The critical angle is $\theta$ for a light going from medium $P$ to medium Q. If speed of light in medium…
The critical angle is $\theta$ for a light going from medium $P$ to medium Q. If speed of light in medium $\mathrm{P}$ is $\mathrm{V}_{\mathrm{P}}$ then speed of light in medium $\mathrm{Q}$ is
$\frac{\mathrm{V}_{\mathrm{p}}}{\sin \theta}$
$\mathrm{V}_{\mathrm{p}} \tan \theta$
$\frac{\sin \theta}{V_p}$
$\mathrm{V}_{\mathrm{P}} \sin \theta$
Solution
For total internal reflection for critical incidence angle $\theta$ the angle of refraction is $\frac{\pi}{2}$.
Using Snell's law:
$\mathrm{n}_{\mathrm{A}} \sin \theta=\mathrm{n}_{\mathrm{B}} \sin \left(\frac{\pi}{2}\right)$, where, $\mathrm{n}_{\mathrm{A}}=$ refractive index of medium $\mathrm{A}$ and $n_B=$ refractive index of medium $B$
and $\mathrm{n}_{\mathrm{B}}=$ refractive index of medium $\mathrm{B}$
$\frac{n_A}{n_B}=\frac{1}{\sin \theta}$
We define refractive index in terms of velocity of light as follows:
$\begin{aligned} & \frac{\mathrm{n}_{\mathrm{A}}}{\mathrm{n}_{\mathrm{B}}}=\frac{\left.\mathrm{V}_{\mathrm{B}} \text { (velocity in medium } \mathrm{B}\right)}{\left.\mathrm{V}_{\mathrm{A}} \text { (velocity in medium } \mathrm{A}\right)} \\ & \therefore \frac{\mathrm{V}_{\mathrm{A}}}{\mathrm{V}_{\mathrm{B}}}=\frac{1}{\sin \theta} \\ & \mathrm{V}_{\mathrm{B}}=\frac{\mathrm{V}_{\mathrm{A}}}{\sin \theta}=\frac{\mathrm{V}}{\sin \theta}\end{aligned}$
.