Question 81: Waves Optics

Question

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

Accepted Answer

To find the wavelength of the refracted light in a medium with a refractive index of 2, we can use the relationship between frequency, wavelength, and the speed of light. The frequency ($f$) of the light remains the same when it enters the medium.
The formula relating frequency, wavelength, and velocity is:
$ v = f \lambda $
Where:
$ v $ is the speed of light in the medium,
$ f $ is the frequency of the light,
$ \lambda $ is the wavelength of the light.
The wavelength of light in the medium ($\lambda_{	ext{medium}}$) can be found using the refractive index ($\mu$) and the wavelength in a vacuum ($\lambda_{	ext{vacuum}}$) as follows:
$ \lambda_{	ext{medium}} = \frac{\lambda_{	ext{vacuum}}}{\mu} $
Given:
Frequency of the light, $ f = 5 	imes 10^{14} \, 	ext{Hz} $
Speed of light in a vacuum, $ v_{	ext{vacuum}} = 3 	imes 10^{8} \, 	ext{m/s} $
Refractive index of the medium, $ \mu = 2 $
First, calculate $\lambda_{	ext{vacuum}}$:
$ \lambda_{	ext{vacuum}} = \frac{v_{	ext{vacuum}}}{f} = \frac{3 	imes 10^{8}}{5 	imes 10^{14}} $
Now calculate $\lambda_{	ext{medium}}$:
$ \lambda_{	ext{medium}} = \frac{\lambda_{	ext{vacuum}}}{2} = \frac{3 	imes 10^{8}}{2 	imes 5 	imes 10^{14}} $
$ \lambda_{	ext{medium}} = 0.3 	imes 10^{-6} \, 	ext{m} $
Converting to nanometers:
$ 0.3 	imes 10^{-6} \, 	ext{m} = 300 \, 	ext{nm} $
Therefore, the wavelength of the refracted light in the medium is
300 nm
.

Answer

400 nm

Answer

300 nm

Answer

600 nm

Answer

500 nm

Waves Optics

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

Choose the correct answer: