Determine the normalized frequency at $0.82 \mu \mathrm{m}$ for a step-index fiber having a $25 \mu \mathrm{m}$ core radius, $n_1=1.48$ and $n_2=1.46$. How many modes propagate in this fiber at $0.82 \mu \mathrm{m}$?

Solution:

Solution: We know that the normalized frequency V is given by,

$ V=\frac{2 \pi a}{\lambda} \sqrt{\left(n_1^2-n_2^2\right)}\\ $

Given

$\quad \alpha=25 \mu \mathrm{m}, \lambda=0.82 \mu \mathrm{m}, n_1=1.48$ and $n_2=1.46$.

$ \therefore \quad V=\frac{2 \pi \times 25}{0.82} \sqrt{(1.48)^2-(1.46)^2}=46.45\\ $

Since the number of modes propagates through the fiber is,

$\frac{V^2}{2}$,

$N=\frac{V^2}{2} \therefore N=\frac{(46.45)^2}{2} \approx 1079$