Question

Light coming out of a point source kept at depth $h$ under the water $\left(\mu \right)$ has only a part of it that escapes into air. What is the radius of the circle through which the 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 $\frac{h}{\sqrt{{\mu }^2-1}}$


From the diagram, we can see that light will escape into air till critical angle ${\theta }_c$, after that total internal reflection starts.

Now applying Snell's law for critical angle,

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

$\Rightarrow \sin {\theta }_c=\frac{1}{\mu }........\left(1\right)$

In $△ABC$

$\sin {\theta }_c=\frac{BC}{AC}=\frac{R}{\sqrt{R^2+h^2}}$

from eq. $\left(1\right)$

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

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

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

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

Hence, option $\left(b\right)$ is correct.