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Question

A pole of length 1.00 m stands half dipped in a swimming pool with water level 50.0 cm higher than the bed. The refractive index of water is 1.33 and sunlight is coming at an angle of 45° with the vertical. Find the length of the shadow of the pole on the bed.

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Solution

Given,
Length of the pole = 1.00 m
Water level of the swimming pool is 50.0 cm higher than the bed.
Refractive index (μ) of water = 1.33



According to the figure, shadow length = BA' = BD + DA'= 0.5 + 0.5 tan r
Using Snell's law:
Now 1.33=sin 45°sin r sin r=11.332=0.53 cos r=1sin2 r=1-(0.53)2=0.85So, tan r=0.6235

Therefore, shadow length of the pole = (0.5)×(1 + 0.6235) = 0.81175 m
= 81.2 cm

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