I know that long focal-length scopes have a smaller fully-illuminated FOV, but I don't have an exact formula. Does anyone have an equation where I can plug in aperture and focal length to get the size of the fully-illuminated FOV? Thanks.
Fully-illuminated field
Started by Jeff Verona, 12/27/2002 11:20PM
Posted 12/27/2002 11:20PM
Opening Post
Posted 12/28/2002 03:36AM
#1
What type of telescope are you asking about? In a Newtonian, the fully illuminated field depends on the aperture, the focal length, the size of the diagonal, and the distance from the center of the diagonal to the focal plane. In a refractor is depends on the aperture, focal length, and the size and placement of the baffles.
Posted 12/28/2002 09:56PM
#2
Thanks for the help. What I'm specifically looking for, I guess, is how large a light cone you can get for a given aperture/focal length combination. For example, suppose I had a 6" f/10 Mak. How large will the image circle be at the visual back? If it's less than 35mm, then it wouldn't make any sense to use a 40mm eyepiece. Anybody have a quick equation for this, or does it depend so much on baffling, etc., that it has to be computed based on the actual design and build of the scope?
Funding Member
