KCET · Physics · Ray Optics
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
- A \(\sqrt{7} \mathrm{~m}\)
- B \(\frac{3}{\sqrt{7}} \mathrm{~m}\)
- C \(3 \mathrm{~m}\)
- D \(\frac{\sqrt{7}}{3} \mathrm{~m}\)
Answer & Solution
Correct Answer
(C) \(3 \mathrm{~m}\)
Step-by-step Solution
Detailed explanation
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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