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

converting 20mw to watt 20000000

Then dividing by 2 after travelling each km

11-10 ----10000000
10-9 ------5000000
9-8 ------2500000
8-7 ------1250000
7-6 ------ 625000
6-5 ------ 312500
5-4 ------ 156250
4-3 ------ 78125
3-2 ------ 39062.5
2-1 ------ 19531.25
1-0 ------ 9765.625