Rays from a point source of light situated at height \(h\) below the liquid surface having refractive index \(\mu\), forms a circular patch of light of radius \(r\) on the surface. The area of the patch is

Consider the figure as shown below:

Area of the light patch \(=\pi r^2\)
We know the patch forms due to total internal reflection. Light rays from source \(S\) are incident at the edge of the circular path at critical angle \(\theta_{\mathrm{C}}\) (as shown at points \(\mathrm{P}\) and \(\mathrm{Q}\) ), the refracted ray exits at angle \(\frac{\pi}{2}\) w.r.t. the normal.
Using Snell's Law
\(
\begin{aligned}
& \therefore \mu\left(\sin \theta_C\right)=1\left(\sin \left(\frac{\pi}{2}\right)\right) \\
& \Rightarrow \sin \theta_C=\frac{1}{\mu}=\frac{r}{\sqrt{h^2+r^2}} \\
& \Rightarrow r^2=\frac{\left(h^2+r^2\right)}{\mu^2} \\
& \Rightarrow r^2=\frac{h^2}{\left(\mu^2-1\right)}
\end{aligned}
\)
\(\therefore\) Area of the patch \(=\left(\frac{\pi h^2}{\mu^2-1}\right)\)