Q86 / 160 · JIPMER 2012 · 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 answer
    B
    $\frac{d(n+\sqrt{2})}{n \sqrt{2}}$
  • C
    $\frac{\sqrt{2} n}{d(n+\sqrt{2})}$
  • D
    $\frac{n d}{d+\sqrt{2 n}}$

Solution

Refractive index $\mu=\frac{\text { Real depth }(d)}{\text { Apparent depth }(x)}$ For 1st liquid, $\sqrt{2}=\frac{d}{x_1}$ $\Rightarrow \quad x_1=\frac{d}{\sqrt{2}}$ Similarly, for 2nd liquid, $n=\frac{d}{x_2}$ $\Rightarrow \quad x_2=\frac{d}{n}$ 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'.

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