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 vessel will be :

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

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

$\frac{h}{2 (\sqrt{2} + 1)}$

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

answer is D.

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Detailed Solution

$A p p a r e n t d e p t h = h (\frac{1}{μ_{1}}) + h (\frac{1}{μ_{2}}) = h (\frac{1}{2 \sqrt{2}}) + h (\frac{1}{\sqrt{2}}) = \frac{3 \sqrt{2} h}{4}$

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