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Originally Posted by Starkler
Its all about photons per unit area.
Consider a 10" f5 and a 5" f5 scope. The exposure time for an extended object will be the same but the image will be smaller on the ccd chip for 5" scope due to its shorter focal length.
If you take an sct of f10, then put a focal reducer in to bring it to f6.3, you are aiming the collected photons for a given object to a smaller number of pixels on the camera.
So, if you keep f-ratio as a constant and vary the aperture of a scope, the photon density per pixel wont change. As aperture increases, the field of view will be decreased due to longer focal length and the image of an object will appear larger.
I notice the same effect with visual observing. For a given exit pupil size, brightness of extended objects appears the same regardless of aperture. Having a great big truss dob means that you can get a closer look at higher mags whilst still maintaining that brightness.
I think thats all correct  |
Geoff, that about the best layman's description I've seen.
The simplified explanation I have in my head from an astrophotography book I've read was along the lines of "aperture affects the exposure time for point sources like stars, and F ratio affects the exposure time for extended sources like nebulae, the moon, terrestrial photos etc".
Of course, the two are related.
Al.