Quote:
Originally Posted by Don Pensack
Yes, if you calculate the field of view per millimeter of focal plane, you'll be able to take the field stop size of the eyepiece and calculate the size of the field of view.
Example:
FL = 1200mm
2" eyepiece inside diameter = 46mm.
True field = 46/1200 * 57.3 = 2.1965 degrees of field=131.8'
So, 46mm = 131.8' of field.
An eyepiece with a 10mm field stop sees 10/46 x 131.8' of field = 28.65' of true field.
Think of the image scale on the telescope's focal plane as being related to its focal length, and an eyepiece acting as a simple magnifier of part of that field.
How much dimmer is the extended image of the background sky in the eyepiece than the apparent illumination at the telescope's focal plane?
Simple.
The True field of that 10mm field stop eyepiece was 28.65'. Let's say the apparent field of the eyepiece was 68 degrees. The "parent" field of the eyepiece is (68x60) 4080'. To get that field width from a 28.65' true field requires 142X. Brightness of the background is 1/(142 squared) = 1/20164 x as bright as the image on the focal plane.
But, stars don't stretch with magnification--they're points. So the background sky is 1/20164 X as bright as the image at the focal plane (about 10.8 magnitudes fainter), but the stars are the same brightness. So the contrast is greatly improved, and we see fainter stars with increased magnification.
Back to the part of the focal plane being viewed:
Assuming two telescopes, one of which has twice the aperture of the other, and having identical focal lengths: the larger scope will put twice the number of photons per square millimeter on the focal plane. The image scale (minutes per millimeter) will be exactly the same, but more photons will be being crammed into that area.
Remember that a star on axis is the result of the entire aperture bringing the image of that star to the focal plane. When you view that star, regardless of what focal length or field width of eyepiece being used, you are using the entire aperture for that on-axis star image. It is true that you are examining only a portion of the focal plane, but the eyepiece does not determine what portion of the primary mirror is used to illuminate anything on the focal plane of the telescope--it merely magnifies that image. So an extended object will get fainter with magnification, but stars do not.
Hope my explanation makes sense. Let me know if it doesn't.
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Okay starting to get the picture.
The maximum diameter of the actual focal plane is also limited by the minor axis of the secondary mirror in a newt?
I am struggling with the idea that the image I see looking down the focus tube (unmagnified) seems a wide angle. But if I am only using a part of the mirror when using eyepices what's the point of all the extra mirror? Does this make sense?
