Total Internal Reflection — Question 3

Question 3

The escaping rays from a submerged point occupy a circular surface region

A point source lies depth h beneath a flat water-air surface. Water index is n greater than one and air index is one. Ignore scattering and examine the first encounter with the surface.

1. Find the radius of the surface circle through which rays can emerge.

2. Evaluate the radius relative to h for n=4/3, and find how that radius changes when the depth doubles.

Question 3 - Solution

The escaping rays from a submerged point occupy a circular surface region

(a) Method and calculation

$$\sin\theta_c=1/n,\quad R=h\tan\theta_c.$$$$\tan\theta_c=\frac{1/n}{\sqrt{1-1/n^2}}\Rightarrow\boxed{R=\frac{h}{\sqrt{n^2-1}}}.$$

(b) Interpretation and further calculation

$$n=4/3\Rightarrow\boxed{R/h=3/\sqrt7=1.13389}.$$$$\boxed{R(2h)=2R(h)}.$$

Physical check

The bounding ray is grazing at emergence. Rays striking beyond this circle exceed the critical angle.