
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 slitwidth 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 fringewidth? 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 sinkxωt. 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 Xaxis is
(a) 0°
(b) 45°
(c) 90°
(d)
cos1 1/3
Answer:
(d)
cos11/3On 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:
13x+13y+13z=c3If
θis the angle between the normal and +X axis, then
cosθ=13⇒θ=cos113
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⇒17×107 m<1λ<14×107 m3×1087×107 Hz<cλ<3×1084×107 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=589×109 mRefractive index of water, μ_{w}= 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=3×108589×109 =5.09×1014 Hz(b)
Let wavelength of sodium light in water be
λw.
We know that
μaμw=λωλa,
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.
⇒11.33=λω589×109⇒ 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, v_{a} = 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 v_{400}.
So,
1.472=3×108v400⇒v400=2.04×108 m/sLet speed of light of wavelength 760 nm in quartz be v_{760}.
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 × 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,
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
5·0×107 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=5×107 mDistance of the screen,
D=1 m(a)
We know that separation between two consecutive maxima = fringe width (β).
That is,
β=λDd …(i)
=5×107×1102 m =5×105 m=0.05 mm(b)
Separation between two consecutive maxima = fringe width
∴
β=1 mm=103 mLet the separation between the sources be ‘d’
Using equation (i), we get:
d’=5×107×1103 ⇒d’=5×104 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 +I22I1I22=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 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) 4I_{0}
Answer:
(b) I_{0}/â€‹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.
Question 16:
A thin transparent sheet is placed in front of a Young’s double slit. The fringewidth will
(a) increase
(b) decrease
(c) remain same
(d) become nonuniform
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 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)
νA>νB>νC(b)
νA<νB<νC(c)
νA=νB=νC(d)
νB=12νA+νC.
Answer:
(c) v_{A}_{ = }v_{B}_{ =} v_{C}_{ }
(d)
νB=12vA+vC
Since the speed of light is a universal constant, v_{A}_{ = }v_{B}_{ =} v_{C} =
3×108m/s.
νB=12vA+vC. This expression also implies that v_{A}_{ = }v_{B}_{ =} v_{C}_{. }
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) v_{A }> v_{B }> v_{C}
(d) v_{B} = (v_{A} + v_{C}_{ })/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 Xaxis. 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
â€‹The wave is travelling along the Xaxis. So, it’ll have planar wavefront perpendicular to the Xaxis.
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.
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
β=λDd
103 m=2.5×λ103⇒λ=12.5 106 m =4×107 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=103 mWavelength of the light used,
λ=5.0×107 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)
=5×107×12×103 =2.5×104 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=0.8×103 mWavelength of the yellow light,
λ=589 nm=589×109 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=589×109×20.8×103 =1.47×103 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 bluegreen light of wavelength 500 nm. The separation between the slits is
2·0×103 m.
Answer:
Given:
Wavelength of the bluegreen light,
λ=500×109 mSeparation between two slits,
d=2×103 m,Let angular separation between the consecutive bright fringes be θ.
Using θ= βD=λDdD=λd, we get: θ=500×1092×103 =250×106 =25×105 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 doubleslit 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=480×109 m and λ2=600×109 mSeparation between the slits,
d=0.25 mm=0.25×103 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 = y_{2} − y_{1}
y2y1=Dy2y1d⇒y2y1=1.50.25×103600×109480×109 y2y1=72×105 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 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} =
mλvDdPosition of the n^{th} bright fringe of red light, y_{r} =
nλrDdFor overlapping, y_{v =} y_{r} .
So, as per the question,
m×400×Dd=n×700×Dd⇒mn=74Therefore, the
7thbright 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,
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
μ1t.
∴ μ1t=λ2⇒t=λ2μ1Hence, the thickness of a plate is
λ2μ1.
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
μ1t.
(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. μ1 t=λ2⇒t=λ2 μ1
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=0.02×103 mWavelength of the light,
λ=620 nm=620×109 mWe know that when we paste a transparent paper in front of one of the slits, then the optical path changes by
μ1t.
And optical path should be changed by λ for the shift of one fringe.
∴ Number of fringes crossing through the centre is
n=μ1tλ =1.451×0.02×103620×109 =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 fringeshift 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=1.964×106 mLet the wavelength of the light used = λ.
Number of fringes shifted is given by
n=μ1tλSo, the corresponding shift in the fringe width equals the number of fringes multiplied by the width of one fringe.
Shift =n × β=μ1tλ×λDd=μ1t×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, μ1t×Dd=λ2Dd⇒λ=μ1 t2 =1.61×1.964×1062 =589.2×109=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 fringewidth? (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=0.5×103 mSeparation between the two slits,
d=0.12 cm=12×104 mThe refractive index of mica, μ_{m} = 1.58 and of polystyrene, μ_{p} = 1.58
Wavelength of the light,
λ=590 nm=590×109 m,Distance between screen and slit, D = 1 m
(a)
We know that fringe width is given by
β=λDd
⇒β =590×109×112×104 =4.9×104 m(b) When both the mica and polystyrene strips are fitted before the slits, the optical path changes by
∆x=μm1 t μp1 t =μmμp t =1.581.55×0.5 103 =0.015×103 m∴ Number of fringes shifted, n =
∆xλ.
⇒n=0.015×103590×109=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=0.43×4.91×104 ∵β=4.91×104 m =0.021 cmOn the other side,
x’=10.43×4.91×104 =0.028 cm
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
∆x=μ11 t μ21 t =μ1μ2 tFor minimum at P_{0}, the path difference should be
λ2.
i.e. ∆x=λ2
So,λ2=μ1μ2t⇒t=λ2μ1μ2Therefore, minimum at point P_{0} is
λ2μ1μ2.
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=0.02×103 mRefractive index of the paper,
μ=1.45.
Wavelength of the light,
λ=600 nm=600×109 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,
I2=49I1We know that
I1I2=a12a22 ∵I∝a2⇒a1a2=32Here, a is the amplitude.
We know that ImaxImin=a1+a22a1a22.⇒ ImaxImin=3+22322 =251⇒Imax:Imin=25 : 1(b)
Number of fringes that will cross through the centre is given by
n=μ1tλ.
⇒n=1.451×0.02×103600×109 =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 fringewidth of the pattern formed on the screen.
Answer:
Given:
Separation between two slits,
d=0.28 mm=0.28×103 mDistance between screen and slit (D) = 48 cm = 0.48 m
Wavelength of the red light,
λa=700 nm in vaccum =700 ×109 mLet the wavelength of red light in water =
λωWe known that refractive index of water (μ_{w} =4/3),
μ_{w} =
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 =525×109×0.480.28×103 =9×104=0.90 mmHence, fringewidth 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=sin1 λ2dwith 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 = âˆ†x.
In âˆ†S_{1}S_{2}X,
sinθ=S1XS1S2=∆xd
⇒∆x=dsinθUsing
θ=sin1 λ2d, we get,
⇒∆x=d×λ2d=λ2Hence, there will be dark fringe at point P_{0} 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 (17E1). 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 fringewidth if the light used has a wavelength of 700 nm.
Answer:
Given:
Separation between two slits,
d’=2d=2 mm=2×103 m(as d = 1 mm)
Wavelength of the light used,
λ=700 nm=700×103 mDistance between the screen and slit (D) = 1.0 m
It is a case of Lloyd’s mirror experiment.
Fringe width, β=λDd’ =700×109×12×103 =0.35×103 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 = 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,
I2=I1×64100⇒I2I1=64100=1625And I2I1=a22a12⇒a2a1=45
We know that ImaxImin=a1+a22a1a22 =5+42542Imax:Imin=81 : 1Hence, the required ratio is 81 : 1.
Question 24:
A double slit S_{1} − S_{2} 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 D_{1} from it and a screen ∑ is placed behind the double slit at a distance D_{2} from it (figure 17E2). The screen ∑ receives only the light reflected by the mirror. Find the fringewidth 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=2D1+D2Fringe width, β=λDd=2D1+D2 λdHence, the required fringe width is
2D1+D2 λd.
Question 25:
White coherent light (400 nm700 nm) is sent through the slits of a Young’s double slit experiment (figure 17E3). 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=0.5×103 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=1×103 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+1×103×0.05×1030.5 =22n+1×106 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 =103×0.5×1030.5 =106 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 (17E4). 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 fringewidth.
Figure
Answer:
From the figure,
AB=BO and AC=CO.
Path difference of the wave front reaching O,
∆x=AB+BOAC+CO =2 ABAC =2 D2+d2DFor dark fringe to be formed at O, path difference should be an odd multiple of
λ2.
So,∆x=2n+1 λ2⇒ 2D2+d2D=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 S_{1} and S_{2}_{,} vibrating in phase, emit light of wavelength
λ. The separation between the sources is
2λ. 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,
S1PS2P=x=2n+1 λ2Thus, we get:
Z2+2λ2Z=2n+1 λ2⇒Z2+4λ2=Z2 2n+12 λ24+2Z 2n+1 λ2⇒Z=4λ22n+1 2λ2/42n+1 λ =16λ22n+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 (17E5) shows three equidistant slits being illuminated by a monochromatic parallel beam of light. Let
BP0AP0=λ/3 and D>>λ. (a) Show that in this case
d=2λD/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 =
λPath difference of wave fronts reaching from A and B is given by
∆xB=BP0AP0=λ3⇒D2+d2D=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 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
CP0AP0=D2+2d2D =D2+8λD3D Using the value of d from part a =D1+8λ3D12DExpanding the value using binomial theorem and neglecting the higher order terms, we get: D1+12×8λ3D+…D
CP0AP0=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 P_{0} 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 I_{po} is the total intensity at P_{0}.
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=2×103 mWavelength of the light,
λ=600 nm=6×107 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=0.5×102 m,path difference,∆x=ydD.⇒∆x=0.5×102×2×1032 =5×106 mSo, the corresponding phase difference is given by
∆ϕ=2π∆xλ=2π×5×1066×107 =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+a2a2 =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/m^{2}.
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) onefourth 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 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=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 onefourth 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=103 mWavelength of the light,
λ=500 nm=5×107 mDistance of the screen,
D=1 mLet 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=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=5×107×14×103 =1.25×104 m∴ The required minimum distance from the central maximum is
1.25×104 m.
Question 32:
The linewidth 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 linewidth 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 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=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 linewidth 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, linewidth = 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 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=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 S_{3}, the path difference is zero. So, the maximum intensity occurs at amplitude = 2a.
(a) When
z=Dλ2d:
The first minima occurs at S_{4}, as shown in figure (a).
With amplitude = 0 on screen ∑_{2}_{, }we get:
lmaxlmin=2a+022a02=1
(b) When
z=Dλd:
The first maxima occurs at S_{4}, as shown in the figure.
With amplitude =
2aon screen ∑_{2}, we get:
lmaxlmin=2a+2a22a2a2=∞(c) When
z=Dλ4d:
The slit S_{4} falls at the midpoint of the central maxima and the first minima, as shown in the figure.
Intensity=lmax2⇒ Amplitude=2a
∴lmaxlmin=2a+2a22a2a2=34
Question 34:
Consider the arrangement shown in the figure (17E7). 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=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 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)
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 S_{3} is given by
∆x=ydD=λD2d×dD=λ2Similarly at S_{4}, path difference,
∆x=ydD=λD2d×dD=λ2
i.e. dark fringes are formed at S3 and S4.
So, the intensity of light at S_{3} and S_{4} 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 S_{3} is given by
∆x=ydD=3λD4d×dD=3λ4Similarly at S_{4}, 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=580×109 mLet the index of refraction of the soap solution be μ.
The condition of minimum reflection of light is 2μd = nλ,
where n is an interger = 1 , 2 , 3 …
⇒μ=nλ2d=2nλ4d =580×109×2n4×11×107 =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=560×109 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 =560×1094×1.4 =107m=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=400×109 m to 700×109 nmRefractive index of water,
μ=1.33The thickness of film,
t=104 cm=106 m
The condition for strong transmission:
2μt=nλ,
where n is an integer.
⇒ λ=2μtn
⇒ λ=2×1.33×106n =2660×109n mPutting 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.
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=400×109 to 750×109 mRefractive index of oil, μ_{oil}, is 1.25 and that of glass, μ_{g}, is 1.50.
The thickness of the oil film,
d=1×104 cm=106m,The condition for the wavelengths which can be completely transmitted through the oil film is given by
λ=2μdn+12 =2×106×1.25×22n+1 =5×106 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 = nλ
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=560×109 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 =1.22×560×109×22×104 =6.832×103 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=620×109mDiameter of the convex lens,
d=8 cm=8×102 mDistance from the lens where light is to be focused,
D=20 cm=20×102 mThe radius of the central bright spot is given by
R=1.22λDd =1.22×620×109×20×1028×102 = 1891×109 m≈1.9×106m∴ Diameter of the central bright spot, 2R = 3.8 × 10^{−6} m.
HC Verma Solutions for Class 11 Physics – Part 1
 Chapter 1 – Introduction to Physics
 Chapter 2 – Physics and Mathematics
 Chapter 3 – Rest and Motion: Kinematics
 Chapter 4 – The Forces
 Chapter 5 – Newton’s Laws of Motion
 Chapter 6 – Friction
 Chapter 7 – Circular Motion
 Chapter 8 – Work and Energy
 Chapter 9 – Center of Mass, Linear Momentum, Collision
 Chapter 10 – Rotational Mechanics
 Chapter 11 – Gravitation
 Chapter 12 – Simple Harmonic Motion
 Chapter 13 – Fluid Mechanics
 Chapter 14 – Some Mechanical Properties of Matter
 Chapter 15 – Wave Motion and Wave on a String
 Chapter 16 – Sound Wave
 Chapter 17 – Light Waves
 Chapter 18 – Geometrical Optics
 Chapter 19 – Optical Instruments
 Chapter 20 – Dispersion and Spectra
 Chapter 21 – Speed of Light
 Chapter 22 – Photometry
HC Verma Solutions for Class 12 Physics – Part 2
 Chapter 23 – Heat and Temperature
 Chapter 24 – Kinetic Theory of Gases
 Chapter 25 – Calorimetry
 Chapter 26 – Laws of Thermodynamics
 Chapter 27 – Specific Heat Capacities of Gases
 Chapter 28 – Heat Transfer
 Chapter 29 – Electric Field and Potential
 Chapter 30 – Gauss’s Law
 Chapter 31 – Capacitors
 Chapter 32 – Electric Current in Conductors
 Chapter 33 – Thermal and Chemical Effects of Electric Current
 Chapter 34 – Magnetic Field
 Chapter 35 – Magnetic Field due to a Current
 Chapter 36 – Permanent Magnets
 Chapter 37 – Magnetic Properties of Matter
 Chapter 38 – Electromagnetic Induction
 Chapter 39 – Alternating Current
 Chapter 40 – Electromagnetic Waves
 Chapter 41 – Electric Current through Gases
 Chapter 42 – Photoelectric Effect and Wave Particle Duality
 Chapter 43 – Bohr’s Model and Physics of the Atom
 Chapter 44 – Xrays
 Chapter 45 – Semiconductors and Semiconductor Devices
 Chapter 46 – The Nucleus
 Chapter 47 – The Special Theory of Relativity
Leave a Reply