Question

A light ray falls on a square glass slab as shown in the diagram. The index of refraction of the glass, if total internal reflection is to occur at the vertical face, is equal to :

• A. $\frac{\left(\sqrt{2}+1\right)}{2}$
• B. $\sqrt{\frac{5}{2}}$
• C. $\frac{3}{2}$
• D. $\sqrt{\frac{3}{2}}$

Answer 1

Answer: (D)

At point A by Snell’s law
$\mu =\frac{sin45^{°}}{sinr}\Rightarrow r=\frac{1}{\mu \sqrt{2}}...\left(i\right)$
At point B, for total internal reflection,
$sin{i}_1=\frac{1}{\mu }$
From figure, $i_1=90^{°}-r$
$\therefore \left(sin90^{°}-r\right)=\frac{1}{\mu }$
$\Rightarrow cosr=\frac{1}{\mu }...\left(ii\right)$
Now $cosr=\sqrt{1-si{n}^{2}r}=\sqrt{1-\frac{1}{2{\mu }^2}}$
$\sqrt{\frac{2{\mu }^2-1}{2{\mu }^2}}$
Squaring both sides and then solving, weget
$\mu =\sqrt{\frac{3}{2}}$