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)

$\sqrt{\frac{3}{2}}$

Answer 2

The correct answer is (D) : $\sqrt{\frac{3}{2}}$
At point A by Snell's law
$\mu=\frac{sin\,45°}{sin\,r}\Rightarrow r=\frac{1}{\mu\sqrt{2}} \,...\left(i\right)$
At point B, for total internal reflection,
$sin\,i_{1}=\frac{1}{\mu}$

From figure, $i_{1} = 90^{?}- r$
$\therefore\left(sin\,90^{?}-r\right)=\frac{1}{\mu}$
$\Rightarrow cos\,r=\frac{1}{\mu}\,...\left(ii\right)$
Now $cos\,r=\sqrt{1-sin^{2}\,r}=\sqrt{1-\frac{1}{2\mu^{2}}}$
$\sqrt{\frac{2\mu ^{2}-1}{2\mu^{2}}}$
Squaring both sides and then solving, weget
$\mu=\sqrt{\frac{3}{2}}$