JEE Physics 2025 Question with Solution

Given: frequency $f = 5 \times 10^{14} \, \text{Hz}$, refractive index $\mu = 2$, and speed of light in air approximately $c = 3 \times 10^8 \, \text{m/s}$.

Find: the wavelength of refracted light in the medium.

The frequency of light remains unchanged when it passes from one medium to another.

First, find the wavelength in air:

$$\lambda_{\text{air}} = \frac{c}{f} = \frac{3 \times 10^8}{5 \times 10^{14}} = 0.6 \times 10^{-6} \, \text{m} = 600 \times 10^{-9} \, \text{m} = 600 \, \text{nm}$$

Now use the relation between wavelength in air and wavelength in the medium:

$$\lambda_{\text{medium}} = \frac{\lambda_{\text{air}}}{\mu}$$

Substituting the values:

$$\lambda_{\text{medium}} = \frac{600 \, \text{nm}}{2} = 300 \, \text{nm}$$

Therefore, the wavelength of the refracted light is $300 \, \text{nm}$. The correct option is A.

Given: $\nu = 5 \times 10^{14} \, \text{Hz}$, $n = 2$, and $c = 3 \times 10^8 \, \text{m/s}$.

Find: wavelength in the medium.

Use

$$v = \nu \lambda$$

and for a medium,

$$v = \frac{c}{n}$$

So,

$$\frac{c}{n} = \nu \lambda_m$$

Hence,

$$\lambda_m = \frac{c}{n\nu}$$

Substitute the values:

$$\lambda_m = \frac{3 \times 10^8}{2 \times 5 \times 10^{14}} = 3 \times 10^{-7} \, \text{m}$$

Convert to nanometres:

$$\lambda_m = 3 \times 10^{-7} \times 10^9 = 300 \, \text{nm}$$

Therefore, the wavelength of the refracted light is $300 \, \text{nm}$.