JEE Physics 2023 Question with Solution

Given: Light passes from air into another medium with angle of incidence $45^\circ$ and angle of refraction $30^\circ$.

Find: The relation between $\lambda_2, \lambda_1$ and $\nu_2, \nu_1$.

Using Snell's law:

$$1 \sin 45^\circ = \mu \sin 30^\circ$$ $$\mu = \sqrt{2}$$

The wavelength in a medium changes inversely with refractive index, so

$$\lambda_2 = \frac{\lambda_1}{\mu} = \frac{\lambda_1}{\sqrt{2}}$$

The frequency does not change when light passes from one medium to another, hence

$$\nu_2 = \nu_1$$

Therefore, the correct option is A.

Given: $n_1 = 1$ for air, $\theta_1 = 45^\circ$, $\theta_2 = 30^\circ$.

Find: $\lambda_2$ and $\nu_2$ in terms of $\lambda_1$ and $\nu_1$.

Snell's law gives:

$$n_1 \sin \theta_1 = n_2 \sin \theta_2$$

Substituting the given values:

$$1 \times \sin 45^\circ = \mu \sin 30^\circ$$ $$\frac{1}{\sqrt{2}} = \mu \times \frac{1}{2}$$ $$\mu = \sqrt{2}$$

Now, refractive index is related to wavelength by:

$$\frac{\lambda_2}{\lambda_1} = \frac{1}{\mu}$$

So,

$$\lambda_2 = \frac{\lambda_1}{\sqrt{2}}$$

Also, frequency remains constant during refraction, therefore:

$$\nu_2 = \nu_1$$

Hence, the refracted wave has wavelength $\frac{1}{\sqrt{2}}\lambda_1$ and frequency $\nu_1$. The correct option is A.