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## Homework Statement

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?

## Homework Equations

a = (10/L)log(P

_{in}/P

_{out})

where L = distance

a= attenuation, units: dB/km

## The Attempt at a Solution

a)

0.30 dB/km = (10/1.2*10

^{-6})log(P

_{in}/P

_{out})

0.30*10

^{-6}/ 10 = log(P

_{in}/P

_{out})

3.0*10

^{-8}= log(P

_{in}/P

_{out})

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

b)

I don't really understand what b) is asking for though and am not really sure where to start.