Question 148 of 295

A hemispherical vessel is completely filled with a liquid of refractive index $\mu$. A small coin is kept at the lowest point $(\mathrm{O})$ of the vessel as shown in 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

A$\sqrt{3}$B$\frac{\sqrt{3}}{2}$C$\frac{3}{2}$D$\sqrt{2}$

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Answer: (D) $\sqrt{2}$

For the rays from coin to reach the point $E$, the refracted rays must grazing the surface, i.e. they must be incident at critical angle $\theta_c$ inside the liquid.
$\mu=\frac{1}{\sin \theta_c}$
$\mu$ is minimum when $\theta_c$ is maximum.
Maximum value of $\theta_c=45^{\circ}$
$\Rightarrow \mu$ has a minimum value of $\sqrt{2}$.