Question

A certain optical fibre has attenuation of 3.5db/km at 850 nm. if 0.5 nW of optical fibre is launched into the fibre, what is the power level in mW after 4 km ?

Answer 1

Attenuation is given by formula

$α=\frac {10} {L}×\log{\frac {P_{in}}{P_{out}}} (1)$

In our case, α=3.5db/km, P_in=0.5 nW, L=4 km

We have

$α×L=14 db (2)$

Using (1) and (2) we get

P_out=19.9µW

Answer

0.01991 mW