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 $\alpha=\sin ^{-1}(\mathrm{NA}) \Rightarrow \alpha=9.88^{\circ}$