Question

A small coin is placed at the bottom of a cylindrical vessel of radius $R$ and height $h$. If a transparent liquid of refractive index $\mu$ completely filled into the cylinder, find the minimum fraction of the area that should be covered in order not to see the coin.

Answer 1

When the rays coming from the coin incident at an angle $\theta \ge {\theta }_c$ will be totally reflected
$\Rightarrow$ Minimum area that should be covered$=A={\pi r}^2$ where
$r=$ radius of the circular aperture $r$ can be found as
$r=h\tan {\theta }_c$
where ${\theta }_c$ can be given by the formula,
$\sin {\theta }_c=\frac{{\mu }_a}{\mu }=\frac{1}{\mu }$
$\Rightarrow \tan {\theta }_c=\frac{1}{\sqrt{{\mu }^2-1}}$
$\Rightarrow r=\frac{h}{\sqrt{{\mu }^2-1}}$
$\therefore A={\pi r}^2=\frac{{\pi h}^2}{{\mu }^2-1}$
The total area of the base $A={\pi R}^2$
$\Rightarrow$ The fraction of base area covered$=\eta =\left(\frac{{\pi h}^2}{{\mu }^2-1}\right)╱{\pi R}^2$
$\Rightarrow \eta =\frac{h^2}{({\mu }^2-1)R^2}$