JEE Physics 2026 Question with Solution

Given: Wavelength in water is $\lambda_{\text{water}} = 540\,\text{nm}$ and refractive index of water is $\mu_{\text{water}} = \frac{4}{3}$. The refractive index of the transparent medium is $\mu = \frac{3}{2}$.

Find: The wavelength of the same light in the second medium.

Frequency of light remains unchanged when it passes from one medium to another, so wavelength changes according to

$$\lambda = \frac{\lambda_0}{\mu}$$

where $\lambda_0$ is the wavelength in vacuum.

For water,

$$\lambda_0 = \mu_{\text{water}} \times \lambda_{\text{water}}$$ $$\lambda_0 = \frac{4}{3} \times 540 = 720\,\text{nm}$$

Now for the second medium,

$$\lambda = \frac{720}{\frac{3}{2}}$$ $$\lambda = 720 \times \frac{2}{3} = 480\,\text{nm}$$

Therefore, the wavelength of the light in the transparent medium is $480\,\text{nm}$. The correct option is A.

Given: The same light passes through two media, so its frequency remains constant.

Find: The wavelength in the medium of refractive index $\frac{3}{2}$.

Since for the same light,

$$\lambda \propto \frac{1}{\mu}$$

we can write

$$\frac{\lambda_2}{\lambda_1} = \frac{\mu_1}{\mu_2}$$

Substituting $\lambda_1 = 540\,\text{nm}$, $\mu_1 = \frac{4}{3}$ and $\mu_2 = \frac{3}{2}$,

$$\lambda_2 = 540 \times \frac{\frac{4}{3}}{\frac{3}{2}}$$ $$\lambda_2 = 540 \times \frac{4}{3} \times \frac{2}{3}$$ $$\lambda_2 = 540 \times \frac{8}{9} = 480\,\text{nm}$$

Therefore, the required wavelength is $480\,\text{nm}$, so the correct option is A.