Math question -) lens coverage.

Nick Zentena · Nov 17, 2005

The question about 150mm versus 135mm enlarger lenses got me thinking. Wide angle lenses that almost cover bigger then intended formats tend to be common. They don't cover at infinity that is. Is it reasonable to assume that the angle of coverage extends? Should I be able to just do the math and work out how close I'd need to focuss to have enough coverage? Does any of this make sense?

Dan Fromm · Nov 17, 2005

Huh?

The lens projects a cone of rays whose intersection with the film plane is a circle. Call the angle of coverage theta. The circle's radius is lens extension * tan(theta).

Do you know how to calculate extension given focal length and magnification?

Double the extension, double the circle's diameter. And so on.

BradS · Nov 17, 2005

Actually, the correct fomula is:

d = 2L Tan (theta / 2)

where:
L is the distance between the lens and the film (== focal length when focused at infinity)

d is the diameter of the image circle

and theta is the so called angle of coverage or covering power.This is a characteristic of the lens. It does not change.

As you can see, the diameter of the image circle increases with the length of the bellows. So, yes, you can figure out how close you'd have to be to getenough coverage.

Nick Zentena · Nov 17, 2005

Thanks. It seemed so simple I had to ask to make sure it wasn't too good to be true.

Dan Fromm · Nov 17, 2005

Brad, thanks for the correction. I always forget to divide the angle by 2.

Math question -) lens coverage.

Nick Zentena

Dan Fromm

BradS

Nick Zentena

Dan Fromm

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