Question

A vessel of depth $2h$ is half-filled with a liquid of refractive index $\sqrt{2}$ in the upper half and with a liquid of refractive index $2\sqrt{2}$ in the lower half. The liquids are immiscible. The apparent depth of the inner surface of the bottom of the vessel will be;

• A. $\frac{3h}{4}\sqrt{2}$
• B. $\frac{h}{\sqrt{2}}$
• C. $\frac{h}{3\sqrt{2}}$
• D. $\frac{h}{2\left(\sqrt{2}+1\right)}$

Answer 1

The correct option is A

$\frac{3h}{4}\sqrt{2}$

Step 1: Given data,

The real Depth of the vessel is $R=2h$.

Refractive index of the upper half, ${\mu }_1$$=\sqrt{2}$

Refractive index of the lower half, ${\mu }_2$$=2\sqrt{2}$

Step 2: Finding the apparent depth

Let the total apparent depth of the vessel $=d$

Let the depth of the upper half $=d_1$

Let the depth of the lower half $=d_2$

Now, By using the formula of apparent depth

Apparent depth $d$ $=$ Real Depth of vessel $/$Refractive Index

$d=\frac{R}{\mu }$

Since, $d=d_1+d_2$

Where, $d_1=$Depth of upper half $/$ Refractive index of the upper half

$d_1=\frac{h}{\sqrt{2}}$

$d_2=$Depth of lower half $/$ Refractive index of the lower half

$d_2=\frac{h}{2\sqrt{2}}$

From the figure, the depth of the upper half and the lower half is $h$ and $h$ respectively, and refractive index of the upper half and the lower half is $\sqrt{2}$ and $2\sqrt{2}$ respectively.

$$\begin{gathered} d=\frac{h}{\sqrt{2}}+\frac{h}{2\sqrt{2}} \\ d=\frac{h}{\sqrt{2}}\left(1+\frac{1}{2}\right) \\ d=\frac{3h}{2\sqrt{2}} \\ d=\frac{3h}{2\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}} \end{gathered}$$

$d=\frac{3h}{4}\sqrt{2}$

Therefore, the apparent depth of the vessel is $\frac{3h}{4}\sqrt{2}$.

Hence, the correct option is A.