JEE Physics 2023 Question with Solution

Given: critical angle is $\theta_c = 45^\circ$ and the speed of light in water medium is $3 \times 10^8 \, \text{m/s}$.

Find: the speed of light in the denser medium.

From the solution, the critical angle relation is

$$\sin \theta_c = \frac{n_2}{n_1}$$

where $n_1$ is the refractive index of the denser medium and $n_2$ is that of the rarer medium.

Using

$$n_1 = \frac{c}{v}$$

the extracted working gives

$$v = \frac{c}{\sin \theta_c} = \frac{3 \times 10^8}{\sin 45^\circ} = 2.12 \times 10^7 \, \text{m/s}$$

Therefore, the speed of light in the denser medium is $2.12 \times 10^7 \, \text{m/s}$ and the correct option is A.