0
$\begingroup$

In this answer I estimated that if a DSLR (digital SLR) camera with an aperture of 40 mm were pointed at a satellite in geostationary orbit with a visual brightness of about +11 magnitude, it would produce roughly 200 e- per second in the camera's silicon sensor.

I did a similar estimate here.

The estimate is only slightly better than order-of-magnitude. I assumed that the spectrum was reflected solar, and the average photon had about 2.5 eV of energy.

Is this about right? Is there any better way to do this starting with a given number meant to be "visual magnitude"?

I understand there are plenty of different definitions of magnitude that involve the specification of a specific band, (e.g. UBV) but when you are handed simply a visual magnitude, how can one estimate number of photons electrons in silicon?

Yes, a precise calculation is quite sophisticated, but here I'm asking only for a good way to estimate. Let's call a silicon CCD's QE 0.8 and flat for example.

$\endgroup$
  • 2
    $\begingroup$ V band magnitude 0 is 3640 Janskys (sorry) which translates to 998.81 photons/s/Angstrom/cm^2 (sorry again, Astronomy units are often stoopid). You can scale to your magnitude and if you have some idea of the collecting area and the bandpass of the filter you can make an estimate. You then need some idea of the extinction of the atmosphere (0.15mag per airmass is a start for V) and reflectivity/transmission of the optics. Be warned, this rabbit hole goes very deep very steeply... $\endgroup$ – astrosnapper Jan 28 at 19:43
  • $\begingroup$ The QE is irrelevant if all you want is incoming photons/pixel. If you want photoelectrons then you need QE, and making it flat is a VeryBadIdea(TM) $\endgroup$ – Carl Witthoft Jan 28 at 20:04
  • $\begingroup$ @CarlWitthoft thank's it's a typo; I mention "e- (carriers)" in the title and 2.5 eV conversion, so it's charge that I'm after. For most of the Sun's visible spectrum the QE of a good silicon silicon CCD, expresses as electron-hole pairs per incoming photon, should be fairly flat. For a bare die with a broadband ARC it could be near 0.8. If you plot with A/W rather than quante per quanta then it's a ramp shape. However for a color camera that's way too high because of the Bayer filter. $\endgroup$ – uhoh Jan 29 at 0:03
  • $\begingroup$ @astrosnapper that's extremely handy & exactly what I'm looking for. My rabbit hole-phobia inspired the sentence at the end re only looking for an estimate. I think your comment, plus the approx wavelength limits of the V band, plus a simple plot of W/m^2/um from some solar power inspired plot of a solar spectrum is all I'd need for an acceptable answer. That plus a reminder (as mentioned above) of the SLR's bayer filter's losses. $\endgroup$ – uhoh Jan 29 at 0:13
  • 2
    $\begingroup$ And the red/IR blocking filter in DSLRs unless it has been removed or is an 'a'stronomy model e.g. Canon 60Da. Here is the Generic Bessell V filter profile filter profile (and many others) $\endgroup$ – astrosnapper Jan 29 at 0:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.