[The_bluester (Paul)]

OK, this one even has my wife semi stumped (Trained as a professional photographer by the RAAF, who at that time even went into the chemical reactions going on in the film/paper to make it all work in film based photography) and she had to mumble about not quite recalling the techincalities of it.

F stop of a camera lens is easy to understand and easy to see why exposure times vary. Stop it down, reduce the apeture, reduce the amount of light falling on the media (Film, sensor, whatever) Open it up, more light, shorter exposure, simple.

How does it work with a scope? I can understand my CPC925 being an F10 system and that it will need exposures of X amount of time at prime focus etc. But how, if I get a fastar setup and convert it to F6 (I think) does it work to reduce exposures required to a fraction of the time when the objective that is gathering the light is precisely the same as before (Except it probably has a larger obstruction from replacing the secondary mirror with the fastar and the body of the camera)

Is it an inverse square thing where the light has had to travel further at F10 from the objective mirror to the camera media than with a fastar or other setup at a faster ratio? And if so, how does that work with folded optics like the 925? Apart from the fact that it just would not work, if I imaged at F2, straight off the objective, versus imaging at the prime focus at F10, the light clearly has not travelled five times as far!





I hate just taking things on faith! I am an ex tech so if anyone has a really detailed and techincal explanation hit me with it and I will go work it out if I do not follow it right away. This matters to me not at all, I am not really into imaging, but it is bugging me and I want to understand so I can forget it again!