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