"f/256" wrote in message

ogers.com...

"jjs" wrote:

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
Very helpful, Guillermo. I'm a bit closer to understanding.

So, finding the Cos power of the lens-design remains problematic. Here's the

lens I'm working with:

http://course1.winona.edu/jstafford/...s1/index2.html There is very

little light fall of even in the corners. In fact there is so little I can

hardly find any. Note the construction of the lens: it covers 4x5" (actually

more than 5x5") and the rear lens is 4.5" in diameter.