Question

A rod of height $h$ is placed in a beaker of same radius. The height of the beaker is thrice its radius. An observer sees the top end of the rod through a pin-hole (see figure). When the beaker is filled with a liquid (R.I. $=\sqrt{\mu}$ ) upto a height $2 h$, he can see the lower end of the rod. Find the value of $\mu$.

Answer 1

The line of sight of the observer remains constant, making an angle of $45^{\circ}$ with the normal.
$\sin \theta =\frac{h}{\sqrt{h^{2}+(2 h)^{2}}} $
$=\frac{1}{\sqrt{5}}$
Refractive index of liquid,
$\sqrt{\mu}=\frac{\sin 45^{\circ}}{\sin \theta}$
$=\frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{5}}}=\sqrt{\left(\frac{5}{2}\right)}$
$\therefore \sqrt{\mu}=\sqrt{2.5} $
$ \Rightarrow \mu=2.5$