Question

A fish looking up through the water sees the outyside world contained in a circular horizon. If the refractive index of water is 4/3 and the fish is 12 cm below the 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

The correct option is B $36/\sqrt{7}$

The refractive index of water is $N=\frac{4}{3}$

The critical angle is $i_c=si{n}^{-1}\frac{1}{n}=si{n}^{-1}\frac{3}{4}$

if $r$ is the radius of the circular horizon and $h$ is the distance of the fish from the surface then from the geometry, $cot{i}_c=\frac{r}{h}$

From trigonometry

$1+co{t}^2{i}_c=cose{c}^2{i}_c$

$1+\frac{h^2}{r^2}=\frac{1}{si{n}^2{i}_c}$

$1+\frac{h^2}{r^2}=\frac{4^2}{3^2}$

$r=\frac{h\sqrt{9}}{\sqrt{7}}=12\times \sqrt{\frac{9}{7}}cm=\frac{36}{\sqrt{7}}cm$