Question

A fish looking up through water sees the outside world contained in a circular horizon. If the fish is $\sqrt{7}\text{cm}$ below the surface of water $(\mu =\frac{4}{3})$, the radius of the circle will be

• A. $2\text{cm}$
• B. $3\text{cm}$
• C. $4\text{cm}$
• D. $5\text{cm}$

Answer 1

The correct option is B $3\text{cm}$


The boundary of the circle will be formed by rays that just graze the surface of water.

For total internal reflection,
$\angle OAF={90}^{\circ }-{\theta }_c$
$={90}^{\circ }-{\sin }^{-1}\left(\frac{1}{\mu }\right)$
$={90}^{\circ }-{\sin }^{-1}\left(\frac{3}{4}\right)$
$={\cos }^{-1}\left(\frac{3}{4}\right)$
Let $\angle OAF=\alpha$, then $\cos \alpha =\frac{3}{4}$
and $\tan \alpha =\frac{\sqrt{7}}{3}\dots \left(1\right)$
In the triangle $AOF$,
$\tan \alpha =\frac{OF}{OA}=\frac{\sqrt{7}}{r}\dots \left(2\right)$
From equation (1) and (2),
$r=3\text{cm}$

Why this question? In this question, we use the principle of reversibility of light rays. Since light rays incident at critical angle from water will just graze the surface after refraction, we can conclude that rays coming from outside parallel to the water surface will have an angle of refraction equal to the critical angle.