Calculate the number of modes at $820 \mathrm{~nm}$ in a graded index fibre having a parabolic index profile $\left(\alpha_1=2\right)$, a $25 \mu \mathrm{m}$ core radius, $n_1=1.48$ and $\mathrm{n}_2=146$. How does this compare to a step-index fiber?

Solution:

$ N=\frac{\alpha_1}{\alpha_1+2} a^2\left(\frac{2 \pi}{\lambda}\right)^2 n_1^2 \Delta \text { where } \Delta=1-\frac{n_2}{n_1}\\ $

Given,

$\alpha_1=2, a=25 \mu \mathrm{m}, \lambda=820 \mathrm{~nm}=0.82 \mu \mathrm{m}, n_1=1.48, n_2=1.46$

$ N=\frac{2}{4}\left(\frac{2 \pi \times 25}{0.82}\right)^2 \times(1.48)^2 \times\left(1-\frac{1.46}{1.48}\right) \approx 543 $

For step index fiber, the number of modes is given by,

$ \begin{aligned} & N=\left(\frac{2 \pi a}{\lambda}\right)^2 \frac{\left(n_1^2-n_2^2\right)}{2} \\ & N=\left(\frac{2 \pi \times 25}{0.82}\right)^2 \times\left(\frac{(1.48)^2-(1.46)^2}{2}\right) \approx 1079 \end{aligned} $