written 3.2 years ago by
thambisrilalith126
• 20
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Solution:
(i) For signal mode propagation, V is given by,
$ \mathrm{V}=\frac{2 \pi a}{\lambda} \sqrt{n_1^2-n_2^2}\\ $
Given $\quad n_1=1.48, \quad n_2=1.47, \lambda=800 \mathrm{~nm}$
Also $\quad \mathrm{V}=2.405$ for single mode
$\therefore \quad a=1.78 \mu \mathrm{m}$.
(ii) $\mathrm{NA}=\sqrt{n_1^2-n_2^2}=0.1717$
(iii) Let $\alpha$ be the acceptance angle, then $\sin \alpha=\mathrm{NA}$ or $\alpha=\sin ^{-1}(\mathrm{NA}) \Rightarrow \alpha=9.88^{\circ}$
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