Quote Originally Posted by Troy Hamon
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In terms of Michael's equation, there is no reason that you should have to add one to the two terms. As far as the physics of the situation, there is no basis for it.........

......... A simple way to look at it is to think of it solely in terms of the light put out by the enlarger. There is a certain quantity of light that is coming through the enlarger lens. That same amount is going to come out of the lens whether the image on the print is covering 4x5 or 8x10 or 16x20. So, by simply taking the ratio of the areas in the prints (or the ratio of the squares of the enlarger heights) you can accurately measure the ratio of exposure times. So...a 4x5 is 20 inches square, an 8x10 is 80 inches square, an 8x10 takes 4 times the exposure of a 4x5. The mathematical version of this is

Tn = To*(An/Ao)

where Tn = new exposure time, To = old exposure time, An = new print area, and Ao = old print area. To alter this equation for enlarger distance, it simply becomes

Tn = To*(Dn^2/Do^2)

where Dn = new enlarger distance and Do = old enlarger distance. Neither of these are new, others above have already provided these results in different forms, but this presentation makes more sense to me, so I hope it is helpful to somebody else.
First of all, the equation that I gave is not really my equation. Second, the "+1" added to the magnification has a physical basis. In fact, it is necessary for the equation to be correct. I tried explaining it in my first answer and I gave references to more detailed derivations than I can give in an internet answer. Do you really think that your mathematical optics by intuition is correcting an equation published by Rudolf Kingslake, who maybe Kodak's best lens designer and is one of the best authors of optics books?

If you had checked the references that I gave, you will find that both Kingslake and Ray use the same numerical example, going from an enlargement of 2x to 4x. Both calculate an exposure time change of x2.8. The paper dial calculator in the Kodak Darkroom Dataguide also gives an exposure change from 10 to 29 s.

What your light spread over paper area argument ignores is refocusing the enlarger lens and the consequent change in the effective f-stop.

Tn = To*(An/Ao) and Tn = To*(Dn^2/Do^2) implies a relation between "enlarger distance" (image distance, I assume -- the term wasn't precisely defined) and print area that simpy isn't correct. If a 4x5 negative is enlarged to 8x10 with a 150 mm lens, the image distance is 450 mm. Switching to a 16x20 print, the image distance becomes 750 mm. The ratio of print areas is 4. The ratio of image distances squared is 2.78. If by "enlarger distance" is meant the total distance from negative to print, the two distances are 675 mm and 937.5 mm. The square of the ratios is 1.92, which still isn't 4.

The correct equations relating image and object distance, focal length and magnfication are given near the top of the lens tutorial: http://www.photo.net/learn/optics/lensTutorial

This numerical examle (same as my previous example) shows the consistency between squaring the image distance and the m+1 squared equation, and the inconsistency with the magnification squared equation. (750 / 450 )^2 = 2.78. (4+1)^2 / (2+1)^2 = 2.78. But 4 ^2 / 2^2 = 4. This should cause worry to those who believe in both the image distance squared and magnification distance squared equations.