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

Question

Philoid · Accepted Answer

Given:

Actual depth of the pin, d = 15 cm
Let apparent depth of the pin = d’ cm
Refractive index of glass, μ = 1.5

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

μ = d / d′

⇒ d’ = d × μ

⇒ d^’ = 15 × 1.5 = 10cm

The distance at which the pin appears to be raised D,

D = d’-d

D = 15-10 = 5cm

So the pin appears to be raised by 5 cm.
For a small angle of incidence, this distance does not depend upon the location of the slab.