Question

A small piece of wood is floating on the surface of a 2.5 m deep lake. Where does the shadow form on the bottom when the sum is just setting? Refractive index of water = 4/3.

Answer 1

Given,
Depth of the lake = 2.5 m
Refractive index (μ) of water = $\frac{4}{3}$
When the sun is just setting, θ is approximately = 90˚



Therefore, incidence angle is 90˚
Using Snell's law:

$\therefore \frac{\sin i}{\sin r}=\frac{{\mathrm{\mu }}_2}{{\mathrm{\mu }}_1}$
$\Rightarrow \frac{1}{\sin r}=\frac{{\displaystyle \frac{4}{3}}}{1}$
$\Rightarrow \sin r=\frac{3}{4}$
$\Rightarrow r=49°$

From the figure, W'O = x is the distance of the shadow
$\mathrm{Thus},\left({\displaystyle \frac{x}{2.5}}\right)=\tan r=1.15$
$\Rightarrow x=2.5\times 1.15=2.8\mathrm{m}$