Question 109 of 295

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 $\sqrt{2}$ . The liquids are immiscible. The apparent depth of the inner surface of the bottom of the vessel will be

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

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Answer: (D) $\dfrac{3\sqrt{2}h}{4}$

$d=\dfrac{h}{\sqrt{2}}+\dfrac{h}{2\sqrt{2}}$
$\Rightarrow d=\dfrac{h}{\sqrt{2}}\times \dfrac{3}{2}=\dfrac{3\sqrt{2}h}{4}$