consider a optical fiber link of 56 km the attenuation is 02

consider a optical fiber link of 56 km. the attenuation is 0.25 db/km. if the transmit power is 1mw what is the received power at the end of the fiber

Solution

attenuation=10 x Log₁₀(P in/P out)

Here attenuation= 0.25 db/km  *56 km

=14db

Hence

14=10 x Log₁₀(P in/P out)

14=10 x Log₁₀(1mW/P out)

1.4=Log₁₀(1mW/P out)

25.118864=1mW/P out

P out=1/25.118864mW

P out=0.039810718 mW