Actually, no math is required. All lenses will paint on the image
plane (where the film or sensor lives) a circular image of some
angular diameter. The center of this circle is always on the optical
axis of the lens; the location and diameter of the image circle are
inherent properties of the lens itself, and do not vary with how the
lens is used. As one would expect, the larger the diameter of the
image circle, the more complex and expensive the lens.
Inside this circle, the image is sharp, lens abberations being
reasonably well corrected. Outside this circle, sharpness (and
illumination) fall off, sometimes slowly, sometimes abruptly,
depending on the optical design of the lens. If the lens is intended
for shift use, the falloff will likely be gradual.
If the lens is to be used with film center always on the optical axis
(that is, unshifted), the diameter of the image circle need not be
much larger than the diagonal dimension of the film (or CCD sensor).
If the lens will be shifted, so the optical axis no longer passes
through the center of the film or sensor, the image circle diameter
must be larger, being large enough that the corner of the film that's
farthest away from the optical axis of the lens remains within the
image circle.
Shift lenses are more expensive than non-shift lenses for two
reasons, then. First, the optics must have a larger image circle, so
that shifted images remain sharp. Second, the lens mechanics must
allow the optics to slide with respect to the film or sensor,
remaining parallel to the image plane, and yet be mechanically solid.
View camera lenses are the extreme case, having image circles out to
+/- 60 degrees:
<http://www.schneideroptics.com/photography/large_format_lenses/super-angulon/>.
For the full story, search for "scheimpflug":
<http://www.trenholm.org/hmmerk/index.html> and
<http://www.columbia.edu/~akj19/scheimpflug.html>.
Joe Gwinn
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