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

Light's frequency is constant across different media. The given frequency is \( f = 5 \times 10^{14} \) Hz. The speed of light in air (approximating vacuum) is \( c = 3 \times 10^8 \) m/s. The wavelength of light in air (\( \lambda_{air} \)) is calculated as: \[ \lambda_{air} = \frac{c}{f} = \frac{3 \times 10^8 \, \text{m/s}}{5 \times 10^{14} \, \text{Hz}} = 0.6 \times 10^{-6} \, \text{m} = 600 \times 10^{-9} \, \text{m} = 600 \, \text{nm} \] With a refractive index of \( \mu = 2 \) for the medium, the wavelength of light in the medium (\( \lambda_{medium} \)) relates to the air wavelength by: \[ \lambda_{medium} = \frac{\lambda_{air}}{\mu} \] Substituting the values yields: \[ \lambda_{medium} = \frac{600 \, \text{nm}}{2} = 300 \, \text{nm} \] Consequently, the wavelength of the refracted light in the medium is 300 nm.