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 = 1.5. Does the answer depend on the location of the slab?
Actual depth of the pin , d=15cm Apparent dept of the pin = d′ Refractive index of glass is μ=1.5 Ratio of actual depth to the apparent depth is equal to the refractive index of the glass i.e. μ=d′d⇒d′=10cm The distance at 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.
