Physics - Geometrical Optics Question with Solution | TestHub

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: $\begin{array}{rl}& \frac{{\mathrm{n}}_{\mathrm{A}}}{{\mathrm{n}}_{\mathrm{B}}}=\frac{{\mathrm{V}}_{\mathrm{B}}\text{ (velocity in medium }\mathrm{B})}{{\mathrm{V}}_{\mathrm{A}}\text{ (velocity in medium }\mathrm{A})}\\ & \therefore \frac{{\mathrm{V}}_{\mathrm{A}}}{{\mathrm{V}}_{\mathrm{B}}}=\frac{1}{\sin \theta }\\ & {\mathrm{V}}_{\mathrm{B}}=\frac{{\mathrm{V}}_{\mathrm{A}}}{\sin \theta }=\frac{\mathrm{V}}{\sin \theta }\end{array}$