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NCERT Class 12 Physics Chapter 10 Wave optics
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Wave optics
Chapter: 10
| Part – II |
EXERCISE
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.
Ans: (a) Given:
λair = 589 nm
c = 3 × 108 m/s
n = 1.33
Here:
𝑓reflected = 𝑓incident = c/ λair
= 3 × 108 m/s / 589 × 10-9 m
= 5.09 × 1014 Hz.
(b) Refracted light:
𝑓reflected = 𝑓incident = 5.09 × 1014 Hz.
vrefracted = c/n = (3 × 108 m/s) / 1.33
= 2.26 × 108 m/s
Wavelength:
λrefracted = vrefracted/𝑓refracted
= (2.26 × 108 m/s) / 5.09 × 1014 Hz
= 444 nm.
2. What is the shape of the wavefront in each of the following cases:
(a) Light diverging from a point source.
Ans: Spherical wavefront with point source as the centre of the sphere.
(b) Light emerging out of a convex lens when a point source is placed at its focus.
Ans: Plane wavefront.
(c) The portion of the wavefront of light from a distant star intercepted by the Earth.
Ans: Plane wavefront.
3. (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)
Ans: Given:
c = 3.0 × 108 m/s
n = 1.5
Here:
v = c/n
= (3.0 × 108 m/s) / 1.5
= 2.0 × 108 m/s.
(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: The speed of light in glass is not independent of the color of light. Since the refractive index of glass is higher for violet light than for red light, the speed of violet light in glass is less than that of red light. Therefore, violet light travels slower in glass than red light.
4. In 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.
Ans: Given:
y4 = 1.2 cm = 1.2 × 10-2 m
D = 1.4 m
d = 0.20 × 10-3 m
Here:
λ = yn × d / n × D
= (1.2 × 10-2 m) × (0.28 × 10-3 m) / 4 × 1.4 m
= 3.36 × 10-6 m2 / 5.6 m
= 6.0 × 10-7 m.
So, λ = 600 nm.
5. In Young’s double-slit experiment using monochromatic light of wavelength, 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: In young’s double – slit experiment two slits are identical and intensity of light coming out of them is exactly same,i.e., Ⅰ1 = Ⅰ2 = Ⅰ.
At a point on the screen where path is different is:
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.
Ans: Given:
λ1 = 650 nm = 6.5 × 10-7 m
D = 60 cm = 0.6 m
d = 1 × 10-3 m
y3 = 3 × (λ × D) / d
= 3 × 3.9 × 10-7 m2 / (1 × 10-3 m)
= 1.17 × 10-3 m
= 1.17 mm.
(b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide?
Ans: nλ1 = mλ2
λ1 = 650 nm = 650 × 10-9 m
λ2 = 520nm = 520 × 10-9 m
n/m = λ2/λ1 = 520 / 650 = 4/5.
𝒙 = 4 × 650 × 10-9 × 0.6 / (1 × 10-3)
= 1.56 × 10-3 m
1.56 mm.
