A hemispherical vessel is completely filled with a liquid of refractive index $\mu$. A small coin is kept at the lowest point $(O)$ of the vessel as shown in the figure. The minimum value of the refractive index of the liquid so that a person can see the coin from point $E$ (at the level of the vessel) is
Answer: (D) $\sqrt2$
$E$ is at the level of the rim, so the light reaching $E$ leaves the liquid surface at grazing emergence (angle of refraction $90^\circ$), at the edge $B$ at the latest.
The ray from $O$ to the edge $B$ goes from the bottom of the hemisphere to the rim: with $C$ the centre of the surface, $CO = CB = R$, so this ray makes $45^\circ$ with the vertical normal.
For it to emerge along the surface:
$$\mu\sin45^\circ = \sin90^\circ \Rightarrow \mu = \sqrt2$$
Any smaller $\mu$ bends the ray upward and it cannot reach $E$.
Solution by Sreeraj P, M.Sc Physics