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A point source of light is kept below the surface of water $\left(n_{w}=4 / 3\right)$ at a depth of $\sqrt{7} \mathrm{~m}$. The radius of the circular bright patch of light noticed on the surface of water is
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Verified Answer
The correct answer is:
$3 \mathrm{~m}$
When the ray of light is incident from water-air interface at critical angle $\left(\theta_{c}\right)$, the refracted ray becomes parallel to the interface.
Hence, the radius of the circular bright patch of light noticed on the surface of water is given by
$$
\begin{aligned}
\mathrm{R} &=\mathrm{h} \tan \theta_{\mathrm{c}} \\
&=\sqrt{7} \frac{\sin \theta_{\mathrm{c}}}{\cos \theta_{\mathrm{c}}}=\sqrt{7} \frac{1 / \mu}{\sqrt{1-\frac{1}{\mu^{2}}}}=\frac{\sqrt{7}}{\sqrt{\mu^{2}-1}}=3 \mathrm{~m}
\end{aligned}
$$
Hence, the radius of the circular bright patch of light noticed on the surface of water is given by
$$
\begin{aligned}
\mathrm{R} &=\mathrm{h} \tan \theta_{\mathrm{c}} \\
&=\sqrt{7} \frac{\sin \theta_{\mathrm{c}}}{\cos \theta_{\mathrm{c}}}=\sqrt{7} \frac{1 / \mu}{\sqrt{1-\frac{1}{\mu^{2}}}}=\frac{\sqrt{7}}{\sqrt{\mu^{2}-1}}=3 \mathrm{~m}
\end{aligned}
$$
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