Correct option: D.$\sqrt{2} < \mu < \sqrt{3}$
d
(d) When light is made incident over face $A B$, refraction occurs as shown below.
From geometry of figure,
$r_1=90-\theta_c$
$i=90-\alpha$
Refractive index $\mu$ of prism is
$\mu=\frac{1}{\sin \theta_c}=\frac{\sin i}{\sin r_1}$
$\Rightarrow \quad \mu =\frac{1}{\sin \theta_c}=\frac{\sin (90-\alpha)}{\sin \left(90-\theta_c\right)}$
$\Rightarrow \quad \mu =\frac{1}{\sin \theta_c}=\frac{\cos x}{\cos \theta_c}$
Also, when incidence is made over face $A C$, refraction occurs as shown below.
Again, from geometry of figure,
At surface $A B$, $TIR$ occurs
$\Rightarrow i > \theta_c$
$\Rightarrow 90-r > \theta_c$
$\Rightarrow 90-\theta_c > r_2$
$\Rightarrow \sin \left(90-\theta_c\right) > \sin r_2$
$\Rightarrow \cos \left(\theta_c\right) > \frac{\sin \alpha}{\mu}$
$\Rightarrow {\left[\therefore \mu=\frac{\sin i}{\sin r_2}=\frac{\sin \alpha}{\sin r_2}\right]}$
$\cos \theta_c > \frac{\sqrt{1-\cos ^2 \alpha}}{\mu}$
$\Rightarrow \cos \theta_c > \frac{\sqrt{1-\mu^2 \cos ^2 \theta_c}}{\mu}$
$\Rightarrow 2 \mu^2 \cos ^2 \theta_c > 1 \Rightarrow 2 \mu^2\left(1-\sin ^2 \theta_c\right) > 1$
$\Rightarrow 2 \mu^2\left(1-\frac{1}{\mu^2}\right) > 1$
$\Rightarrow 2 \mu^2-2 > 1$
$\Rightarrow 2 \mu^2 > 3$
$\Rightarrow \mu > \sqrt{\left(\frac{3}{2}\right)}$
This matches with option (d).
