In the figure shown, for an angle of incidence 45°, at the top surface, the minimum refractive index needed for total internal reflection at vertical face is $\sqrt{\frac{3}{\mathrm{n}}}$ . The value of n is _____ .
 

If $\mathrm{\theta }$ is the angle of reflection, then 

$\begin{array}{l}\frac{\sin {45}^{\circ }}{\sin \mathrm{r}}=\mathrm{\mu }\\ \Rightarrow \sin \mathrm{r}=\frac{1}{\sqrt{2}\mathrm{\mu }} ------ \left(\mathrm{i}\right)\end{array}$
 

$\begin{array}{l}\text{ From figure, }{\mathrm{i}}_1=90-\mathrm{r}\\ \mathrm{for} \mathrm{TIR}, \left({90}^{\circ }-\mathrm{r}\right)>\mathrm{C}\\ \sin \left({90}^{\circ }-\mathrm{r}\right)>\mathrm{sinC}\\ \therefore \sin \left({90}^{\circ }-\mathrm{r}\right)>\frac{1}{\mathrm{\mu }}\\ \Rightarrow \mathrm{cosr}>\frac{1}{\mathrm{\mu }}-------\left(\mathrm{ii}\right)\\ \mathrm{Now}, \mathrm{cosr}=\sqrt{1-{\sin }^2\mathrm{r}}=\sqrt{1-\frac{1}{2{\mathrm{\mu }}^2}}\\ =\sqrt{\frac{2{\mathrm{\mu }}^2-1}{2{\mathrm{\mu }}^2}}----\left(\mathrm{iii}\right)\\ \mathrm{From} \mathrm{equation} \left(\mathrm{ii}\right) \mathrm{and} \left(\mathrm{iii}\right), \\ \sqrt{\frac{2{\mathrm{\mu }}^2-1}{2{\mathrm{\mu }}^2}}>\frac{1}{\mathrm{\mu }}\\ \mathrm{Squaring} \mathrm{both} \mathrm{side} \mathrm{and} \mathrm{then} \mathrm{solving} \mathrm{we} \mathrm{get} \\ \mathrm{\mu }>\sqrt{\frac{3}{2}}\end{array}$