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?
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.