Question

A point source is placed at a depth h below the surface of water (refractive index = μ). (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

Given,
Refractive index is μ

(a)
Let the point source be P, which is placed at a depth of h from the surface of water.
Let us take x as the radius of the circular area.
and let θ_c be the critical angle.



Thus,
$$\begin{gathered} \frac{x}{h}=\tan {\theta }_c \\ \frac{x}{h}=\frac{\sin {\theta }_c}{\sqrt{1-{\sin }^2{\theta }_c}} \\ =\frac{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\mathrm{\mu }$}\right.}}{\sqrt{1-{\displaystyle \frac{1}{{\mathrm{\mu }}^2}}}}\left(\because \sin {\theta }_c=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\mathrm{\mu }$}\right.\right) \\ \frac{x}{h}=\frac{{\displaystyle 1}}{\sqrt{{\mathrm{\mu }}^2-1}} \\ x=\frac{h}{\sqrt{{\mathrm{\mu }}^2-1}} \end{gathered}$$

Clearly from figure, the light escapes through a circular area at a fixed distance r on the water surface, directly above the point source.
That makes a circle, the centre of which is just above P.

(b)
The angle subtended by the radius of the circular area on the point source P:
$\Rightarrow \sin {\mathrm{\theta }}_{\mathrm{c}}=\frac{1}{\mathrm{\mu }}$
$\Rightarrow {\theta }_{\mathrm{c}}={\sin }^{-1}\left(\frac{1}{\mathrm{\mu }}\right)$