Question

A monochromatic light of frequency \( 5 \times 10^{14} \) Hz travelling through air, is incident on a medium of refractive index '2'. Wavelength of the refracted light will be :

• A. 300 nm
• B. 600 nm
• C. 400 nm
• D. 500 nm

Hint:

When light travels from one medium to another, its frequency remains constant, but its speed and wavelength change. The relationship between the wavelength in a medium (\( \lambda_{medium} \)), the wavelength in vacuum (\( \lambda_{vacuum} \)), and the refractive index (\( \mu \)) of the medium is \( \lambda_{medium} = \frac{\lambda_{vacuum}}{\mu} \). First, find the wavelength in air (approximated as vacuum) using the given frequency and the speed of light in vacuum. Then, use the refractive index to find the wavelength in the medium.

Answer 1

The Correct Option is A

The frequency of light remains unchanged when it passes from one medium to another. 
Given frequency \( f = 5 \times 10^{14} \) Hz. 
The speed of light in air (approximately vacuum) is \( c = 3 \times 10^8 \) m/s. 
The wavelength of light in air (\( \lambda_{air} \)) is given by: \[ \lambda_{air} = \frac{c}{f} = \frac{3 \times 10^8 \, \text{m/s}}{5 \times 10^{14} \, \text{Hz}} = 0.6 \times 10^{-6} \, \text{m} = 600 \times 10^{-9} \, \text{m} = 600 \, \text{nm} \] The refractive index of the medium is given as \( \mu = 2 \). The wavelength of light in the medium (\( \lambda_{medium} \)) is related to the wavelength in vacuum (or air) by the refractive index: \[ \lambda_{medium} = \frac{\lambda_{air}}{\mu} \] Substituting the values: \[ \lambda_{medium} = \frac{600 \, \text{nm}}{2} = 300 \, \text{nm} \] The wavelength of the refracted light in the medium is 300 nm.