Question

$ABCD$ is the plane of glass cube. A horizontal beam of light enters the face $AB$ at the grazing incidence. Show that the angle $\theta$ which any ray emerging from $BC$ would make with the normal to $BC$ 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 $BC$ is $\sqrt{a}$, then value of $a$ is ?

Answer 1

Answer: 2.00

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 $BC\le \alpha$ (critical angle)
$90^{\circ }-\alpha \le \alpha$
$\Rightarrow \alpha \ge 45^{\circ }$
$\sin \alpha \ge \sin 45^{\circ }$
$\frac{1}{\mu }\ge \frac{1}{\sqrt{2}}$
$\Rightarrow \mu \le \sqrt{2}$
Maximum value of refractive index $=\sqrt{2}$