Hello Students. Are you Searching for NCERT Solutions for Class 12 Physics Chapter 10? If yes then you are most welcome to NCERTian. Here we have provided you with the complete Question and Answers of Chapter 10: Wave Optics. These solutions are written by expert teachers and faculties keeping the need of students in mind.
|Chapter||10. Wave Optics|
|Category||NCERT Solutions for Class 12|
Class 12 Physics NCERT Solutions are the excellent practice set for students to ace Physics with better understandings and good marks. With the NCERT Solution for Class 12 Physics, students can confidently answer their exam questions and enjoy the best possible marks in the subject. These solutions are prepared keeping in view the latest syllabus and curriculum. In addition, the solutions are designed keeping in mind the requirements of students. Therefore, students can use these solutions for revision or to score high marks in the final examinations. Let’s get started with the solutions of Chapter 10: Wave Optics.
NCERT Solutions for Class 12 Physics Chapter 10
Wave Optics Solutions
Q 10.1) Monochromatic light of wavelength 589 nm is incident from air on a water surface.
What are the wavelength, frequency and speed of (a) reflected, and (b) refracted light? Refractive index of water is 1.33.
Wavelength of incident monochromatic light,
λ = 589 nm = 589 × 10−9 m
Speed of light in air, c = 3 × 108 m/s
Refractive index of water, μ = 1.33
The ray will reflect back in the same medium as that of incident ray. Hence, the wavelength, speed, and frequency of the reflected ray will be the same as that of the incident ray.
Frequency of light is given by the relation,
Hence, the speed, frequency, and wavelength of the reflected light are 3 × 108 m/s, 5.09 ×1014 Hz, and 589 nm respectively.
(b) Frequency of light does not depend on the property of the medium in which it is travelling. Hence, the frequency of the refracted ray in water will be equal to the frequency of the incident or reflected light in air.
Refracted frequency, ν = 5.09 ×1014 Hz
Speed of light in water is related to the refractive index of water as:
Wavelength of light in water is given by the relation,
Hence, the speed, frequency, and wavelength of refracted light are 2.26 ×108 m/s, 444.01nm, and 5.09 × 1014 Hz respectively.
Q 10.2) What is the shape of the wavefront in each of the following cases:
(a) Light diverging from a point source.
(b) Light emerging out of a convex lens when a point source is placed at its focus.
(c) The portion of the wavefront of light from a distant star intercepted by the Earth.
(a) The shape of the wavefront in case of a light diverging from a point source is spherical. The wavefront emanating from a point source is shown in the given figure.
(b) The shape of the wavefront in case of a light emerging out of a convex lens when a point source is placed at its focus is a parallel grid. This is shown in the given figure.
(c) The portion of the wavefront of light from a distant star intercepted by the Earth is a plane.
(a) The refractive index of glass is 1.5. What is the speed of light in glass? Speed of light in vacuum is 3.0 × 108 m s−1)
(b) Is the speed of light in glass independent of the colour of light? If not, which of the two colours red and violet travels slower in a glass prism?
Refractive index of glass, μ = 1.5
Speed of light, c = 3 × 108 m/s
Speed of light in glass is given by the relation,
Hence, the speed of light in glass is 2 × 108 m/s.
(b) The speed of light in glass is not independent of the colour of light. The refractive index of a violet component of white light is greater than the refractive index of a red component. Hence, the speed of violet light is less than the speed of red light in glass. Hence, violet light travels slower than red light in a glass prism.
Q 10.4) In a Young’s double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away.
The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment.
Distance between the slits, d = 0.28 mm = 0.28 × 10−3 m
Distance between the slits and the screen, D = 1.4 m
Distance between the central fringe and the fourth (n = 4) fringe,
u = 1.2 cm = 1.2 × 10−2 m
In case of a constructive interference, we have the relation for the distance between the two fringes as:
n = Order of fringes = 4
λ = Wavelength of light used
Hence, the wavelength of the light is 600 nm.
Q 10.5) In Young’s double-slit experiment using monochromatic light of wavelengthλ, the intensity of light at a point on the screen where path difference is λ, is K units.
What is the intensity of light at a point where path difference is λ /3?
Answer) Let I1 and I2 be the intensity of the two light waves. Their resultant intensities can be obtained as:
ϕ = Phase difference between the two waves
For monochromatic light waves,
Phase difference = path difference
Since path difference = λ,
Phase difference, ϕ = 2π
I’ = K
When path difference
Hence, resultant intensity,
Using equation (1), we can write:
Hence, the intensity of light at a point where the path difference is unit
Q 10.6) A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes in a Young’s double-slit experiment.
(a) Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650 nm.
(b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide?
Wavelength of the light beam, λ1 = 650 nm
Wavelength of another light beam, λ2 = 520 nm
Distance of the slits from the screen = D
Distance between the two slits = d
(a) Distance of the nth bright fringe on the screen from the central maximum is given by the relation,
For third bright fringe, n=3
(b) Let, the nth bright fringe due to wavelength λ2 and (n – 1)th bright fringe due to wavelength λ2 coincide on the screen. The value of n can be obtained by equating the conditions for bright fringes:
Therefore n = 5
Hence, the least distance from the central maximum can be obtained by the relation:
Note: The value of d and D are not given in the question.
Q 10.7) In a double-slit experiment, 0.2° is found to be the angular width of a fringe on a screen placed 1 m away.
The wavelength of light used is 600 nm. What will be the angular width of the fringe if the entire experimental apparatus is immersed in water? Take refractive index of water to be 4/3
Distance of the screen from the slits, D = 1 m
Wavelength of light used, λ1 = 600 nm
Angular width of the fringe in air θ1 = 0.2°
Angular width of the fringe in water= θ2
Refractive index of water,
is the relation between the refractive index and the angular width
Therefore, 0.15° is the reduction in the angular width of the fringe in water.
Q 10.8) What is the Brewster angle for air to glass transition? (Refractive index of glass=1.5.)
Refractive index of glass, μ=1.5
Consider Brewster angle = θ
Following is the relation between the Brewster angle and the refractive index:
Therefore, the Brewster angle for air to glass transition is 56.31°
Q 10.9) Light of wavelength 5000 Armstrong falls on a plane reflecting surface.
What are the wavelength and frequency of the reflected light? For what angle of incidence is the reflected ray normal to the incident ray?
Wavelength of incident light, [λ] = 5000 Armstrong = 5000 x 10-10 m
Speed of light, c =3 x 108 m
Following is the relation for the frequency of incident light:
The wavelength and frequency of incident light is equal to the reflected ray. Therefore, 5000 Armstrong and
6 × 1014 Hz is the wavelength and frequency of the reflected light. When reflected ray is normal to incident ray, the sum of the angle of incidence, ∠i and angle of reflection, ∠r is 90°.
From laws of reflection we know that the angle of incidence is always equal to the angle of reflection
Therefore, 45° is the angle of incidence.
Q 10.10) Estimate the distance for which ray optics is a good approximation for an aperture of 4 mm and wavelength 400 nm.
Fresnel’s distance (ZF) is the distance which is used in ray optics for a good approximation. Following is the relation,
Aperture width, a = 4 mm = 4 × 10-3 m
Wavelength of light, λ = 400 nm = 400 × 10-9 m
Therefore, 40m is the distance for which the ray optics is a good approximation.
Q 10.11) The 6563 Å Hα line emitted by hydrogen in a star is found to be red-shifted by 15 Å.
Estimate the speed with which the star is receding from the Earth.
λ = 6563 Å
Δλ = 15 Å
Since the star is receding, the velocity (v) is negative.
Δλ = – vλ/c
v = – cΔλ/λ
= – (3 x 108) x (15 Å/ 6563 Å)
= – 6.86 x 105 m/s
Q 10.12) Explain how Corpuscular theory predicts the speed of light in a medium, say, water, to be greater than the speed of light in vacuum.
Is the prediction confirmed by experimental determination of the speed of light in water? If not, which alternative picture of light is consistent with experiment?
According to Newton’s Corpuscular theory, velocity of light in the denser medium (water) is greater than the velocity of light in the rarer medium (vacuum). This was experimentally wrong.
At the angle of incidence (i) of the light of velocity v, the angle of refraction is r.
Due to the change in the medium, the change in the velocity of light in water is v
Using Snells law,
c sin i = v sin r ——(1)
The relation between the velocities and the refractive index is
v/c = μ ———(2)
v/c = sin i/sin r = μ ——-(3)
But μ> 1 so v > c is not possible
Huygens wave theory is consistent with the experiment.
Q 10.13) You have learnt in the text how Huygens’ principle leads to the laws of reflection and refraction.
Use the same principle to deduce directly that a point object placed in front of a plane mirror produces a virtual image whose distance from the mirror is equal to the object distance from the mirror.
A circle is drawn from the centre (O) such that it just touches the plane mirror at point O’. According to Huygens’ Principle, XY is the wavefront of incident light.
If the mirror is absent, then a similar wavefront X’Y’ (as XY) would form behind O’ at distance r (as shown in the given figure).
X’ Y’ can be considered as a virtual reflected ray for the plane mirror. Hence, a point object placed in front of the plane mirror produces a virtual image whose distance from the mirror is equal to the object distance (r).
Q 10.14) Let us list some of the factors, which could possibly influence the speed of wave propagation:
(i) nature of the source.
(ii) the direction of propagation.
(iii) the motion of the source and/or observer.
(v) the intensity of the wave. On which of these factors, if any, does (a) the speed of light in a vacuum,
(b) the speed of light in a medium (say, glass or water), depend?
(a) The speed of light in the vacuum does not depend on any of the factors listed.
(b) The speed of light in the medium depends on the wavelength of the light in that medium
Q 10.15) For sound waves, the Doppler formula for frequency shift differs slightly between the two situations:
(i) source at rest; observer moving, and (ii) source moving; observer at rest. The exact Doppler formulas for the case of light waves in vacuum are, however, strictly identical for these situations. Explain why this should be so. Would you expect the formulas to be strictly identical for the two situations in the case of light travelling in a medium?
Answer) The Doppler formula differs slightly between the two situations because the sound waves can travel only through the medium. The motion of the observer relative to the medium is different in both cases. Hence, the doppler formula is different.
Light waves can propagate in vacuum. In the vacuum, the speed of light does not depend on the motion of the observer and the source.
When light travels in the medium, the doppler formula for the two cases will be different.
Q 10.16) In a double-slit experiment using the light of wavelength 600 nm, the angular width of a fringe formed on a distant screen is 0.1°. What is the spacing between the two slits?
Wavelength of the light, λ = 600 nm
The angular width of the fringe formed, θ= 0.10 = 0.1 π/180
Spacing between the slits, d = λ/θ = (600 x 10-9 x 180)/(0.1 x 3.14)
= 108000 x 10-9/0.314
d =3.44 x 10-4 m
Q 10.17) Answer the following questions:
(a) In a single slit diffraction experiment, the width of the slit is made double the original width. How does this affect the size and intensity of the central diffraction band?
(b) In what way is diffraction from each slit related to the interference pattern in a double-slit experiment?
(c) When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the centre of the shadow of the obstacle. Explain why?
(d) Two students are separated by a 7 m partition wall in a room 10 m high. If both light and sound waves can bend around obstacles, how is it that the students are unable to see each other even though they can converse easily?
(e) Ray optics is based on the assumption that light travels in a straight line. Diffraction effects (observed when light propagates through small apertures/slits or around small obstacles) disprove this assumption. Yet the ray optics assumption is so commonly used in an understanding location and several other properties of images in optical instruments. What is the justification?
(a) In a single slit diffraction experiment, if the width of the slit is made double the original width, the size of the central diffraction band reduces to half and the intensity of the band increases four times.
(b) If the width of each slit is comparable to the wavelength of light used, the interference pattern thus obtained in the double-slit experiment is modified by diffraction from each of the two slits.
(c) When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the centre of the shadow of the obstacle. The bright spot is formed due to the constructive interference of the light waves that gets diffracted from the edges of the circular obstacle.
(d) The obstacle bends the waves by a large angle if the wavelength of the wave is comparable with the size of the obstacle. The wavelength of the light wave is much smaller than the size of the wall. Therefore, the diffraction angle is also very small. As a result, the students will not be able to see each other. On the other hand, the size of the wall and the wavelength of the sound wave is comparable. Hence, the diffraction angle is large. Therefore, students can hear each other.
(e) The size of the aperture in the optical instruments are much larger than the wavelength of light. Therefore, the diffraction effect of light is negligible in these instruments. Thus the assumption of light travelling in a straight line can be used in these instruments.
Q 10.18) Two towers on top of two hills are 40 km apart.
The line joining them passes 50 m above a hill halfway between the towers. What is the longest wavelength of radio waves, which can be sent between the towers without appreciable diffraction effects?
Distance between the towers = 40 km
Height of the line joining the hills, d = 50 m
The radial spread of the radio waves must not exceed 50 m
Aperture a = d = 50 m
The hill is located halfway between the towers. Therefore, Fresnel’s distance is Zp = 20 km
Fresnel’s distance can be given by the equation, Zp = a2/λ
λ = a2/Zp
= (50)2/(20 x 103)
= 250/20 = 12.5 x 10-3 m = 12.5 cm
Q 10.19) A parallel beam of light of wavelength 500 nm falls on a narrow slit and the resulting diffraction pattern is observed on a screen 1 m away.
It is observed that the first minimum is at a distance of 2.5 mm from the centre of the screen. Find the width of the slit.
Wavelength of the beam of light, λ = 500 nm
Distance between the slit and the screen, D= 1 m
Distance of the first minimum from the centre of the screen, x = 2.5 mm = 2.5 x 10-3 m
First minima, n = 1
Consider the equation, nλ = xd/D
⇒ d = nλD/x = ( 1 x 500 x 10-9 x 1) /( 2.5 x 10-3 )
= 200 x 10-6 m
Q 10.20) Answer the following questions:
(a) When a low flying aircraft passes overhead, we sometimes notice a slight shaking of the picture on our TV screen. Suggest a possible explanation.
(b) As you have learnt in the text, the principle of linear superposition of wave displacement is basic to understanding intensity distributions in diffraction and interference patterns. What is the justification of this principle?
(a) The weak radar signals from the aircraft interfere with the TV signal received by the antenna.
(b) This is because superposition follows from the linear character of a differential equation that governs wave motion.
Q 10.21) In deriving the single slit diffraction pattern, it was stated that the intensity is zero at angles of nλ/a.
Justify this by suitably dividing the slit to bring out the cancellation.
Let “a” be the width of a single slit. The single slit is further divided into n smaller slits of width a’.
a’ = a/n
Each of the smaller slits should produce zero intensity for the single slit to produce zero intensity.
For this to happen, the angle of diffraction, θ = λ/a’
⇒ θ = λ/(a/n)
or, θ = nλ/a
Therefore, in deriving the single slit diffraction pattern, it was stated that the intensity is zero at angles of nλ/a.
That’s it. These were the solutions of NCERT Class 12 Physics Chapter 10 – Wave Optics. Our team hopes that you have found these solutions helpful for you. If you have any doubt related to this chapter then feel free to comment your doubts below. Our team will try their best to help you with your doubts.