A) \[{{\cos }^{-1}}\sqrt{\mu _{2}^{2}-\mu _{1}^{2}}\]
B) \[{{\sin }^{-1}}\sqrt{\mu _{1}^{2}-\mu _{2}^{2}}\]
C) \[{{\tan }^{-1}}\sqrt{\mu _{1}^{2}-\mu _{2}^{2}}\]
D) \[{{\sec }^{-1}}\sqrt{\mu _{1}^{2}-\mu _{2}^{2}}\]
Correct Answer: B
Solution :
Here the requirement is that \[i>c\] \[\Rightarrow \,\,\sin i>\sin c\,\,\,\Rightarrow \,\,\sin i>\frac{{{\mu }_{2}}}{{{\mu }_{1}}}\] ?..(i) From Snell?s law \[{{\mu }_{1}}=\frac{\sin \alpha }{\sin r}\] ?.(ii) Also in \[\Delta OBA\] \[r+i={{90}^{o}}\] \[\Rightarrow \,\,\,r=(90-i)\] Hence from equation (ii) \[\sin \alpha ={{\mu }_{1}}\sin (90-i)\] \[\Rightarrow \,\,\,\cos i=\frac{\sin \alpha }{{{\mu }_{1}}}\] \[\sin i=\sqrt{1-{{\cos }^{2}}i}\]\[=\sqrt{1-{{\left( \frac{\sin \alpha }{{{\mu }_{1}}} \right)}^{2}}}\] ?.(iii) From equation (i) and (iii) \[\sqrt{1-{{\left( \frac{\sin \alpha }{{{\mu }_{1}}} \right)}^{2}}}>\frac{{{\mu }_{2}}}{{{\mu }_{1}}}\] Þ \[{{\sin }^{2}}\alpha <(\mu _{1}^{2}-\mu _{2}^{2})\] Þ \[\sin \alpha <\sqrt{\mu _{1}^{2}-\mu _{2}^{2}}\] \[{{\alpha }_{\max }}={{\sin }^{-1}}\sqrt{\mu _{1}^{2}-\mu _{2}^{2}}\]You need to login to perform this action.
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