Optical path length change due to absorption

  1. 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. jcsd
  3. Andy Resnick

    Andy Resnick 5,963
    Science Advisor
    Education Advisor

    I got different results from you- we agree until

    0.14 dB power loss means 0.968 mW is *transmitted* through the fiber (with 1 mW initial power)- only 0.03 mW is absorbed. Then, instead of:

    I get a temperature change of 2.3*10^-5 mK (thermally isolating the fiber form the environment). In any case, the reason there is no time dependence is because you started with a time-independent equation: it was ΔT, not T(t).

    Does this help?
     
  4. 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?
     
  5. Andy Resnick

    Andy Resnick 5,963
    Science Advisor
    Education Advisor

    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.
     
  6. Thanks!
     
  7. Except I think its 1.3J
     
  8. Andy Resnick

    Andy Resnick 5,963
    Science Advisor
    Education Advisor

    Oopsy...
     
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