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Measure optical power with spectrometer

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  1. Jun 11, 2013 #1
    Hi all,

    I have only one spectrometer and 2 different light sources. I need to measure the optical power of the light sources.
    The spectrometer has a resolution of 8 nm.
    I measure the spectrum of each light source. And then calculate the optical power by integrating over the spectrum.
    My first light source is a LED with a spectral full-width-at-half-maximum (FWHM) of 24 nm. The optical power caldulated from the measured spectrum should be correct.
    But my second light source is a laser with a spectral FWHM of 2 nm (given by supplier). Then my spectrometer is not able to resolve the spectrum of my laser anymore. But if i still integrate over the measured spectrum, do i still have the "right" value of my laser's optical power?


    Thanks all
     
  2. jcsd
  3. Jun 11, 2013 #2

    Andy Resnick

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    Spectrometers are not a good tool for measuring optical power- that's what power meters are for. The reason is that the efficiency (including the detector) varies over the spectral range; in order to use the spectrometer as a power meter, you first need to calibrate the instrument with a known source (e.g. a calibrated blackbody).

    As for measuring the power output of the laser, why not read the label? Lasers must have that specification listed somewhere on the device.
     
  4. Jun 12, 2013 #3
    Hello Andy,

    My problem is i don't have any powermeter. Of course, i have the laser power from the specification. But I'm setting up a bench to test my laser and i need to be able to read out the optical power (even if only relative value) for different driving conditions.

    To summarize, if i calibrate my spectrometer correctly with a known source. Then the resolution of my spectrometer does not affect my " measurable relative optical power", does it?
     
  5. Jun 12, 2013 #4

    Andy Resnick

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    No, the measurement will still be poor- your spectral resolution is too large as compared to the spectrum of your source. If the laser output stays within the same spectral 'bin', you may be able to get satisfactory results, but some lasers (solid state, mainly) emit different linewidths with different driving powers.

    How much power are you talking about- a few mW? uW? why not get an inexpensive photodiode and use that instead? How precise does this measurement need to be?
     
  6. Jun 12, 2013 #5
    Hi again,

    Yes my laser is a solid-state laser, a semiconductor laser diode. The power emitted by the laser is 1.5W (high power laser diode). As i said, i care only about relative value of optical power, therefore i don't have to couple the whole 1.5W to my measurement device.

    Using a photodiode came also to me as "more natural" approach for power measurement. I'm also considering this solution.

    But as i have "cheap" spectrometer available, i'm just wondering why not?

    Theorically, it should work. But i have doubts that's why i'm asking for help and hope that we could come to some clear conclusion :-)
    As the measured spectrum (S) is the convolution of my laser spectrum and the spectral response function of the spectrometer. If i integrate the measured spectrum (S) over the whole wavelength range, mathematically, the result is a quantity (A) which is proportional to the optical power of the laser. And the constant relating this quantity A and the optical power is defined by the spectrometer characteristics (including the resolution).
    So i deduce that if i want the real value of the optical power, then the spectrometer resolution does matter.
    But if i observe the evolution of A (because i know that it's proportional to my optical power), then i can use my spectrometer to measure A even if the resolution is not enough to resolve the laser spectral line.
    It's clear that i'll not be able to deduce any information about wavelength shift, details and shape of laser spectrum, etc...
    But again, the quantity A is always proportional to the optical power despite the fact that the resolution is not enough to resolve the laser spectral line.
    It's true in the experimental world?

    :confused:
     
  7. Jun 14, 2013 #6

    Claude Bile

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    Had a think about this, and I don't think it works.

    While the integral of

    [tex]\int f_1(\lambda)\tau(\lambda) d\lambda [/tex]

    is indeed proportional to A (f = source spectrum of source 1, tau = instrument response), it tells you nothing about

    [tex]\int f_2(\lambda)\tau(\lambda) d\lambda [/tex]

    unless you know tau.

    A basic thought experiment - what if the two sources are in different parts of the spectrum (one at 400 nm, one at 600 nm say). The instrument response at 400 nm would tell you nothing about the instrument response at 600 nm.

    You would need to know the shape of the spectral instrument response at a minimum.

    Hope this makes sense.

    Claude.
     
    Last edited: Jun 14, 2013
  8. Jun 20, 2013 #7
    Hi Claude,

    I agree with you if i want to measure the absolute value of optical power. But I want only to measure a relative value as explained in the attached file below.

    Hope that clear my issue.

    Regards
     

    Attached Files:

  9. Jun 20, 2013 #8
  10. Jun 21, 2013 #9

    Claude Bile

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    Hi Juliettengo,

    Forgive me, I don't have msword on this pc, but...

    I'm still not convinced it will work, since;

    [tex]\frac{\int f_1(\lambda)\tau(\lambda)d\lambda}{\int f_2(\lambda)\tau(\lambda)d\lambda} \neq \frac{\int f_1(\lambda)d\lambda}{\int f_2(\lambda)d\lambda}[/tex]

    (in general.)

    You can only get useful information re: power if tau is constant.

    I'm open to any suggestions to the contrary, but those are my initial thoughts.

    Claude.
     
  11. Jun 21, 2013 #10
    Hi Claude,

    I completly agree with your formula.
    but i integrate my measured spectrum over the whole wavelength range to get the area under the curve ∫∫f1(λ-λ')tau(λ)dλdλ'.
    This time i attach the file in pdf. Sorry to not type all the equations here.
    Hope that u can see my calculations this time.

    :-)

    @ Johnbbahm: Thanks, it also looks like we will go for a kind of thermal sensor.
     

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