Question

Monochromatic light having a wavelength of 589nm from the air is incident on a water surface. Find the frequency, wavelength and speed of
(i) reflected and
(ii) refracted light?      [1.33 is the Refractive index of water]

Answer 1

Monochromatic light incident having wavelength,\( \lambda = 589 nm = 589 \times 10^{-9}\ m \)

\( Speed\ of\ light\ in\ air,\ c = 3 \times 10^8\ m/s \)

Refractive index of water, \( \mu = 1.33 \)

(i) In the same medium through which incident ray passed the ray will be reflected back.

Therefore the wavelength, speed, and frequency of the reflected ray will be the same as that of the incident ray.

Frequency of light can be found from the relation:

\( v=\frac{c}{\lambda} = \frac{3\times10^{8}}{589\times10^{-9}} = 5.09 × 10^{14} Hz \)

Hence, \( c = 3 \times 10^8\ m/s \), \( 5.09 × 10^{14} Hz \), and \( 589\ nm \) are the speed, frequency, and wavelength of the reflected light.

(b) The frequency of light which is travelling never depends upon the property of the medium. Therefore, the frequency of the refracted ray in water will be equal to the frequency of the incident or reflected light in air.

Refracted frequency, \( v = 5.09 \times 10^{14} Hz \)

Following is the relation between the speed of light in water and the refractive index of the water:

\( v=\frac{c}{\mu} = \frac{3\times10^{8}}{1.33} = 2.26 × 10^8\ m/s \)

Below is the relation for finding the wavelength of light in water:

\( \lambda=\frac{v}{V} = \frac{2.26\times10^{8}}{5.09\times10^{14}} ​= 444.007 × 10^{-9} m = 444.01nm \)

Therefore, \( 444.007 × 10^{-9} m \), \( 444.01\ nm \), and \( 5.09 × 10^{14} Hz \) are the speed, frequency, and the wavelength of the refracted light.

Hence, \( c = 3 \times 10^8\ m/s \), \( 5.09 × 10^{14} Hz \), and \( 589\ nm \) are the speed, frequency, and wavelength of the reflected light.
Therefore, \( 444.007 × 10^{-9} m \), \( 444.01\ nm \), and \( 5.09 × 10^{14} Hz \) are the speed, frequency, and the wavelength of the refracted light.