## Optical Fiber attenuation coefficient

Hi, I am curious to know what is a typical numerical value for the attenuation coefficient of common optical fibre. Please don't include anything strange in the units like dB since I am new to that. Thanks.
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 Google for "optical fiber attenuation" http://www.arcelect.com/Calculating_...d_distance.pdf As one would suspect, it depends on multiple things but the length of one segment seems to fall within the 100s of meters to 10s of Km range (according to this reference anyway).
 Recognitions: Homework Help There are many types of fibers in existence, but I can tell you that the attenuation of most modern Erbium-doped fibers is amazing. We can achieve attenuation coefficients of 4dB per kilometer or even lower (at the transmission wavelengths). I'm not really up to date with the latest manufacturing breakthroughs. You're going to have to get used to the dB scale, because it is the most convenient way to calculate power loss in generally any optical system. To convert between dB and % gain, you do: 10 log (X) = (dB) So for a -4dB gain, you have X = 0.3981 So that means that for every 1 km, you lose 60% of your power. Putting 10 Watts through, you get 3.981 Watts at the other end.

## Optical Fiber attenuation coefficient

So attenuation coefficient here would be 0.6 per km, right?

Mentor
 Quote by teleport So attenuation coefficient here would be 0.6 per km, right?
I'm assuming that the -4dB/km number is for light intensity or power, so you would do the conversion to a ratio like this:

$$-4dB = 10 log \frac{P}{P_0}$$

So $$\frac{P}{P_0} = 10^{-4/10} = 0.398$$
 OK. I'm trying to set up this equation: $$I = I_0e^{-\alpha x}$$. So this $$\alpha$$ attenuation coefficient would be (without dB please)?

Mentor
 Quote by teleport OK. I'm trying to set up this equation: $$I = I_0e^{-\alpha x}$$ . So this $$\alpha$$ attenuation coefficient would be (without dB please)?
I'd have to think about it more to be sure, but my first guess would be that you would want to use the ratio per km that I showed in my post, and fit it to your exponential equation. Something like....

$$I = I_0e^{-\alpha x}$$

$$\frac{I}{I_0} = e^{-\alpha x} = 0.398$$ where $$x = 1km$$

$$ln( 0.398 ) = -\alpha * 1km$$

etc. Does that work? You can test it to see if you plug in 2km, and get 0.398^2 as the intensity ratio....
 It looks good to me. Thanks a lot.
 It also depends on the wavelength of light that you're using. If you operate at 1.55 microns, you'll have the least attenuation.
 But what is the actual relationship to wavelength. Decreases with wavelength? If so, in what way? i.e linearly, or...?
 It's not a simple relationship. You need to look at a graph like this one.
 Thank you. Wow that's very intriguing. Why is it that attenuation is related to specific wavelengths? I do hope the answer does not have to do with QM as most strange things these days are explained by that.
 it also depends on the fiber type, multi-mode will have different dispersion characteristics based on it's modal properties and step index compared to single mode fibers, fiber can be considered a waveguide for photons. some of the more interesting things i guess would be optics basics, polarization, brillouin scattering, raman scattering, four-wave mixing.

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