Calculate minimum intensity of a point source visible to a camera

I have two cameras and know the specified minimum faceplate sensitivity for both:

Camera 1: Minimum faceplate sensitivity 4 x 10-6 Lux
Camera 2: Minimum faceplate sensitivity 2 x 10-7 lux

I also know the minimum intensity (1 x 10-7 W m-2) of a point source (470 nm) visible to Camera 2 at a distance of 0.5 m from the front of the camera (ambient light levels: zero).

I would like to be able to calculate the minimum intensity of a point source visible Camera 1 through the same medium (seawater). (Camera 1 is no longer functional and so cannot be calibrated)

Although there would obviously be a number of factors that would add error into this calculation, is there a way to calculate this (supposing ideal conditions)?

• Is it not simply 20 times greater (so $2\cdot 10^{-6}W/m^2$) – Floris Jul 1 '14 at 13:12
• If that's the case, then great! I didn't want to make a mistake based on such an assumption. – Jesinsky Jul 1 '14 at 13:22
• Just remember not to do this calculation with a spectrally broad source, especially going thru water. "Lux" is a photopic value, and variation in the spectral power emittance (and transmittance to your camera) will throw that linear relationship off. For a narrowband source, no problem :-) – Carl Witthoft Jul 1 '14 at 15:00

So when you consider how important it is to get this answer "as right as possible", you have to know the spectral response of the two cameras. If it is the same, then all is good and your answer is $2 \cdot 10^-6 W / m^2$ . If it is different, then you need to consider whether you expect your point source to be white or colored: because the answer will depend.