24. The ratio of intensities of successive max1ma m the differaction pattern due to single slit is (a) 1 : 4: 9 (b) 1 : 2: 3

4 4 ' 4 9 (c) 1:-2:--2 (d) 1:2:2

9n 25n n n 25. The equations of displacement of two w~ves an~ given as

y 1 = 10sin(3nt + n I 3) Yz = 5(sin 3nt + J:3 cos 3nt), . then what is the ratio of their amplitude? . (a) 1 : 2 (b) 2 : 1 (c) 1 : 1 (d) None ofthese

26 . . Light of wavelength A is incident on a slit of width d. The resJ lting diffraction pattern is observed on a screen at a

distance D. The linear width of the principal maximum is equal to the width of the slit, if D equals

d2 d (a) 2A (b) A

(c) 2A2 . < (d) 2A d ' d

2 7. A glass slab of thickness 8 em contains the same number of waves as 10 em of water when both are traversed by the same monochromatic light. If the refractive index of water is 4/3, the refractive index of glass is (a) 5/4 (b) 3/2 (c) 5/3 (d) 16/15

28. Fringes are obtained with the help of a biprism in the focal plane of an _eyepiece distant 1 m from the slit. A convex lens produces images of the slit in two positions betw.ee:r;J. biprism ~nd eyepiece. Jheqista:r;J.ces,behveen two images of the slit · in two positions are 4.05 x 10-3 m and 2.90 x 10-3 m~espectively. The distance between the slits will be (a) 3.43 x 10-3 m (b) 0.343 rri (c) 0.0343 m (d) 43.3 m

29. In Young's double slit experimel,lt ~ = 10-4 (d =distance . D between slits, D = distance of screen from the slits). At a point P on the screen resultant intensity is equal to the intensity due to · individmihlit I 0 •• Then the distance of point P from the central maximum is (A,= 600QA) (a) 0.5 mm (b) 2 mm (c) 1mm (d) 4mm

. 10 .. In young's double slit experiment, the intensity of the •

0 0

•

0

maxima is I. If the width of each slit is dqubled, the intensity of the r,naxima will be (a) 1!2 . (b) 2 I

\c). ,~I , _ (d) I

May have Mot:e·Qian One CorrectOpti.on

',· 31. A light of wavelength 6000 A in air enters a medium of

refra.ctive inde,x 1.5. Inside the medium, its frequency is v . al}doit~ ~avelength ~sA..: (a) v = 5 x 1014 I:Iz

(c) A= 4oooA (b) v=7.5x1014 Hz

(d) A= 90ooA " 32~ In Yotiri:g's•doubTe slit experiment, the distance betWeen

the two' slits is 0.1 mm and wavelength of light used is 4 x 1 o-7 m: lfthe width of fringe on screen is 4 mm, the

- distance between screen and slit is (a) 0.1 mm (b) 1 em (c) 0.1 em (d) lm

0,

0 3~. yYit~ a. monochromatic light, the fringe width using a

1 ··;; 1·~: _' , : ~~ss'bipJ:~~1U setup in air is 2 mm. Given a ~g = 3 I 2 and ' .. rDik _a P.t.,~ . 5/ ~- ·. {he whole ~~t up is immersed in liquid, then 1 ~.-,.,, _ ~h\\17-f!~: fringe width is , ·:,

' .. kl_i~, r,2 mm- (b) 6 mm (c) 6/5 mm {d) 8/3 mm

o ·· -34. ·In Young's double slit experiment, on interference, ratio of inten~ities of a brighqJand and a dark band is 16 : 1. The 'ratio of amplitudes of interfering waves is (a) 16 · ' (b) 5/3 (c) 4 ,.,L

0

(d) 1/4

35. Light waves travel in vacuum along X-axis. Which of the following may represent the wavefronts? (a) x = a (b) y = a (c) x = a' (d) x + y + z = a

Comprehension Based Questions Passage

In YDSE, intensity of light from two slits is proportional to their width, and also to squar~ of amplitudes of light from the slits, ie,

II rol a2 I2 = (1)2 = ll

At points, where crest of one wave falls on crest of the other and through falls on trough, intensity of light is maximum. Interference is said to be constructive . At points, where crest of one wave falls on trough of the other, resultant intensity of light is minimum. The ·interference is said to be destructive.

, ImaJ<- (a+b)2

Imin- (a-b)2

For sustained interference, the two sources must be coherent; Two independent sources cannot be coherent.

36. The light waves from two coherent sources are represented by 0

.y 1 = a1'sinrot and Y2 = a2 sin(rot + 1t I 2)

' The resultant amplitude will be (a) a1 (b) a2

(c) a 1 +a2 (d•) ~a~+a~ 3?. The ratio of intensities in the interference pattern in the

. above' question will be (a) 9 : 1 (c) 2 : 1

(b) 4 : 1 (d) 1 : 4

38. The amplitudes of light waves from two shts are in the ratio 3 : 1. Widths of slits are in the ratio ca.·)· I :·9 ' · (b) 1:3 (c) 9 : 1 ( (d) 3 :· 1

3'9~ The widths of two slits in YDSE ate in the ' ratio 1 : 4. The ratio of amplitudes of light waves fro'm two slits will be (a) 9 : 1 (b) 4 : 1 (c) 1 : 2 (d) 1 : J2

tO. Two independent sources emit ligP,t waves given by y 1 =a1 sinro1t and Y2 =a2 sin(ro1t+7t/2). Will you observe interference pattern? (a) Yes (b) No (c) Sometimes (d) Cannot say

Assertion and Reason Directions Question No. 41 to 50 are Assertion-Reason type . Each of these contains two Statements : Statement I (Assertion), Statement II (Reason). Each ~f these questions also has four alternative choice, only one of which is correct. You have to select the correct choices from the codes (a) , (b), (c) and (d) given below: (a) If both Assertion and Reason are true and the Reason is

correct explanation of the Assertion. (b) If both Assertion and Reason are true but Reason is not

correct explanation of the Assertion. (c) If AsseJ;;tion is true but Reason is false. (d) If Assertion is false but the Reason is true.


41. Assertion Newton's rings are formed in the reflected system when the space between the lens and the glass plate is filled with a liquid of refractive index greater than that of glass, the central spot of the pattern is dark. Reason The refraction in Newton's ring cases will be from denser to rarer medium and the two interfering rays are reflected under similar conditions.

42. Assertion To observe diffraction of light the size of obstacle/aperture should be of the order of 10-7 m. Reason 10-7 m is the order of wavelength of visible light.

43. Assertion A famous painting was painted by not using brush strokes in the usual manner, but rather a myriad of small colour dots. In this painting the colour you see at any given place on the painting changes as you move away. Reason The angular separation of adjacent dots changes with the distance from the painting.

44. Assertion In Young's experiment, the fringe width for dark fringes is same from that for white fringes. Reason In Young's double slit experiment, when the fringes are performed with a source of white' light, then only black and bright fringes are observed.

45. Assertion Thin films such as soap bubble or a thin layer of oil on water show beautiful colours when illuminated by whi~e light. Reason It happens due to the interference of light reflected from the upper surface of the thin film.

46. Assertion Corpuscular theory fails in explaining the velocities of light in air and water. Reason According to corpuscular theory, light should travel faster in denser media than in rarer media .

47. Assertion When a tiny circular obstacle is placed in the path of light from some distance, a bright spot is seen at the centre of shadow of the obstacle. Reason Destructive interference occur's at the centre of the shadow.

48. Assertion Out of radio waves and microwaves, the former undergoes more diffraction. Reason Radio waves have greater frequency compared to microwaves.

49. Assertion No diffraction is produced in sound waves near a very small opening. Reason For diffraction to take place the aperture of opening should be of the same order as wavelength of the waves.

50. Assertion In Young's experiment, for two coherent

sources, the resultant intensity given by I = 4I 0 cos2 1.. . 2

Reason Ratio of maximum and minimum intensity

I maJ< ( ji; + jf-;y Imin = cji; -ji;F

Previous Year's Questions 51. A mixture of light consisting of wavelength 590 nm and

an unknown wavelength illuminates Young's double slit and gives rise to two overlapping interference patterns on the screen. The central maximum of both light coincide. Further, it is observed that the third bright fringe of known

956 Chapter 24 o Wave Optics

light coincides with the 4'h bright fringe of unknown light. From this data, the wavelength of unknown light is

(AIEEE 2009) (a) 885 .0 nm (c) 776.8 nm

(b) 442.5 nm (d) 393.4 nm

52. A parallel monochromatic beam of light is incident normally on a narrow slit. A diffraction pattern is formed on a screen placed perpendicular to the direction of the incident beam. At the first minimum of the diffraction pattern, the phase difference between the rays corning from the two edges of slit is (UP SEE 2009) (a) zero (b) n

(c) ~ (d) 2n 2

53. What causes chromatic aberration? (UP SEE 2008) (a) Non-paraxial rays (b) Paraxial rays (c) Variation of focal length with colour (d) Difference in radii of curvature of the bounding

surfaces of the lens

54. In an interference experiment, the spacing between successive maxima or minima i.i (UP SEE 2008)

(a) Ad I D

(c) dD!f..

(b) Wid

(d) Ad I 4D

55. In the case of diffraction pattern due to a single slit, angular width of central maximum can be increased by

(Karnataka CET 2008) (a) decreasing the wavelength (b) increasing the slit Width (c) increasing distance between slit and screen (d) decreasing the slit width

56. Both, light and sound waves produce diffraction. It is more difficult to observe diffraction with light waves because (Karnataka CET 2008) (a) light waves do not require medium (b) wavelength of light waves is far too small (c) light waves are transverse in nature (d) speed of light is far greater

57. In an interference experiment using waves of same amplitude, path difference between the waves at a point on the screen is AI 4. The ratio of intensity at this point with that at the central bright fringe is (Karnataka CET 2008) (a) 1 (b) 0.5 (c) 1.5 (d) 2 .0

58. In Young's experiment, the third bright band for light of wavelength 6000 A coincides with the fourth bright hand for another source of light in the same arrangement. Then the wavelength of second source is (Kerala CET 2008)

(a) 36oo A Cb) 4ooo A (c) 5ooo A (d) 45oo A

59. The Young's double slit experiment is performed with blue and with green light of wavelengths 4360 A and 5460 A respectively. If x is th~ distance of 4'h maximum from the central one, then (BVP Engg. 2008) (a) x (blue) = x (green) (b) x (blue) > x (green)

(c) x (blue) < x (green) (d) x(blue) 5460

' x (green) 4360

60. I

61.

62.

63.

The necessary condition for an interference by two sources of light is that (BVP Engg. 2008) (a) two light sources must have the same wave length (b) two point sources should have the same amplitude

and same wavelength (c) two sources should have the same wavelength,

nearly the same amplitude and have a constant phase difference

(d) two point sources should have a randomly varying phase difference

In the Young's double slit experiment, the central maxima is observed to be 10• If one the slits is covered, then the intensity at the central maxima will become (BVP Engg. 2007)

(a) !_g_ (b) J.g_ 2 .J2

(c) Io (d) Io 4

If the polarising angle of a piece of glass for. green light is 54.74°, then the angle of minimum deviation for an equilateral prism made of same glass iS' (Kerala CET 2007) (Given: tan 54.74° = 1.414) (a) 45° (b) 54.7° (c) 60° (d) 30°

In Young's double slit experiment, the intensity at a point

where the path difference is ~(A. being the wavelength 6

oflight used) is I. IfJ0 denotes the maximum intensity, _!_ Io

is equal to (AIEEE 2007)

3 1 (a) 4 (b) .J2

(c) J3 (d) 1

2 2 64. The phenomenon of polarization of electromagnetic

waves proves that the electromagnetic waves are (Gujarat CET 2007)

(a) transverse (b) mechanical (c) longitudinal (d) neither longitudinal nor transverse

65. In the phenomenon of diffraction of light, when blue light is used in the experiment instead of red light, then

(a) fringes will becomes narrower (b) fringes will becomes broader (c) no change in fringe width (d) None of the above

(DCE 2007)

66. Monochromatic light of frequency 6.0 x 1014 Hz is produced by a laser. The power emitted is 2 x 10-3 W. The number of photons emitted, on the average, by the source per second is (UP SEE 2007) (a) 5 x 1015 (b) 5 x 1016

(c) 5 X 1017 (d) 5 X 1014

67. When an unpolarized light of intensity 10 is incident on a polarizing sheet, the intensity of the light which does not get transmitted is (UP SEE 2007)

1 (a) 2Io

(c) zero

(b) ]:_10 4

(d) 10

·'

68. The refractive index of glass is 1.520 for red light and 1.525 for blue light. Let D1 and D2 be angles of minimum deviation for red and blue light respectively in a prism of this glass. Then, (UP SEE 2007)

(a) D1 < D2

(b) D1 = Dz (c) D1 can be less than or greater than D2 depending upon

the angle of prism (d) D 1 > D2

69. In double slit experiment, the distance between two slits is 0.6 mm and these are illirninated with light of wavelength 4800 A. The angular width of first dark fringe on the screen distant 120 em from slits will be (BCECE 2006) (a) 8 x 10-4 rad (b) 6 x 10-4 rad (c) 4 x 10-4 rad (d) 16 x 10-4 rad

70. A lens is made of flint glass (refractive index = 1.5). Which the lens is immersed in a liquid of refractive index 1.25, the focal length (UP SEE 2006)

(a) increases by a factor of 1.25 (b) increases by a factor of 2.5 (c) increases by a factor of 1.2 (d) decreases by a factor of 1.2

71. In Young's double slit experiment, the wavelength of light A. = 4 x 10-7 m and separation between the slits is d = 0.1 mm. If the fringe width is 4 mm, then the separation between the slits and screen will be

(Orissa JEE 2006)

~~l&M'1ff11%Pffftf~~:tl'l:%:;ci'::m'dll1Sfal:i1~if~;::!.~;d'ii

Exercise I 1. (a) 2. (c) 3. (c) 4. (d) 5. (a)

11. (d) 12. (b) 13. (a) 14. (a) 15. (a) 21. (a) 22. (a) 23. (b) 24. (a) 25. (c) 31. (b) 32. (c) 33. (a) 34. (d) 35. (d) 41. (d) 42. (b) 43. (c) 44. (c)· 45. (c) 51. (b) 52. (a) 53. (a) 54. (b) 55. (b) 61. (b) 62. (a) 63. (a) 64. (b) 65. (a) 71. (a) 72. (d) 73. (c) 74. (d) 75. (b)

Exercise II

1. (c) 2. (a) 3. (c) 4. (d) 5. (c) 11. (c) 12. (b) 13. (a) 14. (c) 15. (b) 21. (a) 22. (a) 23. (d) 24. (c) 25. (c) 31. (a,c) 32. (d) 33. (c) 34. (b) 35. (a,c) 41. (d) 42. (a) 43. (a) 44. (c) 45. (c) 51. (b) 52. (d) 53. (c) 54. (b) 55. (d) 61. (c) 62. (d) 63. (a) 64. (a) 65. (a) 71. (b) 72. (d) 73. (c) 74. (b) 75. (b)

,

Chapter 24 • Wave Optics I 957

(a) 100 mm (b) 1 m (c) 106 em (d) 10 A

72. A beam of light of wave~ngth 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 (Karnataka CET 2006) (a) 1.2 em (b) 1.2 mm (c) 2.4 em (d) 2.4 mm

73. In Young's double slit experiment, intensity at a point is (1/ 4) of the maximum intensity angular position of this point is (liT JEE 2005)

(a) sin- 1(A. /d)

(c) sin -l (A. !3d)

(b) sin-1(A. /2d)

(d) sin-1(A./4d)

74. The maximum number of possible interference maxima for slit separation equal to twice the wavelength in Young's double slit experiment is (AIEEE 2004) (a) infinite (b) 5 (c) 3 (d) zero

75. If l 0 is the intensity of the principal maximum in the single slit diffraction pattern, then what will be its intensity when the slit widt~ is doubled? (AIEEE 2004)

(a) :!_g_ (b) 10 2

(c) 4 I 0 (d) 2 I 0

w·

Answers

6. (d) 7. (a) 8. (c) 9. (b) 10. (d) 16. (b) 17. (a) 18. (c) 19. (a) 20. (c) 26. (c) 27. (c) 28. (c) 29. (d) 30. (b) 36. (a) 37. (b) 38. (c) 39. (a) 40. (c) 46. (a) 47. (c) 48. (d) 49. (a) 50. (b) 56. (b) 57. (b) 58. (a) 59. (d) 60. (b) 66. (a) 67. (c) 68. (a) 69. (a) 70. (a) 76. (b) 77. (a) 78. (d) 79. (c) 80. (b)

6. (d) 7. (d) 8. (b) 9. (a) 10. (d) 16. (b) 17. (a) 18. (c) 19. (b) 20. (b) 26. (a) 27. (c) 28. (a) 29. (b) 30. (b) 36. (d) 37. (a) 38. (c) 39. (c) 40. (b) 46. (a) 47 . . (c) 48. (c) 49. (a) 50. (a) 56. (b) 57. (b) 58. (d) 59. (c) 60. (c) 66. (a) 67. (a) 68. (a) 69. (a) 70. (b)


Hints & Solutions Exercise I

1. For a given time, optical path remain constant

· · J.l1X1 = J.lzXz or 1.5 x 2 = J.lz x 2.25

7. As we lmow

and

l.Sx2 2 20 4 1-lz = 2.25 =1.5=15=3

~p =LillA d

~D = d~p A

10-3 X 3 X 10-5

600xl0-9

= 5 em away or towards the slits

From Eqs. (i) and (ii), •

1 pocAoc-

poe~ J.l

. J.l

... (i)

... (ii)

The refractive index of water is greater than air, therefore fringe width will decrease.

B. Here, sin= 8 ={i) So, ~8= ~y

D Angular fringe width 80 = ~8 (Width ~y = p)

8 = l = DA x ~ = !:_ 0 D d D d

80 = 1°= n;/180 rad

and A= 6x10-7 m

d = _!:_ = 180

X 6 X 10-7

8a 7r = 3.44 x 10-5 m ~0.03 mm

9. As field of view is same in both cases

or

n1P1 = nz~z

n1 ( D~1 )= n2( D~z) or

,

10.

A2 =(~~)x4358 A2 = 5904A

For dark fringe,

Here,

or

xd A -=(2m-1)D 2

rj_l----~--:-1 2

0

d m=Sx=-, 2

d d A -·-=(2x5-1)-2 D 2

dz =9A D

,·,.

11. When two cohere!'lt light b~ams of intensities 11 and 12

superimpose, then maximum intensity is cji; +.Fz)2

and minimum intensity is Cji; -ji;)2 • But when two incoherent light beams of intensities 11 and 12 superimpose, then maximum intensity is' (I 1 + 12) and minimum intensity is (11 -12) .

I~ == 51, Imin ~ 31 12. For incoherent waves,

Jmax = nl ·.''

I 32 n=~=-=16

I 2 13. The distance between zeroth order maxima and second

order minima is

Y1 =%+P=%P

3 =-x0.2mm = 0.3 mm

2 :. The distance of second maxima from point P is

. y = (4.8+0.3) mm = 5.1 mm

14. At a point of maxima . . Imax = 410 = 4 wm-2

. . 10 = 1 wm-2

16. As S3, & = s1s3 - S2S3 = o

:.' '. <1> = 27r & =· 0 A

18.

·. 13 = ! 0 + ! 0 + 2-JI0 x I 0 (cos0°)

. . [3 = 4Io The path difference at 54 is

&:' = S1S4 -S2S4 = xd D

d AD A =-X-=-

D 2d 2 , 2n A

!J> =--=1t A 2

I 4 = I 0 +I 0 + 2I 0 cos 1t = 0

_2= 4!0 == [4 0

I max= !1 + [2 + 2ji;f;

= 4! +I +2.,)2!1 xi

= 9! !min= [1 +[2 -2.,)2!1[2

Imm =I

19. The intensity of light reflected from upper surface is

I 1 = !0 x 25% 25

=I0 x -100

= Io 4

The intensity of trans~tted light from upper surface is

I- I - Io- 3Io - 0 4- 4

:. The intensity of reflec;ted light from lower surface is

I = 3I 0 X 50 = 3I 0 2 4 100 8

[max --' cji; +JI;t ._ ' .Imin -: Cjf; -Ji;)2

I.~ JJ¥ + ftFj' Imin- ( fj -P¥J

. (~+HJ -(~-HJ

t; ''. : · .•. ~!

. ' -

··" . ; ,._ ... .... ....

20. The number of fringes on either side of centre C of screen is

n1

= [AE:J"~ [~] = [23 . ~;, = 2~ ,' ~ 0.021 . -uoc; . r

:. Total number of fringes = 2n1 + fringe a:t centre = 2n1 + 1 = 2 x 23 + 1 = 46 + 1= 47

In Young's experiment, the number of fringes sho.uld be ~d. : ,

21.


The position of 30th bright fringe

30AD Y3o=-d-

Now position shift of central fringe is

30 ).J) Yo= - d-

D But we know, y 0 =-(f..L-l)t

d 30AD D -d-=d(f..L-l)t

(f..L-1) = 30 A = 30 X (6000 X 10-10

) = 0.5 t (3 .6 X 10-5)

f..l = 1.5

22. As velocity of light is perpendicular to the wavefront, and light is travelling in vacuum along they-axis, therefore, the wavefront is represented by y = constant.

23. As ~ = ).J) d

:. ~ "" 1 I d

24. Fringe visibility (V) is given by v = 1 max - I min ·· · !max +!min

25. y 1 = a1 cosmt = a1 sin(mt + 90°)

y 2 = a2 sin mt

:. Phase difference = !j> = 90°

R = ~ af + a~ + 2a1 a2 cos !J> = ~ af + a~ :. Resultant intensity II= R2 = (a~ + a~ ) .

26. Light appears to travel along straight lines because its wavelength is very small.

27. For destructive interference, path difference = 2:. 2

,· ' Therefore, phase difference = 27t . 2: = n radian = 180°. A 2

28. Distance of n'h maxima, x =nAp_"" A d

. . xblue < xgreen

29. Distance between ~uccessive fringes = fringe Width

= ~ = AD = 8 x 10-5 x 2 = 0.32 em d 0.05

30. Optical path for 1st ray = n1 L1 Optical path for 2nd ray = n2 L2

:. Phase difference, ~$ = 2 7t &: A

21t = T(nl Ll - n2L2)

31. Wave nature of liglwalone can explain the phenomenon lik.e diffraction.

32. For maximum contrast, ! 1 = !2

33, ~ 1 = ~13 = SOQO X 0.48 = 0.04 A 6000

em

34. 51 P- S2P = ll(A I 2) = odd integral multiple of A/2


35. Let ! 1 = a 2 ,!2 == b2

·· Imax =(a+b)2 =a2+b2 +2ab

= 11 + I2 + 2ji; ji; = Cji; + ji;)2

and

37. At the centre, all colour~ meet in phase, hence central fringe is white.

38. Path difference, x = (SS1 + S10)-(SS2 + S20) If x = n ),, the central fringe at 0 will be bright. If x = (2n- l)A/2, the central fringe at 0 will be dark.

39. Distance between the fringes= fringe width

AD 6x10-7 x2 x = ~ = d-= -O.OS x 10-2

= 24 x 10-4m = 0.24 em

40. When the arrangement is dipped in water,

~· = ~ == -~ = ~x = 0.75x !1 4/3 4

'AD 41. x=(2n-1)2"d-

A= 2xd (2n -1)D

2 X 10-3 X 0.9 X 10-3

(2x 2-1) x 1

= 6 x 10-7 m = 6 x 10-5 em b 2 I b

42. 1max =~=9 or ~=3

lmin (a-bi a-b

or 3a- 3b == a+ b or 2a = 4b;

I a2 4b2

__l_=- =-=-4 : 1 !2 b2 b2

43. Interference fringes are bands on screen XY running parallel to the lengths of slits. Therefore, the locus of fringes is represented correctly by W3 W4•

44. Distance between two successive maxima

A= 0.14 m =10-2m 14 '

v=~= 3 x108

=3x1010 H A 10-2

I a 2 9 45. _j_ =-=-[2 b2 1

a 3

b 1

!max= (a+b)~=(3:1J2

= 4 : l Imin (a-b) 3 1

46. Path difference of 3rd bright fringe from central fringe= 3A, so that phase difference= 3(2n) = 61t rad.

47. When one slit is closed, amplitude becomes half and intensity becomes 1/ 4th

ie, I0 = }:_I or I= 4!0 4

48. ~= AD d

P' =AD' d

~ __ P' = A(D -D') d

3 x 10_5 = AX5x10-2

10-3

or A =: -3

X w-s = 6 X 10-7 m = 6000A 50

49. ~=AD =-A-=2: d d! D e

50. When path difference is A., I max = 4 I = K

When path difference is !: , phase difference, <1> = 1t / 2 4

IR = I1 +I2 +2Nz cos<!>

K =I+I=2I=-

2 51. For a path difference (!l-1)t , the shift is

D X= (!l-1)t d

52. Oil floating on water looks coloured opJy when thickness of oil layer;:;; wavelength of light;:;; 10000 A.

53. Amplitude A 1 and A2 are added as vectors. Angle between these vectors is the phase difference (~1 - ~2 ) between them

.. R= ~A?+Ai+2A1A2 cos(~1 ~ ~2 ) 54. Here, n1 = 12,A1 = 600 nm

n2 = ?, A2 = 400nm

As nr Ar = n2A2

n2

= n1 A1 = 12 x 602_ = 18 A2 400

55. d= 0.1 mm = 10-4, D = 20 em= !m 5

A= 5460A = 5.46 x w-7 m

Angular position of first dark fringe is

6 =..::_=...?:_= 5.46xlo-7

D 2d 2x10-4

= 2.73,x 10-3 rad 180°

= 2.73 X 10-3 X--= 0.156° 1t

56. Here, I1 =I,I2 =4I,e1 =n/2,e2 =n

IA = I 1 +12 +2Nz cos81

= [ +4I +2~1x4J COS1t/2'= 51

I8 = I1 +I2 +~ cos82

= I + 4 I + 2jl x 4 I cos 1t

= SI- 41 =I

IA-IB=5l-l = 4I

I 57. nAr=(n+l)Ab

n + 1 = ~ = 600 = i n Ab 480 5

!=i-l=}_n=4 n 5 4

58. AD ~d 0.06 x 10-2 xl0-3

From ~ = d => A= D = 1

= 6 x 10-7 m= 6000 A 5'9. When white light is used instead of monochromatic light,

the central bright fringe becomes white, while others are coloured. Hence, distinction is rriade.

AD 6000x10-10 x 2 60

' ~=d= 4x1o-3

= 0 .3 x 10-3 m = 0.3 mm

61. From diffraction at a single slit, size of aperture,

A a=--

sinO

3141.59 x lo-Io 18 10-6 -18 a= = x m- 11m sin1°

62. For diffraction to be observed, size of aperture must be of the same order as wavelength of light.

63. llv = 1 and ~ta = 1.003

!:x_ = lla = 1.0003 Aa llv

X = Avn = Aa (n ;- 1)

n + 1 = !:x_ = 1.0003 n Aa

1 1 1 +- = L0003,- = 0.0003

n n

1 104

n=---=-0.0003 3

104

x =An= 6000xl0-7mm x- = 2mm a 3

64. Distance between first and sixth minima S AD

X=--d

nAD 1 x SOOx10-10 x 2 a= -x- = O.S x10-3

d = 2.5 x 10-3 m = 2.5 mm

65. As, a sinO= nA

A= a sinO :. = Ssin30° = 2.Scm n 1

66. Here, A= 6250 A = 6250 X 10-10 m

a= 2 x 10-2 cm = 2 x 10-4m D = SO em = 0.5 m

Chapter 24 • Wave Optics

'd h f 1 . ZAD W1 t o centra mrunmum = --a

2 X 6250 X 10-10 X 0.5 2x10-4

=312.5 x 10-3 em

961

67. In interference , we use two sources while in diffraction, we use light from two points of the same wavefront.

68. From a sin 8 = nA x nAD 1 x6000 x10-10 x2 a-=nA a=--=-=--___:__:------;;--D ' x 4 x 1o-3

= 3 x 10-4 = 0.3 mm

69. asinO =n A

, ax 0.3 x l0-3 xSxl0-3

ax =3A or~~.=-=-------! 3f 3xl

= 5 x 107 m = 5000 A 70. To be invisible in vacuum, 11 of medium must be equal to

11 of vacuum, which is 1.

71. Angular spread on either side is e = ~ = _!_ rad a 5

72. Sound waves cannot be polarised as they are longitudinal. Light waves can be polarised as they are transverse .

73. Intensity of light from C2 = 10

On rotating through 60°, I = [0 cos2 60° (law of Malus)

=l0 GJ =10 / 4

74. As reflected and refracted rays are perpendicular to each other, therefore, iP = i = 60°

"= tani = tan60°= J3 = 1 732 ~ p .

75. The intensity of plane pola1i sed light is = 2a2•

:. Intensity of polarised light from first nicol prism

10 1 2 2 =-= -x 2a =a

2 2 76. Here, e = 9.9° , I= 20cm = 0.2m,s = 66°

c = ?c = ~ = ~ = O.O?Sgcc-1 Is 2 x 66

= 75 gL-1•

77. Along the optic axis, 1-lo = 1-le

a 78. We lmow, Am = - and 11 = tan 0

ll A

A = -a- = A cote m tane a

79. Deviation = iP - r = 20°

Also, iP = r = 90°

Solve to get r = 34°

80. taniP = 11 = J3 :. iP = 60°

I


Exercise II

1. Angular momentum

or

L= nh 21t

U = nhv

w = 2nv::::} v=~ 21t

U= nhro 21t

U = Lw

L = !!_ (J)

2. IR = 11 + 12 + 2.JI1 12 cos~ I0 =I +I +2Icos0°= 4I

When one of the slits is closed, intensity on the same spot =I= I0/4

3. As x = n1 ~1 = n2 ~2 = n1 A1 = n2A2

:. nz = n1 A1 = 60 x 4000 = 40 A2 6000

4. As sound waves are longitudinal, therefore, polarization of sound waves is not possible.

5. If I 0 is intensity of unpolarized light,

then from 1st nicol, I1 = Io 2

From 2nd nicol, I 2 = I1 cos2(90°-60°)

= Io (J3J2

=~I 2, 2 8 °

I _1_ = 37.5% Io

6. From (Jl-l)t = n A.

7. A= 600 nm = 6 x 10-7 m a = 1 mm = 10-3 m, D = 2 m Distance between the first dark fringes on either side of central bright fringe = width of central maximum

2AD 2x6x10-7 x2 a 10-3

= 24 x 10-4 m = 2.4 mm 8. Angular dispersion of central maximum

dispersion of 1st minimum (= 29)

F . 9 n 2x10-3

. 1 rom s1n =-=---,-

a 4x10-3 2 9= 30°

9. As the two bright fringes coincide .. n A1 = (n + 1) A2 ' ·

n+1 A1 7500 5 -n-=~=600o =4 •

2 . .'

1+.!_=~ n=-( · n 4'

10. From~;= w, '· 1'

. tl ' ' ' . - ·, ·, r'

angular

A=~= 0.3xl0-z x2x 1o;}, .,;, ·6 x10-7 m=60ooA. • • '~' ~"" ... ,'" 1 '- ' ~' D -..1 ;T-''. •1~ ··<\ } • • )( .·,~ ~ t

11. X = (n + 1) Ab = n Ar n+l Ar 7.8x10-5 3 -n-=~= 5.2 x 1o-s =2

1 3 1+-=-

b 2 n=2

12. Here, d = 1 mm =10-3 m, A= 6.5x 10-7 m

D =1m

x5 = nA D = 5 X 6.5 X 10-7 X _!_3 d w-

= 32.5 X 10-4 m AD

x 3 = (2n -1)"2d

(2x3-1)x6.5 x 1o-7

2 x 10-3

= 16.25 x 10-4 m x5 - x3 = (32.5 -16.25) 10-4m

= 16.25 x 10-4m = 1.63 mm

13. From x = n A.P._ d D

d1 =7A1d

D dz =7A2d

~=~ d2 A.2

14. When white light is used in a biprism experiment, central spot will be white , while the surronding fringes will be coloured.

15. From I R = I 1 +I 2 + 2ji;l; cos ~

When ~ = 0° ,IR =I+ I+ 2.Jii cos0°= 41

When~= 90° I R I = I+ I + 2.Jii cos 90° = 2 I

I~= 4I = 2 : 1 IR 2I

16. If I0 is intensity of unpolarised light, then intensity of polarised light from 1st polaroid = Iof2 . On rotating through 45°, intensity of light from 2nd polaroid,

I= (!.SL)ccos45o)z = Io ( _.!__ )2 = Io 2 2j2 ' 4

.-. < ~ 25%I0

17. When sourC!=S" are coherent ' ' ' I~ = I1 + I 2 + 2ji;I; cos~

At middle peint of the sereen, q, = 0° ·

IR:, I +I +2.Jii cos0°= 4I When sour~es , are inco,hereJ1t,:·

I~ ·= I 1 +I2 =I +I +21' · I 41· .. -li-=-=2 IR 2I ·,'

18. y 1 = 4sinwt

Yz = 3sin(rot + 1t I 3)

Here, a=4,b=3,<j>=n/3

R = ~a2 +b2 + 2abcos<j>

= ~42 +32 + 2 x 4 x 3cosn / 3

=m =6

'AD . 19. From ~=d

3 (5000x10-10) xl.O 5 x 10- = d

d = 5 x 10-7

=10-4m= 0.1 mm 5 X 10-3

20. Fringe width oc wavelength of light. Therefore fringe will become narrower.

21. S2P=(d2

+b2

)112

=d(1+ :: r2

= d ( 1 + ;:2 J = d + ~: Path difference = S2 P- S1 P

b2 x=d+--d=b2 ! 2d

2d

For missing wavelengths (i2n - 1)!:, = x = bd2

2 2

For

For

b2 n=1 A=-

' d'

2b2

n=2 A= -' 3d

22. Here, w = 31.4 rads-1

:. Time period of revolution,

T = 27t = 2 x 3.14 = 0.2 s (J) 31.4

Energy transmitted/ revolution

= (IA) T = e; A )r = <j> 0 T = 10-

3 X 0.2 = 10-4 J

2 2

23. For 5th dark fringe , 'AD 9A.D x =(2n - 1)--= - -

1 2 d 2d

D ?AD For 7th bright fringe, x 2 = nA. - = --d d

,

1) X 2 - X 1 =(!l-1)t- '

' d

AD [7 -~] = (I!- 1) t B ' d . 2 ' d .

2.5'A t=--- -

(!l - 1) ,'


24. The ratio of intensities of successive maxima is

1 = C2

1tJ = (5~J = C~J = 1 . __±_ . _4_

. 9n2 . 251t2

25. Here, a= 10,b = ~.-52_+_(_5_.[3=3~)2 = 10

~ = 10 = 1 : 1 b 10

26. The linear width of central principal maximum 2 'A D

d If it is equal to width of slit (d), then

2W d2

--=d or D=-d 2'A

27. Let the wavelength of monochromatic light in glass be )...g em, and in water be "-w em.

. ·. Number of waves in 8 em of glass = J!..., and number A.g.

of waves in 10 em of glass = J!.... "-w

8 10 "-w 10 5 - =-or-=-= -'Ag Aw Ag 8 4

c c llg = - and llw = -

vg vw Now,

.&_ = Vw = V Aw = ~ llw Vg VAg 4

5 5 4 5 llg = 4llw = 4 X 3 = 3

28. Distance between the slits, d = ~d1d2 = >f'4-.0-5_x_1_0-:_3:--x-2-.-90- x -10-_-=-3

= 3.427 x w-3 m

29. The resultant intensity at any point P is

1=410 cos2 (%)

10 = 4J0cos2 <j>/ 2

or

If Lll: is the corresponding value of, path difference at P,

then <I> = 2 1t(~x) 'A

As

or

21t 21i: ' ' -=-fix. 3 'A -

,_' ... ~ .... ' .: -- -.x.a -Lll:_'"' r)··~· -·

~=_! xd, 3 'A D

I '

x=-A.- = 6 x 1o-7 =Zx1o-3m _r, 3d I D 3 x 10-4

- X = 2 inm ·: ·' .-,; -.. ' ,' , ",' This is . the'di~~a~~~ ~f poi~tP- fr~m ;e;{t~al ~aximum.

\

964 Chapter 24 * Wave Optics

30. /max= I= 11 + lz + 2 ~G

. 31.

32.

33.

34.

When Wldth of each slit is doubled, intensity from each slit becomes twice ie,

!'1 = 211 and 1'2 = 2!2

I' max = I' = I' 1 + I' 2 + 2~ =2l1 +212 +2J211 x212

= 2(11 +12 +2Ji:xl;) = 21

o C 3 X 108

\, = 6000A, V a =- = · 10 "-a 6000x 10-

=5x1014Hz

A.a 6ooo'A o In water, "-m =- = --- = 4000 A.

~ 1.5

'AD From P=

d

D = ~_:1_ 'A

4x10-3 x0.1x10-3 =1

m 4 X 10-7

p 2 6 Pt =_a = -- = -mm a~l 5/3 5

]max- (a+b)2 - 16 /min- (a-b)2 -l

a+b 4 a-b 1

4a-4b=a+b or3a = 5b

a 5

b 3

35. The wavefront in Y-Z plane travels alongX-axis. Therefore , x = a and x = a' represent such a wavefront.

36. R =~a~+ a~+ 2a1 a2 cos 1t I 2 =~a~+ a~ 37. In the interference pattern,

38.

39.

]max _(a+b)2 _(1+2)2 -9·1 /min- (a-b)2- (1-2)2- .

:: = ~: = G J = 9 : 1.

11 - w1 - 1 - az !

2- w

2 -4-/l

~=1:2 b

40. Two independent sources cannot be coherent. Therefore, no interference pattern would be observed.

41. When the medium between plane-cbli.vex lens and plane gl~ss is rarer than the medium of lens and glass, the central spot of Newton's ring is dark. The darkness of central spot is due to the phase change of 1t which is introduced between the rays·,reflected · from denser to rarer and rarer to denser medium.

42. For diffraction to occur, the si?-e of an obstacle/aperture is comparable to the wavelength of light wave. The order of wavelength of light wave is io-7 m, so diffraction occurs.

43. When we seen the painting which is painted by a myriad of small colour dots near our eyes, scientillating colour of dots are visible due to diffraction of light and when we go away from the painting our eyes blend the dots and we see the different colours. This is due to the change is angular separation of adjacent dots as the distance of the object changes.

44. In Young's experiment fringe width for dark and white fringes are same while in the same experiment, when a white light as a source is used, the central fringe is white around which few coloured frings are observed on either side.

45. The beautiful colours are seen on account of interference of light reflected from the upper and the lower surfaces of the thin films. Since, condition for constructive and destructive interference depends upon the wavelength of light therefore, coloured interference fringes are observed.

46. According to Newton's to corpuscular the01y of light, the light should travel faster in denser mediQ. than in rarer media. It is contrary to present theory of light which explains the light travels faster in air (rarer) and in water (Denser).

47. The waves diffracted from the edges of circular obstacle , placed in the path of light, interfere constructively at the centre of the shadow resulting in the formation of a bright spot.

48. The waves which consist longer wavelength have more diffraction. Since radio waves have greater wavelength than microwaves, hence, radio waves undergo more diffraction than microwaves.

49. The diffraction of sound is only possible when the size of opening should be of the same order as its wavelength and the wavelength of sound is of the order of 1.0 m, hence, for a very small opening no diffraction is produced in sound waves.

50. For two coherent sources,

1=11 +12 +Mcos<jl

Putting,

We have , I= 10 + 10 + 2.JI0 x 10 cos<jl

Simplifying the above expression

I= 210 (1 + cos<jl)

= 210 ( 1 + 2 cos2 % - 1 )

=2I0 x2cos2 1 2

= 410 cos2 1 2

Also, [max = c.ji; +Fz)2

i

I min = ( .ji; - JJ;f

f r I

i

51. As 4th bright fringe of unknown wavelength A' coincides with 3rd btight fringe of known wavelength A= 590 nrn, therefore

4A'D 3AD --=--

d d

A.'= 2A = 2x590 = 442.5nm 4 4

52. The phase difference ($) between the wavelets from the

top edge and the bottom edge of the slit is ljl = 2

1t (d sin 9), . A.

where dis the slit width. The first minima of the diffraction

pattern occurs at sine= 2::_, so ljl = 2

1t (d x !:.) = 21t d A d

53. In chromatic aberration the image formed by a lens has coloured fringes, because the refractive index for different colours is different and hence the focal length of lens for different colours is different. So, the cause of chromatic aberration is the variation of focal length with colour.

54. In an interference experiment the spacing between successive maxima and minima is called the fringe width and is given by

~=DA.Id

55. Angular width of central maximum 2 A.D It can be

a increased by decreasing the slit width (a).

56. It is more difficult to observe diffraction with light waves because wavelength of light waves is far too smaller compared to that of sound waves.

57. From IR = I1 + I2 + 2~I1 I2 cps~

21t A. 1t where ~=-X-=-

58.

A. 4 2 1t

I R = I + I + 2I cos- = 2I 2

At the central bright fringe, I R = 4 I

. . !B._= 3!_ = 0.5 IR I 4I

X= n A!!_= 3 X 6000!!_ = 4 X A!!_ d d d

A = 3 X 6000 = 4500 A 4

59. Position of fourth maxima 4DA

Xo=-d-

or x oc A. : . x (blue) < x (green)

60. Two sources should have the same wavelength, nearly the same amplitude and have a constant phase difference. If the phase difference between two interfering waves does not remain constant, interference pattern will not be sustained.

61. Intensity, I 0 = I1 + I 2 + 2~I1 I 2

ff ~=~=I (~0 then I0 = 4I

When one slit is covered then I 2

I' -I- Io . . , 0- - 4

Chapter 24 • Wave Optics

62. f.l. = taniP = tan54.74°= 1.44 = J2

From ll = sin(A +om) I 2 = )2 sinAl 2

. (60o+Om) r;:;2 . 600 sm = vL.sm-2 2

= J2 x }:_ = ~ = sin 45° 2 v2

om= 90°-60°= 30°

63. ~= 2 1tAx= 2 1tx2:_=~ A. A 6 3

and

1=11 +I2 +2~I1 I2 cos~

I I = I + I + 21 cos~ = 3 I 3

I 0 = I + I + 2 I cos 0° = 4l

I ' 3

I 0 4

965

64. Polarisation proves that light waves are transverse m character.

65. As Ab < /,r

and width of pattern is directly proportional to A, therefore, the fringes will become narrower.

66. Photons are the packets of energy. Power emitted,

Energy of photon,

3 P=2x1o- w

E = hv = 6.6 X 10-34 X 6 X 1014 J h being Planck's constant. Number of photons emitted per second

p n=-

E

2x1o-3 .

6.6 X 10-34 X 6 X 1014

= 5 X 1015

67. Intensity of polarized light = I012 :. Intensity of light which does not get transmitted

= I0 - Iof2 = Iof2

68. D = (f.l.-1) A For blue light Jl is greater than tl1at for red light, so D2 > D1.

69. d=0.6mm=6x10-4m,A=4800x10-10 m

D = 120 em = 1.2 m For dark fringe~ ,

, A.D . A.D x = (2n -1)-- = (2x 1-1)--

, . 2 d 2 d

'· · · 2'x A Angular width = - = -

-' ''

· · .·.·v-· d

r

I


70. Lens-Maker's formula is given by

_!_=( Jl -1)(_!_ _ _!_) f a g Rl Rz

... (i)

where al-lg is refractive index of glass w.r.t. air, R1 and R2 are radii of curvature of two surfaces of lens and f is focal length of the lens. If the lens is immersed is a liquid of refractive index J.l1 then

1 (1 '1) -=(zJlg -1) ---fz R1 Rz

Here zJ.lg is refractive index of glass w.r.t. liquid. Dividing Eq. (i) by Eq. (ii) , we have

f1 (aJlg - 1) -= fz (zJlg -1)

=> !1=(1.5-1_1) / 2 1.25

=> f1 = 0.5 X 1.25 = 2_5 f 2 0.25

Hence, focal length increases by a factor of 2.5. 'AD

71. From 13 = d

D=j3xd ').._

4x10-3 x0.1x10-3 =1

m 4 X 10-7

... (ii)

72. Distance between first dark fringe on either side of central bright fringe = width of central maximum

2f...D 2 X (600 X 10-9 ) X 2 =--=

a 10-3

= 24 x 10-4 m = 2.4 mm

,

73. Distance between first dark fringe on either side of central bright fringe = width of central maximum

2f...D 2 X (600 X 10-9 ) X 2 =--=

a 10-3

= 24 x 10-4 m = 2.4 mm

Here, I= 10

= 11 = 12 and if ~ is the phase difference, 4

then I= I 1 +I2 +2~I1 I 2 cos<j>

· I I I Ml .. _Q_ = _Q_ + _Q_ + 2 _Q_ X _Q_ COS <j> 4 4 4 4 4

1 1 1 1 - = - +-+ 2 X- COS <j> 4 4 4 4

or

1 21t cos<j>=- or <1>=-

2 3 or

If L\x the path difference betwee11 two waves, then

<I> L\x ').._ ').._ 21t ').._ -=-=> L\x=-<j>=-X-.--=-21t ').._ 21t 21t 3 3

For path difference L\x, if angular separation is 8 then d sin 9L\x = 'A I 3

or sm =- or e = sm -. e "- . -1 ( "- J 3d 3d

74. For maxima on the screen dsin8=nA.

Given d = slit-width= 2A. 2A.sin8=nA. 2sin8=n

The maximum value of sin 8 = + 1 .. n=2 :. Eq. (i) is satisfied by- 2,- 1, 0, 1, 2.

75. The intensity of principal maximum in the single slit diffraction pattern does not depend upon the slit width.

24.wave optics

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