The speed of light in media 'A' and 'B' are $$2.0 \times 10^{10}$$ cm s$$^{-1}$$ and $$1.5 \times 10^{10}$$ cm s$$^{-1}$$ respectively. A ray of light enters from the medium $$B$$ to $$A$$ at an incident angle $$\theta$$. If the ray suffers total internal reflection, then

We need to find the condition for total internal reflection when light travels from medium B to medium A. The speed of light in vacuum is $$c = 3.0 \times 10^{10}$$ cm s$$^{-1}$$.

The refractive index of medium A is $$n_A = \frac{c}{v_A} = \frac{3.0 \times 10^{10}}{2.0 \times 10^{10}} = 1.5$$, and that of medium B is $$n_B = \frac{c}{v_B} = \frac{3.0 \times 10^{10}}{1.5 \times 10^{10}} = 2.0$$. Total internal reflection occurs when light travels from a denser medium to a rarer medium and the angle of incidence exceeds the critical angle. Since $$n_B = 2.0 > n_A = 1.5$$, medium B is denser, so light going from B to A can undergo total internal reflection.

At the critical angle $$\theta_c$$, $$\sin \theta_c = \frac{n_A}{n_B} = \frac{1.5}{2.0} = \frac{3}{4}$$ and $$\theta_c = \sin^{-1}\left(\frac{3}{4}\right)$$. Thus, total internal reflection occurs for $$\theta > \sin^{-1}\left(\frac{3}{4}\right)$$. The answer is Option C: $$\theta > \sin^{-1}\left(\frac{3}{4}\right)$$.