Question

In the figure given below, for an angle of incidence ${45}^o$ at the top surface, what is the minimum refractive index need to total internal reflection at vertical face?

• A. $\frac{\sqrt{2}+1}{2}$
• B. $\sqrt{\frac{1}{2}}$
• C. $\sqrt{\frac{3}{2}}$
• D. $\sqrt{2}+1$

Answer 1

Answer: (C)

$\sqrt{\frac{3}{2}}$

At point $A$, by Snell's law,

$\mu =\frac{\sin {45}^o}{\sin r}\Rightarrow \sin r=\frac{1}{\mu \sqrt{2}}$ ....(i)

At point $B$, total internal reflection with $\sin i_1=\frac{1}{\mu }$

From figure, $i_1={90}^o-r$ ....(ii)

$\sin ({90}^o-r)=\frac{1}{\mu }\Rightarrow \cos r=\frac{1}{\mu }$ ....(iii)

Now, $\cos r=\sqrt{1-{\sin }^{2}r}=\sqrt{1-\frac{1}{2{\mu }^2}}=\sqrt{\frac{2{\mu }^2-1}{{\mu }^2}}$

From equation (ii) and (iii), $\frac{1}{\mu }=\sqrt{\frac{2{\mu }^2-1}{2{\mu }^2}}$

Squaring both sides and then solving, we get

$\mu =\sqrt{3/2}$