Q. A light bulb is at a depth of $D$ below the surface of water. An opaque disc of radius $R$ is placed on the surface of water just above the bulb. The bulb is not at all seen through the surface of water, then $(n =$ Refractive index of water)
A) $R = \frac{D}{\sqrt{n^{2} - 1}}$
B) $R > \frac{D}{\sqrt{n^{2} - 1}}$
C) $R < \frac{D}{\sqrt{n^{2} - 1}}$
D) $R = D\sqrt{n^{2} - 1}$
Explanation
(b) $\frac{\sin\theta_{c}}{\sin 90} = \frac{1}{\mu}$
$\sin\theta_{c} = \frac{1}{\mu}$
$\Rightarrow \sin{\theta_{c} = \frac{1}{n}}$
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$\Rightarrow \frac{R}{\sqrt{R^{2} + D^{2}}} = \frac{1}{n} \Rightarrow \frac{R^{2} + D^{2}}{R^{2}} = \frac{n^{2}}{1}$
$\Rightarrow 1 + \frac{D^{2}}{R^{2}} = n^{2} - 1 \Rightarrow R^{2} = \frac{D^{2}}{n^{2} - 1}$
$R = \frac{D}{\sqrt{n^{2} - 1}}$
If the bulb is not at seen through the surface $R$ must be greater than
$\frac{D}{\sqrt{n^{2} - 1}}\ \ \ \ R > \frac{D}{\sqrt{n^{2} - 1}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.