question_answer

In the figure given below, for an angle of incidence \[{{45}^{o}}\] 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 Snells law                 \[\mu =\frac{\sin {{45}^{o}}}{\sin r}\Rightarrow \sin r=\frac{1}{\mu \sqrt{2}}\] ? (i) At point B, total internal reflection with \[\sin {{i}_{1}}=\frac{1}{\mu }\] From figure, \[{{i}_{1}}={{90}^{o}}-r\] ... (ii)                 \[\sin \,\,({{90}^{o}}-r)=\frac{1}{\mu }\] \[\Rightarrow \] \[\cos r=\frac{1}{\mu }\] ... (ii) Now,     \[\cos r=\sqrt{1-{{\sin }^{2}}r}=\sqrt{1-\frac{1}{2{{\mu }^{2}}}}\]                  \[=\sqrt{\frac{2{{\mu }^{2}}-1}{{{\mu }^{2}}}}\] 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 \[\mu =\sqrt{3/2}\]