Optics Question 806

Question: A point light source is moving with a constant velocity v inside a transparent thin spherical shell of radius R, which is filled with a transparent liquid. If at t=0 light source is at the center of the sphere, then at what time a thin dark ring will be visible for an observer outside the sphere. The refractive index of liquid with respect to that of shell is $ \sqrt{2} $ .

Options:

A) $ \frac{R}{\sqrt{2}V} $

B) $ \frac{R}{2V} $

C) $ \frac{R}{3V} $

D) $ \frac{R}{\sqrt{3}V} $

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Answer:

Correct Answer: A

Solution:

[a] This dark ring will be visible if ray from source gets total internal reflection from the spherical shell.

Let the source at any instant be at point P then at point Q ray will be totally reflected if $ \theta $ is equal to or greater than critical angle. If QP is equal to x, then $ z=\cos \theta =\frac{R^{2}+x^{2}-v^{2}t^{2}}{2Rx} $

For $ \theta $ to be minimum $ \frac{dx}{dy}=\frac{2x( 2Rx )-2R( R^{2}+x^{2}-v^{2}t^{2} )}{4R^{2}x^{2}}=0 $

$ \Rightarrow x=\sqrt{R^{2}-v^{2}t^{2}} $

So, $ \cos \theta =\frac{2( R^{2}-v^{2}t^{2} )}{2R\sqrt{R^{2}-v^{2}t^{2}}}=\frac{\sqrt{R^{2}v^{2}t^{2}}}{R} $

For no light come out, $ \sin \theta \ge \frac{1}{\sqrt{2}}\text{ or }\theta \ge 45{}^\circ $

$ \frac{\sqrt{R^{2}-v^{2}t^{2}}}{R}=\frac{1}{\sqrt{2}};$

$\text{ }t=\frac{R}{\sqrt{2}V} $



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