There is a small source of light at some depth below the surface of water (refractive index $= \dfrac{4}{3}$) in a tank of large cross sectional surface area. Neglecting any reflection from the bottom and absorption by water, percentage of light that emerges out of surface is (nearly):
[Use the fact that surface area of a spherical cap of height $h$ and radius of curvature $r$ is $2\pi rh$]
Answer: (C) $17\%$
Only rays within the critical angle $C$ of the vertical escape: $\sin C = \dfrac{3}{4}$, so $\cos C = \dfrac{\sqrt{7}}{4} \approx 0.661$.
These rays fill a cone of half-angle $C$. On a sphere of radius $r$ around the source, the cone cuts a cap of height $h = r(1 - \cos C)$. Fraction of light:
$$\frac{2\pi r\cdot r(1 - \cos C)}{4\pi r^2} = \frac{1 - \cos C}{2} = \frac{1 - 0.661}{2} \approx 0.17$$
About $17\%$.
Solution by Sreeraj P, M.Sc Physics