Rich, Let me take a crack at this. I think the key is to remember that in a linear optical system, angles are preserved. This means that if an object subtends one degree in the sky, it will subtend one degree in a telescope. Here's an example. The moon subtends about 1/2 degree in the sky. For an 8" f7 scope (focal length 56 inches) a 1/2 degree image will be .49 inches tall. The image height for an 8" f4.5 (36 inch focal length) will be .31 inches. An 8" scope gathers 8 inches worth of light. You can spend the light any way you want to. FOR EXTENDED OBJECTS, all eight inches worth of that light is in the image. If, however, you form a smaller image because of a shorter focal length, that image will still contain all the light, but being smaller, it will be brighter. This is not true for stars. They are point sources, and thorough good optical systems of any f ratio their size is determined by aperture. A star in an 8" f7 is the same size as a star in an 8" f4.5. It is determined by diffraction limits. What's the result of this? Extended objects AT THE FOCAL PLANE will be brighter with faster scopes. Stars will be the same brightness independent of f ratio. An 8" scope will show the same magnitude star no matter what the f ratio is. If you run 50 power, the image is the same brightness in the f7 scope as it is in the f4.5 scope. You have magnified the image fifty times OVER WHAT THE EYE SEES in both cases, and you have done it with an 8" scope. The brightness will be the same in both. It is only at the focal plane where the brightness difference exists between the two focal ratios. Remember that power per inch of aperture determines visual brightness. This is th basic principle of what is happening. Other things will come into play (like magnification and non-linearity of the eye) that modify this, but the basic rule remains the same. Still clear as mud? Brent --- Richard Tenney <retenney@yahoo.com> wrote:
My question obviously does belie my complete ignorance of astrophotography and the various methods used to image deep sky objects. When I read that a short focal length scope can photograph a given object X in "half" the time as a longer focal length scope of the same aperture, I see now what they didn't tell me is that the resulting image on the film is also 1/4 the size -- that's a big "duh". So I still fail to see the photographic advantage of a fast scope, unless what is imaged is huge (e.g., the veil nebula) and image size isn't all that important, right? Am I thinking straight here? At any rate, I'll quit pestering the list with my ignorance and go check out a few books on astrophotography and get a clue how this all works.
-Rich
--- Chuck Hards <chuckhards@yahoo.com> wrote:
It's not different. It's identical, given identical apertures. You're confused because you're using eyepiece projection and making the image identical.
You get a shorter exposure in an f/4 scope than an f/8 scope of identical aperture because the image is only 1/4 as large at the focal plane. If you make the images of identical size through eyepiece projection, they are of the same brightness because they are now the same size (cover the same area). You are now no longer comparing an f/4 to an f/8. You've made them identical with the eyepiece, as far as brightness goes.
C.
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