CBSE Class 12 Physics Chapter 10: Wave Optics NCERT Solutions

Given:

Incident light wavelength, \lambda = 589 \text{ nm} = 589 \times 10^{-9} \text{ m}

Speed of light in air, c = 3 \times 10^8 \text{ m/s}

Refractive index of water, \mu = 1.33

(a) Reflected Light:

When light reflects off a surface, it bounces back into the same medium. Therefore, the properties of the reflected light are identical to those of the incident light.

The frequency of light is determined by the source and does not change with reflection. The frequency (\nu) is calculated as:

\nu = \frac{c}{\lambda} = \frac{3 \times 10^8 \text{ m/s}}{589 \times 10^{-9} \text{ m}} \approx 5.09 \times 10^{14} \text{ Hz}

Thus, for the reflected light:

• Speed = Speed of light in air = $3\; \backslash times\; 10^8\; \backslash text\{\; m/s\}$
• Frequency = Frequency of incident light = $5.09\; \backslash times\; 10^\{14\}\; \backslash text\{\; Hz\}$
• Wavelength = Wavelength of incident light = $589\; \backslash text\{\; nm\}$

(b) Refracted Light:

When light refracts, it enters a different medium. The frequency of the light remains unchanged because it is determined by the source.

Frequency of refracted light = $\backslash nu\; =\; 5.09\; \backslash times\; 10^\{14\}\; \backslash text\{\; Hz\}$

The speed of light in water ($v$) is related to the speed of light in air ($c$) and the refractive index of water ($\backslash mu$) by the formula $v\; =\; \backslash frac\{c\}\{\backslash mu\}$.

$v\; =\; \backslash frac\{3\; \backslash times\; 10^8\; \backslash text\{\; m/s\}\}\{1.33\}\; \backslash approx\; 2.26\; \backslash times\; 10^8\; \backslash text\{\; m/s\}$

The wavelength of light in water ($\backslash lambda\text{'}$) can be calculated using the formula $v\; =\; \backslash nu\; \backslash lambda\text{'}$.

$\backslash lambda\text{'}\; =\; \backslash frac\{v\}\{\backslash nu\}\; =\; \backslash frac\{2.26\; \backslash times\; 10^8\; \backslash text\{\; m/s\}\}\{5.09\; \backslash times\; 10^\{14\}\; \backslash text\{\; Hz\}\}\; \backslash approx\; 4.44\; \backslash times\; 10^\{-7\}\; \backslash text\{\; m\}\; =\; 444\; \backslash text\{\; nm\}$

Therefore, for the refracted light:

• Speed = $2.26\; \backslash times\; 10^8\; \backslash text\{\; m/s\}$
• Frequency = $5.09\; \backslash times\; 10^\{14\}\; \backslash text\{\; Hz\}$
• Wavelength = $444\; \backslash text\{\; nm\}$

The shape of a wavefront depends on the nature and position of the source of light.

1. Light diverging from a point source:

When light originates from a point source, it spreads out uniformly in all directions. At any given instant, all points at the same distance from the source will have the same phase. Therefore, the wavefront emanating from a point source is spherical. The center of the sphere is the point source itself.

2. Light emerging out of a convex lens when a point source is placed at its focus:

According to the properties of a convex lens, when a point source of light is placed at its principal focus, the rays emerging from the lens become parallel to the principal axis. A collection of parallel rays constitutes a plane wavefront. Thus, the wavefront emerging from the convex lens in this case is a plane wavefront.

3. The portion of the wavefront of light from a distant star intercepted by the Earth:

A star is an extremely distant source of light. Light rays coming from such a distant source are practically parallel by the time they reach Earth. Even though the light originates from a point source (the star), due to the immense distance, the curvature of the spherical wavefront becomes negligible over the small area intercepted by the Earth. Hence, the portion of the wavefront from a distant star that is intercepted by the Earth can be approximated as a plane wavefront.