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} $
Show Answer
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} $
