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Do someone knows the units of the spectra provided here ? It seems obvious enough that it's said nowhere, but even Wikipedia and other sites are quite blurry on this point.

So, is it power ($W$), radiance($\:\rm{W/m^2sr}$), or something else ?

Thanks !

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  • $\begingroup$ The README file provided with the samples seems to give all the information you might need. What precisely do you not understand? $\endgroup$
    – Marek
    Commented Dec 1, 2010 at 17:46
  • $\begingroup$ It doesn't specify in what units the numbers in the other files are. But I suppose knowing the apparatus that did the measurement can help resolve that issue. $\endgroup$
    – Raskolnikov
    Commented Dec 1, 2010 at 17:54
  • $\begingroup$ Yeah, I meant READMEs included with other measurements. But then I noticed that different apparatuses were used in those cases and also different format of data so it's not really very useful. $\endgroup$
    – Marek
    Commented Dec 1, 2010 at 20:20

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Usually, the spectrum is in arbitrary unit, proportional to the power, the radiance or whatever. The information the spectrum provides is the relative radiance of various wavelengths which is contained in the spectrum. If you want the radiance of one wavelength band, you simply multiply the relevant band of a normalized spectrum (of integral 1) by the total radiance of your source.

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    $\begingroup$ At the price of those tools, one could expect something better than relative results, and as a matter of fact, the spec (cs.unc.edu/Research/stc/FAQs/spectroradiometer/705spec.pdf) says it can mesure luminance, radiance, and some other things. Anyway, I e-mailed the author, maybe he'll answer.. $\endgroup$
    – Calvin1602
    Commented Dec 2, 2010 at 8:10
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From the author :

Hi,

The data is in W/sr/m2.

Best,

Jussi

So, radiance. Thanks anyway :)

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  • $\begingroup$ It should be a radiance/nm. So you have to take into account the 2 nm spectral window width. $\endgroup$
    – Frédéric Grosshans
    Commented Dec 3, 2010 at 16:49
  • $\begingroup$ @Frédéric : Yes, sure, it's the integral of the spectrum which is in w/sr/m². $\endgroup$
    – Calvin1602
    Commented Dec 4, 2010 at 0:53
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Most spectrometers work by spatially separating light into its component frequencies by means of a diffraction grating, and measuring that with a CCD array. That means measuring the intensity*, or power per unit area, the units of which are watts per square meter. However, there is usually an unknown scaling factor involved, so don't expect the values to actually be in W/m². I would guess that the values in these files are normalized so that a value of 1 is the intensity at which the corresponding CCD pixel saturates.

*Strictly speaking, you are measuring the intensity present in one range of frequencies, but it's simpler and usually justified to just treat it as one frequency.

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