A small pin fixed on a table top is viewed from above from a distance of 50 cm.

Actual depth of the pin, d = 15 cm

Apparent dept of the pin $=d^{\prime}$

Refractive index of glass, $\mu=1.5$

Ratio of actual depth to the apparent depth is equal to the refractive index of glass, i.e.

$\mu=\frac{d}{d^{\prime}}$

$\therefore d^{\prime}=\frac{d}{\mu}$

$=\frac{15}{1.5}=10 \mathrm{~cm}$

The distance at which the pin appears to be raised $=d^{\prime}-d$

$=15-10=5 \mathrm{~cm}$

For a small angle of incidence, this distance does not depend upon the location of the slab.