Question

Light coming out of a point source kept at depth ${}^{\prime }{h}^{\prime }$ under water($\mu$) has only a part of it that escapes into air as shown in the figure. what is the radius of the circle through which light ray escapes.

• A.
  $\frac{h}{\mu }$
• B.
  $\frac{h}{\sqrt{{\mu }^2-1}}$
• C.
  $\frac{h\mu }{\sqrt{{\mu }^2-1}}$
• D.
  None

Answer 1

The correct option is B

$\frac{h}{\sqrt{{\mu }^2-1}}$

As we can see from the given figure, rays under less than ${\theta }_c$ only escapes into air and rays beyond ${\theta }_c$ undergoes total internal reflection.

Using Snell's law at the interface

$\mu \sin {\theta }_c=1\times \sin {90}^{\circ }$

$\sin {\theta }_c=\frac{1}{\mu }$

From the given figure,

$\sin {\theta }_c=\frac{R}{\sqrt{{\left(R\right)}^2+{\left(h\right)}^2}}$

$\Rightarrow \frac{1}{\mu }=\frac{R}{\sqrt{{\left(R\right)}^2+{\left(h\right)}^2}}$

$\Rightarrow \frac{1}{{\mu }^2}=\frac{R^2}{R^2+h^2}$

$\Rightarrow R^2+h^2=R^2{\mu }^2$

$\Rightarrow R^2({\mu }^2-1)=h^2$

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

Hence, option (b) is the correct answer.

Why this question ? To understand the concept of total internal reflection and the circle having radius of curvature $R=\frac{h}{\sqrt{{\mu }^2-1}}$ is known as circle of illuminance.