- #1
elegysix
- 406
- 15
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
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