$A$ jar of height $h$ is filled with a transparent liquid of refractive index $\mu$. At the centre of the jar on the bottom surface is a dot. Find the minimum diameter of a disc,such that when placed on the top surface symmetrically about the centre,the dot is invisible.

(D) Suppose the required minimum diameter of the given disc is $d$. For light rays emanating from the point-like object $O$ and incident on the water surface from inside,if the angle of incidence $i \geq C$,then the object $O$ will not be seen by an observer outside the water due to total internal reflection. (Where $C$ is the critical angle for water to air).
Assume the angle $i$ in the figure is equal to $C$.
According to the formula for the critical angle:
$\sin C = \frac{1}{\mu}$
Since $i = C$,we have $\sin i = \frac{1}{\mu}$.
From the geometry of the figure:
$\tan i = \frac{d/2}{h}$
$\therefore \frac{d}{2} = h \tan i$
$\therefore d = 2h \tan i$ ... $(1)$
Now,since $\sin i = \frac{1}{\mu}$,we find $\cos i$:
$\cos i = \sqrt{1 - \sin^2 i} = \sqrt{1 - \frac{1}{\mu^2}} = \frac{\sqrt{\mu^2 - 1}}{\mu}$
Therefore,$\tan i = \frac{\sin i}{\cos i} = \frac{1/\mu}{\sqrt{\mu^2 - 1}/\mu} = \frac{1}{\sqrt{\mu^2 - 1}}$.
Substituting this into equation $(1)$:
$d = \frac{2h}{\sqrt{\mu^2 - 1}}$.