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(B) The image formation by a mirror (either plane or spherical) does not depend on the medium.The object $P$ is at a height $h$ above the mirror. Ther

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$\frac{2h}{\mu}$

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(B) The image formation by a mirror (either plane or spherical) does not depend on the medium.The object $P$ is at a height $h$ above the mirror. Therefore,the image $P'$ of $P$ will be formed at a distance $h$ below the mirror.Let $d$ be the depth of the liquid in the tank.The distance of object $P$ from the surface of the liquid is $(d - h)$. The apparent depth of $P$ as seen by the observer $O$ is $x_1 = \frac{d - h}{\mu}$.The distance of the image $P'$ from the surface of the liquid is $(d + h)$. The apparent depth of the image $P'$ as seen by the observer $O$ is $x_2 = \frac{d + h}{\mu}$.The apparent distance between $P$ and its image is the difference between their apparent depths:Apparent distance $= x_2 - x_1 = \frac{d + h}{\mu} - \frac{d - h}{\mu} = \frac{d + h - d + h}{\mu} = \frac{2h}{\mu}$.

$A$ plane mirror is placed at the bottom of a tank containing a liquid of refractive index $\mu$. $P$ is a small object at a height $h$ above the mirror. An observer $O$,vertically above $P$ outside the liquid,sees $P$ and its image in the mirror. The apparent distance between these two will be

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