The critical angle is theta for light going f | vedclass

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Question

(A) For total internal reflection at the critical angle $	heta$,the angle of refraction is $\frac{\pi}{2}$.Using Snell's law at the interface between

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$\frac{V_{P}}{\sin 	heta}$

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(A) For total internal reflection at the critical angle $	heta$,the angle of refraction is $\frac{\pi}{2}$.Using Snell's law at the interface between medium $P$ and medium $Q$:$n_{P} \sin 	heta = n_{Q} \sin \left(\frac{\pi}{2}ight)$Since $\sin \left(\frac{\pi}{2}ight) = 1$,we have $n_{P} \sin 	heta = n_{Q}$.This implies $\frac{n_{P}}{n_{Q}} = \frac{1}{\sin 	heta}$.The refractive index $n$ is inversely proportional to the speed of light $V$ in that medium $(n = \frac{c}{V})$,so $\frac{n_{P}}{n_{Q}} = \frac{V_{Q}}{V_{P}}$.Equating the two expressions:$\frac{V_{Q}}{V_{P}} = \frac{1}{\sin 	heta}$Therefore,the speed of light in medium $Q$ is $V_{Q} = \frac{V_{P}}{\sin 	heta}$.

The critical angle is $\theta$ for light going from medium $P$ to medium $Q$. If the speed of light in medium $P$ is $V_{P}$,then the speed of light in medium $Q$ is:

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