View Full Version : Lenses & Maths

Pfeiffer Duckett
13-Feb-2008, 01:00
I have a mathematical problem that I am having trouble with. It isn't very hard, I've just forgotten my schooling.

Here it goes: Lets say a lens only has coverage for 2x3 inch film when focused at infinity. Using the pythagorean theorem you can get a reasonable idea of what the diameter of the image circle would be: (the square root of two squared plus 3 squared so 3.6056). Divide that if half to get the radius, and using trigonometry and the theoretical length of the lens combined with the radius of the image circle, you should be able to find the angle of view.

Let us then say that you want the lens to cover 4x5. Using the pythagorean theorem again, you find that the image circle needs to be 6.4031 inches in diameter or, alternatively, have a 3.2016 inch radius. Using the angle of view that was found in the last problem and the radius, you should be able to find out how far away the lens needs to be from the plane of focus to cover the larger film format, and thereby find out how close the subject would have to be for the image to be both in focus and cover the film.

I want to see how far away from the film a lens with a focal length of 100mm which covers 6x9 cm would have to be to cover the 4x5 inch format- with bonus points for a theoretical distance of how close the subject would have to be.

My problem is that I have forgotten most of my trigonometry in leu of photo theory, and I sold my TI-83 for film long ago. If somebody who retained this information could help me out on this I would be very grateful.


Ole Tjugen
13-Feb-2008, 02:56
You don't need pythagoras to find the circle of coveraage at a given extension, as long as you know the coverage at one extension!

Calling extension D and image circles C1 and C2,

your D2 is simply D2 = D1 * (C2 / C1)

Since the angle is constant, the diameter varies linearly with extension. So you would need 177.59mm extension in the given case, assuming extension at infinity focus with a 100mm lens is 100mm.

Then, assuming a perfect simple lens (zero thickness and all principal points in the same place), 1/F=1/u+1/v.

1/100= 1/177.59 + 1/v, so

v= 1(1/100-1/177.59), or 228.88mm.

That's your minimum distance.

Jim Jones
13-Feb-2008, 07:30
I have complete faith in Ole's encyclopedic knowledge and mathematical ability. However, there is a caveat. Some lenses perform well in macro photography: others fail dismally. Macro performance in most lenses on 2x3 cameras has been neglected to optimize their use in general photography. One should improvise a check on their macro ability before investing much in adapting them to larger cameras.

Pfeiffer Duckett
13-Feb-2008, 17:30
Great, Thanks!

16-Feb-2008, 12:13
I agree with Ole except:

1) That subject distance is the maximum, not the minimmum

2) Your values for the 4x5 & "2x3" diagonals are suspect, particularly if you mean the 6x9cm nominal format. I use 4x5 => 154mm diag, and 6x9 => 100mm diag.

3) Your lens may have extra coverage on the "2x3" format. Per Schneider, a 100mm Xenar covers a 116mm diagonal. That lens will cover 4x5 with an image distance of 132mm, corresponding to a subject distance of 407mm

100 / 116 = X / 154
X = 100 / 116 * 154 = 132

Ole's simple lens formula then gives the 407mm value which is almost exactly 16 inches and means that you minimum image magnification is 0.325 so your subject had better be smaller than about 15 inches.

Other lens designs cover better or less well.

Ole Tjugen
16-Feb-2008, 12:39
I agree with Ole except:

1) That subject distance is the maximum, not the minimmum

Right. That's the maximum distance, or minimum extension. I got a bit muddled there.

For the rest, you are correct. I tried to make as few assumptions as possible, and used the givens from the original question. As for "other designs", I have a 100mm Symmar bought for moderately close work on 4x5". :)