A) \[\frac{\sqrt{3}+1}{2}\]
B) \[\frac{\sqrt{2}+1}{2}\]
C) \[\sqrt{\frac{3}{2}}\]
D) \[\sqrt{\frac{7}{6}}\]
Correct Answer: C
Solution :
Key Idea: For total internal reflection angle of incidence should be greater than critical angle. For total internal reflection to take place, angle of incidence > critical angle \[i.e.,i>C\] \[or\theta >C\] \[or\sin \theta >\sin C\] \[but\sin C=\frac{1}{\mu }\] and from figure, \[\theta ={{90}^{o}}-r\] So, \[\sin ({{90}^{o}}-r)>\frac{1}{\mu }\] \[i.e.,\mu >\frac{1}{\cos r}...(i)\] From Snell?s law, \[\frac{\sin \,{{45}^{o}}}{\sin \,r}=\mu \] \[\Rightarrow \sin r=\frac{1}{\sqrt{2}\,\,\mu }\] \[\therefore \cos r=\sqrt{1-{{\sin }^{2}}\,r}=\sqrt{1-\frac{1}{2{{\mu }^{2}}}}\] Thus Eq. (i) becomes \[\mu >\frac{1}{\sqrt{1-\frac{1}{2{{\mu }^{2}}}}}\] \[\therefore {{\mu }^{2}}=\frac{1}{1-\frac{1}{2{{\mu }^{2}}}}\] \[or{{\mu }^{2}}-\frac{1}{2}=1\] \[or\mu =\sqrt{\frac{3}{2}}\]
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