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A plane mirror in placed at bottom of a tank containing a liquid of refractive index $μ .$ P is a small object at a height ' $h$ ' above the mirror. An observer "O" vertically above P outside the liquid observes P & its image in the mirror. The apparent distance between these two will be

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$\frac{2 h}{μ - 1}$

$2 μ h$

$h (1 + \frac{1}{μ})$

$\frac{2 h}{μ}$

answer is B.

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Detailed Solution

Let x be the depth of point P from surface

App. depth of point P from surface, $x = \frac{x}{μ}$

App. depth of image of P from surface $= \frac{x + 2 h}{μ}$

So, separation between two $= \frac{x + 2 h}{μ} - \frac{x}{μ} \Rightarrow \frac{2 h}{μ}$

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