A glass fibre exhibits material dispersion given by $\left|\lambda^2 \frac{d^2 n}{d \lambda^2}\right|$ of 0.025. Determine the material dispersion parameter at a wavelength of $0.85 \mu \mathrm{m}$, and estimate the pulse broadening per kilometer with a spectral width of $20 \mathrm{~nm}$ at this wavelength.

Solution:

We know that the pulse broadening is given by,

$ \Delta \tau=\mathrm{D}_{\text {mat }}(\lambda) \mathrm{L} \Delta \lambda\\ $

where $D_{\text {mat }}(\lambda)$ defines material dispersion and is given by

$ \mathrm{D}_{\mathrm{mat}}(\lambda)=\frac{\lambda}{c}\left[\frac{d^2 n}{d \lambda^2}\right]=\frac{1}{\lambda c}\left[\lambda^2 \frac{d^2 n}{d \lambda^2}\right] $

Given,

$\quad \lambda=0.85 \mu \mathrm{m},\left[\lambda^2 \frac{d^2 n}{d \lambda^2}\right]=0.025$ and $c=3 \times 10^5 \mathrm{~km} / \mathrm{s}$

$\therefore \quad D_{\text {mat }}(\lambda)=98.1 \mathrm{ps} \mathrm{nm}^{-1} \mathrm{~km}^{-1} \quad$

Also given spectral width $\Delta \lambda=20 \mathrm{~nm}, \mathrm{~L}=1 \mathrm{~km}$

$\therefore$ Pulse brodening $\Delta \tau=98.1 \times 1 \times 20=1960 \mathrm{ps} \mathrm{km}^{-1}=1.96 \mathrm{~ns} \mathrm{~km}^{-1}$