Q 12-09-139JEE MainJEE Main 2022 (26 Jun, Shift 2)Medium
A small bulb is placed at the bottom of a tank containing water to a depth of $\sqrt7\ \text{m}$. The refractive index of water is $\dfrac43$. The area of the surface of water through which light from the bulb can emerge out is $x\pi\ \text{m}^2$. The value of $x$ is ______.
Numerical value type. Enter your answer.
Answer: 9
Light emerges within a circle whose edge corresponds to the critical angle:
$$r = \frac{h}{\sqrt{n^2-1}} = \frac{\sqrt7}{\sqrt{16/9 - 1}} = \frac{\sqrt7}{\sqrt7/3} = 3\ \text{m}$$
Area $= \pi r^2 = 9\pi\ \text{m}^2$, so $x = 9$.
Solution by Sreeraj P, M.Sc Physics