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. |
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