question_answer

An optical fibre consists of core of m­1 surrounded by a cladding of m2 < m1. A beam of light enters from air at an angle a with axis of fibre. The highest a for which ray can be travelled through fibre is

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 \[\sin \alpha <\sqrt{\mu _{1}^{2}-\mu _{2}^{2}}\]            \[{{\alpha }_{\max }}={{\sin }^{-1}}\sqrt{\mu _{1}^{2}-\mu _{2}^{2}}\]