$A$ small pin fixed on a table top is viewed from above from a distance of $50 \, cm$. By what distance would the pin appear to be raised if it is viewed from the same point through a $15 \, cm$ thick glass slab held parallel to the table? Refractive index of glass $= 1.5$. Does the answer depend on the location of the slab?

(5 CM) The actual depth of the pin within the glass slab is $d = 15 \, cm$.
The apparent depth of the pin is given by $d^{\prime} = \frac{d}{\mu}$,where $\mu = 1.5$ is the refractive index of the glass.
$d^{\prime} = \frac{15}{1.5} = 10 \, cm$.
The shift or the distance by which the pin appears to be raised is given by $\Delta d = d - d^{\prime}$.
$\Delta d = 15 \, cm - 10 \, cm = 5 \, cm$.
Since the shift depends only on the thickness of the slab and its refractive index,the answer does not depend on the location of the slab between the pin and the observer.