Maybe you can help me:
I understand that the amount of photons that will hit a sensor will depend on the aperture (more aperture meaning more photons, shorter exposures) and on the f-ratio (longer f ratio meaning fewer photons, longer exposures).
What are the formulas to quantify this flux of photons?
What would allow for shorter exposures, a 5.5" F6 scope, or a 6.1" F5.5 scope?
Putting aside a few stupendously incorrect answers, there are a number of good comments here. I'd like to expand a bit on how this stuff works. Right up front you need to understand that there is a difference in the radiometry of extended sources (pretty much everything but stars) and extended sources. In normal, every day photography, we only deal with extended sources. In the night sky, we deal with both stars, which are point sources, and extended sources--such as nebula, galaxies, and solar system objects. The equations from my Etendue post comparing two systems are valid for extended objects where the irradiance in the focal plane varies only as the inverse of the focal ratio squared. That means that for any given object, an 8 m telescope operating at say F/4 will produce an image that has exactly the same irradiance in the focal plane as an 80 mm, F/4 telescope.
That is an amazing result! The 8 m telescope is 100 times larger with 10,000 times more collecting area so how can that be? The answer lies in how the light is spread out in the image plane. Remember that the 8m scope has a focal length that is 100 times larger so the image will also be 100 times larger in the focal plane. That means that even though the large scope gathers 10,000 times more photons, it spreads them out over 10,000 times more area in the focal plane. That will illuminate the focal plane with the same irradiance (in W/m2) for each of the two scopes. The big (and I mean really big) difference will be the image scale. That big scope produces a much bigger image than the image produced by the 80 mm scope even though they are at the exact same irradiance.
When it comes to stars, things are different because diffraction, rather than first order properties, determines the size of the point spread function. First, forget about the atmosphere and think of a scope in a vacuum. The maximum irradiance of a star image from a perfect telescope varies with the (diameter/focal ratio)2 = (D/F)2. That means that for any given focal ratio, a larger telescope will pick up fainter stars than a smaller telescope. Double the size and you quadruple peak stellar irradiance.
So what happens when we throw in the atmosphere? When light passes through the atmosphere it adds a time varying deviation to the inbound plane wavefront from a star. That means the instantaneous star image will no longer be a perfect Airy pattern and that the form of the pattern will vary with time. When we take a long time exposure of a star, we record the time averaged pattern, which will always be larger than what we would get in a vacuum. That means that the light energy from a star will be smeared over a larger area on the sensor and we'll record a lower peak signal compared to what we might expect. A star image will still not behave like an extended source but it won't vary perfectly as (D/F)2 either. It will be close...just not perfect and how imperfect will depend on bad the seeing might be. You can see this effect for yourself by imaging an area of the sky dense in stars and carefully comparing the faintest stars detectable on nights of good seeing verses nights with poor seeing. So, a larger diameter telescope will always produce star images with higher peak irradiance in the focal plane than a smaller scope with the same focal ratio even when you consider atmospheric seeing.
Remember that so far I've only mentioned the effect on the optical irradiance in the focal plane. To get at the signal you measure, you then have to fold in the sensor characteristics. The size of the pixels, the responsivity (Q.E.) and the exposure length will determine the ultimate signal level that you'll record. Remember that larger pixels always record a higher signal but if you make the pixels too big, you risk under-sampling the image and throwing away detail. However, this is where sampling plays its most important role in this whole thing. Realize that if you take two telescopes of different focal ratios and use equal sampling in object space, the signal strength will be exactly the same! That's why I'm content with my slow F/10.8 system for imaging. It produces exactly the same signal as an F/5 system that uses a camera with the exact same angular sampling relative to the sky. I've attached a slide to show how this works. Just remember that the two scopes have to have different cameras--you can't just remove the camera from the f/10.8 and plop it onto the F/5 scope (that's equal sampling in image space.) You only get the same signal when the sampling relative to the sky is constant (assuming the QE of the two cameras is equal.)
Finally, to directly answer your question, the larger faster F/5.5, 6.1" scope will produce around 19% more irradiance in the focal plane for extended objects than the F/6, 5.5" scope. It will also image slightly fainter stars for the same camera and exposure time. That's not a huge difference. In my view, differences like this don't become significant until they reach a level of around 30% - 50% since other factors can may overwhelm their relative importance. Portability, weight, optical quality, available accessories and stuff like that may be more important than that 19%. On the other hand, differences of 200% - 300% really get my attention and that's where I'm much more willing to put my efforts. Everything being equal, I will personally always opt for the larger aperture.
Edited by jhayes_tucson, 19 July 2019 - 10:47 AM.