1. Refraction at first face $AB$

Let the angle of incidence on face $AB$ be $\theta$, and the angle of refraction inside the prism at face $AB$ be $r_1$.

By Snell's law: $\sin \theta = \mu \sin r_1$ so, $r_1 = \sin^{-1}\!\left(\frac{\sin\theta}{\mu}\right)$

2. Condition at second face $AC$

Let the angle of incidence at the second face inside the prism be $r_2$. For a prism, $r_1 + r_2 = A$ Hence, $r_2 = A - r_1$

For the ray to emerge from face $AC$, it must not suffer total internal reflection. Thus, $r_2 \le i_c$ where the critical angle $i_c$ for glass-air interface is $i_c = \sin^{-1}\left(\frac{1}{\mu}\right)$

Therefore, $A - r_1 \le \sin^{-1}\left(\frac{1}{\mu}\right)$

This gives $r_1 \ge A - \sin^{-1}\left(\frac{1}{\mu}\right)$

3. Convert this into condition on $\theta$

Using $\sin \theta = \mu \sin r_1$ we get $\sin \theta \ge \mu \sin\left(A - \sin^{-1}\left(\frac{1}{\mu}\right)\right)$

Hence, $\theta \ge \sin^{-1}\left[\mu \sin\left(A - \sin^{-1}\left(\frac{1}{\mu}\right)\right)\right]$

So the ray will be transmitted through face $AC$ provided $\theta$ is greater than this limiting value.

Thus the matching option is: $\boxed{\theta > \sin^{-1}\left[\mu \sin\left(A - \sin^{-1}\left(\frac{1}{\mu}\right)\right)\right]}$

4. Check options

• A and B involve $\cos^{-1}$, which does not arise here.
• D has the wrong inequality sign.
• C matches the derived result.

Therefore, the correct option is C.