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, 

\[\frac{x}{h} = \tan   \theta_c \] 

\[\frac{x}{h} = \frac{\sin  \theta_c}{\sqrt{1 - \sin^2 \theta_c}}\] 

\[           = \frac{\frac{1}{\mu}}{\sqrt{1 - \frac{1}{\mu^2}}}  \left( \because \sin  \theta_c = \frac{1}{\mu} \right)\] 

\[\frac{x}{h} = \frac{1}{\sqrt{\mu^2 - 1}}\] 

\[x = \frac{h}{\sqrt{\mu^2 - 1}}\]
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:
\[\Rightarrow \sin \theta_c = \frac{1}{\mu}\]
\[\Rightarrow \theta_c = \sin^{- 1} \left( \frac{1}{\mu} \right)\]