JEE Main 2025 — Wave Optics Question with Solution

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 (${\lambda }_{\text{medium}}$) can be found using the refractive index ($\mu$) and the wavelength in a vacuum (${\lambda }_{\text{vacuum}}$) as follows:

${\lambda }_{\text{medium}}=\frac{{\lambda }_{\text{vacuum}}}{\mu }$

Given:

Frequency of the light, $f=5\times 10^{14}\text{Hz}$

Speed of light in a vacuum, $v_{\text{vacuum}}=3\times 10^8\text{m/s}$

Refractive index of the medium, $\mu =2$

First, calculate ${\lambda }_{\text{vacuum}}$:

${\lambda }_{\text{vacuum}}=\frac{v_{\text{vacuum}}}{f}=\frac{3\times 10^8}{5\times 10^{14}}$

Now calculate ${\lambda }_{\text{medium}}$:

${\lambda }_{\text{medium}}=\frac{{\lambda }_{\text{vacuum}}}{2}=\frac{3\times 10^8}{2\times 5\times 10^{14}}$

${\lambda }_{\text{medium}}=0.3\times 10^{-6}\text{m}$

Converting to nanometers:

$0.3\times 10^{-6}\text{m}=300\text{nm}$

Therefore, the wavelength of the refracted light in the medium is 300 nm.