
Q1 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. Ans: 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
(a) 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,
v = c / λ
= 3 x 10^{8} / 589 x 10^{9}
= 5 .09 x 10^{14} Hz
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.
So Refracted frequency, ν = 5.09 ×1014 Hz
Speed of light in water is related to the refractive index of water as:
v = c / μ
v = 3 x 10^{8 }/ 1.33 = 2.26 x 10^{8 }m/s
Wavelength of light in water is given by the relation,
λ = v / v
v = 2.26 x 10^{8}^{ }/ 5 .09 x 10^{14} = 444.007 x 10^{9} m
= 444.01 nm
Hence, the speed, frequency, and wavelength of refracted light are 2.26 ×10^{8} m/s, 444.01nm, and 5.09 × 10^{14} Hz respectively.
Q2 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. Ans: (a) Light diverging from a point source.
The geometrical shape of the wavefront in case of light diverging from a point source will be diverging spherical wavefront . It can be shown as,
(b) Light emerging out of a convex lens when a point source is placed at its focus.
When a point source is placed at the focus of a convex lens, the rays emerging from the lens are parallel. Therefore, the wavefront must be plane, as shown in figure,
(c) The portion of the wavefront of light from a distant star intercepted by the Earth.
The source of light (i.e. star) is very far off (i.e. at infinity), the wavefront intercepted by the earth will be a plane wavefront, as shown above.
Q3 (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 × 10 8 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? Ans: (a) It is given that,
Refractive index of glass, μ = 1.5
We know, Speed of light, c = 3 × 10^{ 8} m/s
So, speed of light in glass can be calculated by the relation, v= c/µ = 3x10^{8} / 1.5 = 2x10^{8 }
Hence, the speed of light in glass is 2 × 10^{ 8} m/s.
(b) No, the refractive index and the speed of light in a medium depend on wavelength i.e. colour of light.
We know that μ _{v} > μ_{ r} . Therefore, v _{voilet} < v _{red} . Hence voilet component of white light travels slower than the red component in a glass prism.
Q4 In a Young’s doubleslit 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. Ans: Here it is given that,
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
For constructive interference, the distance between the two fringes is given by relation: u = nλ D/d
where, n = Order of fringes
wavelength of the light can be given as: λ = ud/nD = 1.2x10^{2}x0.28x10^{3}/4x1.4 = 6x10^{7} = 600 nm
Hence, the wavelength of the light is 6 x 10^{ 7} m.
Q5 In Young’s doubleslit experiment using monochromatic light of wavelength λ, 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? Ans: Let I_{1} and I_{2} be the intensity of the two light waves.
For monochromatic light waves, I_{1} = I_{2}
If, Ø = Phase difference between the two waves
Then, their resultant intensities can be obtained by the relation:
Phase difference of the light is given by: 2¶/λ x Path Difference
When path difference = λ,
Phase difference, Ø = 2¶
∴ I' = 2I_{1} +2I_{1} = 4I_{1}
Given, I’ = K
∴ I_{1} = K/4
When path difference = λ/3
Phase difference, ø = 2¶/3
Hence, resultant intensity is given by,
Using equation (1), we can write:
I_{R} = I_{1} = K/4
Hence, the intensity of light with path difference of λ/3 is K/4 units.
Q6 A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes in a Young’s doubleslit 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? Ans: Given that,
Wavelength of the light beam, λ_{1} = 650 nm
Wavelength of another light beam, λ_{2} = 520 nm
Distance of the slits from the screen= D
Distance between the two slits = d
(a) distance of the third bright fringe on the screen from the central maximum
Distance of the n th bright fringe on the screen from the central maximum is given by the relation, x = nλ_{1} (D/d)
For third bright fringe, n = 3
∴ x = 3 x 650 D/d = 1950 (D/d) nm
(b) Least distance from the central maximum
Let the n th bright fringe due to wavelength, λ_{2} and (n − 1)^{th} bright fringe due to wavelength coincide on the screen. We can equate the conditions for bright fringes as:
nλ_{2} = (n  1) λ_{1}
520n = 650n  650
650 = 130n
∴ n = 5
Hence, the least distance from the central maximum can be obtained by the relation:
x = nλ_{2} D/d = 5 x 520 D/d = 2600 D/d nm
Q7 In a doubleslit experiment the angular width of a fringe is found to be 0.2° 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. Ans: Given that,
Distance of the screen from the slits, D = 1 m
Wavelength of light used, λ_{1} =600 nm
Angular width of the fringe in air, θ_{1} = 0.2°
Angular width of the fringe in water = θ,
Refractive index of water, µ = 4/3
Refractive index is related to angular width as: µ = θ_{1}/ θ_{2}
θ_{2} = 3/4 θ_{1} = 3/4 x 0.2 = 0.15
Therefore, the angular width of the fringe in water will reduce to 0.15°.
Q8 What is the Brewster angle for air to glass transition? (Refractive index of glass = 1.5.) Ans: Given that,
Refractive index of glass, µ = 1.5
Brewster angle = θ
Brewster angle is related to refractive index as: tanθ = µ, θ = tan^{1}(1.5) = 56.31°
Therefore, the Brewster angle for air to glass transition is 56.31°.
Q9 Light of wavelength 5000 Å 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? Ans: It is given that,
Wavelength of incident light, λ = 5000 Å = 5000 × 10 ^{10} m
We know that, Speed of light, c = 3 × 10^{ 8} m
Frequency of incident light is given by the relation, v = c/λ = 3x10^{8}/5000x10^{10} = 6x10^{14} Hz
The wavelength and frequency of incident light is the same as that of reflected ray. Hence, the wavelength of reflected light is 5000 Å and its frequency is 6 × 10^{ 14 }Hz. When reflected ray is normal to incident ray, the sum of the angle of incidence, ang. i and angle of reflection, ang. r is 90°.
Q10 Estimate the distance for which ray optics is good approximation for an aperture of 4 mm and wavelength 400 nm. Ans: Fresnel’s distance (Z_{f} ) is the distance for which the ray optics is a good approximation. It is given by the relation, Z_{f} = a^{2}/λ
Where,
Aperture width, a = 4 mm = 4 ×10 ^{3} m
Wavelength of light, λ = 400 nm = 400 × 10_{ }^{9} m
Z_{f} = (4x10^{3})^{2}/400x10^{9} = 40 m
Therefore, the distance for which the ray optics is a good approximation is 40 m.
Q11 The 6563 Å H2 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. Ans: Given that,
Wavelength of H_{2} line emitted by hydrogen, λ = 6563 Å = 6563 × 10^{10} m.
Star’s redshift can be obtain by, (λ'  λ) = 15 Å = 15 x 10^{10} m
We know that, Speed of light, c = 3 x 10^{8} m/s
Let the velocity of the star receding away from the Earth be v. Since the star is receding away, hence its velocity v is negative.
The red shift is related with velocity as: λ'  λ = v/c λ
∴ v = c/λ x (λ'  λ) = 3 x 10^{8} x 15 x 10^{10} / 6563 x 10^{10} = 6.87 x 10^{5} m/s
Therefore, the speed with which the star is receding away from the Earth is 6.87 × 10^{5} m/s.
Q12 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? Ans: No; Wave theory
Newton’s corpuscular theory of light states that when light corpuscles strike the interface of two media from a rarer (air) to a denser (water) medium, the particles experience forces of attraction normal to the surface. Hence, the normal component of velocity increases while the component along the surface remains unchanged.
Hence, we can write the expression: c sin i = v sin r ... (i)
Where,
i = Angle of incidence
r = Angle of reflection
c = Velocity of light in air
v = Velocity of light in water
We have the relation for relative refractive index of water with respect to air as: µ = v/c
Hence, equation (i) reduces to v/c = sin i/sin r = µ ... (ii)
But, μ > 1
Hence, it can be inferred from equation (ii) that v > c. This is not possible since this prediction is opposite to the experimental results of c > v. The wave picture of light is consistent with the experimental results.
Q13 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. Ans: Let an object at O be placed in front of a plane mirror MO’ at a distance r (as shown in the given figure).
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).
Q14 Let us list some of the factors, which could possibly influence the speed of wave propagation: (i) Nature of the source. (ii) Direction of propagation. (iii) Motion of the source and/or observer. (iv) Wave length. (v) Intensity of the wave. On which of these factors, if any, does (a) The speed of light in vacuum, (b) The speed of light in a medium (say, glass or water), depend? Ans: (a) Speed of light in vacuum is an absolute constant, according to Einstein’s theory of relativity. It does not depend upon any of the factors listed above or any other factor.
(b) The speed of light in a medium like water or glass
(i) does not depend upon the nature of the source.
(ii) does not depend upon the direction of propagation, when the medium is isotropic.
(iii) does not depend upon motion of the source w.r.t the medium, but depends on motion of the observer relative to the medium.
(iv) depends on wavelength of light, being lesser for shorter wavelength and viceversa.
(v) does not depend upon intensity of light.
Q15 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 case of light travelling in a medium? Ans: Sound waves require a material medium for propagation. That is why situation (i) and (ii) are not identical physically though relative motion between the source and the observer is the same in the two cases. Infact, relative motion of the observer relative to the medium is different in the two situations. That is why Doppler’s formulae for sound are different in the two cases.
For light waves travelling in vacuum, there is nothing to distinguish between the two situations. That is why the formulae are strictly identical.
For light propagating in a medium, situations (i) and (ii) are not identical. The formulae governing the two situations would obviously be different.
Q16 In doubleslit experiment using light of wavelength 600 nm, the angular width of a fringe formed on a distant screen is 0.1o. What is the spacing between the two slits? Ans: It is given that,
Wavelength of light used, λ = 6000 nm = 600 × 10 ^{9} m
Angular width of fringe, θ = 0.1° = 0.1 x ¶/180 = 3.14/1800 rad
Angular width of a fringe is related to slit spacing (d) as: d = λ/θ = 600 x 10^{9}/3.14/1800 = 3.44 x 10^{4} m
Therefore, the spacing between the slits is 3.44 x 10 4 m.
Q17 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 doubleslit 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 understanding location and several other properties of images in optical instruments. What is the justification? Ans: (a) When width (a) of single slit is made double, the half angular width of central maximum which is λ/a, reduces to half. The intensity of central maximum will become 4 times. This is because area of central diffraction band would become 1/4th.
(b) If width of each slit is of the order of λ, then interference pattern in the double slit experiment is modified by the diffraction pattern 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. This is because light waves are diffracted from the edge of the circular obstacle, which interferes constructively at the centre of the shadow. This constructive interference produces a bright spot.
(d) Bending of waves by obstacles by a large angle is possible when the size of the obstacle is comparable to the wavelength of the waves. On the one hand, the wavelength of the light waves is too small in comparison to the size of the obstacle. Thus, the diffraction angle will be very small. Hence, the students are unable to see each other. On the other hand, the size of the wall is comparable to the wavelength of the sound waves. Thus, the bending of the waves takes place at a large angle. Hence, the students are able to hear each other.
(e) The ray optics assumption is used in understanding location and several other properties of images in optical instruments. This is because typical sizes of aperture involved in ordinary optical instruments are much larger than the wavelength of light. Therefore, diffraction or bending of waves is of no significance.
Q18 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? Ans: Distance between the towers, d = 40 km
Height of the line joining the hills, d = 50 m.
Thus, the radial spread of the radio waves should not exceed 50 km.
Since the hill is located halfway between the towers, Fresnel’s distance can be obtained as: Z _{P} = 20 km = 2 × 10 ^{4} m
Aperture can be taken as, a = d = 50 m
Fresnel’s distance is given by the relation, Z_{p} = a^{2}/λ
Where, λ = Wavelength of radio waves
Therefore, λ = a^{2}/Z_{p} = (50)^{2} / 2x10^{4} = 1250x10^{4} = 0.1250 m = 12.5 cm
Therefore, the wavelength of the radio waves is 12.5 cm.
Q19 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. Ans: Wavelength of light beam, λ = 500 nm = 500 × 10 ^{9} m
Distance of the screen from the slit, D = 1 m
For first minima, n = 1
Distance between the slits = d
Distance of the first minimum from the centre of the screen can be obtained as:
x = 2.5 mm = 2.5 × 10 ^{3} m
It is related to the order of minima as:
d = nλD/x = 1x500x10^{9}x1 / 2.5x10^{3} = 2x10^{4} = 0.2 mm
Therefore, the width of the slit is 0.2 mm.
Q20 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? Ans: (a) A low flying aircraft reflects the T.V. signal. The slight shaking on the T.V. screen may be due to interference between the direct signal and the reflected signal.
(b) The principle of linear superposition of wave displacement is essential to our understanding of intensity distributions and interference patterns. This is because superposition follows from the linear character of a differential equation that governs wave motion. If y_{ 1} and y _{2} are the solutions of the second order wave equation, then any linear combination of y_{ 1 }and y _{2} will also be the solution of the wave equation.
Q21 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. Ans: Consider that a single slit of width d is divided into n smaller slits.
Therefore, Width of each slit, d' = d/n
Angle of diffraction can be calculated by,
θ = d/d' λ /d = λ/d'
Now, each of these infinitesimally small slit sends zero intensity in direction θ. Hence, the combination of these slits will give zero intensity.