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

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

Answer 1

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

At point A, by Snell’s law
$\mu =\frac{sin45}{sinr}\Rightarrow sinr=\frac{1}{\mu \sqrt{2}}....\left(i\right)$
At point B, for total internal reflection $sin{i}_1=\frac{1}{\mu }$
From figure, $i_1=90-r$
$\therefore sin({90}^{\circ }-r)=\frac{1}{\mu }$
$\Rightarrow cosr=\frac{1}{\mu }...\left(ii\right)$
Now $cosr=\sqrt{1-si{n}^{2}r}=\sqrt{1-\frac{1}{2{\mu }^2}}$
$=\sqrt{\frac{2{\mu }^2-1}{2{\mu }^2}}....\left(iii\right)$
From equation (ii) and (iii) $\frac{1}{\mu }=\sqrt{\frac{2{\mu }^2-1}{2{\mu }^2}}$
Squaring both side and then solving we get $\mu =\sqrt{\frac{3}{2}}$