Question

A light ray fall on a square glass slab shown in the diagram the refractive index of the glass , if total internal reflection is to occur at the vertical face is equal to :

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

Answer 1

Answer: (D)

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

At the point $A$-

$\mu \mathrm{Sin}\left(i_1\right)=\mathrm{Sin}{45}^{\circ }$

$\mathrm{Sin}\left(i_1\right)=\frac{1}{\sqrt{2}\mu }$

Now, $i_2={90}^{\circ }-i_1$

So, at the point $B$ due to total internal reflection-

$\mathrm{Sin}\left(i_2\right)=\frac{1}{\mu }$

$\mathrm{Sin}({90}^{\circ }-i_1)=\frac{1}{\mu }$

$\mathrm{Cos}\left(i_1\right)=\frac{1}{\mu }$

$\sqrt{1-{\mathrm{Sin}}^2\left(i_1\right)}=\frac{1}{\mu }$

$1-\frac{1}{2{\mu }^2}=\frac{1}{{\mu }^2}$

$1=\frac{1}{{\mu }^2}\left(1+\frac{1}{2}\right)$

${\mu }^2=\frac{3}{2}$

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