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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.
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:
\[Now 1 . 33 = \frac{\sin 45^\circ }{\sin r}\]
\[ \Rightarrow \sin r = \frac{1}{1 . 33\sqrt{2}} = 0 . 53\]
\[ \Rightarrow \cos r = \sqrt{1 -\sin^2 r}\]
\[ = \sqrt{1 - (0 . 53 )^2} = 0 . 85\]
\[So, \tan r = 0 . 6235\]
Therefore, shadow length of the pole = (0.5)
\[\times\] (1 + 0.6235) = 0.81175 m
= 81.2 cm
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