Physics - Geometrical Optics Question with Solution | TestHub

Velocity of light in medium $${\mathrm{V}}_{\text{med }}=\frac{3\mathrm{cm}}{0.2\mathrm{ns}}=\frac{3\times 10^{-2}\mathrm{m}}{0.2\times 10^{-9}\mathrm{s}}=1.5\mathrm{m}/\mathrm{s}$$ Refractive index of the medium $$\mu =\frac{V_{\text{air }}}{V_{\text{med }}}=\frac{3\times 10^8}{1.5}=2\mathrm{m}/\mathrm{s}$$ As $\mu =\frac{1}{\sin C}$ $$\therefore \sin \mathrm{C}=\frac{1}{\mu }=\frac{1}{2}=30^{\circ }$$ Condition of TIR is angle of incidence i must be greater than critical angle. Hence ray will suffer TIR in case of $(B)\left(i=40^{\circ }>30^{\circ }\right)$ only.