Question

A point source is placed at a depth h below the surface of water (refractive index $=\mu$).
(a) Show that light escapes through a circular area on the water surface with its centre directly above the point source.
(b) Find the angle subtended by a radius of the area on the source.

Answer 1

a) Let, $x=$ radius of the circular area

$\frac{x}{h}=\tan C$ (where C is the critical angle)

$\Rightarrow \frac{x}{h}=\frac{\sin C}{\sqrt{1-{\sin }^{2}C}}=\frac{1/\mu }{\sqrt{1-\frac{1}{{\mu }_2}}}$ (because $\sin C=1/\mu$)

$\Rightarrow \frac{x}{h}=\frac{1}{\sqrt{{\mu }^2-1}}$ or $x=\frac{h}{\sqrt{{\mu }^2-1}}$

So, light escapes through a circular area on the water surface directly above the point source.

b) Angle subtained by a radius of the area on the source, $C={\sin }^{-1}\left(1/\mu \right)$