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

Answer: 1.32

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