Assignment 5, solutions:

Problem 1:

At l_free = 830 nm we have (d²b/dw²)(w/l)d = ~1000 ps/nm = 1 s/m.

D(1/v_g)= (d²b/dw²)Dw
(d²b/dw²)(w/l)11000m = (1s/m).
(d²b/dw²) = (9.09*10^-5 s/m)(l/w) = (9.09*10^-5 s/m)(l²/(2pc)) = (2.087*10^-25/(2p))s²/m

Here  Df = 2(c/(2nl)) =  (3 10⁸)/(3.5 * 200 10^-6) Hz = 4.286 10¹¹ Hz between the highest and lowest mode.

Dw = 2pDf. 

D(1/v_g) = (d²b/dw²)Dw = (2.087*10^-25)(s²/m) *44.286 10¹¹/s = 8.9*10^-14(s/m)
|D(v_g)| = v_g²D(1/v_g) = (c²/n²)*8.9*10^-14(s/m) = 657m/s

The change in the pulse width t over a distance d is then given by

Dt = (d²b/dw²)Dwd.

Dt   = [(2.087*10^-25/(2p))s²/m]*2pDfd = (8.95*10^-14s/m) *d

The pulses is 10 ns wide.  If the pulse spreading is to be kept below 20%, we need Dt < 2ns,
or d < 2*10^-9 s /(8.95*10^-14 s/m) = 22.3 km.