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field reduction factor

Started by darryls, 12/26/2011 06:49PM
Posted 12/26/2011 06:49PM Opening Post
What is the "field reduction factor"?

It was used (on a webpage) to compute the effective FL of a telescope/DSLR combo. Such as "the telescope's FL is 540mm so with the 40d's "field reduction factor" the effective FL is 848mm.

Thanks!
Darryl
Posted 12/26/2011 08:01PM #1
That would be a comparison to a 35mm film full frame. At prime focus the focal length is that of the telescope. The field of view of the DSLR is calculated using the chip size and scope focal length.
Posted 12/27/2011 07:43AM #2
The reduction factor really means nothing in astronomical use. When DSLR's were first being sold, it was used to help a buyer translate how a lens designed for a full fralme 35 mm camera would work on a DSLR.

A standard 35 mm frame has a diagonal of 55 mm. A "normal" lens--the 55 mm lens that was traditionally the main lens sold with a 35 mm film camera--is designed to cover just this area. The perspective, etc. was designed to look like it does to the regular human eye-brain vision system. However, the DSLR sensor only covered a portion of that full frame. What it did not "see" was considered a "crop." It was as if you had taken the center section out of the full 35 mm frame. When you do this, and blow it up to the full size 8x10 print, then you are in effect, magnifying the central subject. The finished picture would look just as it might had you used a longer telephoto lens on the original picture. The 1.6 factor (Common in many DSLR's) is the difference between the normal lens and the telephoto lens focal length.

Alex