1. Find refractive indices of the two media

The refractive index is $n = \frac{c}{v}$ where $c$ is the speed of light in vacuum and $v$ is the speed in the medium.

Since only the ratio matters, $n_A = \frac{c}{2.0\times 10^{10}}, \qquad n_B = \frac{c}{1.5\times 10^{10}}$

Because $1.5\times 10^{10} < 2.0\times 10^{10}$, medium $B$ has larger refractive index than medium $A$. So light is going from denser to rarer medium, which is the required condition for total internal reflection.

2. Condition for total internal reflection

For total internal reflection, $\theta > \theta_c$ where $\theta_c$ is the critical angle given by $\sin \theta_c = \frac{n_A}{n_B}$

Now, $\frac{n_A}{n_B} = \frac{c/(2.0\times 10^{10})}{c/(1.5\times 10^{10})} = \frac{1.5}{2.0} = \frac{3}{4}$

Hence, $\sin \theta_c = \frac{3}{4}$ so $\theta_c = \sin^{-1}\left(\frac{3}{4}\right)$

3. Apply TIR condition

Therefore, for total internal reflection, $\theta > \sin^{-1}\left(\frac{3}{4}\right)$

4. Check options

• A: $\theta = \sin^{-1}(3/4)$ → this is only the critical angle, not TIR.
• B: $\theta > \sin^{-1}(2/3)$ → not the exact condition.
• C: $\theta < \sin^{-1}(3/4)$ → wrong.
• D: $\theta > \sin^{-1}(3/4)$ → correct.

Therefore, the correct option is D.