question_answer

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\[\sqrt{2}\].        

A) \[\frac{R}{\sqrt{2}V}\]

B) \[\frac{R}{2V}\]

C) \[\frac{R}{3V}\]

D) \[\frac{R}{\sqrt{3}V}\]

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 \[\frac{dx}{dy}=\frac{2x\left( 2Rx \right)-2R\left( {{R}^{2}}+{{x}^{2}}-{{v}^{2}}{{t}^{2}} \right)}{4{{R}^{2}}{{x}^{2}}}=0\] \[\Rightarrow x=\sqrt{{{R}^{2}}-{{v}^{2}}{{t}^{2}}}\] So, \[\cos \theta =\frac{2\left( {{R}^{2}}-{{v}^{2}}{{t}^{2}} \right)}{2R\sqrt{{{R}^{2}}-{{v}^{2}}{{t}^{2}}}}=\frac{\sqrt{{{R}^{2}}{{v}^{2}}{{t}^{2}}}}{R}\] For no light come out, \[\sin \theta \ge \frac{1}{\sqrt{2}}\text{ or }\theta \ge \text{45}{}^\circ \] \[\frac{\sqrt{{{R}^{2}}-{{v}^{2}}{{t}^{2}}}}{R}=\frac{1}{\sqrt{2}};\text{    }t=\frac{R}{\sqrt{2}V}\]