Question

A small bulb is placed at the bottom of a tank containing water to a depth of 80 cm. What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is 1.33. (Consider the bulb to be a point source.)

Answer 1

Given: Actual depth of bulb $d_1=0.8m$

The refractive index of water is $\mu =1.33$

Circle's radius, $R=\frac{AC}{2}$

We know that,

$\mu =\frac{sin{90}^{\circ }}{sini}$

$\Rightarrow i={48.75}^{\circ }$

Now, in the $\Delta OBC$,

$tani=\frac{OC}{OB}=\frac{R}{d_1}\Rightarrow R=0.91m$

Area,

$\pi R^2=\pi \times (0.91{)}^2=2.61m^2$