Q.
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 :
Solution:
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
$\mu=\sqrt{\frac{3}{2}}$
