Question

In the figure given below, for an angle of incidence $45^{\circ }$ at the top surface, what is the minimum refractive index needed for total internal reflection at the 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)

At point $A$ , by Snell's law

$\mu =\frac{\sin 45^{\circ }}{\sin r}\Rightarrow \sin r=\frac{1}{\mu \sqrt{2}}\dots$ (i)
At point $B$, total internal reflection with $\sin i_1=\frac{1}{\mu }$
From figure, $i_1=90^{\circ }-r\dots$ (ii)
$\sin \left(90^{\circ }-r\right)=\frac{1}{\mu }\Rightarrow \cos r=\frac{1}{\mu }\dots .$ (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
$\mu =\sqrt{\frac{3}{2}}$