Optics Question 958
Question: If the signal is transmitted from an optical fibre core of Refractive Index (RI) $ \sqrt{\frac{12}{5}} $ to an another optical fibre with RI of core and cladding as 1.8 and 1.2 respectively, then the maximum angle of acceptance for 2nd optical fibre is
Options:
A) $ 30{}^\circ $
B) $ 45{}^\circ $
C) $ 60{}^\circ $
D) $ {{\sin }^{-1}}( \frac{2}{\sqrt{3}} ) $
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Answer:
Correct Answer: C
Solution:
[c] Idea In optical fibre maximum angle of accepts is given by
$ \sin \theta =\frac{\sqrt{n _{1}^{2}-n _{2}^{2}}}{n _{0}} $
Maximum angle of acceptance is given by
$ \sin \theta =\frac{\sqrt{n _{1}^{2}-n _{2}^{2}}}{n _{0}}=\frac{\sqrt{{{1.8}^{2}}-{{1.2}^{2}}}}{\sqrt{12/5}} $
$ =\sqrt{(1.8-1.2)(1.8+1.2)}\times \sqrt{\frac{5}{12}} $
$ \Rightarrow $ $ \sin \theta =\sqrt{0.6\times 3}\times \sqrt{\frac{5}{12}}=\frac{\sqrt{3}}{2} $
$ \Rightarrow $ $ \theta ={{\sin }^{-1}}( \frac{\sqrt{3}}{2} ) $
$ \Rightarrow $ $ \theta =60{}^\circ $
BETA
