Question

A vessel of depth $2h$ is half filled with a liquid of refractive index $2\sqrt{2}$ ​ and the upper half with another liquid of refractive index $2$. The liquids are immiscible. The apparent depth of an object which has a height $\frac{h}{2}$ from the bottom of the vessel inside the liquid as seen from the top of the vessel is (Take $h=\sqrt{2}\text{m})$

Answer 1

Let, the apparent height of the object is $h_a$. So, we can write,
$h_a=h_{a1}+h_{a2}....\left(1\right)$

Since, the object is being look from the top, so apparent height due to the liquid having refractive index $2\sqrt{2}$ is
$h_{a1}=\frac{h_{r1}}{{\mu }_1}=\frac{h}{2\times 2\times \sqrt{2}}$

Similarly, $h_{a2}=\frac{h_{r2}}{{\mu }_2}=\frac{h}{\sqrt{2}}$

From equation $\left(1\right)$,
$h_a=\frac{h}{2\times 2\times \sqrt{2}}+\frac{h}{\sqrt{2}}$

$\Rightarrow h_a=\frac{h}{\sqrt{2}}[\frac{1}{4}+1]$

$\therefore h_a=\frac{5h}{4\sqrt{2}}$

Taking $h=\sqrt{2}\text{m}$ gives, $h_a=1.25\text{m}$