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 depth of the pin = d’

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’

Therefore  ,d’ = d/μ

= 15/1.5 = 10 cm

The distance at which the pin appears to be raised = d’ – d = 15 – 10 = 5 cm

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