Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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 !

share|improve this question
    
The README file provided with the samples seems to give all the information you might need. What precisely do you not understand? –  Marek Dec 1 '10 at 17:46
    
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. –  Raskolnikov Dec 1 '10 at 17:54
    
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. –  Marek Dec 1 '10 at 20:20

3 Answers 3

up vote 0 down vote accepted

From the author :

Hi,

The data is in W/sr/m2.

Best,

Jussi

So, radiance. Thanks anyway :)

share|improve this answer
    
It should be a radiance/nm. So you have to take into account the 2 nm spectral window width. –  Frédéric Grosshans Dec 3 '10 at 16:49
    
@Frédéric : Yes, sure, it's the integral of the spectrum which is in w/sr/m². –  Calvin1602 Dec 4 '10 at 0:53

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.

share|improve this answer

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.

share|improve this answer
2  
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.. –  Calvin1602 Dec 2 '10 at 8:10

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.