A $6 \mathrm{~km}$ optical link consists of multimode step-index fiber with a core refractive index of 1.5 and a relative refractive index difference of $1 \%$. Estimate the delay difference between the slowest and fastest modes at the fiber output.

Solution:

We know that in the case of step index fiber, the fiber output. $n_2$ is less than the core refractive index $n_1$.

$ n_1=1.5 \text { and } n_1-n_2=1 \% \text { of } n_1=\frac{1}{100} \times 1.5=.015\\ $

$ n_2=1.5-0.015=1.485\\ $

Now the pulse dispersion for step-index fiber is,

$ \Delta T=\frac{n_1 L}{c}\left(\frac{n_1}{n_2}-1\right)\\ $

where L is the length of optical fiber, and c is the velocity of light.

$ \begin{aligned} \therefore \quad \Delta T & =\frac{1.5 \times 6}{3 \times 10^5}\left(\frac{1.5}{1.485}-1\right) \quad\left(\because c=3 \times 10^5 \mathrm{~km} / \mathrm{sec}\right) \\\\ & =30.3 \times 10^{-3} \mathrm{sec}=30.3 \mathrm{~m} \mathrm{sec}\\ \end{aligned} $

Thus, the delay difference between the slowest and fastest modes at the fiber output is $30.3 \mathrm{~m}$ sec.