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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
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Verified Answer
The correct answer is:
$\frac{12 \times 3}{\sqrt{7}}$
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 $\sin i_{c}=\frac{1}{\mu}$
$$
\begin{aligned}
\therefore r &=h \frac{\frac{1}{\mu}}{\sqrt{1-\frac{1}{\mu^{2}}}}=\frac{h}{\sqrt{\mu^{2}-1}}=\frac{12}{\sqrt{\frac{16}{9}-1}} \\
&=\frac{12 \times 3}{\sqrt{7}} \mathrm{~cm}
\end{aligned}
$$
or $r=h \frac{\sin i_{c}}{\cos i_{c}}$ or $r=h \frac{\sin i_{c}}{\sqrt{1-\sin ^{2} i_{c}}}$
But $\sin i_{c}=\frac{1}{\mu}$
$$
\begin{aligned}
\therefore r &=h \frac{\frac{1}{\mu}}{\sqrt{1-\frac{1}{\mu^{2}}}}=\frac{h}{\sqrt{\mu^{2}-1}}=\frac{12}{\sqrt{\frac{16}{9}-1}} \\
&=\frac{12 \times 3}{\sqrt{7}} \mathrm{~cm}
\end{aligned}
$$
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