chapter 10 wave optics

7/21/2019 Chapter 10 Wave Optics

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Question 10.1:

Monochromatic light of wavelength 589 nm is incident from air on a water surface. What

are the wavelength, frequency and speed of (a) reflected, and (b) refracted light?

Refractive index of water is 1.33.

Answer

Wavelength of incident monochromatic light,

λ = 589 nm = 589 × 10−9

m

Speed of light in air, c = 3 × 108 m/s

Refractive index of water, μ = 1.33

The ray will reflect back in the same medium as that of incident ray. Hence, the

wavelength, speed, and frequency of the reflected ray will be the same as that of the

incident ray.

Frequency of light is given by the relation,

Hence, the speed, frequency, and wavelength of the reflected light are 3 × 108

m/s, 5.09

×1014

Hz, and 589 nm respectively.

Frequency of light does not depend on the property of the medium in which it is

travelling. Hence, the frequency of the refracted ray in water will be equal to the

frequency of the incident or reflected light in air.

Refracted frequency, ν = 5.09 ×1014

Hz

Speed of light in water is related to the refractive index of water as:



Wavelength of light in water

Hence, the speed, frequency,

444.01nm, and 5.09 × 10

14

Question 10.2:

What is the shape of the wav

Light diverging from a point

Light emerging out of a conv

The portion of the wavefront

Answer

The shape of the wavefront iThe wavefront emanating fr

is given by the relation,

and wavelength of refracted light are 2.26 ×108

z respectively.

efront in each of the following cases:

source.

ex lens when a point source is placed at its focu

of light from a distant star intercepted by the E

case of a light diverging from a point source im a point source is shown in the given figure.

m/s,

s.

rth.

spherical.



The shape of the wavefront isource is placed at its focus i

The portion of the wavefront plane.

Question 10.3:

The refractive index of glassvacuum is 3.0 × 108 m s−1)

Is the speed of light in glass icolours red and violet travels

Answer

Refractive index of glass, μ

Speed of light, c = 3 × 108 m

Speed of light in glass is giv

case of a light emerging out of a convex lenss a parallel grid. This is shown in the given figu

of light from a distant star intercepted by the E

is 1.5. What is the speed of light in glass? Spee

ndependent of the colour of light? If not, whichslower in a glass prism?

1.5

s

n by the relation,

hen a pointe.

rth is a

of light in

of the two



Hence, the speed of light in

The speed of light in glass is

The refractive index of a violindex of a red component. Hlight in glass. Hence, violet l

Question 10.4:

In a Young’s double-slit exp placed 1.4 m away. The distafringe is measured to be 1.2experiment.

Answer

Distance between the slits, d

Distance between the slits an

Distance between the central

u = 1.2 cm = 1.2 × 10−2 m

In case of a constructive inte

two fringes as:

Where,

n = Order of fringes = 4

lass is 2 × 108 m/s.

not independent of the colour of light.

et component of white light is greater than the rnce, the speed of violet light is less than the spght travels slower than red light in a glass pris

riment, the slits are separated by 0.28 mm andnce between the central bright fringe and the fom. Determine the wavelength of light used in t

= 0.28 mm = 0.28 × 10−3 m

d the screen, D = 1.4 m

fringe and the fourth (n = 4) fringe,

ference, we have the relation for the distance b

efractiveed of red.

he screen isurth brighte

tween the



λ = Wavelength of light used

∴

Hence, the wavelength of the

Question 10.5:

In Young’s double-slit experintensity of light at a point o

intensity of light at a point w

Answer

Let I 1 and I 2 be the intensityobtained as:

Where,

= Phase difference betwee

For monochromatic light wa

Phase difference =

light is 600 nm.

iment using monochromatic light of wavelengtthe screen where path difference is λ, is K unithere path difference is λ /3?

f the two light waves. Their resultant intensitie

the two waves

es,

λ, thes. What is the

s can be



Since path difference = λ,

Phase difference,

Given,

I ’ = K

When path difference ,

Phase difference,

Hence, resultant intensity,

Using equation (1), we can

Hence, the intensity of light

Question 10.6:

A beam of light consisting o

interference fringes in a You

Find the distance of the thirdwavelength 650 nm.

What is the least distance frothe wavelengths coincide?

Answer

rite:

t a point where the path difference is is

two wavelengths, 650 nm and 520 nm, is used

g’s double-slit experiment.

bright fringe on the screen from the central ma

m the central maximum where the bright fringe

units.

to obtain

imum for

due to both



Wavelength of the light bea

Wavelength of another light

Distance of the slits from the

Distance between the two sli

Distance of the nth bright frinrelation,

Let the nth bright fringe due t

wavelength coincide on th

Hence, the least distance fro

Note: The value of d and D a

Question 10.7:

,

eam,

screen = D

s = d

ge on the screen from the central maximum is g

wavelength and (n − 1)th bright fringe due

e screen. We can equate the conditions for brig

the central maximum can be obtained by the r

re not given in the question.

iven by the

to

t fringes as:

elation:



In a double-slit experiment t placed 1 m away. The wavelof the fringe if the entire expindex of water to be 4/3.

Answer

Distance of the screen from t

Wavelength of light used,

Angular width of the fringe i

Angular width of the fringe i

Refractive index of water,

Refractive index is related to

Therefore, the angular width

Question 10.8:

What is the Brewster angle f

Answer

e angular width of a fringe is found to be 0.2°ngth of light used is 600 nm. What will be the

erimental apparatus is immersed in water? Take

he slits, D = 1 m

water =

angular width as:

of the fringe in water will reduce to 0.15°.

r air to glass transition? (Refractive index of gl

n a screenngular widthrefractive

ass = 1.5.)



Refractive index of glass,

Brewster angle = θ

Brewster angle is related to r

Therefore, the Brewster angl

Question 10.9:

Light of wavelength 5000 Å

and frequency of the reflectenormal to the incident ray?

Answer

Wavelength of incident light,

Speed of light, c = 3 × 108 m

Frequency of incident light i

The wavelength and frequenHence, the wavelength of ref

When reflected ray is normalangle of reflection, is 90

According to the law of reflereflection. Hence, we can wr

efractive index as:

e for air to glass transition is 56.31°.

falls on a plane reflecting surface. What are the

d light? For what angle of incidence is the refle

λ = 5000 Å = 5000 × 10

−10

m

given by the relation,

y of incident light is the same as that of reflectlected light is 5000 Å and its frequency is 6 × 1

to incident ray, the sum of the angle of inciden.

ction, the angle of incidence is always equal toite the sum as:

wavelength

ted ray

d ray.014 Hz.

ce, and

he angle of



Therefore, the angle of incid

Question 10.10:

Estimate the distance for whiand wavelength 400 nm.

Answer

Fresnel’s distance ( Z F ) is theis given by the relation,

Where,

Aperture width, a = 4 mm =

Wavelength of light, λ = 400

Therefore, the distance for w

Question 10.11:

The 6563 Å line emittedEstimate the speed with whi

Answer

nce for the given condition is 45°.

ch ray optics is good approximation for an aper

distance for which the ray optics is a good appr

×10−3 m

nm = 400 × 10−9 m

hich the ray optics is a good approximation is 4

by hydrogen in a star is found to be red shiftedh the star is receding from the Earth.

ure of 4 mm

oximation. It

m.

y 15 Å.



Wavelength of line emitt

λ = 6563 Å

= 6563 × 10−10 m.

Star’s red-shift,

Speed of light,

Let the velocity of the star re

The red shift is related with

Therefore, the speed with wm/s.

Question 10.12:

Explain how Corpuscular thgreater than the speed of lighdetermination of the speed oconsistent with experiment?

Answer

ed by hydrogen,

ceding away from the Earth be v.

elocity as:

ich the star is receding away from the Earth is

ory predicts the speed of light in a medium, sayt in vacuum. Is the prediction confirmed by explight in water? If not, which alternative picture

.87 × 105

, water, to beerimentalof light is



No; Wave theory

Newton’s corpuscular theoryof two media from a rarer (aiforces of attraction normal toincreases while the compone

Hence, we can write the exp

… (i)

Where,

i = Angle of incidence

r = Angle of reflection

c = Velocity of light in air

v = Velocity of light in water

We have the relation for rela

Hence, equation (i) reduces t

But, > 1

Hence, it can be inferred fro prediction is opposite to the

The wave picture of light is

Question 10.13:

You have learnt in the text hrefraction. Use the same prina plane mirror produces a virobject distance from the mirr

Answer

of light states that when light corpuscles striker) to a denser (water) medium, the particles expthe surface. Hence, the normal component of vt along the surface remains unchanged.

ession:

ive refractive index of water with respect to air

o

equation (ii) that v > c. This is not possible sixperimental results of c > v.

onsistent with the experimental results.

w Huygens’ principle leads to the laws of refleciple to deduce directly that a point object placetual image whose distance from the mirror is eqor.

the interfaceerienceelocity

as:

ce this

ction andd in front ofual to the



Let an object at O be placedthe given figure).

A circle is drawn from the ceAccording to Huygens’ Prin

If the mirror is absent, then adistance r (as shown in the g

can be considered as aobject placed in front of thethe mirror is equal to the obj

Question 10.14:

Let us list some of the factor propagation:

Nature of the source.

n front of a plane mirror MO’ at a distance r (a

ntre (O) such that it just touches the plane mirriple, XY is the wavefront of incident light.

similar wavefront X’Y’ (as XY) would form bven figure).

virtual reflected ray for the plane mirror. Hencelane mirror produces a virtual image whose disct distance (r ).

, which could possibly influence the speed of

shown in

r at point O’.

hind O’ at

, a pointtance from

ave



Direction of propagation.

Motion of the source and/or

Wave length.

Intensity of the wave.

On which of these factors, if

The speed of light in

The speed of light in

Answer

Thespeed of light in a vacuuis not affected by the motiondoes not affect the speed of l

Out of the listed factors, thelight in that medium.

Question 10.15:

For sound waves, the Dopplsituations: (i) source at rest;The exact Doppler formulasidentical for these situations.formulas to be strictly identimedium?

Answer

No

bserver.

any, does

acuum,

a medium (say, glass or water), depend?

i.e., 3 × 108 m/s (approximately) is a universaof the source, the observer, or both. Hence, theight in a vacuum.

peed of light in a medium depends on the wave

r formula for frequency shift differs slightly bet bserver moving, and (ii) source moving; obserfor the case of light waves in vacuum are, howeExplain why this should be so. Would you expal for the two situations in case of light travelli

l constant. Itgiven factor

length of

ween the twoer at rest.er, strictlyct theg in a



Sound waves can propagatescientifically identical becauin the two situations. Hence,same.

In case of light waves, sound

are identical because the spethe motion of the source. Whidentical because the speed o

Question 10.16:

In double-slit experiment usiformed on a distant screen is

Answer

Wavelength of light used, λ

Angular width of fringe,

Angular width of a fringe is

Therefore, the spacing betwe

Question 10.17:

Answer the following questi

nly through a medium. The two given situatione the motion of an observer relative to a mediuthe Doppler formulas for the two situations can

can travel in a vacuum. In a vacuum, the above

d of light is independent of the motion of the oen light travels in a medium, the above two casf light depends on the wavelength of the mediu

g light of wavelength 600 nm, the angular wid0.1º. What is the spacing between the two slits?

6000 nm = 600 × 10−9 m

elated to slit spacing (d ) as:

en the slits is .

ns:

s are notis different

ot be the

two cases

server ands are not.

h of a fringe



In a single slit diffraction experiment, the width of the slit is made double the original

width. How does this affect the size and intensity of the central diffraction band?

In what way is diffraction from each slit related to the interference pattern in a double-slit

experiment?

When a tiny circular obstacle is placed in the path of light from a distant source, a brightspot is seen at the centre of the shadow of the obstacle. Explain why?

Two students are separated by a 7 m partition wall in a room 10 m high. If both light and

sound waves can bend around obstacles, how is it that the students are unable to see each

other even though they can converse easily.

Ray optics is based on the assumption that light travels in a straight line. Diffraction

effects (observed when light propagates through small apertures/slits or around small

obstacles) disprove this assumption. Yet the ray optics assumption is so commonly used

in understanding location and several other properties of images in optical instruments.

What is the justification?

Answer

In a single slit diffraction experiment, if the width of the slit is made double the original

width, then the size of the central diffraction band reduces to half and the intensity of the

central diffraction band increases up to four times.

The interference pattern in a double-slit experiment is modulated by diffraction from each

slit. The pattern is the result of the interference of the diffracted wave from each slit.

When a tiny circular obstacle is placed in the path of light from a distant source, a bright

spot is seen at the centre of the shadow of the obstacle. This is because light waves are

diffracted from the edge of the circular obstacle, which interferes constructively at the

centre of the shadow. This constructive interference produces a bright spot.

Bending of waves by obstacles by a large angle is possible when the size of the obstacle

is comparable to the wavelength of the waves.

On the one hand, the wavelength of the light waves is too small in comparison to the size

of the obstacle. Thus, the diffraction angle will be very small. Hence, the students are

unable to see each other. On the other hand, the size of the wall is comparable to the

wavelength of the sound waves. Thus, the bending of the waves takes place at a large

angle. Hence, the students are able to hear each other.

The justification is that in ordinary optical instruments, the size of the aperture involved is

much larger than the wavelength of the light used.



Question 10.18:

Two towers on top of two hila hill halfway between the to

can be sent between the towe

Answer

Distance between the towers,

Height of the line joining the

Thus, the radial spread of the

Since the hill is located halfas:

Z P = 20 km = 2 × 104 m

Aperture can be taken as:

a = d = 50 mFresnel’s distance is given b

Where,

λ = Wavelength of radio wav

Therefore, the wavelength o

ls are 40 km apart. The line joining them passeswers. What is the longest wavelength of radio

rs without appreciable diffraction effects?

d = 40 km

hills, d = 50 m.

radio waves should not exceed 50 km.

ay between the towers, Fresnel’s distance can

the relation,

es

the radio waves is 12.5 cm.

50 m aboveaves, which

e obtained



Question 10.19:

A parallel beam of light of wdiffraction pattern is observeis at a distance of 2.5 mm fro

Answer

Wavelength of light beam, λ

Distance of the screen from t

For first minima, n = 1

Distance between the slits =

Distance of the first minimu

x = 2.5 mm = 2.5 × 10−3 m

It is related to the order of m

Therefore, the width of the sl

Question 10.20:

Answer the following questi

When a low flying aircraft p

a slight shaking of the pictur

avelength 500 nm falls on a narrow slit and thed on a screen 1 m away. It is observed that the fm the centre of the screen. Find the width of th

= 500 nm = 500 × 10−9 m

he slit, D = 1 m

from the centre of the screen can be obtained

nima as:

its is 0.2 mm.

ns:

sses overhead, we sometimes notice

on our TV screen. Suggest a possible explanat

resultingirst minimumslit.

s:

ion.



As you have learnt in the texis basic to understanding inteWhat is the justification of t

Answer

Weak radar signals sent by a by the antenna. As a result, taircraft passes overhead, wescreen.

The principle of linear super

understanding of intensity disuperposition follows from twave motion. If y1 and y2 arelinear combination of y1 and

Question 10.21:

In deriving the single slit difangles of nλ/a. Justify this b

Answer

Consider that a single slit of

Width of each slit,

Angle of diffraction is given

, the principle of linear superposition of wave dnsity distributions in diffraction and interferencis principle?

low flying aircraft can interfere with the TV sige TV signals may get distorted. Hence, when a

sometimes notice a slight shaking of the picture

osition of wave displacement is essential to ou

tributions and interference patterns. This is bece linear character of a differential equation thatthe solutions of the second order wave equatio2 will also be the solution of the wave equation

raction pattern, it was stated that the intensity isuitably dividing the slit to bring out the cance

idth d is divided into n smaller slits.

by the relation,

isplacemente patterns.

nals receivedlow flyingon our TV

ausegoverns, then any.

zero atllation.



Now, each of these infinitesicombination of these slits wi

ally small slit sends zero intensity in direction ll give zero intensity.

θ . Hence, the

chapter 10 wave optics

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