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 \[\mu =\sqrt{3/2}\]You need to login to perform this action.
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