"jjs" wrote in message
...

"Richard Knoppow" wrote in message
om...

[...]
One of the inherent properties of lenses which produce
orthographic images is that the illumination falls off with image
angle. For a "normal" design lens this fall off is approximately
proportional to cos^4 theta where theta is the "half angle" of the
image point. There are designs of lenses which have improved
illumination but these are not better than about cos^3 theta, better,
but there is still fall off. [...]

Care to help an innumerate? Taking the 38mm Biogon as an example, what's
the
light fall-off in terms of F-stops?

If you consider the light at the center of the image circle to have a
magnitude of 1 (units do not matter because we will be dealing with ratios),
the number of stops of fall off at an angle Theta degrees off of the center
of the center of the image would be:

Fall-off in Stops = 3.322 x Log(1 / Cos^4 of the angle Theta)

If you assume the fall off of the lens design is Cos^3 instead, the formula
would obviously be:

Fall-off in Stops = 3.322 x Log(1 / Cos^3 of the angle Theta)

For instance, the fall off at a point 30 degrees off of the center,
considering a lens with Cos^4 fall off, would be:

Fall-off in Stops = 3.322 x Log(1 / Cos^4 ( 30 ))
Fall-off in Stops = 3.322 x Log(1 / (0.866^4))
Fall-off in Stops = 3.322 x Log(1 / 0.5625 )
Fall-off in Stops = 3.322 x Log(1.77777)
Fall-off in Stops = 3.322 x 0.2498
Fall-off in Stops = 0.83 stops

Guillermo