Question

A fish looking from within water sees the outside world through a circular horizon. If the fish is $\sqrt{7}m$ below the surface of water, then the radius of the circular horizon will be

• A. $1m$
• B. $2m$
• C. $3m$
• D. $4m$

Answer 1

The correct option is C $3m$

Retracing the rays of light reaching the fish from outside, we see that there is maximum angle of incidence (called critical angle $i_c$) for which the refracted ray just graces the surface of water.


i.e Angle of refraction $r={90}^{\circ }$
So, applying Snell’s law, $m{u}_d\times \sin {\theta }_c={\mu }_r\times \sin {90}^{\circ }$
$\Rightarrow \sin {\theta }_c=\frac{{\mu }_r}{{\mu }_d}$
Rays incident at angle greater than ${\theta }_c$ will suffer Total Internal Reflection (TIR) and thus will not go outside.

Using the principle of reversibility of light, we can say the light rays from outside the ocean can reach the fish’s eye at a maximum angle of ${\theta }_c$. So the fish will see the outside world through a circle of radius AC.


To determine radius AC, $\sin {\theta }_c=\frac{{\mu }_r}{{\mu }_d}=\frac{AC}{OA}$
i.e $\frac{1}{\mu }=\frac{r}{\sqrt{r^2+h^2}}$
From this, we get $r=\frac{h}{\sqrt{{\mu }^2-1}}$
Substituting $h=\sqrt{7}$ and $\mu =\frac{4}{3}$,
$r=3m$