Question

As shown in the figure, light ray $P$ enters slab at an angle $60^{\circ}$ with normal and inside the slab, light ray $Q$ suffers total internal reflection. Find the minimum refractive index of the slab.

Answer 1

At point $A$,
$1 \sin 60^{\circ}=\mu \sin r $
$\Rightarrow \sin r=\frac{\sqrt{3}}{2 \mu}$
At point $B$,
$\sin \left(90^{\circ}-r\right)>\sin C$
$\Rightarrow \cos r>\frac{1}{\mu}$
$\Rightarrow \sqrt{1 \frac{3}{4 \mu^{2}}}>\frac{1}{\mu}$
$\Rightarrow 1-\frac{3}{4 \mu^{2}}>\frac{1}{\mu^{2}}$
$\Rightarrow 1>\frac{7}{4 \mu^{2}}$
$\Rightarrow \mu^{2}>\frac{7}{4} $
$\Rightarrow \mu > 1.32$