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