Need Help with Optical Formulas for Macro

How do I calculate the focus distance (object to film plane or sensor) from a known magnification and focal length?

The practical application of this calculation is to determine if the working height of a copy stand is appropriate to copy a given size film negative or slide with a given lens on a given sensor size. I have been looking at the optical formulas until my eyes are bleeding, and I can't work it out.

For example if I want to copy a 6x7 cm negative with an APS-C camera, I can calculate the magnification needed. To just fill the height of the APS-C sensor (15.6mm) with the height of the 6x7 negative (56mm), the resulting magnification is 15.6/56=0.28, right?

If I want to use a 100mm enlarger lens on a bellows at 0.28x magnification, it should be possible to calculate the distance between the negative and the camera sensor, right?

Solve this equation: 1/100=1/lens to subject + 1/lens to sensor and the results are 457 for lens to subject and 128 for lens to sensor but....
There is a but because these distances are from the front nodal plane of the lens and rear nodal plane of the lens which I wouldn't know where they are.

Chan Tran said:

Solve this equation: 1/100=1/lens to subject + 1/lens to sensor and the results are 457 for lens to subject and 128 for lens to sensor but....
There is a but because these distances are from the front nodal plane of the lens and rear nodal plane of the lens which I wouldn't know where they are.

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Ignoring for a minute the uncertainty introduced by not knowing the nodal planes of the lens, where does magnification enter into your equation? And how can you solve for either lens-to-sensor distance or lens-to-subject distance without already knowing the distance of at least one of them?

runswithsizzers said:

where does magnification enter into your equation?

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Hi, looks like he forgot to say that magnification is the ratio of the two lens distances.

Something I found useful 20-some years back were a couple of short optical summaries posted on a particular web site I frequented. They seem to have disappeared from there but can still be found here:

https://graflex.org/lenses/lens-faq.html
https://www.graflex.org/lenses/photographic-lenses-tutorial.html

Your specific questions are covered in the faq, see Q4 and Q5.

I was gonna explain how to (approximately) deal with the issue of the lens' nodal points, but on a quick review of the faq I see that this is also covered: first paragraph of the "Technical notes." At least for the front nodal point. He doesn't seem to mention it, but for the rear nodal point you can generally turn the lens backwards and repeat the procedure. Some lenses may not be able to form such an image when reversed (it can be internal to the lens), but it will generally work for enlarging lenses.

Mr Bill said:

Hi, looks like he forgot to say that magnification is the ratio of the two lens distances.

Something I found useful 20-some years back were a couple of short optical summaries posted on a particular web site I frequented. They seem to have disappeared from there but can still be found here:

https://graflex.org/lenses/lens-faq.html
https://www.graflex.org/lenses/photographic-lenses-tutorial.html

Your specific questions are covered in the faq, see Q4 and Q5.

I was gonna explain how to (approximately) deal with the issue of the lens' nodal points, but on a quick review of the faq I see that this is also covered: first paragraph of the "Technical notes." At least for the front nodal point. He doesn't seem to mention it, but for the rear nodal point you can generally turn the lens backwards and repeat the procedure. Some lenses may not be able to form such an image when reversed (it can be internal to the lens), but it will generally work for enlarging lenses.

Click to expand...

In Q4 and Q5, I see terms for "So" - distance from front principal point to subject (object), and for "Si" - distance from rear principal point to film (image) plane. I don't know either. And, let's say I don't have a lens, so I can't measure So or Si. What I want to do is calulate the focusing distance at a given magnification for various focal length lenses before actually buying any lenses.

But if I did know values for So and Si, would it be correct to say that the sum of So + Si is the total distance from the subject to the sensor? If so, I can use the magnification formulas in Q5 to solve for So and Si, and then add them together to get the total focusing distance, right? (Apparently, that is what Chan Tran did. Solving for So and Si, I get the same numbers as Chan Tran.) But if there is still some unknown distance between So and Si then the calculated focusing distance might be incorrect by some unknown amount (?)

OK, I just used the equations "So = (Mf + f)/M" and "Si = Mf + f" to solve for my 100mm enlarger lens and a magnification I have actually used. When I add So and Si, the sum of the calculated numbers is close to my measured focusing distance, so that seems to work.

Thanks!

runswithsizzers said:

But if I did know values for So and Si, would it be correct to say that the sum of So + Si is the total distance from the subject to the sensor?

Click to expand...

runswithsizzers said:

But if there is still some unknown distance between So and Si then the calculated focusing distance might be incorrect by some unknown amount (?)

Click to expand...

Yes, approximately, and yes.

The unknown distance happens due to the two nodal points not being in the same position. But for a typical enlarging lens they are generally fairly close together. To find them, approximately, you could hold the lens so as to focus an image of the sun onto a surface. At the same time measure one focal length from the image and place a mark on the side of the lens. Then turn the lens backwards and repeat. If there is a space between the marked points then this distance gets added to the calculated distance. If, instead, the two points have overlapped, then you subtract the overlapping distance.

I pointed out that some lenses, for example wide angle designs for SLRs, may not be able to form an image of the sun when reversed. In such a case you could move a light source closer to the lens until the image comes forward from the front element, then calculate a corrected approximate position.

runswithsizzers said:

Ignoring for a minute the uncertainty introduced by not knowing the nodal planes of the lens, where does magnification enter into your equation? And how can you solve for either lens-to-sensor distance or lens-to-subject distance without already knowing the distance of at least one of them?

Click to expand...

I rewite the equation:
1/100= 1/distance from lens to subject + 1/0.28*distance from lens to subject. Because you knew the maginfication..
The equation would be a quadratic equation.

Regarding post #1,

“If I want to use a 100mm enlarger lens on a bellows at 0.28x magnification, it should be possible to calculate the distance between the negative and the camera sensor, right?”

The distance d from the object to the image as a function of magnification and focal length is:

For a thin lens, d = ((m + 1)^2)f/m

For a real-world thick lens, d = ((m + 1)^2))f/m + nodal distance

The usual problem with this is not knowing the nodal distance. That’s generally the case with roll-film and digital cameras. The makers of large-format camera, enlarger, and process lenses usually provided the nodal distances.

A more-practical approach is to adjust the position of the camera and lens relative to the subject until the particular subject dimension chosen fills the corresponding dimension in the camera’s finder.

Ian C said:

Regarding post #1,

“If I want to use a 100mm enlarger lens on a bellows at 0.28x magnification, it should be possible to calculate the distance between the negative and the camera sensor, right?”

The distance d from the object to the image as a function of magnification and focal length is:

For a thin lens, d = ((m + 1)^2)f/m

For a real-world thick lens, d = ((m + 1)^2))f/m + nodal distance

The usual problem with this is not knowing the nodal distance. That’s generally the case with roll-film and digital cameras. The makers of large-format camera, enlarger, and process lenses usually provided the nodal distances.

A more-practical approach is to adjust the position of the camera and lens relative to the subject until the particular subject dimension chosen fills the corresponding dimension in the camera’s finder.

Click to expand...

Thanks for that!

I was hoping to find a more elegant way of calculating the focused distance rather than solving for So and Si, then finding theif sum. Both calculation methods result in the same distance.