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.


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