Question

In fig For an angle of incidence ${45}^{\circ }$ at top surface what is the minimum refractive index needed for total internal reflection at vertical face AD?

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

Answer 1

Answer: (C)

The correct option is C $\sqrt{3/2}$

AT point of $A$ by snell's

$\begin{array}{c}\mu =\frac{\sin {45}^0}{\sin r}\\ \Rightarrow \sin r=\frac{1}{\mu \sqrt{2}}..........\left(1\right)\end{array}$

From figure,

$\begin{array}{c}{i}_1=90-r\\ \therefore \sin \left({90}^0-r\right)=\frac{1}{\mu }\\ Now,\\ \Rightarrow \cos r=\frac{1}{\mu }..........\left(2\right)\\ \cos r=\sqrt{1-{\sin }^{2}r}=\sqrt{1-\frac{1}{2{\mu }^2}}\\ =\sqrt{\frac{2{\mu }^2-1}{2{\mu }^2}}...........\left(3\right)\\ fromequation\left(2\right)and\left(3\right)\\ \frac{1}{\mu }=\sqrt{\frac{2{\mu }^2-1}{2{\mu }^2}}\end{array}$

Squaring both sides then solving we get $=\sqrt{\frac{3}{2}}$