Irradiance measurement procedure

  1. Here's what I'm thinking. The sun is too bright to measure directly with our equipment. If I calibrate without a filter, and capture the reference spectrum with and without the filter, then I can model how the filter changes the spectrum. This way I can capture the sun spectrum with the filter, and transform the data as if I had not used it.

    What I mean to say is, if
    [itex]I_{r}(\lambda)=Y(\lambda)[/itex]
    and
    [itex]I_{r+f}(\lambda)=G(\lambda)Y(\lambda)[/itex]
    then is it valid to argue that
    [itex]I_{s}(\lambda)=\frac{I_{s+f}(\lambda)}{G(\lambda)}[/itex]

    Or is it that [itex]G(\lambda)[/itex] is dependent on I?

    where
    [itex]I_{r}[/itex] is the irradiance of the reference
    [itex]I_{r+f}[/itex] is the irradiance of the reference measured through a filter
    [itex]I_{s}[/itex] is the irradiance of the sample
    [itex]I_{s+f}[/itex] is the irradiance of the sample measured through the same filter
     
  2. jcsd
  3. Andy Resnick

    Andy Resnick 5,751
    Science Advisor

    As a first approximation, that approach is fine. The main shortcomings with this approach are 1) if G(λ) << 1 (the filter transmits very little light at some wavelengths) and/or 2) if your detector is a coarse-grained spectrometer (say a color camera). Problem (1) introduces error by amplifying noise, and problem (2) means that the measured spectrum is a convolution, not a multiplication, so the 'division' step is actually a deconvolution.

    The filter transmittance should not vary with intensity unless it has been specifically designed to do so (e.g. a saturable absorber).
     
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