Question

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

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

Answer 1

The correct option is B $\sqrt{\frac{3}{2}}$


$\mu =\frac{\sin {45}^{\circ }}{\sin r}$
$\Rightarrow \sin r=\frac{1}{\mu \sqrt{2}}\dots \dots \left(i\right)$
At point $Q$, for total internal reflection,
$\sin i^{{}^{\prime }}=\frac{1}{\mu }$
From figure, $i^{{}^{\prime }}={90}^{\circ }-r$
$\therefore \sin ({90}^{\circ }-r)=\frac{1}{\mu }$
$\Rightarrow \cos r=\frac{1}{\mu }\dots \dots \left(ii\right)$
Now, $\cos r=\sqrt{1-{\sin }^{2}r}=\sqrt{(1-\frac{1}{2{\mu }^2})}$
$=\sqrt{\left(\frac{2{\mu }^2-1}{2{\mu }^2}\right)}\dots \dots \left(iii\right)$
From Eqs. (ii) and(iii),
$\frac{1}{\mu }=\sqrt{\frac{2{\mu }^2-1}{2{\mu }^2}}$
$\Rightarrow \mu =\sqrt{\frac{3}{2}}$