A plane mirror is placed at the bottom of a tank containing a liquid of refractive index ' $\mu$ ', ' p ' is…
A plane mirror is placed at the bottom of a tank containing a liquid of refractive index ' $\mu$ ', ' p ' is a small object at a height ' $h$ ' above the mirror. An observer ' $O$ ' vertically above ' $p$ ' outside the liquid sees ' $p$ ' and the image in a mirror. The apparent distance between these two will be
$2 \mu \mathrm{~h}$
$\frac{2 h}{\mu}$
$\frac{2 \mathrm{~h}}{(\mu-1)}$
$\mathrm{h}\left(1+\frac{1}{\mu}\right)$
Solution
Let the apparent depth of P be $\mathrm{x}_1$ and the apparent depth of the image of P be $\mathrm{x}_2$.
$\therefore \quad \mathrm{x}_1=\frac{\mathrm{d}-\mathrm{h}}{\mu}$ and $\mathrm{x}_2=\frac{\mathrm{d}+\mathrm{h}}{\mu}$
$\therefore \quad$ Apparent distance between P and its image is
$\begin{aligned}
\dot{x}_2-x_1 & =\frac{d+h}{\mu}-\frac{(d-h)}{\mu} \\
& =\frac{d+h-d+h}{\mu}=\frac{2 h}{\mu}
\end{aligned}$
