HC Verma Solutions for Class 11 Physics Chapter 17 – Light Waves

Page No 379:

Question 1:

Is the colour of 620 nm light and 780 nm light same? Is the colour of 620 nm light and 621 nm light same? How many colours are there in white light?

Answer:

White light is a composition of seven colours. These are violet, indigo, blue, green, yellow, orange and red, collectively known as VIBGYOR. Spectrum of white light consists of  sever colour bands. Each band consists of some range of wavelengths or frequencies.
For orange colour : (590 nm to 620 nm)
For red colour: (620 nm to 780 nm)

So, the colour of 620 nm and 780 nm lights may be different. But the colour of 620 nm light and 621 nm light is same.

Question 2:

The wavelength of light in a medium is

λ=λ0/μ, where

λis the wavelength in vacuum. A beam of red light

λ0=720 nmenters water. The wavelength in water is

λ=λ0/μ=540 nm. To a person under water, does this light appear green?

Answer:

Colour of light will depend only on the frequency of light and not on the wavelength of the light. So, light will appear red to an observer under water.

Question 3:

Will the diffraction effects from a slit be more or less clearly visible if the slit-width is increased?

Answer:

The width of the central band is inversely proportional to the slit width. So, as the width of the slit is increased, the central band will become less wider and further bands will start merging in them. Hence, diffraction effects will be visible less clearly.

Question 4:

If we put a cardboard (say 20 cm × 20 cm) between a light source and our eyes, we can’t see the light. But when we put the same cardboard between a sound source and out ear, we hear the sound almost clearly. Explain.

Answer:

Light waves have the property of travelling in a straight line, unlike sound waves. When we put a cardboard between the light source and our eyes, the light waves are obstructed by the cardboard and cannot reach our eyes, which doesn’t happen when the cardboard is inserted between sound source and our ear.

Question 5:

TV signals broadcast by a Delhi studio cannot be directly received at Patna, which is about 1000 km away. But the same signal goes some 36000 km away to a satellite, gets reflected and is then received at Patna. Explain.

Answer:

To receive TV signals transmitted from Delhi in Patna directly, one has to use antennas of great height, which will cost much. On the other hand, transmission of signals with the help of satellites requires only high frequency waves and can be done easily.

Question 6:

Can we perform Young’s double slit experiment with sound waves? To get a reasonable “fringe pattern”, what should be the order of separation between the slits? How can the bright fringes and the dark fringes be detected in this case?

Answer:

Young’s double slit experiment can be performed with sound waves, as the sound waves also show interference pattern. To get a reasonable “fringe pattern”, the separation of the slits should be of the order of the wavelength of the sound waves used.
In this experiment, the bright and dark fringes can be detected by measuring the intensity of sound using a microphone or any other detector.

Question 7:

Is it necessary to have two waves of equal intensity to study interference pattern? Will there be an effect on clarity if the waves have unequal intensity?

Answer:

Interference pattern can be studied with waves of unequal intensity.
Ι=Ι1+Ι2+2Ι1×Ι2cosϕ,
where

ϕ=phase difference.
In case of waves of unequal intensities, the contrast will not be clear, as the minima will not be completely dark.

Question 8:

Can we conclude from the interference phenomenon whether light is a transverse wave or a longitudinal wave?

Answer:

The interference pattern can be produced by any two coherent waves moving in the same direction. It cannot be concluded from the interference phenomenon that light is a transverse wave, as sound waves that are longitudinal in nature also interfere.

Question 9:

Why don’t we have interference when two candles are placed close to each other and the intensity is seen on a distant screen? What happens if the candles are replaced by laser sources?

Answer:

In order to get interference, the sources should be coherent, i.e. they should emit wave of the same frequency and a stable phase difference. Two candles that are placed close to each other are distinct and cannot be considered as coherent sources. Two independent sources cannot be coherent. So, two different laser sources will also not serve the purpose.

Question 10:

If the separation between the slits in a Young’s double slit experiment is increased, what happens to the fringe-width? If the separation is increased too much, will the fringe pattern remain detectable?

Answer:

The fringe width in Young’s double slit experiment depends on the separation of the slits.

χ=λDd,
where
λ=wavelengthχ=fringe widthD=distance between slits and screend=separation between slitsOn increasing d, fringe width decreases. If the separation is increased too much, the fringes will merge with each other and the fringe pattern won’t be detectable.

Question 11:

Suppose white light falls on a double slit but one slit is covered by a violet filter (allowing λ = 400 nm). Describe the nature of the fringe pattern observed.

Answer:

The violet filter will allow only violet light to pass through it. Now, if the double slit experiment is performed with the white light and violet light, the fringe pattern will not be the same as obtained by just using white light as the source. To have interference pattern, the light waves entering from the slits should be monochromatic. So, in this case, the violet light will superimpose with only violet light (of wavelength 400 nm) in such a way that the bright bands will be of violet colour and the minima will be completely dark.

Question 1:

Light is
(a) a wave phenomenon
(b) a particle phenomenon
(c) both a particle and a wave phenomenon

Answer:

(c) both a particle and a wave phenomenon

Light shows photoelectric effect and Compton effect, which depicts its particle nature. It also shows interference and diffraction, which depicts the wave nature of light.

Question 2:

The speed of light depends
(a) on elasticity of the medium only
(b) on inertia of the medium only
(c) on elasticity as well as inertia
(d) neither on elasticity nor on inertia

Answer:

(d) neither on elasticity nor on inertia

The speed of light in any medium depends on the refractive index of that medium, which is an intensive property. Hence, speed of light is not affected by the elasticity and inertia of the medium.

Question 3:

The equation of a light wave is written as

y=A sinkx-ωt. Here, y represents
(a) displacement of ether particles
(b) pressure in the medium
(c) density of the medium
(d) electric field

Answer:

(d) electric field

Light consists of mutually perpendicular electric and magnetic fields. So, the equation of a light wave is represented by its field vector.

Question 4:

Which of the following properties shows that light is a transverse wave?
(a) Reflection
(b) Interference
(c) Diffraction
(d) Polarization

Answer:

(d) Polarization

Reflection, interference and diffraction are the phenomena shown by both transverse waves and longitudinal waves. Polarization is the phenomenon shown only by transverse waves.

Question 5:

When light is refracted into a medium,
(a) its wavelength and frequency increase
(b) its wavelength increases but frequency remains unchanged
(c) its wavelength decreases but frequency remains unchanged
(d) its wavelength and frequency decrease

Answer:

(c) its wavelength decreases but frequency remains unchanged

Frequency of a light wave, as it travels from one medium to another, always remains unchanged, while wavelength decreases.

Decrease in the wavelength of light entering a medium of refractive index

μis given by
λΜ=λμ,where  λΜ=wavelength in medium             λ=wavelength in vacuum             μ=refractive index

Question 6:

When light is refracted, which of the following does not change?
(a) Wavelength
(b) Frequency
(c) Velocity
(d) Amplitude

Answer:

(b) Frequency

Frequency of a light wave doesn’t change on changing the medium of propagation of light.

Question 7:

An amplitude modulated (AM) radio wave bends appreciably round the corners of a 1 m × 1 m board but a frequency modulated (FM) wave only bends negligibly. If the average wavelengths of the AM and FM waves are

λa and λf,
(a)

λa>λf(b)

λa=λf(c)

λa<λf(d) We don’t have sufficient information to evaluate the relation of

λa and λf.

Answer:

(a)

λa>λfAn electromagnetic wave bends round the corners of an obstacle if the size of the obstacle is comparable to the wavelength of the wave. An AM wave has less frequency than an FM wave. So, an AM wave has a higher wavelength than an FM wave and it bends round the corners of a 1 m 

×1m board.
λ

Question 8:

Which of the following sources provides the best monochromatic light?
(a) A candle
(b) A bulb
(c) A mercury tube
(d) A laser

Answer:

(d) A laser

Among the given sources, laser is the best coherent source providing monochromatic light with constant phase difference.

Question 9:

The wavefronts of a light wave travelling in vacuum are given by x + y + z = c. The angle made by the direction of propagation of light with the X-axis is
(a) 0°
(b) 45°
(c) 90°
(d)

cos-1 1/3

Answer:

(d)

cos-11/3On writing the given equation in the plane equation form lx + my + nz = p,
where l2 + m2 + n2 = 1 and p>0, we get:

13x+13y+13z=c3If

θis the angle between the normal and +X axis, then

cosθ=13⇒θ=cos-113

Question 10:

The wavefronts of light coming from a distant source of unknown shape are nearly
(a) plane
(b) elliptical
(c) cylindrical
(d) spherical

Answer:

(a) plane

Wave travelling from a distant source always has plane wavefront.

Question 11:

The inverse square law of intensity (i.e. the intensity

∞1r2) is valid for a
(a) point source
(b) line source
(c) plane source
(d) cylindrical source

Answer:

(a) point source

Intensity of a point source obeys the inverse square law.
Intensity of light at distance r from the point source is given by

I=S/4πr2,
where S is the source strength.

Question 12:

Two sources are called coherent if they produce waves
(a) of equal wavelength
(b) of equal velocity
(c) having same shape of wave front
(d) having a constant phase difference

Answer:

(d) having a constant phase difference
For light waves emitted by two sources of light to remain coherent, the initial phase difference between waves should remain constant in time. If the phase difference changes continuously or randomly with time, then the sources are incoherent.

Question 13:

When a drop of oil is spread on a water surface, it displays beautiful colours in daylight because of
(a) disperson of light
(b) reflection of light
(c) polarization of light
(d) interference of light

Answer:

(d) interference of light

Interference effect is produced by a thin film ( coating of a thin layer of a translucent material on a medium of different refractive index which allows light to pass through it))In the present case, oil floating on water forms a thin film on the surface of water, leading to the display of beautiful colours in daylight because of the interference of sunlight.

Page No 380:

Question 1:

Find the range of frequency of light that is visible to an average human being

400 nm <λ<700 nm.

Answer:

Given:
Range of wave length is

400 nm<λ<700 nmWe know that frequency is given by

f=cλ,

where c=speed of light=3×108 m/s          f is the frequency
          λ is the wavelength
We can write wavelength as:

1700 nm<1λ<1400 nm⇒17×10-7 m<1λ<14×10-7 m3×1087×10-7 Hz<cλ<3×1084×10-7 Hz⇒4.3×1014 Hz<cλ<7.5×1014 Hz⇒4.3×1014 Hz<f<7.5×1014 HzHence, frequency of the range of light that is visible to an average human being is

4.3×1014 Hz to7.5×1014 Hz.

Question 2:

The wavelength of sodium light in air is 589 nm. (a) Find its frequency in air. (b) Find its wavelength in water (refractive index = 1.33). (c) Find its frequency in water. (d) Find its speed in water.

Answer:

Given:
Wavelength of sodium light in air,

λa=589 nm=589×10-9 mRefractive index of water, μw= 1⋅33
We know that

f=cλ,

where c=speed of light=3×108 m/s          f  = frequency
λ = wavelength
(a) Frequency in air,

fair=cλa

fair=3×108589×10-9      =5.09×1014 Hz(b)
Let wavelength of sodium light in water be

λw.
We know that

μaμw=λωλa,
where μa is the refractive index of air which is equal to 1 and
λw is the wavelength of sodium light in water.

⇒11.33=λω589×10-9⇒      λω=443 nm(c) Frequency of light does not change when light travels from one medium to another.

∴ fω=fa               =5.09×1014 Hz(d) Let the speed of sodium light in water be

νω     and speed in air, va = c.
Using

μaμω=νωνa, we get:

νω=μacμω                    =3×1081.33=2.25×108 m/s

Question 3:

The index of refraction of fused quartz is 1⋅472 for light of wavelength 400 nm and is 1⋅452 for light of wavelength 760 nm. Find the speeds of light of these wavelengths in fused quartz.

Answer:

Given:
Refractive index of fused quartz for light of wavelength 400 nm is 1.472.
And refractive index of fused quartz for light of wavelength 760 nm is 1.452.
We known that refractive index of a material is given by
μ =

Speed of light in vacuumSpeed of light in the material=cvLet speed of light for wavelength 400 nm in quartz be v400.
So,

1.472=3×108v400⇒v400=2.04×108 m/sLet speed of light of wavelength 760 nm in quartz be v760.
Again,

1.4521=3×108ν760

⇒v760=2.07×108 m/s

Question 4:

The speed of yellow light in a certain liquid is 2⋅4 × 108 m s−1. Find the refractive index of the liquid.

Answer:

Given:
Speed of yellow light in liquid (vL)= 2⋅4 × 108 m s−1
And  speed of yellow light in air speed = va
Let μL be the refractive index of the liquid
Using,

μL=speed of light in vaccumvelocity of light in the given medium=cvL

μL=3×1082.4×108=1.25Hence, the required refractive index is 1.25.

Question 5:

Two narrow slits emitting light in phase are separated by a distance of 1⋅0 cm. The wavelength of the light is

5·0×10-7 m. The interference pattern is observed on a screen placed at a distance of 1⋅0 m. (a) Find the separation between consecutive maxima. Can you expect to distinguish between these maxima? (b) Find the separation between the sources which will give a separation of 1⋅0 mm between consecutive maxima.

Answer:

Given:
Separation between two narrow slits, d = 1 cm = 10−2 m
Wavelength of the light,

λ=5×10-7 mDistance of the screen,

D=1 m(a)
We know that separation between two consecutive maxima =  fringe width (β).
That is,

β=λDd    …(i)
=5×10-7×110-2 m =5×10-5 m=0.05 mm(b)
Separation between two consecutive maxima = fringe width

β=1 mm=10-3 mLet the separation between the sources be ‘d’
Using equation (i), we get:

d’=5×10-7×110-3 ⇒d’=5×10-4 m=0.50 mm.

Question 14:

Two coherent sources of different intensities send waves which interfere. The ratio of maximum intensity to the minimum intensity is 25. The intensities of the sources are in the ratio
(a) 25 : 1
(b) 5 : 1
(c) 9 : 4
(d) 625 : 1

Answer:

(c) 9 : 4

Ratio of maximum intensity and minimum intensity is given by

ImaxImin=I1 +I22I1-I22=251⇒I1=3 and I2=2⇒I1=9 and I2=4
Then,

I1I2=94

Question 15:

The slits in a Young’s double slit experiment have equal width and the source is placed symmetrically with respect to the slits. The intensity at the central fringe is I0. If one of the slits is closed, the intensity at this point will be
(a) I0
(b) I0/4
(c) I0/2
(d) 4I0

Answer:

(b) I0/​4

Total intensity coming from the source is Iwhich is present at the central maxima. In case of two slits, the intensity is getting distributed between the two slits and for a single slit, the amplitude of light coming from the slit is reduced to half  which leads to 1/4th of intensity.

Question 16:

A thin transparent sheet is placed in front of a Young’s double slit. The fringe-width will
(a) increase
(b) decrease
(c) remain same
(d) become non-uniform

Answer:

(c) remain same

On the introduction of a transparent sheet in front of one of the slits, the fringe pattern will shift slightly but the width will remain the same.

Question 17:

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 unchanged
(d) there will be no fringe

Answer:

(a) the fringe width will decrease

As fringe width is proportional to the wavelength and wavelength of light is inversely proportional to the refractive index of the medium,
Here,

λΜ=λ/ηλΜ=wavelength in mediumλ=wavelength in vacuumη=refractive index of mediumHence, fringe width decreases when Young’s double slit experiment is performed under water.

Question 1:

A light wave can travel
(a) in vacuum
(b) in vacuum only
(c) in a material medium
(d) in a material medium only

Answer:

(a) in vacuum
(c) in a material medium

Light is an electromagnetic wave that can travel through vacuum or any optical medium.

Question 2:

Which of the following properties of light conclusively support the wave theory of light?
(a) Light obeys the laws of reflection.
(b) Speed of light in water is smaller than its speed in vacuum.
(c) Light shows interference.
(d) Light shows photoelectric effect.

Answer:

(b) Speed of light in water is smaller than its speed in vacuum.
(c) Light shows interference.

Snell’s Law, which states that the speed of light reduces on moving from a rarer to a denser medium, can be concluded from Huygens’ wave theory and interference of light waves is based on the wave properties of light.

Question 3:

When light propagates in vacuum, there is an electric field as well as a magnetic field. These fields
(a) are constant in time
(b) have zero average value
(c) are perpendicular to the direction of propagation of light.
(d) are mutually perpendicular

Answer:

(b) have zero average value
(c) are perpendicular to the direction of propagation of light
(d) are mutually perpendicular

Light is an electromagnetic wave that propagates through its electric and magnetic field vectors, which are mutually perpendicular to each other, as well as to the direction of propagation of light. The average value of both the fields is zero.

Question 4:

Huygens’ principle of secondary wavelets may be used to
(a) find the velocity of light in vacuum
(b) explain the particle behaviour of light
(c) find the new position of a wavefront
(d) explain Snell’s Law

Answer:

(c) find the new position of a wavefront
(d) explain Snell’s Law

Huygen’s wave theory explains the origin of points for the new wavefront proceeding successively. It also explains the variation in speed of light on moving from one medium to another, i.e. it proves Snell’s Law.

Question 5:

Three observers A, B and C measure the speed of light coming from a source to be νA, νB and νC. A moves towards the source and C moves away from the source at the same speed. B remains stationary. The surrounding space is vacuum everywhere.
(a)

νA>νB>νC(b)

νA<νB<νC(c)

νA=νB=νC(d)

νB=12νA+νC.

Answer:

(c) vAvB = vC  
(d) 

νB=12vA+vC

Since the speed of light is a universal constant, vAvB = vC =

3×108m/s.

νB=12vA+vC. This expression also implies that vAvB = vC.
1

νs=12

Question 6:

Suppose the medium in the previous question is water. Select the correct option(s) from the list given in that question.

Answer:

(a) vv> vC
(d) vB = (vA + vC )/2

In any other medium, the speed of light is given by

v=c/η, where η is the refractive index of the medium and according to Doppler effect,  for an observer moving towards the source ,speed of light appears to be  more than the other two cases. On the other hand, it will be least when the observer is moving away from the source.

Question 7:

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

Answer:

(a) x = c

​The wave is travelling along the X-axis. So, it’ll have planar wavefront perpendicular to the X-axis.

Question 8:

If the source of light used in a Young’s double slit experiment is changed from red to violet,
(a) the fringes will become brighter
(b) consecutive fringes will come closer
(c) the intensity of minima will increase
(d) the central bright fringe will become a dark fringe

Answer:

(b) consecutive fringes will come closer

Fringe width,

β=λD/d.
Wavelength of red light is greater than wavelength of violet light; so, the fringe width will reduce.

Question 9:

A Young’s double slit experiment is performed with white light.
(a) The central fringe will be white.
(b) There will not be a completely dark fringe.
(c) The fringe next to the central will be red.
(d) The fringe next to the central will be violet.

Answer:

(a) The central fringe will be white.
(b) There will not be a completely dark fringe.
(d) The fringe next to the central will be violet.

The superposition of all the colours at the central maxima gives the central band a white colour. As we go from the centre to corner, the fringe colour goes from violet to red. There will not be a completely dark fringe, as complete destructive interference does not take place.

Question 10:

Four light waves are represented by
(i)

y=a1 sin ωt(ii)

y=a2 sin ωt+ε(iii)

y=a1 sin 2ωt(iv)

y=a2 sin 2ωt+ε.
Interference fringes may be observed due to superposition of
(a) (i) and (ii)
(b) (i) and (iii)
(c) (ii) and (iv)
(d) (iii) and (iv)

Answer:

(a) (i) and (ii)
(d) (iii) and (iv).

The waves are travelling with the same frequencies and varying by constant phase difference.

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Question 6:

The separation between the consecutive dark fringes in a Young’s double slit experiment is 1⋅0 mm. The screen is placed at a distance of 2⋅5 m from the slits and the separation between the slits is 1⋅0 mm. Calculate the wavelength of light used for the experiment.

Answer:

Given:
Separation between consecutive dark fringes = fringe width (β) = 1 mm = 10−3 m
Distance between screen and slit (D) = 2.5 m
The separation between slits (d) = 1 mm = 10−3 m
Let the wavelength of the light used in experiment be λ.
We know that

β=λDd

10-3 m=2.5×λ10-3⇒λ=12.5 10-6 m      =4×10-7 m=400 nmHence, the wavelength of light used for the experiment is 400 nm.

Question 7:

In a double slit interference experiment, the separation between the slits is 1⋅0 mm, the wavelength of light used is 5⋅0 × 10−7 m and the distance of the screen from the slits is 1⋅0 m. (a) Find the distance of the centre of the first minimum from the centre of the central maximum. (b) How many bright fringes are formed in one centimetre width on the screen?

Answer:

Given:
Separation between the two slits,

d=1 mm=10-3 mWavelength of the light used,

λ=5.0×10-7 mDistance between screen and slit,

D=1 m(a) The distance of the centre of the first minimum from the centre of the central maximum, x =

width of central maxima2That is,

x=β2=λD2d    …(i)
=5×10-7×12×10-3 =2.5×10-4 m=0.25 mm(b) From equation (i),
fringe width,

β=2×x=0.50 mmSo, number of bright fringes formed in one centimetre (10 mm) =

100.50=20.

Question 8:

In a Young’s double slit experiment, two narrow vertical slits placed 0⋅800 mm apart are illuminated by the same source of yellow light of wavelength 589 nm. How far are the adjacent bright bands in the interference pattern observed on a screen 2⋅00 m away?

Answer:

Given:
Separation between two narrow slits,

d=0.8 mm=0.8×10-3 mWavelength of the yellow light,

λ=589 nm=589×10-9 mDistance between screen and slit,

D=2.0 mSeparation between the adjacent bright bands = width of one dark fringe
That is,

β=λDd    …(i)

⇒β=589×10-9×20.8×10-3      =1.47×10-3 m      =1.47 mm.Hence, the adjacent bright bands in the interference pattern are 1.47 mm apart.

Question 9:

Find the angular separation between the consecutive bright fringes in a Young’s double slit experiment with blue-green light of wavelength 500 nm. The separation between the slits is

2·0×10-3 m.

Answer:

Given:
Wavelength of the blue-green light,

λ=500×10-9 mSeparation between two slits,

d=2×10-3 m,Let angular separation between the consecutive bright fringes be θ.

Using θ= βD=λDdD=λd, we get:     θ=500×10-92×10-3       =250×10-6       =25×10-5 radian or 0.014°Hence, the angular separation between the consecutive bright fringes is 0.014 degree. 

Question 10:

A source emitting light of wavelengths 480 nm and 600 nm is used in a double-slit interference experiment. The separation between the slits is 0⋅25 mm and the interference is observed on a screen placed at 150 cm from the slits. Find the linear separation between the first maximum (next to the central maximum) corresponding to the two wavelengths.

Answer:

Given:
Wavelengths of the source of light,

λ1=480×10-9 m and  λ2=600×10-9 mSeparation between the slits,

d=0.25 mm=0.25×10-3 mDistance between screen and slit,

D=150 cm=1.5 mWe know that the position of the first maximum is given by
y=λDdSo, the linear separation between the first maximum (next to the central maximum) corresponding to the two wavelengths = y2y1

y2-y1=Dy2-y1d⇒y2-y1=1.50.25×10-3600×10-9-480×10-9     y2-y1=72×10-5 m =0.72 mm

Question 11:

White light is used in a Young’s double slit experiment. Find the minimum order of the violet fringe

λ=400 nmwhich overlaps with a red fringe

λ=700 nm.

Answer:

Let the separation between the slits be d and distance between screen from the slits be D.
Suppose, the mth bright fringe of violet light overlaps with the nth bright fringe of red light.
Now, the position of the mth bright fringe of violet light, yv =

mλvDdPosition of the nth bright fringe of red light, yr =

nλrDdFor overlapping, yv = yr .
So, as per the question,

m×400×Dd=n×700×Dd⇒mn=74Therefore, the

7thbright fringe of violet light overlaps with the 4th bright fringe of red light.
It can also be seen that the 14th violet fringe will overlap with the 8th red fringe.
Because,

mn=74=148

Question 12:

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

Answer:

Given:
The refractive index of the plate is

μ.
Let the thickness of the plate be ‘t’ to produce a change in the optical path difference of

λ2.
We know that optical path difference is given by

μ-1t.

∴ μ-1t=λ2⇒t=λ2μ-1Hence, the thickness of a plate is

λ2μ-1.

Question 13:

A plate of thickness t made of a material of refractive index µ is placed in front of one of the slits in a double slit experiment. (a) Find the change in the optical path due to introduction of the plate. (b) What should be the minimum thickness t which will make the intensity at the centre of the fringe pattern zero? Wavelength of the light used is

λ. Neglect any absorption of light in the plate.

Answer:

Given:
Refractive index of the plate is μ.
The thickness of the plate is t.
Wavelength of the light is λ.
(a)
When the plate is placed in front of the slit, then the optical path difference is given by

μ-1t.

(b) For zero intensity at the centre of the fringe pattern, there should be distractive interference at the centre.
So, the optical path difference should be =

λ2

i.e. μ-1 t=λ2⇒t=λ2 μ-1

Question 14:

A transparent paper (refractive index = 1⋅45) of thickness 0⋅02 mm is pasted on one of the slits of a Young’s double slit experiment which uses monochromatic light of wavelength 620 nm. How many fringes will cross through the centre if the paper is removed?

Answer:

Given:
Refractive index of the paper, μ = 1.45
The thickness of the plate,

t=0.02 mm=0.02×10-3 mWavelength of the light,

λ=620 nm=620×10-9 mWe know that when we paste a transparent paper in front of one of the slits, then the optical path changes by

μ-1t.
And optical path should be changed by λ for the shift of one fringe.
∴ Number of fringes crossing through the centre is

n=μ-1tλ  =1.45-1×0.02×10-3620×10-9  =14.5Hence, 14.5 fringes will cross through the centre if the paper is removed.

Question 15:

In a Young’s double slit experiment, using monochromatic light, the fringe pattern shifts by a certain distance on the screen when a mica sheet of refractive index 1⋅6 and thickness 1⋅964 micron (1 micron = 10−6 m) is introduced in the path of one of the interfering waves. The mica sheet is then removed and the distance between the screen and the slits is doubled. It is found that the distance between the successive maxima now is the same as the observed fringe-shift upon the introduction of the mica sheet. Calculate the wavelength of the monochromatic light used in the experiment.

Answer:

Given:
Refractive index of the mica sheet,μ = 1.6
Thickness of the plate,

t=1.964 micron=1.964×10-6 mLet the wavelength of the light used = λ.
Number of fringes shifted is given by

n=μ-1tλSo, the corresponding shift in the fringe width equals the number of fringes multiplied by the width of one fringe.
Shift =n × β=μ-1tλ×λDd=μ-1t×Dd     …(i)As per the question, when the distance between the screen and the slits is doubled,
i.e.

D’=2D,
fringe width,

β=λD’d=λ2DdAccording to the question, fringe shift in first case = fringe width in second case.

So, μ-1t×Dd=λ2Dd⇒λ=μ-1 t2      =1.6-1×1.964×10-62      =589.2×10-9=589.2 nmHence, the required wavelength of the monochromatic light is 589.2 nm.

Question 16:

A mica strip and a polystyrene strip are fitted on the two slits of a double slit apparatus. The thickness of the strips is 0⋅50 mm and the separation between the slits is 0⋅12 cm. The refractive index of mica and polystyrene are 1.58 and 1.55, respectively, for the light of wavelength 590 nm which is used in the experiment. The interference is observed on a screen at a distance one metre away. (a) What would be the fringe-width? (b) At what distance from the centre will the first maximum be located?

Answer:

Given:
The thickness of the strips =

t1=t2=t=0.5 mm=0.5×10-3 mSeparation between the two slits,

d=0.12 cm=12×10-4 mThe refractive index of mica, μm = 1.58 and of polystyrene, μp = 1.58
Wavelength of the light,

λ=590 nm=590×10-9 m,Distance between screen and slit, D = 1 m

(a)
We know that fringe width is given by
β=λDd

⇒β =590×10-9×112×10-4        =4.9×10-4 m(b) When both the mica and polystyrene strips are fitted before the slits, the optical path changes by

∆x=μm-1 t- μp-1 t      =μm-μp t      =1.58-1.55×0.5 10-3      =0.015×10-3 m∴ Number of fringes shifted, n =

∆xλ.

⇒n=0.015×10-3590×10-9=25.43∴ 25 fringes and 0.43th of a fringe.
⇒ In which 13 bright fringes and 12 dark fringes and 0.43th of a dark fringe.
So, position of first maximum on both sides is given by
On one side,

x=0.43×4.91×10-4   ∵β=4.91×10-4 m      =0.021 cmOn the other side,

x’=1-0.43×4.91×10-4   =0.028 cm

Question 17:

Two transparent slabs having equal thickness but different refractive indices µ1 and µ2 are pasted side by side to form a composite slab. This slab is placed just after the double slit in a Young’s experiment so that the light from one slit goes through one material and the light from the other slit goes through the other material. What should be the minimum thickness of the slab so that there is a minimum at the point P0 which is equidistant from the slits?

Answer:

Given:
Refractive index of the two slabs are µ1 and µ2.
Thickness of both the plates is t.
When both the strips are fitted, the optical path changes by

∆x=μ1-1 t- μ2-1 t      =μ1-μ2 tFor minimum at P0, the path difference should be

λ2.

i.e. ∆x=λ2

So,λ2=μ1-μ2t⇒t=λ2μ1-μ2Therefore, minimum at point P0 is

λ2μ1-μ2.

Question 18:

A thin paper of thickness 0⋅02 mm having a refractive index 1⋅45 is pasted across one of the slits in a Young’s double slit experiment. The paper transmits 4/9 of the light energy falling on it. (a) Find the ratio of the maximum intensity to the minimum intensity in the fringe pattern. (b) How many fringes will cross through the centre if an identical paper piece is pasted on the other slit also? The wavelength of the light used is 600 nm.

Answer:

Given:
The thickness of the thin paper,

t=0.02 mm=0.02×10-3 mRefractive index of the paper,

μ=1.45.
Wavelength of the light,

λ=600 nm=600×10-9 m(a)
Let the intensity of the source without paper = I1
and intensity of source with paper =I2
Let a1 and a2 be corresponding amplitudes.
As per the question,

I2=49I1We know that

I1I2=a12a22      ∵I∝a2⇒a1a2=32Here, a is the amplitude.

We know that  ImaxImin=a1+a22a1-a22.⇒ ImaxImin=3+223-22             =251⇒Imax:Imin=25 : 1(b)
Number of fringes that will cross through the centre is given by

n=μ-1tλ.

⇒n=1.45-1×0.02×10-3600×10-9       =0.45×0.02×1046=15

Question 19:

A Young’s double slit apparatus has slits separated by 0⋅28 mm and a screen 48 cm away from the slits. The whole apparatus is immersed in water and the slits are illuminated by red light

λ=700 nm in vacuum. Find the fringe-width of the pattern formed on the screen.

Answer:

Given:
Separation between two slits,

d=0.28 mm=0.28×10-3 mDistance between screen and slit (D) = 48 cm = 0.48 m
Wavelength of the red light,

λa=700 nm in vaccum =700 ×10-9 mLet the wavelength of red light in water =

λωWe known that refractive index of water (μw =4/3),
μw =

Speed of light in vacuumSpeed of light in the water

So, μw=vavω=λaλω⇒43=λaλω⇒λω= 3λa4=3×7004=525 nmSo, the fringe width of the pattern is given by

β=λωDd   =525×10-9×0.480.28×10-3   =9×10-4=0.90 mmHence, fringe-width of the pattern formed on the screen is 0.90 mm.

Question 20:

A parallel beam of monochromatic light is used in a Young’s double slit experiment. The slits are separated by a distance d and the screen is placed parallel to the plane of the slits. Slow that if the incident beam makes an angle

θ=sin-1 λ2dwith the normal to the plane of the slits, there will be a dark fringe at the centre P0 of the pattern.

Answer:

Let the two slits are S1 and S2 with separation d as shown in figure.

 

The wave fronts reaching P0 from S1 and S2 will have a path difference of S1X = ∆x.
In ∆S1S2X,

sinθ=S1XS1S2=∆xd

⇒∆x=dsinθUsing

θ=sin-1 λ2d, we get,

⇒∆x=d×λ2d=λ2Hence, there will be dark fringe at point P0 as the path difference is an odd multiple of

λ2.

Question 21:

A narrow slit S transmitting light of wavelength

λis placed a distance d above a large plane mirror, as shown in the figure (17-E1). The light coming directly from the slit and that coming after the reflection interfere at a screen ∑ placed at a distance D from the slit. (a) What will be the intensity at a point just above the mirror, i.e. just above O? (b) At what distance from O does the first maximum occur?

Figure

Answer:

(a) The phase of a light wave reflecting from a surface differs by ‘

π’ from the light directly coming from the source.

Thus, the wave fronts reaching just above the mirror directly from the source and after reflecting from the mirror have a phase difference of

π, which is the condition of distractive interference. So, the intensity at a point just above the mirror is zero.

(b) Here, separation between two slits is

2d.
Wavelength of the light is

λ.
Distance of the screen from the slit is

D.
Consider that the bright fringe is formed at position y. Then,
path difference,

∆x=y×2dD=nλ.
After reflection from the mirror, path difference between two waves is

λ2.

⇒y×2dD=λ2+nλFor first order, put n=0.⇒y=λD4d

Page No 382:

Question 22:

A long narrow horizontal slit is paced 1 mm above a horizontal plane mirror. The interference between the light coming directly from the slit and that after reflection is seen on a screen 1⋅0 m away from the slit. Find the fringe-width if the light used has a wavelength of 700 nm.

Answer:

Given:
Separation between two slits,

d’=2d=2 mm=2×10-3 m(as d = 1 mm)
Wavelength of the light used,

λ=700 nm=700×10-3 mDistance between the screen and slit (D) = 1.0 m
It is a case of Lloyd’s mirror experiment.

Fringe width, β=λDd’                  =700×10-9×12×10-3                  =0.35×10-3 m=0.35 mmHence, the width of the fringe is 0.35 mm.

Question 23:

Consider the situation of the previous problem. If the mirror reflects only 64% of the light energy falling on it, what will be the ratio of the maximum to the minimum intensity in the interference pattern observed on the screen?

Answer:

Given:
The mirror reflects 64% of the energy or intensity of light.
Let intensity of source = I1.
And intensity of light after reflection from the mirror = I2.
Let a1 and a2 be corresponding amplitudes of intensity I1 and I2.
According to the question,

I2=I1×64100⇒I2I1=64100=1625And I2I1=a22a12⇒a2a1=45

We know that ImaxImin=a1+a22a1-a22                         =5+425-42Imax:Imin=81 : 1Hence, the required ratio is 81 : 1.

Question 24:

A double slit S1S2 is illuminated by a coherent light of wavelength

λ. The slits are separated by a distance d. A plane mirror is placed in front of the double slit at a distance D1 from it and a screen ∑ is placed behind the double slit at a distance D2 from it (figure 17-E2). The screen ∑ receives only the light reflected by the mirror. Find the fringe-width of the interference pattern on the screen.

Figure

Answer:

Given:
Separation between the two slits = d
Wavelength of the coherent light =λ
Distance between the slit and mirror is D1.
Distance between the slit and screen is D2.
Therefore,
apparent distance of the screen from the slits,

D=2D1+D2Fringe width, β=λDd=2D1+D2 λdHence, the required fringe width is

2D1+D2 λd.

Question 25:

White coherent light (400 nm-700 nm) is sent through the slits of a Young’s double slit experiment (figure 17-E3). The separation between the slits is 0⋅5 mm and the screen is 50 cm away from the slits. There is a hole in the screen at a point 1⋅0 mm away (along the width of the fringes) from the central line. (a) Which wavelength(s) will be absent in the light coming from the hole? (b) Which wavelength(s) will have a strong intensity?

Figure

Answer:

Given: Separation between two slits,

d=0.5 mm=0.5×10-3 mWavelength of the light,

λ=400 nm to 700 nmDistance of the screen from the slit,

D=50 cm=0.5 m,
Position of hole on the screen,

yn=1 mm=1×10-3 m(a) The wavelength(s) will be absent in the light coming from the hole, which will form a dark fringe at the position of hole.

yn=2n+1λn2Dd, where n = 0, 1, 2, …

⇒λn=22n+1 yndD       =22n+1×10-3×0.05×10-30.5       =22n+1×10-6 m        =22n+1×103 nm For n=1,λ1=23×1000 =667 nmFor n=2,λ2=25×1000=400 nmThus, the light waves of wavelength 400 nm and 667 nm will be absent from the light coming from the hole.

(b) The wavelength(s) will have a strong intensity, which will form a bright fringe at the position of the hole.

So, yn=nλnDd⇒λn=yndnDFor n=1, λ1=yndD     =10-3×0.5×10-30.5     =10-6 m=1000 nm.But 1000 nm does not fall in the range 400 nm − 700 nm.

Again, for n=2,λ2=ynd2D=500 nmSo, the light of wavelength 500 nm will have strong intensity.

Question 26:

Consider the arrangement shown in the figure (17-E4). The distance D is large compared to the separation d between the slits. (a) Find the minimum value of d so that there is a dark fringe at O. (b) Suppose d has this value. Find the distance x at which the next bright fringe is formed. (c) Find the fringe-width.

Figure

Answer:


From the figure,

AB=BO and AC=CO.
Path difference of the wave front reaching O,

∆x=AB+BO-AC+CO     =2 AB-AC     =2 D2+d2-DFor dark fringe to be formed at O, path difference should be an odd multiple of

λ2.

So,∆x=2n+1 λ2⇒ 2D2+d2-D=2n+1 λ2⇒D2+d2=D+2n+1 λ4⇒D2+d2=D2+2n+12λ216+2n+1 λD2Neglecting,

2n+12λ216, as it is very small, we get:

d=2n+1 λD2For minimum d, putting, n=0dmin=λD2, we get:
Thus, for

dmin=λD2there is a dark fringe at O.

Question 27:

Two coherent point sources S1 and S2, vibrating in phase, emit light of wavelength

λ. The separation between the sources is

2λ. Consider a line passing through S2 and perpendicular to the line S1 S2. What is the smallest distance from S2 where a minimum intensity occurs?

Answer:

Let P be the point of minimum intensity.
For minimum intensity at point P,

S1P-S2P=x=2n+1 λ2Thus, we get:

Z2+2λ2-Z=2n+1 λ2⇒Z2+4λ2=Z2 2n+12 λ24+2Z 2n+1 λ2⇒Z=4λ2-2n+1 2λ2/42n+1 λ      =16λ2-2n+14 2n+1 λ                … (i) When n=0,        Z=15λ4            n=-1,    Z=-15λ4            n=1,       Z=7λ12            n=2 ,      Z=-9λ20Thus,

Z=7λ12is the smallest distance for which there will be minimum intensity.

Question 28:

Figure (17-E5) shows three equidistant slits being illuminated by a monochromatic parallel beam of light. Let

BP0-AP0=λ/3 and D>>λ. (a) Show that in this case

d=2λD/3. (b) Show that the intensity at P0 is three times the intensity due to any of the three slits individually.

Figure

Answer:

(a) Given:
Wavelength of light =

λPath difference of wave fronts reaching from A and B is given by

∆xB=BP0-AP0=λ3⇒D2+d2-D=λ3⇒D2+d2= D2+λ29 +2λD3We will neglect the term

λ29, as it has a very small value.

∴d=2λD3(b) To calculating the intensity at P0, consider the interference of light waves coming from all the three slits.
Path difference of the wave fronts reaching from A and C is given by

CP0-AP0=D2+2d2-D               =D2+8λD3-D   Using the value of d from part a               =D1+8λ3D12-DExpanding the value using binomial theorem and neglecting the higher order terms, we get:               D1+12×8λ3D+…-D

CP0-AP0=4λ3So, the corresponding phase difference between the wave fronts from A and C is given by

ϕc=2π∆xCλ=2π×4λ3λ⇒ϕc=8π3 or 2π+2π3⇒ϕc=2π3     …(1)Again, ϕB=2π ∆xBλ⇒ϕB=2πλ3λ=2π3    …(2)So, it can be said that light from B and C are in the same phase, as they have the same phase difference with respect to A.

Amplitude of wave reaching P0 is given by

A=2a2+a2+2a×acos2π3       =4a2+a2+2a23∴lpo=K 3 r2=3 Kr2=3lHere, I is the intensity due to the individual slits and Ipo is the total intensity at P0.
Thus, the resulting amplitude is three times the intensity due to the individual slits.

Question 29:

In a Young’s double slit experiment, the separation between the slits = 2⋅0 mm, the wavelength of the light = 600 nm and the distance of the screen from the slits = 2⋅0 m. If the intensity at the centre of the central maximum is 0⋅20 W m−2, what will be the intensity at a point 0⋅5 cm away from this centre along the width of the fringes?

Answer:

Given:
Separation between the slits,

d=2 mm=2×10-3 mWavelength of the light,

λ=600 nm=6×10-7 mDistance of the screen from the slits, D = 2⋅0 m

Imax=0.20 W/m2For the point at a position y=0.5 cm=0.5×10-2 m,path difference,∆x=ydD.⇒∆x=0.5×10-2×2×10-32         =5×10-6 mSo, the corresponding phase difference is given by

∆ϕ=2π∆xλ=2π×5×10-66×10-7    =50π3=16π+2π3or ∆ϕ=2π3So, the amplitude of the resulting wave at point y = 0.5 cm is given by

A=a2+a2+2a2 cos 2π3  =a2+a2-a2 =aSimilarly, the amplitude of the resulting wave at the centre is 2a.
Let the intensity of the resulting wave at point y = 0.5 cm be I.

Since IImax=A22a2, we have:I0.2=A24a2=a24a2⇒I=0.24=0.05 W/m2Thus, the intensity at a point 0.5 cm away from the centre along the width of the fringes is 0.05 W/m2.

Question 30:

In a Young’s double slit interference experiment, the fringe pattern is observed on a screen placed at a distance D from the slits. The slits are separated by a distance d and are illuminated by monochromatic light of wavelength

λ. Find the distance from the central point where the intensity falls to (a) half the maximum, (b) one-fourth the maximum.

Answer:

Given:
Separation between the two slits = d
Wavelength of the light =

λDistance of the screen =

D(a) When the intensity is half the maximum:
Let Imax be the maximum intensity and I be the intensity at the required point at a distance y from the central point.
So,

I=a2+a2+2a2cosϕHere,

ϕis the phase difference in the waves coming from the two slits.
So,

I=4a2cos2ϕ2

⇒IImax=12⇒4a2cos2ϕ24a2=12⇒cos2ϕ2=12⇒cosϕ2=12⇒ϕ2=π4⇒ϕ=π2Corrosponding path difference, ∆x=λ4⇒y=∆xDd=λD4d(b) When the intensity is one-fourth of the maximum:

IImax=14⇒4a2cos2ϕ2=14⇒cos2 ϕ2=14⇒cosϕ2=12⇒ϕ2=π3So, corrosponding path difference, ∆x=λ3and position, y=∆xDd=λD3d .

Question 31:

In a Young’s double slit experiment,

λ=500 nm, d=1·0 mm and D=1·0 m. Find the minimum distance from the central maximum for which the intensity is half of the maximum intensity.

Answer:

Given:
Separation between the two slits,

d=1 mm=10-3 mWavelength of the light,

λ=500 nm=5×10-7 mDistance of the screen,

D=1 mLet Imax be the maximum intensity and I be the intensity at the required point at a distance y from the central point.
So,

I=a2+a2+2a2cosϕHere,

ϕis the phase difference in the waves coming from the two slits.
So,

I=4a2cos2ϕ2

⇒IImax=12⇒4a2cos2ϕ24a2=12⇒cos2ϕ2=12⇒cosϕ2=12⇒ϕ2=π4⇒ϕ=π2Corrosponding path difference, ∆x=14⇒y=∆xDd=λD4d

⇒y=5×10-7×14×10-3      =1.25×10-4 m∴ The required minimum distance from the central maximum is

1.25×10-4 m.

Question 32:

The line-width of a bright fringe is sometimes defined as the separation between the points on the two sides of the central line where the intensity falls to half the maximum. Find the line-width of a bright fringe in a Young’s double slit experiment in terms of

λ, d and D where the symbols have their usual meanings.

Answer:

Given:
Separation between two slits = d
Wavelength of the light =

λDistance of the screen =

DLet Imax be the maximum intensity and I be half the maximum intensity at a point at a distance y from the central point.
So,

I=a2+a2+2a2cosϕHere,

ϕis the phase difference in the waves coming from the two slits.
So,

I=4a2cos2ϕ2

⇒IImax=12⇒4a2cos2ϕ24a2=12⇒cos2ϕ2=12⇒cosϕ2=12⇒ϕ2=π4⇒ϕ=π2Corrosponding path difference, ∆x=14⇒y=∆xDd=λD4dThe line-width of a bright fringe is defined as the separation between the points on the two sides of the central line where the intensity falls to half the maximum.
So, line-width = 2y

=2Dλ4d=Dλ2dThus, the required line width of the bright fringe is

Dλ2d.

Page No 383:

Question 33:

Consider the situation shown in the figure. The two slits S1 and S2 placed symmetrically around the central line are illuminated by a monochromatic light of wavelength λ. The separation between the slits is d. The light transmitted by the slits falls on a screen ∑1 placed at a distance D from the slits. The slit S3 is at the central line and the slit S4 is at a distance z from S3. Another screen ∑2 is placed a further distance D away from ∑1. Find the ratio of the maximum to minimum intensity observed on ∑2 if z is equal to
(a)

z=λD2d
(b)

λDd(c)

λD4dFigure

Answer:

Given:
Separation between the two slits = d
Wavelength of the light =

λDistance of the screen =

DThe fringe width (β) is given by

β=λDd.
At S3, the path difference is zero. So, the maximum intensity occurs at amplitude = 2a.
(a) When

z=Dλ2d:
The first minima occurs at S4, as shown in figure (a).
With amplitude = 0 on screen ∑2, we get:

lmaxlmin=2a+022a-02=1
(b) When

z=Dλd:
The first maxima occurs at S4, as shown in the figure.

With amplitude =

2aon screen ∑2, we get:

lmaxlmin=2a+2a22a-2a2=∞(c) When

z=Dλ4d:

The slit S4 falls at the mid-point of the central maxima and the first minima, as shown in the figure.

Intensity=lmax2⇒ Amplitude=2a

∴lmaxlmin=2a+2a22a-2a2=34

Question 34:

Consider the arrangement shown in the figure (17-E7). By some mechanism, the separation between the slits S3 and S4 can be changed. The intensity is measured at the point P, which is at the common perpendicular bisector of S1S2 and S2S4. When

z=Dλ2d, the intensity measured at P is I. Find the intensity when z is equal to

a Dλd,      b 3Dλ2d       and        c 2Dλd.Figure

Answer:

Given:
Fours slits S1, S2, S3 and S4.
The separation between slits S3 and S4 can be changed.
Point P is the common perpendicular bisector of S1S2 and S3S4.

(a)

For z=λDd:
The position of the slits from the central point of the first screen is given by

y=OS3=OS4=z2=λD2dThe corresponding path difference in wave fronts reaching S3 is given by

∆x=ydD=λD2d×dD=λ2Similarly at S4, path difference,

∆x=ydD=λD2d×dD=λ2

i.e. dark fringes are formed at S3 and S4.
So, the intensity of light at S3 and S4 is zero. Hence, the intensity at P is also zero.

(b)

For z=3λD2dThe position of the slits from the central point of the first screen is given by

y=OS3=OS4=z2=3λD4dThe corresponding path difference in wave fronts reaching S3 is given by

∆x=ydD=3λD4d×dD=3λ4Similarly at S4, path difference,

∆x=ydD=3λD4d×dD=3λ4Hence, the intensity at P is I.

(c) Similarly, for

z=2Dλd,
the intensity at P is 2I.

Question 35:

A soap film of thickness 0.0011 mm appears dark when seen by the reflected light of wavelength 580 nm. What is the index of refraction of the soap solution if it is known to be between 1.2 and 1.5?

Answer:

Given:
Thickness of soap film, d =0.0011 mm = 0.0011 × 10−3 m
Wavelength of light used,

λ=580 nm=580×10-9 mLet the index of refraction of the soap solution be μ.
The condition of minimum reflection of light is 2μd = ,
where n is an interger = 1 , 2 , 3 …

⇒μ=nλ2d=2nλ4d     =580×10-9×2n4×11×10-7     =5.82n44=0.1322n
As per the question, μ has a value between 1.2 and 1.5. So,

n=5So, μ=0.132×10=1.32Therefore, the index of refraction of the soap solution is 1.32.

Question 36:

A parallel beam of light of wavelength 560 nm falls on a thin film of oil (refractive index = 1.4). What should be the minimum thickness of the film so that it strongly reflects the light?

Answer:

Given:
Wavelength of  light used,

λ=560×10-9 mRefractive index of the oil film,

μ=1.4Let the thickness of the film for strong reflection be t.
The condition for strong reflection is

2μt=2n+1λ2⇒t=2n+1λ4μwhere n is an integer.
For minimum thickness, putting n = 0, we get:

t=λ4μ     =560×10-94×1.4     =10-7m=100 nmTherefore, the minimum thickness of the oil film so that it strongly reflects the light is 100 nm.

Question 37:

A parallel beam of white light is incident normally on a water film 1.0 × 10−4 cm thick. Find the wavelengths in the visible range (400 nm − 700 nm) which are strongly transmitted by the film. Refractive index of water = 1.33.

Answer:

Given,
Wavelength of  light used,

λ=400×10-9 m to 700×10-9 nmRefractive index of water,

μ=1.33The thickness of film,

t=10-4 cm=10-6 m
The condition for strong transmission:

2μt=nλ,
where n is an integer.

⇒ λ=2μtn

⇒ λ=2×1.33×10-6n       =2660×10-9n mPutting n = 4, we get, λ1 = 665 nm.
Putting n = 5, we get, λ2 = 532 nm.
Putting n = 6, we get, λ3 = 443 nm.
Therefore, the wavelength (in visible region) which are strongly transmitted by the film are 665 nm, 532nm and 443 nm.

Question 38:

A glass surface is coated by an oil film of uniform thickness 1.00 × 10−4 cm. The index of refraction of the oil is 1.25 and that of the glass is 1.50. Find the wavelengths of light in the visible region (400 nm − 750 nm) which are completely transmitted by the oil film under normal incidence.

Answer:

Given:
Wavelength of  light used,

λ=400×10-9 to 750×10-9 mRefractive index of oil, μoil, is 1.25 and that of glass, μg, is 1.50.
The thickness of the oil film,

d=1×10-4 cm=10-6m,The condition for the wavelengths which can be completely transmitted through the oil film is given by

λ=2μdn+12  =2×10-6×1.25×22n+1  =5×10-6 2n+1 m⇒ λ=50002n+1 nmWhere n is an integer.
For wavelength to be in visible region i.e (400 nm to 750 nm)
When n = 3, we get,

λ=50002×3+1 =50007=714.3 nmWhen, n = 4, we get,

λ=50002×4+1  =50009=555.6 nmWhen, n = 5, we get,

λ=50002×5+1  =500011=454.5 nmThus the wavelengths of light in the visible region (400 nm − 750 nm) which are completely transmitted by the oil film under normal incidence are 714 nm, 556 nm, 455  nm.

Question 39:

Plane microwaves are incident on a long slit of width 5.0 cm. Calculate the wavelength of the microwaves if the first diffraction minimum is formed at θ = 30°.

Answer:

Given:
Width of the slit, b = 5.0 cm
First diffraction minimum is formed at θ = 30°.
For the diffraction minima, we have:
bsinθ =
For the first minima, we put n = 1.

5×sin30°=1×λ⇒λ=52=2.5 cmTherefore, the wavelength of the microwaves is 2.5 cm.

Question 40:

Light of wavelength 560 nm goes through a pinhole of diameter 0.20 mm and  falls on a wall at a distance of 2.00 m. What will be the radius of the central bright spot formed on the wall?

Answer:

Given:
Wavelength of the light used,

λ=560 nm=560×10-9 mDiameter of the pinhole, d = 0.20 mm = 2 × 10−4 m
Distance of the wall, D = 2m
We know that the radius of the central bright spot is given by

R=1.22λDd   =1.22×560×10-9×22×10-4   =6.832×10-3 m or=0.683 cmHence, the diameter 2R of the central bright spot on the wall is 1.37 cm.

Question 41:

A convex lens of diameter 8.0 cm is used to focus a parallel beam of light of wavelength 620 nm. If the light is focused at a distance of 20 cm from the lens, what would be the radius of the central bright spot formed?

Answer:

Given:
Wavelength of the light used,

λ=620 nm=620×10-9mDiameter of the convex lens,

d=8 cm=8×10-2 mDistance from the lens where light is to be focused,

D=20 cm=20×10-2 mThe radius of the central bright spot is given by

R=1.22λDd  =1.22×620×10-9×20×10-28×10-2   = 1891×10-9 m≈1.9×10-6m∴ Diameter of the central bright spot, 2R = 3.8 × 10−6 m.

HC Verma Solutions for Class 11 Physics – Part 1

HC Verma Solutions for Class 12 Physics – Part 2

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