Q115 / 120 · VITEEE 2010 · Physics

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 $12 \mathrm{~cm}$ below the surface of water, the radius of the circle in centimetre is

  • A
    $\frac{12 \times 3}{\sqrt{5}}$
  • B
    $12 \times 3 \times \sqrt{5}$
  • Correct answer
    C
    $\frac{12 \times 3}{\sqrt{7}}$
  • D
    $12 \times 3 \times \sqrt{7}$

Solution

Here, $\tan i_{c}=\frac{r}{h}$ or $r=h \tan i_{c}$ or $r=h \frac{\sin i_{c}}{\cos i_{c}}$ or $r=h \frac{\sin i_{c}}{\sqrt{1-\sin ^{2} i_{c}}}$ But $\sin i_{c}=\frac{1}{\mu}$ $$ \begin{aligned} \therefore r &=h \frac{\frac{1}{\mu}}{\sqrt{1-\frac{1}{\mu^{2}}}}=\frac{h}{\sqrt{\mu^{2}-1}}=\frac{12}{\sqrt{\frac{16}{9}-1}} \\ &=\frac{12 \times 3}{\sqrt{7}} \mathrm{~cm} \end{aligned} $$

Hence, the correct answer is option: 'C'.

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