9.16: A small pin fixed on a tabletop 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. Does the answer depend on the location of the slab?
Hint: Shift = \(d(1-\frac{1}{\mu})\)
Step: Find the shift due to slab.
Let the apparent depth of the pin = d’
Refractive index of glass, = 1.5
The refractive index of the glass is given by:
The distance by which the pin appears to be raised = d’ — d = 15 — 10 = 5cm
For a small angle of incidence, this distance does not depend upon the location of the slab.
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