question_answer

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 [DCE 2002]

A)            \[\frac{\sqrt{2}+1}{2}\]

B)            \[\sqrt{\frac{3}{2}}\]

C)            \[\sqrt{\frac{1}{2}}\]

D)            \[\sqrt{2}+1\]

Correct Answer: B

Solution :

                   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, \[{{i}_{1}}=90-r\]                    \[\therefore \sin (90{}^\circ -r)=\frac{1}{\mu }\]                    \[\Rightarrow \cos r=\frac{1}{\mu }\]         .....(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 \[\mu =\sqrt{\frac{3}{2}}\]