In a particular fibre, the attenuation coefficient for Rayleigh scattering is measured to be 0.30 dB/km at 1.20 μm.
(a) How large would it be expected to be at 0.80 μm?
(b) Ignoring other sources of attenuation, if a signal of a certain initial power was sent over a distance of 50 km of the fibre, what would be the power in the signal at the other end if the wavelength was 0.80 μm, as a fraction of the power that would be received at 1.20 μm?
a = (10/L)log(Pin/Pout)
where L = distance
a= attenuation, units: dB/km
The Attempt at a Solution
0.30 dB/km = (10/1.2*10-6)log(Pin/Pout)
0.30*10-6 / 10 = log(Pin/Pout)
3.0*10-8 = log(Pin/Pout)
a = (10/0.8*10-6)*3.0*10-8
a = 0.375 dB/km - this seems reasonable, as the distance is shorter, attenuation should be higher
I don't really understand what b) is asking for though and am not really sure where to start.