Question

A fish looking up through the water sees the outside world, contained in a circular horizon. If the refractive index of water is $4/3$ and the fish is $12cm$ below the water surface, the radius of this circle (in cm) is :

• A. $36\sqrt{7}$
• B. $36/\sqrt{7}$
• C. $36\sqrt{5}$
• D. $4\sqrt{5}$

Answer 1

Answer: (B)

$36/\sqrt{7}$

If you wonder why there would be circle then try to recall critical angle, if the incident angle is beyond that particular angle there would be the total internal reflection, so the fish won't be able to see the outside world.

So, as to obtain the radius. We need to take the last light ray into consideration which can enter the eyeball of fish. So, here we'll use critical angle.

Using Snell's law,

$\frac{sini}{sinr}=\frac{{\mu }_a}{{\mu }_w}$

$r={90}^o\therefore sinr=1$

$sini=\frac{1}{4/3}$

$=\frac{3}{4}$

$\Rightarrow i={48.59}^o$

$tani=\frac{3}{\sqrt{7}}$

Using trigonometry,

$tani=\frac{a}{12}$

$\Rightarrow a=12\times tan48.59$

$\Rightarrow a=\frac{36}{\sqrt{7}}$