• # question_answer 11) In the figure shown, for an angle of incidence ${{45}^{o}},$at the top surface, what is the minimum refractive index needed for total internal reflection at vertical face:-                                                    A)  $\frac{\sqrt{2}+1}{2}$B)  $\sqrt{\frac{3}{2}}$C)  $\sqrt{\frac{1}{2}}$D)  $\sqrt{2}+1$

At point A, by Snell's law $\mu =\frac{\sin 45}{\sin r}\Rightarrow \sin r=\frac{1}{\mu \sqrt{2}}$         ??.(i) At point B, for total internal reflection $\sin {{i}_{1}}=\frac{1}{\mu }$ From figure, ${{i}_{1}}=90-r$ $\therefore$   $sin\,({{90}^{o}}-r)=\frac{1}{\mu }$ $\Rightarrow$   $\cos \,\,r=\frac{1}{\mu }$            ???.(ii) Now $\cos r=\sqrt{1-{{\sin }^{2}}r}=\sqrt{1-\frac{1}{2{{\mu }_{2}}}}$ $=\sqrt{\frac{2{{\mu }^{2}}-1}{2{{\mu }^{2}}}}$ ??(iii) From equation (ii) and (iii) $\frac{1}{\mu }=\sqrt{\frac{2{{\mu }^{2}}-1}{2{{\mu }^{2}}}}$ Squaring both side and then solving we get $\mu =\sqrt{\frac{3}{2}}$