Physics - Geometrical Optics Question with Solution | TestHub

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 $$\begin{array}{rl}& \therefore \mu \left(\sin {\theta }_C\right)=1\left(\sin \left(\frac{\pi }{2}\right)\right)\\ & \Rightarrow \sin {\theta }_C=\frac{1}{\mu }=\frac{r}{\sqrt{h^2+r^2}}\\ & \Rightarrow r^2=\frac{\left(h^2+r^2\right)}{{\mu }^2}\\ & \Rightarrow r^2=\frac{h^2}{\left({\mu }^2-1\right)}\end{array}$$ $\therefore$ Area of the patch $=\left(\frac{\pi h^2}{{\mu }^2-1}\right)$