Any calculation for exposure as enlargement changes?

"...and why the physical aperture of a lens is larger to achieve the same F-stop (which is a ratio) for a longer focal length..."

Just to clarify Troy's excellent post, the size of the physical aperture doesn't appear in the f-stop calculation, but the diameter of the entrance pupil (the appearance of the aperture, as seen from in front of the lens) does. I'm sure that Troy knows that, and I hope that he doesn't mind me mentioning it.

Best,
Helen

Hi Helen,

You are right, and I'm sorry to have been somewhat opaque in my presentation. Optically, the aperture size (from which F-stops are calculated) is the size of the light-gathering element. Meanwhile we simultaneously have a mechanical aperture that controls the amount of light entering the lens that is transmitted to the film plane. As photographers, we tend to think of the mechanical device as 'the' aperture, since that's where the aperture-based exposure control happens.

I've probably made this even more confusing...short version...Helen is right.

And now to wander way off topic, I've wondered a bit about the front element size of some of the newer generation long lenses. The latest Nikon 600 f4 doesn't look like it is as large in the front element as the older versions. In fact, according to the data sheets, it is 6.6 inches for the new ones and 6.9 inches for the older ones. Is this because they have made the actual front element smaller? In order to do so, they must either change the focal length or the maximum aperture... It used to be that Nikon (and Canon I assume, though I'm not as familiar with their products) produced their super telephotos very close to the published specs, while Tamron, for example, produced a 280 mm instead of a 300 mm. This put them within legal tolerances for their product labeling, but also reduced the size, and presumably expense, of their lenses. In my irrelevant musings, I've wondered whether Nikon has made a similar adjustment in their latest 600. Since I've never touched one of these lenses and am not likely to, I'm free to wonder unfettered by details like the actual size of the lenses, which may not have changed in any real fashion.

Random thoughts from my troubled mind...

Troy, thanks for the clarification. I asked this question way back in this thread.

Troy Hamon said:

..........

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.

Click to expand...

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.

Slightly off topic:

Troy wrote
"Is this because they have made the actual front element smaller? In order to do so, they must either change the focal length or the maximum aperture..."

Troy,

The diameter of the front element does not directly determine the maximum aperture - ie it isn't the number that goes into the f-stop ratio. The diameter of the entrance pupil, with the lens wide open, does. The front element is not the entrance pupil, of course.

Different lens designs of the same focal length, same maximum aperture and same coverage can have different front element diameters, but they will all have the same entrance pupil diameter.

Back on topic:
Michael wrote
"What your light spread over paper area argument ignores is refocusing the enlarger lens and the consequent change in the effective f-stop."

I agree with Michael on this one, and apologies to him if I suggested otherwise. If you want the correct answer this must be taken into consideration.

Best,
Helen

Conservation of light, i.e., what area is the light spread over, is certainly pertinent. The other physical factor is that as the lens is refocused the distance of the entrance pupil from the film changes. For the small print with the small enlargement factor, the light is spread out over a small area, but the lens is focused farther from the negative and so the entrance pupil is farther from the film and therefore captures a smaller fraction of the light that is emitted by the film. This effect is usually termed "effective aperture". This is still conservation of light -- as the entrance pupil is farther from the film, some of the light rays no longer enter the pupil.

The effective aperture change when the lens is far from the film is also why one must correct exposure when taking a closeup photo. This is most commonly encountered by large format photographers ("bellows extension"). (Cameras with through-the-lens metering do the correction automatically.) In that application, the equation is highly accurate.

For the enlarging application, the equation is implicitly making some assumptions about how the film emits light, which of course depends on the illumination system of the enlarger. So the equation might break down in extreme cases, such as point source enlargers or very wide apertures. It should be most accurate for diffusion enlargers with the lens at reasonable apertures for making a print.

For someone who wants a calculator/computer implementation of this, there is a paper dial "computer" in the Kodak Darkroom Dataguide that gives the change in exposure time going from one magnification to another. You can probably find this inexpensively on eBay or internet used book sites.

Hi Michael,

My apologies. As you astutely pointed out, I do in general prefer intuitive mathematics. I am also aware that there are times when the simplified version of an equation loses its transparency. Perhaps that is what has happened in this instance. As you suggest, I certainly have not read the references you provide.

I understand the change in effective aperture, and apologize for my denseness in not recognizing it as you stated it originally. I had never considered it as a factor with enlargements, but I can see how it would apply. I am interested a bit because I've used my intuitive approach for some time and have never had to recalculate exposure, perhaps I ought to go back and compare some of my prints of different sizes for tonal variation.

On the other hand, perhaps I won't, since it seems to be working...

Happy printing.

I find the following explanation a little more succinct and clear. I realize this is just a restate of what's already been posted, but perhaps in more clear, laymen's terms. I can't remember where I found this, but i've been using it for about 5 years and works for me, unless you make a dramatic change to the photo's aspect ratio

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The exposure needed to make a print is proportional to the picture’s surface area. Thus, if you change print size, you can eliminate the need for new test strips by calculating the change in area mathematically. A pocket calculator facilitates this.

1. Divide the new print size by the old one to get the change in width.
2. Multiply this answer by itself—square it—to get the change in area.
3. Finally, multiply the original exposure time by the change in area to get the new exposure time.

For example, if you were adjusting from a 4x5 to a 16x20 print size:

Old print width = 5 inches
Old exposure time = 4 seconds
New print width = 20 inches

New width / old width = 20 / 5 = 4.0 (change in width)
4.0 * 4.0 = 16.0 (change in area)
Change in area x original exposure = 16.0 x 4 = 64

New exposure time = 64 seconds

Matt5791 said:

One question I have is:

Say I decide that 30 seconds exposure is about right for my 5X5 print I am about to make.

Is there any calculation to enable me to decide on comparable exposure if I increase the size of the enlargement to say 8X8?

Click to expand...

Matt:
I found this chart Enlarging Ratios to be extremely useful. I believe that it is what you are looking for. I have used it myself and it works very well.

Hope that this helps,
Kent

johnnywalker said:

Is there any reason why you can't simply calculate it
using the square inches of the print? If a 4x5 (20 sq. in)
print takes 20 seconds, then an 8x10 (80 sq in) will take
80/20 = 4 times as long at the same f stop.

Click to expand...

Perhaps this matter has already been mentioned. As
the negative's distance from the baseboard increases
and the focus is adjusted by bringing the lens closer
to the negative, the speed of the lens increases.
Just as one would expect from a camera lens.
So, with your example the time might be
3.5 times as long.

I also recommend the EM-10. If calibrated it can
be used as a densitometer. Same exposures are
obtained by adjusting the F stop. Dan