Accepted Answer
After one meter, 3.4% of the light is gone ... either soaked up in the fiber material or escaped from it. So only (100 - 3.4) = 96.6% of the light remains, to go on to the next meter.After the second meter, 96.6% of what entered it emerges from it, and that's 96.6% of 96.6% of the original signal that entered the beginning of the fiber.==> After 2 meters, the intensity has dwindled to (0.966)² of its original level.It's that exponent of ' 2 ' that corresponds to the number of meters that the lighthas traveled through.==> After 'x' meters of fiber, the remaininglight intensity is (0.966) ^x-powerof its original value.If you shine 1,500 lumens into the front of the fiber, then after 'x' meters of cable, you'll have (1,500) · (0.966)^xlumens of light remaining. =========================================The genius engineers in the fiber design industry would not handle it this way.When they look up the 'attenuation' of the cable in the fiber manufacturer'scatalog, it would say "15dB per 100 meters".What does that mean ? Break it down: 15dB in 100 meters is 0.15dB per meter.Now, watch this:Up at the top, the problem told us that the loss in 1 meter is 3.4% . We applied super high mathematics to that and calculated that 96.6% remains, or 0.966.Look at this ==> 10 log(0.966) = -0.15 <== loss per meter, in dB .Armed with this information, the engineer ... calculating the loss in 'x' meters offiber cable, doesn't have to mess with raising numbers to powers. All he has todo is say ...-- 0.15 dB loss per meter -- 'x' meters of cable-- 0.15x dB of loss.If 'x' happens to be, say, 72 meters, then the loss is (72) (0.15) = 10.8 dB .and 10 ^ (-10.8/10) = 10 ^ -1.08 = 0.083 = 8.3% <== That's how much light he'll have left after 72 meters, and all he had to do was a simple multiplication.Sorry. Didn't mean to ramble on. But I do stuff like this every day.
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