Question

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

• A. $1.72$
• B. $1.52$
• C. $1.32$
• D. $1.12$

Answer 1

The correct option is C $1.32$

$\alpha =90-C$

$\beta =180-90-\alpha$

$\beta =90-\alpha$

$\beta =C$

So r = 90 - C

& i = ${60}^0$ (given)

${\mu }_1=1$

${\mu }_2=\mu$

So from snells formula

${\mu }_{1}sini={\mu }_{2}sinr$

$1sin{60}^0=\mu sin(90-C)$

$\frac{\sqrt{3}}{2}=\mu cosC$

$\frac{\sqrt{3}}{2}=\mu \sqrt{1-\frac{1}{{\mu }^2}}$

$\frac{3}{4}={\mu }^2(1-\frac{1}{{\mu }^2})$

$\frac{3}{4}={\mu }^2-1$

$\frac{7}{4}={\mu }^2$

$\frac{\sqrt{7}}{2}=\mu \Rightarrow \mu =1.3287$