Problem 7.6.4: Modal dispersion of a multimode fibre

$a=25\mu \mathrm{m}$, ${\lambda }_0=0.85\mu \mathrm{m}$, $L=1\mathrm{km}=10^3\mathrm{m}$.

and the number of modes is about $V^2/2=1004\approx 1000$.

Axial ray: $L{n}_1/c=4.933\mu \mathrm{s}$. Critical ray: $L{n}_1/(c\sin {\theta }_c)=L{n}_1^2/(c{n}_2)=5.001\mu \mathrm{s}$. The difference is

Over $2\mathrm{km}$ $\Delta T=135\mathrm{ns}$, so $B_{\max }=1/\Delta T=7.4Mb/s$ ($BL=14.8Mb/s\mathrm{km}$).

$\Delta =0.020/1.480=0.01351$:

$225\mathrm{ps}$ over $2\mathrm{km}$, and $B_{\max }=4.4Gb/s$: $600$ times better than the step profile.

$V<2.405$ requires

The ratio of the two spreads is $\frac{(n_1-n_2)/n_2}{{\Delta }^2/8}=\frac{0.01370}{2.283\times 10^{-5}}=600$, close to $8/\Delta =592$, since the step-index spread is $\approx L{n}_1\Delta /c$.