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 surface, the radius of this circle in cm is _____.

• A. $36/\sqrt{5}$
• B. $4\sqrt{5}$
• C. $9$
• D. $36/\sqrt{7}$

Answer 1

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

Since, the world is seen in the form of a circle, it implies, that the rays after refraction at the interface of the two media, must be going parallel to the surface.

Applying Snell's law at the interface of the two media, we have,

${\mu }_{1}Sini={\mu }_{2}Sinr$

Given, ${\mu }_1={\mu }_{water}=\frac{4}{3}$

We know, ${\mu }_2={\mu }_{air}=1$

From the figure,

$i=\theta$ and $r={90}^{\circ }$

$\Rightarrow \frac{4}{3}Sin\theta =1$

$Sin\theta =\frac{3}{4}$

From the figure,

$Sin\theta =\frac{r}{\sqrt{r^2+h^2}}$

Given, $h=12cm$

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

Squaring both sides, we have

$\frac{r^2}{r^2+{12}^2}=\frac{9}{16}$
$\frac{r^2+144}{r^2}=\frac{16}{9}$

$\frac{144}{r^2}=\frac{16}{9}-1\Rightarrow \frac{144}{r^2}=\frac{7}{9}$

$r^2=\frac{144\times 9}{7}$

$r=\frac{36}{\sqrt{7}}cm$

Hence, the correct answer is OPTION D.