Q152 / 180 · KCET 2007 · Physics
A vessel of height $2 d$ is half-filled with a liquid of refractive index $\sqrt{2}$ and the other half with a liquid of refractive index $n$ (the given liquids are immiscible). Then the apparent depth of the inner surface of the bottom of the vessel (neglecting the thickness of the bottom of the vessel) will be
- A$\frac{n}{d(n+\sqrt{2})}$
- Correct answerB$\frac{d(n+\sqrt{2})}{n \sqrt{2}}$
- C$\frac{\sqrt{2} n}{d(n+\sqrt{2})}$
- D$\frac{n d}{d+\sqrt{2 n}}$
Solution
Refractive index
$$
\mu=\frac{\text { Real depth }(d)}{\text { Apparent depth }(x x)}
$$
For 1st liquid, $\sqrt{2}=\frac{d}{x_{1}}$
$\Rightarrow \quad x_{1}=\frac{d}{\sqrt{2}}$
Similarly, for 2 nd liquid,
$$
n=\frac{d}{x_{2}} \Rightarrow x_{2}=\frac{d}{n}
$$
Total apparent depth $=x_{1}+x_{2}$
$$
=\frac{d}{\sqrt{2}}+\frac{d}{n}=\frac{d(n+\sqrt{2})}{n \sqrt{2}}
$$
Hence, the correct answer is option: 'B'.