Q 12-09-022NEETJEE MainMedium
A point source of light is placed $4$ m below the surface of water ($\mu = \dfrac{4}{3}$). The radius of the circle on the surface through which light emerges is
Answer: (A) $\dfrac{12}{\sqrt{7}}$ m
At the edge of the circle, light meets the surface at the critical angle: $\sin C = \dfrac{3}{4}$, $\tan C = \dfrac{3}{\sqrt{7}}$.
$$r = h\tan C = \frac{12}{\sqrt{7}} \approx 4.5\ \text{m}$$
Solution by Sreeraj P, M.Sc Physics