Also called "The circle of full illumination" or "the 100 %
illuminated area".
This area is located at the focal plane. It defines the area in which the entire area of the primary mirror or lens is used (minus a possible obstruction).
In a Newtonian it is usually a circular area centered on the optical axis, but it may also be de-centered due to a wrong placement of the secondary (i.e. when the secondary offset is omitted).
In a Newtonian the size of the fully illuminated area is determined by the size of the secondary. It can be calculated by the following formula :
There are two interesting aspects concerning the existence of the fully illuminated area, the first is obvious :
- Outside the fully illuminated area NOT the whole (minus obstruction) surface of the primary is used. This leads to vignetting, a loss of resolution and possibly increased aberrations.
The second is :
- Within the area of full illumination the whole surface of the primary is projected PLUS an area that spreads around the primary.
The second is not unimportant, because in many telescopes it is possible to look past the edge of the telescope out of the tube (as seen from the focal point). It is thus possible for stray light to reach the focal plane !. If this happens, contrast is severely reduced resulting is a big performance loss. It is thus of vital interest to make sure that no stray light can enter past the edge of the primary !. This simple fact is often overlooked in the cheaper commercial telescopes.
How to choose the radius of the fully illuminated area
For imaging in the primary focal point, this is easy. It should be so big that it can accommodate the whole area of the imaging device. Though for photography you may want to compromise a bit.
For visual use, it can be calculated with the following formula:
The relation between the apparent FOV in the eyepiece and the true FOV in the night sky is the magnification (AFOV = m * FOV). If you want a certain apparent FOV to be evenly illuminated all over, then divide this AFOV by the magnification you'll get with the eyepiece in question. Put the result in the above formula and presto !. Example: f = 2000 mm, the focal length of an eyepiece is 10 mm, the magnification is thus 200 times. If in this eyepiece you'll want an AFOV of 45 degrees, then the true FOV will be 0.225 degrees. The size of k is then: 2000 * tan (0.1125) = 3.9 mm
For low magnifications the size of k will be quite large, especially for long focus telescopes. So a compromise must be made.