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Question
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?
Solution
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'"`
∴ 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.
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