
#1
Nov1313, 07:22 PM

P: 269

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
Nov1413, 08:04 AM

Sci Advisor
P: 5,468

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 coarsegrained 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). 


Register to reply 
Related Discussions  
UV and irradiance  General Physics  3  
Actual earth measurement contradicts measurement predicted by special relativity  Special & General Relativity  93  
Irradiance, UV, measurement  General Physics  2  
computing irradiance  Advanced Physics Homework  0  
irradiance  Advanced Physics Homework  10 