Question

A fish in a lake at a depth h sees the outside world through a circular aperture at the surface. The radius of the aperture is

• A. $\frac{\mu }{\sqrt{h^2-1}}$
• B. $\mu \sqrt{h^2-1}$
• C. $h\sqrt{{\mu }^2-1}$
• D. $\frac{h}{\sqrt{{\mu }^2-1}}$

Answer 1

The correct option is D $\frac{h}{\sqrt{{\mu }^2-1}}$

Referring figure, $tanC=\frac{r}{h}$

Where, $C=$ critical angle

$r=htanC=h\left(sinC/cosC\right)=\frac{hsinC}{\sqrt{1-si{n}^{2}C}}$

We know, $sinC=\frac{1}{\mu }$

$\therefore r=\frac{h/\mu }{\sqrt{1-\frac{1}{{\mu }^2}}}=\frac{h}{\sqrt{{\mu }^2-1}}$