Hey, so I am trying to figure out the optical path length change of an optical fiber due to absorption. I'll post what I have so far and let me know if I have done anything wrong or if you have suggestions.(adsbygoogle = window.adsbygoogle || []).push({});

so the optical path length is

[itex]L' = nL[/itex]

(L' is optical path length, n is refractive index and L is the length)

the change in length is

[itex]dL' = L dn + n dL \approx L_0 n_0 \beta \Delta T + n_0 L_0 \alpha \Delta T [/itex]

Where [itex]\beta=1.28\times10^{-5} /^{\circ}C[/itex] is the change in refractive index with temperature

[itex] \alpha = 5.5\times10^{-7}/^{\circ}C[/itex] is the thermal expansion coefficient

[itex] L_0 = 700\text{m}[/itex] is the original length and [itex] n_0 = 1.4585 [/itex] is the original refractive index

this all gives [itex]dL' = (14.763\times 10^{-3} m/^{\circ} C)\Delta T [/itex]

For the temperature change I know the attenuation is 0.2dB/km so for 700m that is 0.14dB and if the output is 0dBm (1mW) then this gives an absorbed amount of power of 0.14dB or 0.9683mW

(this next part is where I am a little unclear)

The change in temperature would be equal to

[itex] \Delta T = \frac{\text{Power Dissapated}}{\text{Thermal Conductivity}\times\text{Length}} = \frac{P}{TC \times L_0} [/itex]

the thermal conductivity of fused silica is 1.3W/mK

This would give a temperature change of

[itex] \Delta T = 1.064 \times 10^{-6} C[/itex]

But there is a problem here, time is not involved, my system is running for close to 12 hours and should be constantly increasing in temperature (minus the dissipated heat but I am assuming that is not much of a factor as the cable is insulated)

How do I factor in time to determine the rate of temperature change which would then give me the change in length per hour?

Any help is greatly appreciated, Cheers!

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# Optical path length change due to absorption

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