1. Find refractive indices of the two media

The refractive index is $n=\frac{c}{v}$ where $c=3\times 10^8\,\text{m s}^{-1}$.

For $M_1$: $n_1=\frac{3\times 10^8}{1.5\times 10^8}=2$

For $M_2$: $n_2=\frac{3\times 10^8}{2.0\times 10^8}=1.5=\frac{3}{2}$

So, $M_1$ is optically denser than $M_2$.

2. Use the formula for critical angle

Critical angle is defined when light goes from denser to rarer medium: $\sin C=\frac{n_{\text{rarer}}}{n_{\text{denser}}}$

Thus, $\sin C=\frac{n_2}{n_1}=\frac{\tfrac{3}{2}}{2}=\frac{3}{4}$

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

3. Match with the given options

We need to check which option is equivalent to $\sin^{-1}(3/4)$.

Using a right triangle: if $\sin C=\frac{3}{4}$ then take opposite $=3$, hypotenuse $=4$. So adjacent side is $\sqrt{4^2-3^2}=\sqrt{16-9}=\sqrt{7}$

Therefore, $\tan C=\frac{3}{\sqrt{7}}$ which gives $C=\tan^{-1}\left(\frac{3}{\sqrt{7}}\right)$

So the correct option is A.

4. Check other options briefly

• B: $\tan^{-1}(2/3)$ is not equal to $\sin^{-1}(3/4)$
• C: $\cos^{-1}(3/4)$ means $\cos C=3/4$, not correct
• D: $\sin^{-1}(2/3)$ is not correct since $\sin C=3/4$

Therefore, the answer is A.