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#### 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.

#### Page No 379:

#### Question 2:

The wavelength of light in a medium is $\lambda ={\lambda}_{0}/\mu $, where $\lambda $ is the wavelength in vacuum. A beam of red light $\left({\lambda}_{0}=720\mathrm{nm}\right)$ enters water. The wavelength in water is $\lambda ={\lambda}_{0}/\mu =540\mathrm{nm}$. 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.

#### Page No 379:

#### 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.

#### Page No 379:

#### 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.

#### Page No 379:

#### 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.

#### Page No 379:

#### 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.

#### Page No 379:

#### 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.

${\rm I}={{\rm I}}_{\mathit{1}}+{{\rm I}}_{\mathit{2}}+\mathit{2}\sqrt{\left[{{\rm I}}_{\mathit{1}}\times {{\rm I}}_{\mathit{2}}\right]}\mathit{cos}\left(\varphi \right)$ ,

where $\varphi =\mathrm{phase}\mathrm{difference}$.

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

#### Page No 379:

#### 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.

#### Page No 379:

#### 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.

#### Page No 379:

#### 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.

$\chi =\frac{\lambda D}{d}$,

where

$\lambda =\mathrm{wavelength}\phantom{\rule{0ex}{0ex}}\chi =\mathrm{fringe}\mathrm{width}\phantom{\rule{0ex}{0ex}}D=\mathrm{distance}\mathrm{between}\mathrm{slits}\mathrm{and}\mathrm{screen}\phantom{\rule{0ex}{0ex}}d=\mathrm{separation}\mathrm{between}\mathrm{slits}$

On 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.

#### Page No 379:

#### 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.

#### Page No 379:

#### 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.

#### Page No 379:

#### 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.

#### Page No 379:

#### Question 3:

The equation of a light wave is written as $y=A\mathrm{sin}\left(kx-\omega t\right)$. 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.

#### Page No 379:

#### 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.

#### Page No 379:

#### 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 $\mu $ is given by

${\lambda}_{{\rm M}}=\frac{\lambda}{\mu},\phantom{\rule{0ex}{0ex}}where{\lambda}_{{\rm M}}=wavelengthinmedium\phantom{\rule{0ex}{0ex}}\lambda =wavelengthinvacuum\phantom{\rule{0ex}{0ex}}\mu =refractiveindex$

#### Page No 379:

#### 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.

#### Page No 379:

#### 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 ${\lambda}_{a}\mathrm{and}{\lambda}_{f}$,

(a) ${\lambda}_{a}>{\lambda}_{f}$

(b) ${\lambda}_{a}={\lambda}_{f}$

(c) ${\lambda}_{a}<{\lambda}_{f}$

(d) We don't have sufficient information to evaluate the relation of ${\lambda}_{a}\mathrm{and}{\lambda}_{f}$.

#### Answer:

(a) ${\lambda}_{a}>{\lambda}_{f}$

An 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 $\times $ 1m board.

λ

#### Page No 379:

#### 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.

#### Page No 379:

#### 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) ${\mathrm{cos}}^{-1}\left(1/\sqrt{3}\right)$

#### Answer:

(d) ${\mathrm{cos}}^{-1}\left(1/\sqrt{3}\right)$

On writing the given equation in the plane equation form *lx + my + nz *= p,

where *l*^{2} + *m*^{2} + *n*^{2} = 1 and p>0, we get:

$\frac{1}{\sqrt{3}}x+\frac{1}{\sqrt{3}}y+\frac{1}{\sqrt{3}}z=\frac{c}{\sqrt{3}}$

If $\theta $ is the angle between the normal and +X axis, then

$\mathrm{cos}\theta =\frac{1}{\sqrt{3}}\phantom{\rule{0ex}{0ex}}\Rightarrow \theta ={\mathrm{cos}}^{-1}\left(\frac{1}{\sqrt{3}}\right)$

#### Page No 379:

#### 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.

#### Page No 379:

#### Question 11:

The inverse square law of intensity (i.e. the intensity $\infty \frac{1}{{r}^{2}}$) 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

$I=S/\left(4{\mathrm{\pi r}}^{2}\right)$ ,

where *S* is the source strength.

#### Page No 379:

#### 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.

#### Page No 379:

#### 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 $\left(400\mathrm{nm}\lambda 700\mathrm{nm}\right)$.

#### Answer:

Given:

Range of wave length is

$400\mathrm{nm}\lambda 700\mathrm{nm}$

We know that frequency is given by $f=\frac{c}{\lambda}$,

$\mathrm{where}\mathrm{c}=\mathrm{speed}\mathrm{of}\mathrm{light}=3\times {10}^{8}\mathrm{m}/\mathrm{s}$

* f* is the frequency

* λ* is the wavelength

We can write wavelength as:

$\frac{1}{700\mathrm{nm}}\frac{1}{\lambda}\frac{1}{400\mathrm{nm}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{7\times {10}^{-7}\mathrm{m}}\frac{1}{\lambda}\frac{1}{4\times {10}^{-7}\mathrm{m}}\phantom{\rule{0ex}{0ex}}\frac{3\times {10}^{8}}{7\times {10}^{-7}}\mathrm{Hz}\frac{c}{\lambda}\frac{3\times {10}^{8}}{4\times {10}^{-7}}\mathrm{Hz}\phantom{\rule{0ex}{0ex}}\Rightarrow 4.3\times {10}^{14}\mathrm{Hz}\frac{c}{\lambda}7.5\times {10}^{14}\mathrm{Hz}\phantom{\rule{0ex}{0ex}}\Rightarrow 4.3\times {10}^{14}\mathrm{Hz}f7.5\times {10}^{14}\mathrm{Hz}$

Hence, frequency of the range of light that is visible to an average human being is $4.3\times {10}^{14}\mathrm{Hz}\mathrm{to}7.5\times {10}^{14}\mathrm{Hz}$.

#### Page No 380:

#### 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, ${\lambda}_{\mathrm{a}}=589\mathrm{nm}=589\times {10}^{-9}\mathrm{m}$

Refractive index of water, μ_{w}= 1⋅33

We know that $f=\frac{c}{\lambda}$,

$\mathrm{where}c=\mathrm{speed}\mathrm{of}\mathrm{light}=3\times {10}^{8}\mathrm{m}/\mathrm{s}$

*f* = frequency

*λ* = wavelength

(a) Frequency in air, ${f}_{\mathrm{air}}=\frac{c}{{\lambda}_{\mathrm{a}}}$

${f}_{\mathrm{air}}=\frac{3\times {10}^{8}}{589\times {10}^{-9}}\phantom{\rule{0ex}{0ex}}=5.09\times {10}^{14}\mathrm{Hz}$

(b)

Let wavelength of sodium light in water be ${\lambda}_{\mathrm{w}}$.

We know that

$\frac{{\mathrm{\mu}}_{\mathrm{a}}}{{\mathrm{\mu}}_{\mathrm{w}}}=\frac{{\lambda}_{\mathrm{\omega}}}{{\lambda}_{\mathrm{a}}}$,

where μ_{a} is the refractive index of air which is equal to 1 and

*λ*_{w} is the wavelength of sodium light in water.

$\Rightarrow \frac{1}{1.33}=\frac{{\lambda}_{\mathrm{\omega}}}{589\times {10}^{-9}}\phantom{\rule{0ex}{0ex}}\Rightarrow {\lambda}_{\mathrm{\omega}}=443\mathrm{nm}$

(c) Frequency of light does not change when light travels from one medium to another.

$\therefore {f}_{\mathrm{\omega}}={f}_{\mathrm{a}}\phantom{\rule{0ex}{0ex}}=5.09\times {10}^{14}\mathrm{Hz}$

(d) Let the speed of sodium light in water be ${\nu}_{\mathrm{\omega}}$

and speed in air, *v*_{a} = c.

Using $\frac{{\mu}_{\mathrm{a}}}{{\mu}_{\mathrm{\omega}}}=\frac{{\nu}_{\mathrm{\omega}}}{{\nu}_{\mathrm{a}}}$, we get:

${\nu}_{\mathrm{\omega}}=\frac{{\mu}_{\mathrm{a}}c}{{\mu}_{\mathrm{\omega}}}\phantom{\rule{0ex}{0ex}}=\frac{3\times {10}^{8}}{\left(1.33\right)}=2.25\times {10}^{8}\mathrm{m}/\mathrm{s}$

#### Page No 380:

#### 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 = $\frac{\mathrm{Speed}\mathrm{of}\mathrm{light}\mathrm{in}\mathrm{vacuum}}{\mathrm{Speed}\mathrm{of}\mathrm{light}\mathrm{in}\mathrm{the}\mathrm{material}}=\frac{c}{v}$

Let speed of light for wavelength 400 nm in quartz be *v*_{400}.

So,

$1.472=\frac{3\times {10}^{8}}{{v}_{400}}\phantom{\rule{0ex}{0ex}}\Rightarrow {v}_{400}=2.04\times {10}^{8}\mathrm{m}/\mathrm{s}\phantom{\rule{0ex}{0ex}}$

Let speed of light of wavelength 760 nm in quartz be *v*_{760}.

Again, $\frac{1.452}{1}=\frac{3\times {10}^{8}}{{\nu}_{760}}$

$\Rightarrow {v}_{760}=2.07\times {10}^{8}\mathrm{m}/\mathrm{s}$

#### Page No 380:

#### Question 4:

The speed of yellow light in a certain liquid is 2⋅4 × 10^{8} m s^{−1}. Find the refractive index of the liquid.

#### Answer:

Given:

Speed of yellow light in liquid (*v*_{L}*)*= 2⋅4 × 10^{8} m s^{−1}

And speed of yellow light in air speed* = v*_{a}

Let μ_{L} be the refractive index of the liquid

Using, ${\mathrm{\mu}}_{\mathrm{L}}=\frac{\mathrm{speed}\mathrm{of}\mathrm{light}\mathrm{in}\mathrm{vaccum}}{\mathrm{velocity}\mathrm{of}\mathrm{light}\mathrm{in}\mathrm{the}\mathrm{given}\mathrm{medium}}=\frac{c}{{v}_{\mathrm{L}}}$

${\mathrm{\mu}}_{\mathrm{L}}=\frac{3\times {10}^{8}}{\left(2.4\right)\times {10}^{8}}=1.25\phantom{\rule{0ex}{0ex}}$

Hence, the required refractive index is 1.25.

#### Page No 380:

#### Question 5:

Two narrow slits emitting light in phase are separated by a distance of 1⋅0 cm. The wavelength of the light is $5\xb70\times {10}^{-7}\mathrm{m}$. 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, $\lambda =5\times {10}^{-7}\mathrm{m}\phantom{\rule{0ex}{0ex}}$

Distance of the screen, $D=1\mathrm{m}$

(a)

We know that separation between two consecutive maxima = fringe width (*β*).

That is, $\beta =\frac{\lambda D}{d}$ ...(i)

$=\frac{5\times {10}^{-7}\times 1}{{10}^{-2}}\mathrm{m}\phantom{\rule{0ex}{0ex}}=5\times {10}^{-5}\mathrm{m}=0.05\mathrm{mm}$

(b)

Separation between two consecutive maxima = fringe width

∴ $\beta =1\mathrm{mm}={10}^{-3}\mathrm{m}$

Let the separation between the sources be *'d'*

Using equation (i), we get:

$d\text{'}=\frac{5\times {10}^{-7}\times 1}{{10}^{-3}}\phantom{\rule{0ex}{0ex}}\Rightarrow d\text{'}=5\times {10}^{-4}\mathrm{m}=0.50\mathrm{mm}.$

#### Page No 380:

#### 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

$\frac{{I}_{max}}{{I}_{min}}=\frac{{\left(\sqrt{{I}_{1}}+\sqrt{{I}_{2}}\right)}^{2}}{{\left(\sqrt{{I}_{1}}-\sqrt{{I}_{2}}\right)}^{2}}=\frac{25}{1}\phantom{\rule{0ex}{0ex}}\Rightarrow \sqrt{{I}_{1}}=3and\sqrt{{I}_{2}}=2\phantom{\rule{0ex}{0ex}}\Rightarrow {I}_{1}=9and{I}_{2}=4\phantom{\rule{0ex}{0ex}}$

Then,

$\frac{{I}_{1}}{{I}_{2}}=\frac{9}{4}$

#### Page No 380:

#### 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 *I*_{0}. If one of the slits is closed, the intensity at this point will be

(a) *I*_{0}

(b) *I*_{0}/4

(c) *I*_{0}/2

(d) 4*I*_{0}

#### Answer:

(b) *I*_{0}/4

Total intensity coming from the source is *I*_{0 }which 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.

#### Page No 380:

#### 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.

#### Page No 380:

#### 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, ${\lambda}_{{\rm M}}=\lambda /\eta \phantom{\rule{0ex}{0ex}}{\lambda}_{{\rm M}}=\mathrm{wavelength}\mathrm{in}\mathrm{medium}\phantom{\rule{0ex}{0ex}}\lambda =\mathrm{wavelength}\mathrm{in}\mathrm{vacuum}\phantom{\rule{0ex}{0ex}}\eta =\mathrm{refractive}\mathrm{index}\mathrm{of}\mathrm{medium}$

Hence, fringe width decreases when Young's double slit experiment is performed under water.

#### Page No 380:

#### 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.

#### Page No 380:

#### 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.

#### Page No 380:

#### 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.

#### Page No 380:

#### 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.

#### Page No 380:

#### Question 5:

Three observers *A*, *B* and *C* measure the speed of light coming from a source to be ν_{A}, ν_{B} and ν_{C}. *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) ${\nu}_{A}>{\nu}_{B}>{\nu}_{C}$

(b) ${\nu}_{A}<{\nu}_{B}<{\nu}_{C}$

(c) ${\nu}_{A}={\nu}_{B}={\nu}_{C}$

(d) ${\nu}_{B}=\frac{1}{2}\left({\nu}_{A}+{\nu}_{C}\right)$.

#### Answer:

(c) *v _{A}_{ = }v_{B}_{ =} v_{C}_{ }*

(d) ${\nu}_{B}=\frac{1}{2}\left({v}_{A}+{v}_{C}\right)$

*v*

_{A}

_{ = }

*v*

_{B}

_{ =}

*v*

_{C}

_{ }= $3\times {10}^{8}$ m/s.

${\nu}_{B}=\frac{1}{2}\left({v}_{A}+{v}_{C}\right)$. This expression also implies that

*v*

_{A}

_{ = }

*v*

_{B}

_{ =}

*v*

_{C}

_{. }

1

#### Page No 380:

#### 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) *v _{A }*>

*v*

_{B }> v_{C}_{ }(d)

*v*(

_{B}_{ }=*v*)/2

_{A}+ v_{C}_{ }In any other medium, the speed of light is given by $v=c/\eta ,\mathrm{where}\mathrm{\eta}\mathrm{is}\mathrm{the}\mathrm{refractive}\mathrm{index}\mathrm{of}\mathrm{the}\mathrm{medium}$ 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.

#### Page No 380:

#### 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.

#### Page No 380:

#### 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, $\beta =\lambda D/d$.

Wavelength of red light is greater than wavelength of violet light; so, the fringe width will reduce.

#### Page No 380:

#### 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.

#### Page No 380:

#### Question 10:

Four light waves are represented by

(i) $y={a}_{1}\mathrm{sin}\omega t$

(ii) $y={a}_{2}\mathrm{sin}\left(\omega t+\epsilon \right)$

(iii) $y={a}_{1}\mathrm{sin}2\omega t$

(iv) $y={a}_{2}\mathrm{sin}2\left(\omega t+\epsilon \right)$.

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.

#### Page No 381:

#### 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

$\beta =\frac{\lambda D}{d}$

${10}^{-3}\mathrm{m}=\frac{2.5\times \lambda}{{10}^{-3}}\phantom{\rule{0ex}{0ex}}\Rightarrow \lambda =\frac{1}{2.5}{10}^{-6}\mathrm{m}\phantom{\rule{0ex}{0ex}}=4\times {10}^{-7}\mathrm{m}=400\mathrm{nm}$

Hence, the wavelength of light used for the experiment is 400 nm.

#### Page No 381:

#### 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, $d=1\mathrm{mm}={10}^{-3}\mathrm{m}$

Wavelength of the light used, $\lambda =5.0\times {10}^{-7}\mathrm{m}\phantom{\rule{0ex}{0ex}}$

Distance between screen and slit, $D=1\mathrm{m}$

(a) The distance of the centre of the first minimum from the centre of the central maximum, *x* = $\frac{\mathrm{width}\mathrm{of}\mathrm{central}\mathrm{maxima}}{2}$

That is, $x=\frac{\beta}{2}=\frac{\lambda D}{2d}$ ...(i)

$=\frac{5\times {10}^{-7}\times 1}{2\times {10}^{-3}}\phantom{\rule{0ex}{0ex}}=2.5\times {10}^{-4}\mathrm{m}=0.25\mathrm{mm}$

(b) From equation (i),

fringe width, $\beta =2\times x=0.50\mathrm{mm}$

So, number of bright fringes formed in one centimetre (10 mm) = $\frac{10}{0.50}=20$.

#### Page No 381:

#### 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, $d=0.8\mathrm{mm}=0.8\times {10}^{-3}\mathrm{m}$

Wavelength of the yellow light, $\lambda =589\mathrm{nm}=589\times {10}^{-9}\mathrm{m}\phantom{\rule{0ex}{0ex}}$

Distance between screen and slit, $D=2.0\mathrm{m}$

Separation between the adjacent bright bands = width of one dark fringe

That is, $\beta =\frac{\lambda D}{d}$ ...(i)

$\Rightarrow \beta =\frac{589\times {10}^{-9}\times 2}{0.8\times {10}^{-3}}\phantom{\rule{0ex}{0ex}}=1.47\times {10}^{-3}\mathrm{m}\phantom{\rule{0ex}{0ex}}=1.47\mathrm{mm}.$

Hence, the adjacent bright bands in the interference pattern are 1.47 mm apart.

#### Page No 381:

#### 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 $2\xb70\times {10}^{-3}\mathrm{m}$.

#### Answer:

Given:

Wavelength of the blue-green light, $\lambda =500\times {10}^{-9}\mathrm{m}\phantom{\rule{0ex}{0ex}}$

Separation between two slits, $d=2\times {10}^{-3}\mathrm{m},$

Let angular separation between the consecutive bright fringes be *θ.*

$\mathrm{Using}\theta =\frac{\mathit{}\beta}{D}=\frac{\lambda D}{dD}=\frac{\lambda}{d},\text{weget:}\phantom{\rule{0ex}{0ex}}\theta =\frac{500\times {10}^{-9}}{2\times {10}^{-3}}\phantom{\rule{0ex}{0ex}}=250\times {10}^{-6}\phantom{\rule{0ex}{0ex}}=25\times {10}^{-5}\mathrm{radian}\mathrm{or}0.014\xb0$

Hence, the angular separation between the consecutive bright fringes is 0.014 degree.^{ }

#### Page No 381:

#### 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,

${\lambda}_{1}=480\times {10}^{-9}\mathrm{m}\mathrm{and}{\lambda}_{2}=600\times {10}^{-9}\mathrm{m}\phantom{\rule{0ex}{0ex}}$

Separation between the slits, $d=0.25\mathrm{mm}=0.25\times {10}^{-3}\mathrm{m}$

Distance between screen and slit, $D=150\mathrm{cm}=1.5\mathrm{m}$

We know that the position of the first maximum is given by

$y=\frac{\lambda D}{d}$

So, the linear separation between the first maximum (next to the central maximum) corresponding to the two wavelengths = *y*_{2}* *− *y*_{1}

${y}_{2}-{y}_{1}=\frac{D\left({y}_{2}-{y}_{1}\right)}{d}$

$\Rightarrow {y}_{2}-{y}_{1}=\frac{1.5}{0.25\times {10}^{-3}}\left(600\times {10}^{-9}-480\times {10}^{-9}\right)\phantom{\rule{0ex}{0ex}}{y}_{2}-{y}_{1}=72\times {10}^{-5}\mathrm{m}=0.72\mathrm{mm}\phantom{\rule{0ex}{0ex}}$

#### Page No 381:

#### Question 11:

White light is used in a Young's double slit experiment. Find the minimum order of the violet fringe $\left(\lambda =400\mathrm{nm}\right)$ which overlaps with a red fringe $\left(\lambda =700\mathrm{nm}\right)$.

#### Answer:

Let the separation between the slits be *d* and distance between screen from the slits be *D*.

Suppose, the *m*^{th} bright fringe of violet light overlaps with the *n*^{th} bright fringe of red light.

Now, the position of the *m*^{th} bright fringe of violet light, *y*_{v} = $\frac{m{\lambda}_{v}D}{d}$

Position of the *n*^{th} bright fringe of red light, *y*_{r} = $\frac{n{\lambda}_{r}D}{d}$

For overlapping, *y*_{v =} *y*_{r} .

So, as per the question,

$\frac{m\times 400\times \mathrm{D}}{d}=\frac{n\times 700\times \mathrm{D}}{d}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{m}{n}=\frac{7}{4}$

Therefore, the ${7}^{\mathrm{th}}$ bright fringe of violet light overlaps with the 4^{th} bright fringe of red light.

It can also be seen that the 14^{th} violet fringe will overlap with the 8^{th} red fringe.

Because, $\frac{m}{n}=\frac{7}{4}=\frac{14}{8}$

#### Page No 381:

#### 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 $\mu $.

Let the thickness of the plate be '*t'* to produce a change in the optical path difference of $\frac{\lambda}{2}$.

We know that optical path difference is given by $\left(\mathrm{\mu}-1\right)t$.

$\therefore \left(\mathrm{\mu}-1\right)t=\frac{\lambda}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow t=\frac{\lambda}{2\left(\mathrm{\mu}-1\right)}$

Hence, the thickness of a plate is $\frac{\lambda}{2\left(\mathrm{\mu}-1\right)}$.

#### Page No 381:

#### 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 $\lambda $. 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 $\left(\mathrm{\mu}-1\right)t$.

(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 = $\frac{\lambda}{2}$

$\mathrm{i}.\mathrm{e}.\left(\mu -1\right)t=\frac{\lambda}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow t=\frac{\lambda}{2\left(\mu -1\right)}$

#### Page No 381:

#### 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, $t=0.02\mathrm{mm}=0.02\times {10}^{-3}\mathrm{m}$

Wavelength of the light, $\lambda =620\mathrm{nm}=620\times {10}^{-9}\mathrm{m}\phantom{\rule{0ex}{0ex}}$

We know that when we paste a transparent paper in front of one of the slits, then the optical path changes by $\left(\mathrm{\mu}-1\right)t$.

And optical path should be changed by *λ** *for the shift of one fringe.

∴ Number of fringes crossing through the centre is

$n=\frac{\left(\mathrm{\mu}-1\right)t}{\lambda}\phantom{\rule{0ex}{0ex}}=\frac{\left(1.45-1\right)\times 0.02\times {10}^{-3}}{620\times {10}^{-9}}\phantom{\rule{0ex}{0ex}}=14.5$

Hence, 14.5 fringes will cross through the centre if the paper is removed.

#### Page No 381:

#### 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, $t=1.964\mathrm{micron}=1.964\times {10}^{-6}\mathrm{m}$

Let the wavelength of the light used = λ.

Number of fringes shifted is given by

$n=\frac{\left(\mathrm{\mu}-1\right)t}{\lambda}$

So, the corresponding shift in the fringe width equals the number of fringes multiplied by the width of one fringe.

$\mathrm{Shift}=n\times \beta \phantom{\rule{0ex}{0ex}}=\frac{\left(\mu -1\right)t}{\lambda}\times \frac{\lambda D}{d}\phantom{\rule{0ex}{0ex}}=\frac{\left(\mu -1\right)t\times D}{d}...\left(\mathrm{i}\right)$

As per the question, when the distance between the screen and the slits is doubled,

i.e. $D\text{'}=2D$,

fringe width, $\beta =\frac{\lambda D\text{'}}{d}=\frac{\lambda 2D}{d}$

According to the question, fringe shift in first case = fringe width in second case.

$\mathrm{So},\frac{\left(\mathrm{\mu}-1\right)t\times \mathrm{D}}{d}=\frac{\lambda 2\mathrm{D}}{d}\phantom{\rule{0ex}{0ex}}\Rightarrow \lambda =\frac{\left(\mu -1\right)t}{2}\phantom{\rule{0ex}{0ex}}=\frac{\left(1.6-1\right)\times \left(1.964\right)\times {10}^{-6}}{2}\phantom{\rule{0ex}{0ex}}=589.2\times {10}^{-9}=589.2\mathrm{nm}$

Hence, the required wavelength of the monochromatic light is 589.2 nm.

#### Page No 381:

#### 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 = ${t}_{1}={t}_{2}=t=0.5\mathrm{mm}=0.5\times {10}^{-3}\mathrm{m}$

Separation between the two slits,$\mathit{}d=0.12\mathrm{cm}=12\times {10}^{-4}\mathrm{m}$

The refractive index of mica, μ_{m} = 1.58 and of polystyrene, μ_{p} = 1.58

Wavelength of the light, $\mathrm{\lambda}=590\mathrm{nm}=590\times {10}^{-9}\mathrm{m},$

Distance between screen and slit, *D* = 1 m

(a)

We know that fringe width is given by

$\beta =\frac{\lambda D}{d}$

$\Rightarrow \beta =\frac{590\times {10}^{-9}\times 1}{12\times {10}^{-4}}\phantom{\rule{0ex}{0ex}}=4.9\times {10}^{-4}\mathrm{m}$

(b) When both the mica and polystyrene strips are fitted before the slits, the optical path changes by

$\u2206x=\left({\mathrm{\mu}}_{\mathrm{m}}-1\right)t-\left({\mathrm{\mu}}_{\mathrm{p}}-1\right)t\phantom{\rule{0ex}{0ex}}=\left({\mathrm{\mu}}_{\mathrm{m}}-{\mathrm{\mu}}_{\mathrm{p}}\right)t\phantom{\rule{0ex}{0ex}}=\left(1.58-1.55\right)\times \left(0.5\right)\left({10}^{-3}\right)\phantom{\rule{0ex}{0ex}}=\left(0.015\right)\times {10}^{-3}\mathrm{m}$

∴ Number of fringes shifted, *n* = $\frac{\u2206x}{\lambda}$.

$\Rightarrow n=\frac{0.015\times {10}^{-3}}{590\times {10}^{-9}}=25.43$

∴ 25 fringes and 0.43^{th} of a fringe.

⇒ In which 13 bright fringes and 12 dark fringes and 0.43^{th} of a dark fringe.

So, position of first maximum on both sides is given by

On one side,

$x=\left(0.43\right)\times 4.91\times {10}^{-4}\left(\because \beta =4.91\times {10}^{-4}\mathrm{m}\right)\phantom{\rule{0ex}{0ex}}=0.021\mathrm{cm}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

On the other side,

$x\text{'}=\left(1-0.43\right)\times 4.91\times {10}^{-4}\phantom{\rule{0ex}{0ex}}=0.028\mathrm{cm}$

#### Page No 381:

#### Question 17:

Two transparent slabs having equal thickness but different refractive indices µ_{1} and µ_{2} 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 *P*_{0} which is equidistant from the slits?

#### Answer:

Given:

Refractive index of the two slabs are µ_{1} and µ_{2}.

Thickness of both the plates is* t*.

When both the strips are fitted, the optical path changes by

$\u2206x=\left({\mathrm{\mu}}_{1}-1\right)t-\left({\mathrm{\mu}}_{2}-1\right)t\phantom{\rule{0ex}{0ex}}=\left({\mathrm{\mu}}_{1}-{\mathrm{\mu}}_{2}\right)t\phantom{\rule{0ex}{0ex}}$

For minimum at *P*_{0}, the path difference should be $\frac{\lambda}{2}$.

$\mathrm{i}.\mathrm{e}.\u2206x=\frac{\lambda}{2}$

$\mathrm{So},\frac{\lambda}{2}=\left({\mathrm{\mu}}_{1}-{\mathrm{\mu}}_{2}\right)t\phantom{\rule{0ex}{0ex}}\Rightarrow t=\frac{\lambda}{2\left({\mathrm{\mu}}_{1}-{\mathrm{\mu}}_{2}\right)}$

Therefore, minimum at point *P*_{0} is $\frac{\lambda}{2\left({\mathrm{\mu}}_{1}-{\mathrm{\mu}}_{2}\right)}$.

#### Page No 381:

#### 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, $t=0.02\mathrm{mm}=0.02\times {10}^{-3}\mathrm{m}$

Refractive index of the paper, $\mathrm{\mu}=1.45$.

Wavelength of the light, $\lambda =600\mathrm{nm}=600\times {10}^{-9}\mathrm{m}$

(a)

Let the intensity of the source without paper = *I*_{1}

and intensity of source with paper =*I*_{2}

Let *a*_{1} and *a*_{2} be corresponding amplitudes.

As per the question,

${I}_{2}=\frac{4}{9}{I}_{1}$

We know that

$\frac{{I}_{1}}{{I}_{2}}=\frac{{{a}_{1}}^{2}}{{{a}_{2}}^{2}}\left(\because I\propto {a}^{2}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{a}_{1}}{{a}_{2}}=\frac{3}{2}$

Here, *a* is the amplitude.

$\mathrm{We}\mathrm{know}\text{that}\frac{{I}_{\mathrm{max}}}{{I}_{\mathrm{min}}}=\frac{{\left({a}_{1}+{a}_{2}\right)}^{2}}{{\left({a}_{1}-{a}_{2}\right)}^{2}}.\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{I}_{\mathrm{max}}}{{I}_{\mathrm{min}}}=\frac{{\left(3+2\right)}^{2}}{{\left(3-2\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{25}{1}\phantom{\rule{0ex}{0ex}}\Rightarrow {I}_{\mathrm{max}}:{I}_{\mathrm{min}}=25:1$

(b)

Number of fringes that will cross through the centre is given by $n=\frac{\left(\mu -1\right)t}{\lambda}$.

$\Rightarrow n=\frac{\left(1.45-1\right)\times 0.02\times {10}^{-3}}{600\times {10}^{-9}}\phantom{\rule{0ex}{0ex}}=\frac{0.45\times 0.02\times {10}^{4}}{6}=15$

#### Page No 381:

#### 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 $\left(\lambda =700\mathrm{nm}\mathrm{in}\mathrm{vacuum}\right)$. Find the fringe-width of the pattern formed on the screen.

#### Answer:

Given:

Separation between two slits, $d=0.28\mathrm{mm}=0.28\times {10}^{-3}\mathrm{m}\phantom{\rule{0ex}{0ex}}$

Distance between screen and slit (*D*) = 48 cm = 0.48 m

Wavelength of the red light, ${\lambda}_{a}=700\mathrm{nm}\mathrm{in}\mathrm{vaccum}=700\times {10}^{-9}\mathrm{m}$

Let the wavelength of red light in water = ${\lambda}_{\omega}$

We known that refractive index of water (μ_{w} =4/3),

μ_{w} = $\frac{\mathrm{Speed}\mathrm{of}\mathrm{light}\mathrm{in}\mathrm{vacuum}}{\mathrm{Speed}\mathrm{of}\mathrm{light}\mathrm{in}\mathrm{the}\mathrm{water}}$

$\mathrm{So},{\mathrm{\mu}}_{\mathrm{w}}=\frac{{v}_{a}}{{v}_{\omega}}=\frac{{\lambda}_{a}}{{\lambda}_{\omega}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{4}{3}=\frac{{\lambda}_{a}}{{\lambda}_{\omega}}\phantom{\rule{0ex}{0ex}}\Rightarrow {\lambda}_{\omega}=\frac{3{\lambda}_{a}}{4}=\frac{3\times 700}{4}=525\mathrm{nm}$

So, the fringe width of the pattern is given by

$\mathrm{\beta}=\frac{{\lambda}_{\omega}D}{d}\phantom{\rule{0ex}{0ex}}=\frac{525\times {10}^{-9}\times \left(0.48\right)}{\left(0.28\right)\times {10}^{-3}}\phantom{\rule{0ex}{0ex}}=9\times {10}^{-4}=0.90\mathrm{mm}$

Hence, fringe-width of the pattern formed on the screen is 0.90 mm.

#### Page No 381:

#### 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 $\mathrm{\theta}={\mathrm{sin}}^{-1}\left(\frac{\lambda}{2d}\right)$ with the normal to the plane of the slits, there will be a dark fringe at the centre *P*_{0} of the pattern.

#### Answer:

Let the two slits are S_{1} and S_{2} with separation *d* as shown in figure.

The wave fronts reaching P_{0}_{ }from S_{1} and S_{2} will have a path difference of S_{1}X = ∆*x*.

In ∆S_{1}S_{2}X, $\mathrm{sin}\theta =\frac{{\mathrm{S}}_{1}\mathrm{X}}{{\mathrm{S}}_{1}{\mathrm{S}}_{2}}=\frac{\u2206x}{d}$

$\Rightarrow \u2206x=d\mathrm{sin}\theta \phantom{\rule{0ex}{0ex}}$

Using $\mathrm{\theta}={\mathrm{sin}}^{-1}\left(\frac{\lambda}{2d}\right)$, we get,

$\Rightarrow \u2206x=d\times \frac{\lambda}{2d}=\frac{\lambda}{2}$

Hence, there will be dark fringe at point P_{0} as the path difference is an odd multiple of $\frac{\lambda}{2}$.

#### Page No 381:

#### Question 21:

A narrow slit *S* transmitting light of wavelength $\lambda $is 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 '$\pi $' 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 *$\pi $,* 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 $\lambda $.

Distance of the screen from the slit is $D$.

Consider that the bright fringe is formed at position *y. *Then,

path difference, $\u2206x=\frac{y\times 2d}{D}=n\lambda $.

After reflection from the mirror, path difference between two waves is $\frac{\lambda}{2}$.

$\Rightarrow \frac{y\times 2d}{D}=\frac{\lambda}{2}+n\lambda \phantom{\rule{0ex}{0ex}}\mathrm{For}\mathrm{first}or\mathrm{der},\mathrm{put}n=0.\phantom{\rule{0ex}{0ex}}\Rightarrow y=\frac{\lambda D}{4d}$

#### 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, $d\text{'}=2d=2\mathrm{mm}=2\times {10}^{-3}\mathrm{m}$ (as *d* = 1 mm)

Wavelength of the light used, $\lambda =700\mathrm{nm}=700\times {10}^{-3}\mathrm{m}\phantom{\rule{0ex}{0ex}}$

Distance between the screen and slit (*D*) = 1.0 m

It is a case of Lloyd's mirror experiment.

$\mathrm{Fringe}\mathrm{width},\beta =\frac{\mathrm{\lambda D}}{d\text{'}}\phantom{\rule{0ex}{0ex}}=\frac{700\times {10}^{-9}\times 1}{2\times {10}^{-3}}\phantom{\rule{0ex}{0ex}}=0.35\times {10}^{-3}\mathrm{m}=0.35\mathrm{mm}$

Hence, the width of the fringe is 0.35 mm.

#### Page No 382:

#### 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 = *I*_{1}.

And intensity of light after reflection from the mirror = *I*_{2}.

Let *a*_{1} and *a*_{2} be corresponding amplitudes of intensity *I*_{1} and *I*_{2}.

According to the question,

${I}_{2}=\frac{{I}_{1}\times 64}{100}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{I}_{2}}{{I}_{1}}=\frac{64}{100}=\frac{16}{25}\phantom{\rule{0ex}{0ex}}\mathrm{And}\frac{{I}_{2}}{{I}_{1}}=\frac{{{a}_{2}}^{2}}{{{a}_{1}}^{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{a}_{2}}{{a}_{1}}=\frac{4}{5}$

$\mathrm{We}\mathrm{know}\text{that}\frac{{I}_{\mathrm{max}}}{{I}_{\mathrm{min}}}=\frac{{\left({a}_{1}+{a}_{2}\right)}^{2}}{{\left({a}_{1}-{a}_{2}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{{\left(5+4\right)}^{2}}{{\left(5-4\right)}^{2}}\phantom{\rule{0ex}{0ex}}{I}_{\mathrm{max}}:{I}_{\mathrm{min}}=81:1$

Hence, the required ratio is 81 : 1.

#### Page No 382:

#### Question 24:

A double slit *S*_{1} − *S*_{2} is illuminated by a coherent light of wavelength $\lambda $. The slits are separated by a distance *d*. A plane mirror is placed in front of the double slit at a distance *D*_{1} from it and a screen ∑ is placed behind the double slit at a distance *D*_{2} 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 *D*_{1}.

Distance between the slit and screen is *D*_{2}.

Therefore,

apparent distance of the screen from the slits,

$D=2{D}_{1}+{D}_{2}\phantom{\rule{0ex}{0ex}}\mathrm{Fringe}\mathrm{width},\beta =\frac{\lambda D}{d}=\frac{\left(2{D}_{1}+{D}_{2}\right)\lambda}{d}\phantom{\rule{0ex}{0ex}}$

Hence, the required fringe width is $\frac{\left(2{D}_{1}+{D}_{2}\right)\lambda}{d}$.

#### Page No 382:

#### 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,

$d=0.5\mathrm{mm}=0.5\times {10}^{-3}\mathrm{m}\phantom{\rule{0ex}{0ex}}$

Wavelength of the light, $\lambda =400\mathrm{nm}\mathrm{to}700\mathrm{nm}$

Distance of the screen from the slit, $D=50\mathrm{cm}=0.5\mathrm{m}$,

Position of hole on the screen, ${y}_{n}=1\mathrm{mm}=1\times {10}^{-3}\mathrm{m}$

(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.

${y}_{n}=\frac{\left(2n+1\right){\lambda}_{n}}{2}\frac{D}{d}$ , where *n* = 0, 1, 2, ...

$\Rightarrow {\lambda}_{n}=\frac{2}{\left(2n+1\right)}\frac{{y}_{n}d}{D}\phantom{\rule{0ex}{0ex}}=\frac{2}{\left(2n+1\right)}\times \frac{{10}^{-3}\times 0.05\times {10}^{-3}}{0.5}\phantom{\rule{0ex}{0ex}}=\frac{2}{\left(2n+1\right)}\times {10}^{-6}\mathrm{m}\phantom{\rule{0ex}{0ex}}=\frac{2}{\left(2n+1\right)}\times {10}^{3}\mathrm{nm}$

$\mathrm{For}n=1,\phantom{\rule{0ex}{0ex}}{\lambda}_{1}=\left(\frac{2}{3}\right)\times 1000=667\mathrm{nm}\phantom{\rule{0ex}{0ex}}\mathrm{For}n=2,\phantom{\rule{0ex}{0ex}}{\lambda}_{2}=\left(\frac{2}{5}\right)\times 1000=400\mathrm{nm}$

Thus, 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.

$\mathrm{So},{y}_{n}=n{\lambda}_{n}\frac{D}{d}\phantom{\rule{0ex}{0ex}}\Rightarrow {\lambda}_{n}={y}_{n}\frac{d}{nD}\phantom{\rule{0ex}{0ex}}\mathrm{For}n=1,\phantom{\rule{0ex}{0ex}}{\lambda}_{1}={y}_{n}\frac{d}{D}\phantom{\rule{0ex}{0ex}}={10}^{-3}\times \left(0.5\right)\times \frac{{10}^{-3}}{0.5}\phantom{\rule{0ex}{0ex}}={10}^{-6}\mathrm{m}=1000\mathrm{nm}.$

But 1000 nm does not fall in the range 400 nm − 700 nm.

$\mathrm{Again},\mathrm{for}n=2,\phantom{\rule{0ex}{0ex}}{\lambda}_{2}={y}_{n}\frac{d}{2D}=500\mathrm{nm}$

So, the light of wavelength 500 nm will have strong intensity.

#### Page No 382:

#### 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, $\mathrm{AB}=\mathrm{BO}\mathrm{and}\mathrm{AC}=\mathrm{CO}$.

Path difference of the wave front reaching O,

$\u2206x=\left(\mathrm{AB}+\mathrm{BO}\right)-\left(\mathrm{AC}+\mathrm{CO}\right)\phantom{\rule{0ex}{0ex}}=2\left(\mathrm{AB}-\mathrm{AC}\right)\phantom{\rule{0ex}{0ex}}=2\left(\sqrt{{D}^{2}+{d}^{2}}-D\right)$

For dark fringe to be formed at O, path difference should be an odd multiple of $\frac{\lambda}{2}$.

$\mathrm{So},\u2206x=\left(2n+1\right)\frac{\lambda}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 2\left(\sqrt{{D}^{2}+{d}^{2}}-D\right)=\left(2n+1\right)\frac{\lambda}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \sqrt{{D}^{2}+{d}^{2}}=D+\left(2n+1\right)\frac{\lambda}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow {D}^{2}+{d}^{2}={D}^{2}+{\left(2n+1\right)}^{2}\frac{{\lambda}^{2}}{16}+\left(2n+1\right)\frac{\lambda D}{2}$

Neglecting, ${\left(2n+1\right)}^{2}\frac{{\lambda}^{2}}{16}\phantom{\rule{0ex}{0ex}}$, as it is very small, we get:

$d=\left(\sqrt{2n}+1\right)\frac{\lambda D}{2}\phantom{\rule{0ex}{0ex}}\mathrm{For}\mathrm{minimum}d,\mathrm{putting},n=0\phantom{\rule{0ex}{0ex}}{d}_{\mathrm{min}}=\sqrt{\left(\frac{\lambda D}{2}\right)}$, we get:

Thus, for ${d}_{\mathrm{min}}=\sqrt{\left(\frac{\lambda D}{2}\right)}$ there is a dark fringe at O.

#### Page No 382:

#### Question 27:

Two coherent point sources *S*_{1} and *S*_{2}_{,} vibrating in phase, emit light of wavelength $\lambda $. The separation between the sources is $2\lambda $. Consider a line passing through *S*_{2} and perpendicular to the line *S*_{1} *S*_{2}. What is the smallest distance from *S*_{2} where a minimum intensity occurs?

#### Answer:

Let P be the point of minimum intensity.

For minimum intensity at point P,

${S}_{1}\mathrm{P}-{S}_{2}\mathrm{P}=x=\left(2n+1\right)\frac{\lambda}{2}$

Thus, we get:

$\sqrt{{\mathrm{Z}}^{2}+{\left(2\lambda \right)}^{2}}-\mathrm{Z}=\left(2n+1\right)\frac{\lambda}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{Z}}^{2}+4{\lambda}^{2}={\mathrm{Z}}^{2}{\left(2n+1\right)}^{2}\frac{{\lambda}^{2}}{4}+2\mathrm{Z}\left(2n+1\right)\frac{\lambda}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{Z}=\frac{4{\lambda}^{2}-\left(2n+1\right)2{\lambda}^{2}/4}{\left(2n+1\right)\lambda}\phantom{\rule{0ex}{0ex}}=\frac{16{\lambda}^{2}-\left(2n+1\right)}{4\left(2n+1\right)\lambda}...\left(i\right)$

$\text{When}n=0,\text{Z}=\frac{15\lambda}{4}\phantom{\rule{0ex}{0ex}}n=-1,\mathrm{Z}=\frac{-15\lambda}{4}\phantom{\rule{0ex}{0ex}}n=1,\mathrm{Z}=\frac{7\lambda}{12}\phantom{\rule{0ex}{0ex}}n=2,\text{Z}=\frac{-9\lambda}{20}$

Thus, $\mathrm{Z}=\frac{7\lambda}{12}$ is the smallest distance for which there will be minimum intensity.

#### Page No 382:

#### Question 28:

Figure (17-E5) shows three equidistant slits being illuminated by a monochromatic parallel beam of light. Let $B{P}_{0}-A{P}_{0}=\lambda /3\mathrm{and}D\lambda $. (a) Show that in this case $d=\sqrt{2\lambda D/3}$. (b) Show that the intensity at *P*_{0} is three times the intensity due to any of the three slits individually.

Figure

#### Answer:

(a) Given:

Wavelength of light = $\lambda $

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

$\u2206{x}_{B}={\mathrm{BP}}_{0}-{\mathrm{AP}}_{0}=\frac{\lambda}{3}\phantom{\rule{0ex}{0ex}}\Rightarrow \sqrt{{D}^{2}+{d}^{2}}-D=\frac{\lambda}{3}\phantom{\rule{0ex}{0ex}}\Rightarrow {D}^{2}+{d}^{2}={D}^{2}+\frac{{\lambda}^{2}}{9}+\frac{2\lambda D}{3}$

We will neglect the term $\frac{{\lambda}^{2}}{9}$, as it has a very small value.

$\therefore d=\sqrt{\frac{\left(2\lambda D\right)}{3}}$

(b) To calculating the intensity at P_{0}_{, }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

${\mathrm{CP}}_{0}-{\mathrm{AP}}_{0}=\sqrt{{D}^{2}+{\left(2d\right)}^{2}}-D\phantom{\rule{0ex}{0ex}}=\sqrt{{D}^{2}+\frac{8\lambda D}{3}}-D\left(\mathrm{Using}\mathrm{the}\mathrm{value}\mathrm{of}d\mathrm{from}\mathrm{part}a\right)\phantom{\rule{0ex}{0ex}}=D{\left\{1+\frac{8\lambda}{3D}\right\}}^{\frac{1}{2}}-D\phantom{\rule{0ex}{0ex}}\mathrm{Expanding}\mathrm{the}\mathrm{value}\mathrm{using}\mathrm{binomial}\mathrm{theorem}\mathrm{and}\mathrm{neglecting}\text{the}\mathrm{higher}\mathrm{order}\mathrm{terms},\mathrm{we}\mathrm{get}:\phantom{\rule{0ex}{0ex}}D\left\{1+\frac{1}{2}\times \frac{8\lambda}{3D}+...\right\}-D\phantom{\rule{0ex}{0ex}}$

${\mathrm{CP}}_{0}-{\mathrm{AP}}_{0}=\frac{4\mathrm{\lambda}}{3}$

So, the corresponding phase difference between the wave fronts from A and C is given by

${\varphi}_{c}=\frac{2\pi \u2206{x}_{C}}{\lambda}=\frac{2\pi \times 4\lambda}{3\lambda}\phantom{\rule{0ex}{0ex}}\Rightarrow {\varphi}_{c}=\frac{8\pi}{3}\mathrm{or}\left(2\pi +\frac{2\pi}{3}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow {\varphi}_{c}=\frac{2\pi}{3}...\left(1\right)\phantom{\rule{0ex}{0ex}}\mathrm{Again},{\varphi}_{\mathrm{B}}=\frac{2\pi \u2206{x}_{B}}{\lambda}\phantom{\rule{0ex}{0ex}}\Rightarrow {\varphi}_{\mathrm{B}}=\frac{2\pi \lambda}{3\lambda}=\frac{2\pi}{3}...\left(2\right)$

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 P_{0} is given by

$A=\sqrt{{\left(2a\right)}^{2}+{a}^{2}+2a\times a\mathrm{cos}\left(\frac{2\pi}{3}\right)}\phantom{\rule{0ex}{0ex}}=\sqrt{4{a}^{2}+{a}^{2}+2{a}^{2}\sqrt{3}}\phantom{\rule{0ex}{0ex}}\therefore {l}_{po}=\mathrm{K}{\left(\sqrt{3}r\right)}^{2}=3\mathrm{K}{r}^{2}=3l$

Here, *I* is the intensity due to the individual slits* *and *I*_{po} is the total intensity at P_{0}.

Thus, the resulting amplitude is three times the intensity due to the individual slits.

#### Page No 382:

#### 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, $d=2\mathrm{mm}=2\times {10}^{-3}\mathrm{m}\phantom{\rule{0ex}{0ex}}$

Wavelength of the light, $\lambda =600\mathrm{nm}=6\times {10}^{-7}\mathrm{m}$

Distance of the screen from the slits, *D *= 2⋅0 m

${I}_{\mathrm{max}}=0.20\mathrm{W}/{\mathrm{m}}^{2}\phantom{\rule{0ex}{0ex}}\mathrm{For}\mathrm{the}\mathrm{point}\mathrm{at}\mathrm{a}\mathrm{position}y=0.5\mathrm{cm}=0.5\times {10}^{-2}\mathrm{m},\phantom{\rule{0ex}{0ex}}\text{p}\mathrm{ath}\mathrm{difference},\u2206x=\frac{yd}{D}.\phantom{\rule{0ex}{0ex}}\Rightarrow \u2206x=\frac{0.5\times {10}^{-2}\times 2\times {10}^{-3}}{2}\phantom{\rule{0ex}{0ex}}=5\times {10}^{-6}\mathrm{m}$

So, the corresponding phase difference is given by

$\u2206\varphi =\frac{2\pi \u2206x}{\lambda}=\frac{2\pi \times 5\times {10}^{-6}}{6\times {10}^{-7}}\phantom{\rule{0ex}{0ex}}=\frac{50\pi}{3}=16\pi +\frac{2\pi}{3}\phantom{\rule{0ex}{0ex}}\mathrm{or}\u2206\mathrm{\varphi}=\frac{2\pi}{3}$

So, the amplitude of the resulting wave at point *y* = 0.5 cm is given by

$A=\sqrt{{a}^{2}+{a}^{2}+2{a}^{2}\mathrm{cos}\left(\frac{2\pi}{3}\right)}\phantom{\rule{0ex}{0ex}}=\sqrt{{a}^{2}+{a}^{2}-{a}^{2}}=a\phantom{\rule{0ex}{0ex}}$

Similarly, the amplitude of the resulting wave at the centre is 2*a**.*

Let the intensity of the resulting wave at point *y* = 0.5 cm be* I*.

$\mathrm{Since}\frac{I}{{I}_{\mathrm{max}}}=\frac{{A}^{2}}{{\left(2a\right)}^{2}},\mathrm{we}\mathrm{have}:\phantom{\rule{0ex}{0ex}}\frac{I}{0.2}=\frac{{A}^{2}}{4{a}^{2}}=\frac{{a}^{2}}{4{a}^{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow I=\frac{0.2}{4}=0.05\mathrm{W}/{\mathrm{m}}^{2}$

Thus, the intensity at a point 0.5 cm away from the centre along the width of the fringes is 0.05 W/m^{2}.

#### Page No 382:

#### 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 $\lambda $. 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 = $\lambda $

Distance of the screen = $D$

(a) When the intensity is half the maximum:

Let *I*_{max} be the maximum intensity and *I* be the intensity at the required point at a distance *y* from the central point.

So, $I={a}^{2}+{a}^{2}+2{a}^{2}\mathrm{cos}\varphi $

Here, *$\varphi $* is the phase difference in the waves coming from the two slits.

So, $I=4{a}^{2}{\mathrm{cos}}^{2}\left(\frac{\varphi}{2}\right)$

$\Rightarrow \frac{I}{{I}_{\mathrm{max}}}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{4{a}^{2}{\mathrm{cos}}^{2}\left({\displaystyle \frac{\varphi}{2}}\right)}{4{a}^{2}}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{cos}}^{2}\left(\frac{\varphi}{2}\right)=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{cos}\left(\frac{\varphi}{2}\right)=\frac{1}{\sqrt{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\mathrm{\varphi}}{2}=\frac{\mathrm{\pi}}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow \varphi =\frac{\mathrm{\pi}}{2}\phantom{\rule{0ex}{0ex}}\mathrm{Corrosponding}\mathrm{path}\mathrm{difference},\u2206x=\frac{\lambda}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow y=\frac{\u2206xD}{d}=\frac{\lambda D}{4d}$

(b) When the intensity is one-fourth of the maximum:

$\frac{I}{{I}_{\mathrm{max}}}=\frac{1}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow 4{a}^{2}{\mathrm{cos}}^{2}\left(\frac{\varphi}{2}\right)=\frac{1}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{cos}}^{2}\left(\frac{\varphi}{2}\right)=\frac{1}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{cos}\left(\frac{\varphi}{2}\right)=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\varphi}{2}=\frac{\pi}{3}\phantom{\rule{0ex}{0ex}}\mathrm{So},\mathrm{corrosponding}\mathrm{path}\mathrm{difference},\u2206x=\frac{\lambda}{3}\phantom{\rule{0ex}{0ex}}\mathrm{and}\mathrm{position},y=\frac{\u2206xD}{d}=\frac{\lambda D}{3d}.$

#### Page No 382:

#### Question 31:

In a Young's double slit experiment, $\lambda =500\mathrm{nm},\mathrm{d}=1\xb70\mathrm{mm}\mathrm{and}D=1\xb70\mathrm{m}$. 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, $d=1\mathrm{mm}={10}^{-3}\mathrm{m}\phantom{\rule{0ex}{0ex}}$

Wavelength of the light, $\lambda =500\mathrm{nm}=5\times {10}^{-7}\mathrm{m}\phantom{\rule{0ex}{0ex}}$

Distance of the screen, $D=1\mathrm{m}\phantom{\rule{0ex}{0ex}}$

Let *I*_{max} be the maximum intensity and *I* be the intensity at the required point at a distance *y* from the central point.

So, $I={a}^{2}+{a}^{2}+2{a}^{2}\mathrm{cos}\varphi $

Here, *$\varphi $* is the phase difference in the waves coming from the two slits.

So, $I=4{a}^{2}{\mathrm{cos}}^{2}\left(\frac{\varphi}{2}\right)$

$\Rightarrow \frac{I}{{I}_{\mathrm{max}}}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{4{a}^{2}{\mathrm{cos}}^{2}\left({\displaystyle \frac{\varphi}{2}}\right)}{4{a}^{2}}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{cos}}^{2}\left(\frac{\varphi}{2}\right)=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{cos}\left(\frac{\varphi}{2}\right)=\frac{1}{\sqrt{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\mathrm{\varphi}}{2}=\frac{\mathrm{\pi}}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow \varphi =\frac{\mathrm{\pi}}{2}\phantom{\rule{0ex}{0ex}}\mathrm{Corrosponding}\mathrm{path}\mathrm{difference},\u2206x=\frac{1}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow y=\frac{\u2206xD}{d}=\frac{\lambda D}{4d}$

$\Rightarrow y=\frac{5\times {10}^{-7}\times 1}{4\times {10}^{-3}}\phantom{\rule{0ex}{0ex}}=1.25\times {10}^{-4}\mathrm{m}$

∴ The required minimum distance from the central maximum is $1.25\times {10}^{-4}\mathrm{m}$.

#### Page No 382:

#### 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 $\lambda $, *d* and *D* where the symbols have their usual meanings.

#### Answer:

Given:

Separation between two slits = *d*

Wavelength of the light = $\lambda \phantom{\rule{0ex}{0ex}}$

Distance of the screen = $D$

Let *I*_{max} be the maximum intensity and *I* be half the maximum intensity at a point at a distance *y* from the central point.

So, $I={a}^{2}+{a}^{2}+2{a}^{2}\mathrm{cos}\varphi $

Here, *$\varphi $* is the phase difference in the waves coming from the two slits.

So, $I=4{a}^{2}{\mathrm{cos}}^{2}\left(\frac{\varphi}{2}\right)$

$\Rightarrow \frac{I}{{I}_{\mathrm{max}}}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{4{a}^{2}{\mathrm{cos}}^{2}\left({\displaystyle \frac{\varphi}{2}}\right)}{4{a}^{2}}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{cos}}^{2}\left(\frac{\varphi}{2}\right)=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{cos}\left(\frac{\varphi}{2}\right)=\frac{1}{\sqrt{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\mathrm{\varphi}}{2}=\frac{\mathrm{\pi}}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow \varphi =\frac{\mathrm{\pi}}{2}\phantom{\rule{0ex}{0ex}}\mathrm{Corrosponding}\mathrm{path}\mathrm{difference},\u2206x=\frac{1}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow y=\frac{\u2206xD}{d}=\frac{\lambda D}{4d}$

The 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 = 2*y*

$=2\frac{D\lambda}{4d}=\frac{D\lambda}{2d}$

Thus, the required line width of the bright fringe is $\frac{D\lambda}{2d}$.

#### Page No 383:

#### Question 33:

Consider the situation shown in the figure. The two slits *S*_{1} and *S*_{2} 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 *S*_{3} is at the central line and the slit *S*_{4} is at a distance *z* from *S*_{3}. 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) $z=\frac{\lambda \mathrm{D}}{2d}$

(b) $\frac{\lambda \mathrm{D}}{d}$

(c) $\frac{\lambda \mathrm{D}}{4d}$

Figure

#### Answer:

Given:

Separation between the two slits = *d*

Wavelength of the light =$\lambda $

Distance of the screen = $D$

The fringe width (*β*) is given by $\beta =\frac{\lambda D}{d}$.

At S_{3}, the path difference is zero. So, the maximum intensity occurs at amplitude = 2*a*.

(a) When $z=\frac{D\lambda}{2d}$:

The first minima occurs at *S*_{4}, as shown in figure (a).

With amplitude = 0 on screen ∑_{2}_{, }we get:

$\frac{{l}_{max}}{{l}_{min}}=\frac{{\left(2a+0\right)}^{2}}{{\left(2a-0\right)}^{2}}=1$

(b) When $z=\frac{D\lambda}{d}$:

The first maxima occurs at S_{4}, as shown in the figure.

With amplitude = $2a$ on screen ∑_{2}, we get:

$\frac{{l}_{\mathrm{max}}}{{l}_{\mathrm{min}}}=\frac{{\left(2a+2a\right)}^{2}}{{\left(2a-2a\right)}^{2}}=\infty \phantom{\rule{0ex}{0ex}}$

(c) When $z=\frac{D\lambda}{4d}$:

The slit S_{4} falls at the mid-point of the central maxima and the first minima, as shown in the figure.

$\mathrm{Intensity}=\frac{{l}_{\mathrm{max}}}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{Amplitude}=\sqrt{2}a$

$\therefore \frac{{l}_{\mathrm{max}}}{{l}_{\mathrm{min}}}=\frac{{\left(2a+\sqrt{2}a\right)}^{2}}{{\left(2a-\sqrt{2}a\right)}^{2}}=34$

#### Page No 383:

#### Question 34:

Consider the arrangement shown in the figure (17-E7). By some mechanism, the separation between the slits *S*_{3} and *S*_{4} can be changed. The intensity is measured at the point *P,* which is at the common perpendicular bisector of *S*_{1}*S*_{2} and *S*_{2}*S*_{4}. When $z=\frac{D\lambda}{2d}$, the intensity measured at *P* is *I*. Find the intensity when *z* is equal to

$\left(\mathrm{a}\right)\frac{D\lambda}{d},\left(\mathrm{b}\right)\frac{3D\lambda}{2d}\mathrm{and}\left(\mathrm{c}\right)\frac{2D\lambda}{d}.$

Figure

#### Answer:

Given:

Fours slits S_{1}, S_{2}, S_{3} and S_{4}.

The separation between slits S_{3} and S_{4} can be changed.

Point P is the common perpendicular bisector of S_{1}S_{2} and S_{3}S_{4}.

(a) $\mathrm{For}z=\frac{\lambda D}{d}$:

The position of the slits from the central point of the first screen is given by

$y={\mathrm{OS}}_{3}={\mathrm{OS}}_{4}=\frac{z}{2}=\frac{\lambda D}{2d}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

The corresponding path difference in wave fronts reaching S_{3} is given by

$\u2206x=\frac{yd}{D}=\frac{\lambda D}{2d}\times \frac{d}{D}=\frac{\lambda}{2}$

Similarly at S_{4}, path difference, $\u2206x=\frac{yd}{D}=\frac{\lambda D}{2d}\times \frac{d}{D}=\frac{\lambda}{2}$

$\mathrm{i}.\mathrm{e}.\mathrm{dark}\mathrm{fringes}\mathrm{are}\mathrm{formed}\mathrm{at}{\mathrm{S}}_{3}\mathrm{and}{\mathrm{S}}_{4}\phantom{\rule{0ex}{0ex}}$.

So, the intensity of light at S_{3} and S_{4} is zero. Hence, the intensity at P is also zero.

(b) $\mathrm{For}z=\frac{3\lambda D}{2d}$

The position of the slits from the central point of the first screen is given by

$y={\mathrm{OS}}_{3}={\mathrm{OS}}_{4}=\frac{z}{2}=\frac{3\lambda D}{4d}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

The corresponding path difference in wave fronts reaching S_{3} is given by

$\u2206x=\frac{yd}{D}=\frac{3\lambda D}{4d}\times \frac{d}{D}=\frac{3\lambda}{4}$

Similarly at S_{4}, path difference, $\u2206x=\frac{yd}{D}=\frac{3\lambda D}{4d}\times \frac{d}{D}=\frac{3\lambda}{4}$

Hence, the intensity at P is *I*.

(c) Similarly, for $z=\frac{2D\lambda}{d}$,

the intensity at P is 2*I*.

#### Page No 383:

#### 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, $\lambda =580\mathrm{nm}=580\times {10}^{-9}\mathrm{m}$

Let the index of refraction of the soap solution be μ.

The condition of minimum reflection of light is 2μ*d* = *nλ*,

where *n *is an interger = 1 , 2 , 3 ...

*$\Rightarrow \mathrm{\mu}=\frac{n\lambda}{2d}=\frac{2n\lambda}{4d}\phantom{\rule{0ex}{0ex}}=\frac{580\times {10}^{-9}\times \left(2n\right)}{4\times 11\times {10}^{-7}}\phantom{\rule{0ex}{0ex}}=\frac{5.8\left(2n\right)}{44}=0.132\left(2n\right)$*

As per the question, μ has a value between 1.2 and 1.5. So,

$n=5\phantom{\rule{0ex}{0ex}}\mathrm{So},\mathrm{\mu}=0.132\times 10=1.32$

Therefore, the index of refraction of the soap solution is 1.32.

#### Page No 383:

#### 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, $\mathrm{\lambda}=560\times {10}^{-9}\mathrm{m}$

Refractive index of the oil film, $\mathrm{\mu}=1.4$

Let the thickness of the film for strong reflection be *t.*

The condition for strong reflection is

$2\mathrm{\mu}t=\left(2n+1\right)\frac{\lambda}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow t=\left(2n+1\right)\frac{\lambda}{4\mathrm{\mu}}$

where *n* is an integer.

For minimum thickness, putting *n* = 0, we get:

$t=\frac{\lambda}{4\mathrm{\mu}}\phantom{\rule{0ex}{0ex}}=\frac{560\times {10}^{-9}}{4\times 1.4}\phantom{\rule{0ex}{0ex}}={10}^{-7}\mathrm{m}=100\mathrm{nm}$

Therefore, the minimum thickness of the oil film so that it strongly reflects the light is 100 nm.

#### Page No 383:

#### 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, $\mathrm{\lambda}=400\times {10}^{-9}\mathrm{m}\mathrm{to}700\times {10}^{-9}\mathrm{nm}$

Refractive index of water, $\mathrm{\mu}=1.33$

The thickness of film, *$t={10}^{-4}\mathrm{cm}={10}^{-6}\mathrm{m}$*

The condition for strong transmission: $2\mathrm{\mu}t=n\lambda $,

where n is an integer.

$\Rightarrow \lambda =\frac{2\mathrm{\mu}t}{n}$

$\Rightarrow \lambda =\frac{2\times 1.33\times {10}^{-6}}{n}\phantom{\rule{0ex}{0ex}}=\frac{2660\times {10}^{-9}}{n}\mathrm{m}$

Putting* n* = 4, we get, λ_{1} = 665 nm.

Putting* n* = 5, we get, λ_{2} = 532 nm.

Putting *n* = 6, we get, λ_{3} = 443 nm.

Therefore, the wavelength (in visible region) which are strongly transmitted by the film are 665 nm, 532nm and 443 nm.

#### Page No 383:

#### 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, $\lambda =400\times {10}^{-9}\mathrm{to}750\times {10}^{-9}\mathrm{m}$

Refractive index of oil, μ_{oil}, is 1.25 and that of glass, μ_{g}, is 1.50.

The thickness of the oil film, $d=1\times {10}^{-4}\mathrm{cm}={10}^{-6}\mathrm{m},$

The condition for the wavelengths which can be completely transmitted through the oil film is given by

$\mathrm{\lambda}=\frac{2\mathrm{\mu}d}{\left(n+{\displaystyle \frac{1}{2}}\right)}\phantom{\rule{0ex}{0ex}}=\frac{2\times {10}^{-6}\times \left(1.25\right)\times 2}{\left(2n+1\right)}\phantom{\rule{0ex}{0ex}}=\frac{5\times {10}^{-6}}{\left(2n+1\right)}\mathrm{m}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{\lambda}=\frac{5000}{\left(2n+1\right)}\mathrm{nm}$

Where *n* is an integer.

For wavelength to be in visible region i.e (400 nm to 750 nm)

When *n* = 3, we get,

$\lambda =\frac{5000}{\left(2\times 3+1\right)}\phantom{\rule{0ex}{0ex}}=\frac{5000}{7}=714.3\mathrm{nm}$

When, *n* = 4, we get,

$\lambda =\frac{5000}{\left(2\times 4+1\right)}\phantom{\rule{0ex}{0ex}}=\frac{5000}{9}=555.6\mathrm{nm}$

When, *n* = 5, we get,

$\lambda =\frac{5000}{\left(2\times 5+1\right)}\phantom{\rule{0ex}{0ex}}=\frac{5000}{11}=454.5\mathrm{nm}$

Thus 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.

#### Page No 383:

#### 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:

*b*sin*θ* = *nλ*

For the first minima, we put *n* = 1.

$5\times \mathrm{sin}30\xb0=1\times \mathrm{\lambda}\phantom{\rule{0ex}{0ex}}\Rightarrow \lambda =\frac{5}{2}=2.5\mathrm{cm}$

Therefore, the wavelength of the microwaves is 2.5 cm.

#### Page No 383:

#### 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, $\mathrm{\lambda}=560\mathrm{nm}=560\times {10}^{-9}\mathrm{m}$

Diameter 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

$R=1.22\frac{\lambda D}{d}\phantom{\rule{0ex}{0ex}}=1.22\times \frac{560\times {10}^{-9}\times 2}{2\times {10}^{-4}}\phantom{\rule{0ex}{0ex}}=6.832\times {10}^{-3}\mathrm{m}\mathrm{or}=0.683\mathrm{cm}$

Hence, the diameter 2*R** *of the central bright spot on the wall is 1.37 cm.

#### Page No 383:

#### 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, $\mathrm{\lambda}=620\mathrm{nm}=620\times {10}^{-9}\mathrm{m}\phantom{\rule{0ex}{0ex}}$

Diameter of the convex lens, $d=8\mathrm{cm}=8\times {10}^{-2}\mathrm{m}$

Distance from the lens where light is to be focused, $D=20\mathrm{cm}=20\times {10}^{-2}\mathrm{m}$

The radius of the central bright spot is given by

$R=1.22\frac{\lambda D}{d}\phantom{\rule{0ex}{0ex}}=1.22\times \frac{620\times {10}^{-9}\times 20\times {10}^{-2}}{8\times {10}^{-2}}\phantom{\rule{0ex}{0ex}}=1891\times {10}^{-9}\mathrm{m}\approx 1.9\times {10}^{-6}\mathrm{m}$

∴ Diameter of the central bright spot, 2*R* = 3.8 × 10^{−6} m.

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