NCERT Solutions Wave Optics

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Wave Optics Solutions

Wave Optics: Solutions and Explanations

Question 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.

Solution:

Concept: The speed of light in a medium is given by \( v = \frac{c}{n} \) where \( c \) is the speed of light in vacuum and \( n \) is the refractive index of the medium. The frequency of light remains unchanged during reflection and refraction, but the wavelength changes in the refractive medium.

(a) Reflected Light:

The wavelength and speed of the reflected light will be the same as the incident light since it remains in the same medium (air).

  • Wavelength: 589 nm
  • Speed: \( c = 3 \times 10^8 \) m/s
  • Frequency: \( f = \frac{c}{\lambda} = \frac{3 \times 10^8 \, \text{m/s}}{589 \times 10^{-9} \, \text{m}} = 5.09 \times 10^{14} \, \text{Hz} \)

(b) Refracted Light:

The speed of light in water: \( v = \frac{c}{n} = \frac{3 \times 10^8 \, \text{m/s}}{1.33} = 2.26 \times 10^8 \, \text{m/s} \)

The wavelength in water: \( \lambda' = \frac{\lambda}{n} = \frac{589 \, \text{nm}}{1.33} = 443 \, \text{nm} \)

  • Wavelength: 443 nm
  • Speed: \( 2.26 \times 10^8 \, \text{m/s} \)
  • Frequency: \( 5.09 \times 10^{14} \, \text{Hz} \) (unchanged)

Question 10.2

What is the shape of the wavefront in each of the following cases:

  1. Light diverging from a point source.
  2. Light emerging out of a convex lens when a point source is placed at its focus.
  3. The portion of the wavefront of light from a distant star intercepted by the Earth.

Solution:

Concept: A wavefront is a surface over which an optical wave has a constant phase. The shape of the wavefront depends on the source and the medium.
  1. For light diverging from a point source, the wavefronts are spherical.
  2. For light emerging out of a convex lens with a point source at its focus, the wavefronts are plane.
  3. For light from a distant star intercepted by the Earth, the wavefronts are also plane due to the large distance.

Question 10.3

(a) The refractive index of glass is 1.5. What is the speed of light in glass? (Speed of light in vacuum is \( 3.0 \times 10^8 \) 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?

Solution:

Concept: The speed of light in a medium is given by \( v = \frac{c}{n} \). The refractive index varies with wavelength (dispersion), so different colours of light travel at different speeds in the same medium.

(a) Speed of light in glass:

\( v = \frac{c}{n} = \frac{3 \times 10^8 \, \text{m/s}}{1.5} = 2 \times 10^8 \, \text{m/s} \)

(b) The speed of light in glass is not independent of the colour of light. Violet light travels slower than red light in a glass prism because the refractive index for violet is higher than for red.

Question 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.

Solution:

Concept: In Young's double-slit experiment, the fringe spacing (distance between adjacent bright or dark fringes) is given by \( \beta = \frac{\lambda D}{d} \) where \( \lambda \) is the wavelength of light, \( D \) is the distance between the slits and the screen, and \( d \) is the separation between the slits.

The distance between the central bright fringe and the fourth bright fringe is \( 4\beta \). Thus,

\( 4\beta = 1.2 \, \text{cm} = 1.2 \times 10^{-2} \, \text{m} \)

\( \beta = \frac{1.2 \times 10^{-2} \, \text{m}}{4} = 3 \times 10^{-3} \, \text{m} \)

Using the fringe spacing formula:

\( \beta = \frac{\lambda D}{d} \Rightarrow \lambda = \frac{\beta d}{D} = \frac{3 \times 10^{-3} \, \text{m} \times 0.28 \times 10^{-3} \, \text{m}}{1.4 \, \text{m}} = 6 \times 10^{-7} \, \text{m} = 600 \, \text{nm} \)

Question 10.5

In Young’s double-slit experiment using monochromatic light of wavelength \( \lambda \), the intensity of light at a point on the screen where path difference is \( \lambda \) is K units. What is the intensity of light at a point where path difference is \( \lambda/3 \)?

Solution:

Concept: The intensity in an interference pattern depends on the path difference \( \Delta x \) between the two waves and is given by \( I = I_0 \cos^2 \left( \frac{\pi \Delta x}{\lambda} \right) \).

The intensity at path difference \( \Delta x = \lambda \) is given as K units. Thus,

\( K = I_0 \cos^2 \left( \frac{\pi \lambda}{\lambda} \right) = I_0 \cos^2 (\pi) = I_0 \cdot 0 = 0 \, \text{units} \)

For path difference \( \Delta x = \lambda/3 \):

\( I = I_0 \cos^2 \left( \frac{\pi \lambda/3}{\lambda} \right) = I_0 \cos^2 \left( \frac{\pi}{3} \right) = I_0 \left( \frac{1}{2} \right)^2 = \frac{I_0}{4} \)

Since \( K = I_0 \), the intensity at path difference \( \lambda/3 \) is \( \frac{K}{4} \).

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