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Optical path length change due to absorption 
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#1
Dec213, 08:02 PM

P: 28

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.
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! 


#2
Dec313, 08:51 AM

Sci Advisor
P: 5,510

I got different results from you we agree until
Does this help? 


#3
Dec313, 06:18 PM

P: 28

Oh thanks for the correction on the first part, that was a silly mistake.
The amount of energy absorbed by the fiber is instantaneous and continuous right. So if the system absorbs 0.03mW of the optical power, there should be a way to make an approximation an figure out the increase in temperature over a period of 12 hours, right? 


#4
Dec413, 08:54 AM

Sci Advisor
P: 5,510

Optical path length change due to absorption
Sure: 0.03 mW * 12 hrs = 2.6 J. Now all you need is the specific heat of the fiber and anything in thermal contact with the fiber.



#5
Dec413, 07:07 PM

P: 28

Thanks!



#6
Dec413, 08:32 PM

P: 28

Except I think its 1.3J



#7
Dec513, 10:12 AM

Sci Advisor
P: 5,510




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