A ray of light is incident at an angle of 60⁰ on the face AD of a transparent slab. What should be the refractive index of the material of the slab so that total internal reflection occurs at face AB of the slab?

By Snell's law,

$\begin{array}{rcl}\mathrm{\mu } \sin \mathrm{r}& =& \sin {60}^{\mathrm{o}}=\frac{\sqrt{3}}{2}\\ \sin \mathrm{r}& =& \frac{\sqrt{3}}{2\mathrm{\mu }} .... \left(1\right)\end{array}$

The refracted ray will be totally reflected at E if it is incident at an angle equal to or greater than the critical angle $i_c$ Which is given by

$\sin {\mathrm{i}}_{\mathrm{c}}=\frac{1}{\mathrm{\mu }} ....\left(2\right)$

It follows from the figure that$\mathrm{r}=90° - {\mathrm{i}}_{\mathrm{c}}$, 

Therefore,
$\begin{array}{rcl}\sin \mathrm{r}& =& \sin (90°-{\mathrm{i}}_{\mathrm{c}})=\cos {\mathrm{i}}_{\mathrm{c}}\\ \sin \mathrm{r}& =& {(1-{\sin }^2{\mathrm{i}}_{\mathrm{c}})}^{\frac{1}{2}} ....\left(3\right)\end{array}$

Using (1) and (2) in (3), we get
$\begin{array}{rcl}\frac{\sqrt{3}}{2\mathrm{\mu }}& =& {\left(1-\frac{1}{{\mathrm{\mu }}^2}\right)}^{\frac{1}{2}}\\ & \Rightarrow & \frac{3}{4{\mathrm{\mu }}^2}=1-\frac{1}{{\mathrm{\mu }}^2}\\ & \Rightarrow & \mathbf{\mu }\mathbf{=}\frac{\sqrt{\mathbf{7}}}{\mathbf{2}}\end{array}$