question_answer

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

A) \[30{}^\circ \]

B) \[45{}^\circ \]

C) \[60{}^\circ \]

D) \[{{\sin }^{-1}}\left( \frac{2}{\sqrt{3}} \right)\]

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

\[\sin \theta =\frac{\sqrt{n_{1}^{2}-n_{2}^{2}}}{{{n}_{0}}}=\frac{\sqrt{{{1.8}^{2}}-{{1.2}^{2}}}}{\sqrt{12/5}}\]

\[=\sqrt{(1.8-1.2)(1.8+1.2)}\times \sqrt{\frac{5}{12}}\]\[\Rightarrow \]\[\sin \theta =\sqrt{0.6\times 3}\times \sqrt{\frac{5}{12}}=\frac{\sqrt{3}}{2}\]\[\Rightarrow \]   \[\theta ={{\sin }^{-1}}\left( \frac{\sqrt{3}}{2} \right)\]\[\Rightarrow \]      \[\theta =60{}^\circ \]

TEST Edge Questions based on optical are frequently asked in examination students are advised to study important terms like cladding principle of optical fibre etc.