A pole is vertically submerged in swimming pool, such that it gives a length of shadow $2.15\ \text{m}$ within water when sunlight is incident at an angle of $30^\circ$ with the surface of water. If swimming pool is filled to a height of $1.5\ \text{m}$, then the height of the pole above the water surface in centimetres is ______. ($n_w=\frac43$)
Numerical value type. Enter your answer.
Answer: 50
The angle of incidence is $60^\circ$. $\sin r=\dfrac{\sin60^\circ}{4/3}=0.65\Rightarrow\tan r\approx0.854$.
The part of the pole above water (height $h$) throws the shadow edge $h\cot30^\circ=\sqrt3h$ from the pole on the water surface; the refracted ray then shifts a further $1.5\tan r\approx1.28\ \text{m}$ before reaching the bottom:
$$\sqrt3h+1.28=2.15\ \Rightarrow\ h\approx0.50\ \text{m}=50\ \text{cm}$$
Solution by Sreeraj P, M.Sc Physics