Optics Question 127
Question: 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
Options:
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}} $
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Answer:
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-{{( \frac{\sin \alpha }{{\mu _{1}}} )}^{2}}} $ ..(iii) From equation (i) and (iii) $ \sqrt{1-{{( \frac{\sin \alpha }{{\mu _{1}}} )}^{2}}}>\frac{{\mu _{2}}}{{\mu _{1}}} $
Therefore $ {{\sin }^{2}}\alpha <(\mu _{1}^{2}-\mu _{2}^{2}) $
Therefore $ \sin \alpha <\sqrt{\mu _{1}^{2}-\mu _{2}^{2}} $
$ {\alpha _{\max }}={{\sin }^{-1}}\sqrt{\mu _{1}^{2}-\mu _{2}^{2}} $
