question_answer

In the figure given below, for an angle of incidence \[\text{45}{}^\circ \] at the top surface, what is the minimum refractive index need to total internal reflection at vertical face?

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

B)  \[\sqrt{\frac{1}{2}}\]

C) \[\sqrt{\frac{3}{2}}\]                                     

D) \[\sqrt{2}+1\]

Correct Answer: C

Solution :

At point A, by Shell?s law \[\mu =\frac{\sin {{45}^{0}}}{\sin r}\Rightarrow \sin r=\frac{1}{\mu \sqrt{2}}\]                   ?(i) At point B, total internal reflection with \[{{\operatorname{sini}}_{1}}=\frac{1}{\mu }\] From figure,       \[{{i}_{1}}={{90}^{o}}-r\]                               ?(ii) \[\sin ({{90}^{o}}-r)=\frac{1}{\mu }\Rightarrow \cos r=\frac{1}{\mu }\]                   ?(ii) \[\mu =\frac{\sin {{45}^{\circ }}}{\sin r}\Rightarrow \sin r=\frac{1}{\mu \sqrt{2}}\]                            ? (i) At point B, total internal reflection with \[\sin {{i}_{1}}=\frac{1}{\mu }\] From figure,          \[{{i}_{1}}={{90}^{\circ }}-r\]                      ... (ii) \[\sin ({{90}^{\circ }}-r)=\frac{1}{\mu }\Rightarrow \cos r=\frac{1}{\mu }\]                            ... (iii) Now,  \[\cos r=\sqrt{1-{{\sin }^{2}}r}=\sqrt{1-\frac{1}{2{{\mu }^{2}}}}=\sqrt{\frac{2{{\mu }^{2}}-1}{{{\mu }^{2}}}}\] From Eq. (ii) and (iii), \[\frac{1}{\mu }=\sqrt{\frac{2{{\mu }^{2}}-1}{2{{\mu }^{2}}}}\] Squaring both sides and then solving, we get \[\mu =\sqrt{3/2}\]