Hi. I'm hoping to confirm some assumptions I've made about the physics of reflectors and eyepieces. I'm not a complete beginner but will be buying my first decent telescope, and I'm comparing bigger, faster scopes with smaller, slower ones.
Can anyone help me determine whether the following assumptions are accurate?
A reflector observes a fixed field of view in its parabolic main mirror and throws that entire portion of sky onto its secondary mirror, which reflects the image into an eyepiece. For any given aperture, a mirror's FOV is determined by its focal length. A slower telescope (flatter mirror, higher f-ratio) concentrates more resolving power into a narrower FOV than a faster telescope (more concave mirror, lower f-ratio). So a telescope's f-ratio defines its maximum or "natural" FOV.
It would be convenient if parabolic mirrors were flexible, and we could adjust their focal length to apply the entire power of the mirror to different fields of view/magnifications. Sadly, this engineering marvel doesn't exist, so magnification is instead achieved by cropping into the full image observed by the mirror and magnifying the result with an eyepiece lens. A higher powered eyepiece means a more aggressive crop of the natural FOV of the mirror. (The eyepiece has its own FOV too, but my understanding is that this describes the size of its image in relation to the eye, not the FOV in terms of degrees of sky.)
I assume the diameter of the focuser (whether 1.25" or 2") more or less defines the edges of this natural potential FOV of the main mirror as it projects into the eyepiece. So when we use an eyepiece with a diameter which is substantially less than the diameter of the focuser, we crop into the image observed by the main mirror by the same factor.
A slower 6" reflector at f/8 is going to see a narrower FOV than a 10" f/5. The 10" is capable of collecting more light, but due to its fast mirror and corresponding wider FOV, it spreads that power across more of the sky. If the 10" had the same f-ratio as the 6" it would indeed produce a brighter image of the same FOV as the 6", but instead, we must crop into its natural FOV to see a FOV equivalent to the slower 6".
Following this logic, for a FOV that the 6" f/8 is capable of observing, its performance is in fact more or less the same as the 10" f/5, because although the 10" mirror is bigger, to achieve the narrower FOV we crop into the main mirror and we're now seeing about 6" of the complete 10" potential of the glass. The only way a 10" f/5 can be said to be more "powerful" than a 6" f/8 is in wider FOVs, which the 6" isn't capable of observing. I understand that these wider FOVs are more suitable for DSOs, but this logic implies that objects preferring a tighter FOV like planets will perform more or less the same in both telescopes.
The telescope I'm considering (Skywatcher collapsible 10" dobs) sports a 2" focuser but comes out of the box with eyepieces that are substantially smaller than that: 25mm and 10mm. 25mm is about 1", or half the diameter of the focuser, so with a 25mm eyepiece, if my assumptions about how the optics work are accurate, we're only observing about half the diameter of the 10" mirror. If a 6" reflector also comes with a 2" focuser, then a low power eyepiece approaching the 2" diameter (40mm+) will capture the same light and produce a similar image as the 25mm on the 10" f/5.
So in summary, the fast 10" is only a useful upgrade to the 6" f/8 if I also want to buy a low powered eyepiece (perhaps a 42mm) to observe the wide view afforded by the faster mirror, but for planets, it won't perform better than the slow 6".
Thanks! I'd really appreciate knowing if these assumptions are accurate or if I've missed something obvious.