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(B) Let the total depth of the liquid in the tank be $H$. The object $P$ is at a height $h$ above the mirror,so its depth from the liquid surface is $

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

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(B) Let the total depth of the liquid in the tank be $H$. The object $P$ is at a height $h$ above the mirror,so its depth from the liquid surface is $(H-h)$.The apparent depth of the object $P$ as seen by the observer $O$ is $d_1 = \frac{H-h}{\mu}$.The image of the object $P$ formed by the plane mirror is at a distance $h$ below the mirror. The total depth of this image from the liquid surface is $(H+h)$.The apparent depth of the image as seen by the observer $O$ is $d_2 = \frac{H+h}{\mu}$.The apparent distance between the object $P$ and its image is the difference between their apparent depths:$	ext{Apparent distance} = d_2 - d_1 = \frac{H+h}{\mu} - \frac{H-h}{\mu} = \frac{H+h-H+h}{\mu} = \frac{2h}{\mu}$.

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