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?
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.