Question

$A B C D$ is the plane of glass cube. A horizontal beam of light enters the face $A B$ at the grazing incidence. Show that the angle $\theta$ which any ray emerging from $B C$ would make with the normal to $B C$ is given by $\sin \theta=\cot \alpha$ where $\alpha$ is the critical angle. The greatest value that the refractive index of glass may have if any of the light is to emerge from $B C$ is $\sqrt{a}$, then value of $a$ is ?

Answer 1

Grazing incidence means $i =90^{\circ}$
At point $E$,
$1 \times \sin 90^{\circ}=\mu \sin \alpha $
$\Rightarrow \mu=\frac{1}{\sin \alpha}$
At point $F$,
$\mu \sin \left(90^{\circ}-\alpha\right)=1 \times \sin \theta$
$\sin \theta=\mu \cos \alpha=\frac{\cos \alpha}{\sin \alpha}$
$\sin \theta=\cot \alpha$
Now for ray to emerge from $BC$,
Angle of incidence on $BC \leq \alpha$ (critical angle)
$90^{\circ}-\alpha \leq \alpha $
$\Rightarrow \alpha \geq 45^{\circ}$
$\sin \alpha \geq \sin 45^{\circ}$
$\frac{1}{\mu} \geq \frac{1}{\sqrt{2}} $
$\Rightarrow \mu \leq \sqrt{2}$
Maximum value of refractive index $=\sqrt{2}$