Question

A fish is situated at a depth ${}^{\prime }{h}^{\prime }$ inside a lake. Calculate the radius of the circular aperture through which the fish sees the outside world. $R.I$. of water $=\mu$

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

Answer 1

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

From the above fig. if $C$ is critical angle, then

$\tan C=\frac{r}{h}⟹r=h\times \tan C⟹r=h\times \frac{\sin C}{\cos C}$

$⟹r=h\times \frac{\sin C}{\sqrt{1-{\sin }^{2}C}}$ [as ${\sin }^{2}a+{\cos }^{2}a=1$]

We know that,

$\sin C=\frac{1}{\mu }$

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