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# Thread: Calculating Bellows Extension from Magnification Factors? Anyone?

1. ## Calculating Bellows Extension from Magnification Factors? Anyone?

It's embarrassing to ask such a basic question for me but.. I just spent a good amount of time wading through the 'net trying to find what I'm looking for but can't seem to locate it (here either!) - Does ANYONE know a simple formula for deriving the effective focal length (bellows draw) from magnification and nominal focal length ALONE?

So here's the deal - Let's say I'm trying to make a photograph of an object I know I need a magnification of 10x on in terms of what I want the aerial image height (fancy talk for image size) on the negative to be. One of the lenses I have at my disposal is a 105mm componon - but I think that's going to be way too long. Basic rule of thumb tells me that at 1:1, the focal length will be 210mm, and at 2:1 twice that, it will be 420mm, at 4:1, 840mm, 8:1 it will be 1680mm... and 10:1 would be SOMEWHERE in the region of 2400mm (7.5 feet) I'm guessing. So in that case I would try to find myself a 50mm lens to keep the bellows extension around 1.2m since it's more convenient...

So fine - I can work through this manually as above... but I have a dozens of such scenarios to work out so I'm wondering if there's a good simple formula that can help me automate my shot planning... (?)

Thanks if anyone can help...

2. Why would you wish to do this?

3. making photographs of small objects...

4. Bellows extensions when using various lenses may require exposure compensation. There is a formula for figuring it out (forgot the formula) but when I started with LF, I sat down with a calculator and figured it out from F stops to length and marked a small retractable ruler that I keep with my light meter. I just measure the distance from the lensboard to the film plane and see what increase is needed. You should be able to see how the image will appear on the ground glass. Lens selection will be governed by the maximum length of your bellows. How small are the objects?

http://www.jeffreyglasser.com/

5. In direct answer to your question, you can almost certainly find the formula on the front of the LFPF homepage or with a second of googling.

You'll find that most lenses have huge aberrations when focused that close. Getting to 1x with good quality requires a macro lens and even then it's going to start looking bad probably by 4x. At 10x, you're going to have mush from any lens not designed for use at that range, i.e. microscope objective lenses. If you're using normal lenses for very high magnification, you can somewhat avoid this problem by using the lenses reversed because they are generally well-corrected for the rear image plane being very close and the front image plane being further away. If you want a 10x shot, set it up like a 0.1x shot but swap the film and subject locations.

Don't forget also that diffraction gets worse: a 2x bellows extension to reach 1x magnification will double the effective focal length and add two stops to your effective aperture. Go to 5x or 10x and you'll find that diffraction means you need to be shooting at much larger apertures than you're used to so the DOF will be practically nil.

If you want to do ultra-macro and despite this being APUG, I strongly suggest looking into the newer digital techniques like focus-stacking. It also helps that there are lenses available for 35mm that are specifically designed to reach about 6x magnification. Right tool(s) for the job and all that.

6. The lens' focal length is independent of its position between film and subject.

Extension can be thought as starting from two different positions: the film plane and the lens' infinity position.

The distance from the film plane to the lens' rear node when the lens is focused at infinity is the lens' focal length.

The distance from the film plane to the lens' rear node when magnification (m) > 0 (m = 0 means the subject is really, really far away) is f * (1 + m). By subtraction, extension from the infinity position is f * m. f is the lens focal length.

For lenses commonly used on LF cameras (not retrofocus, not telephoto) both nodes (front and rear) are close to the diaphragm.

7. well thanks for the responses but I dare say few people have read the question. I'm really not concerned with exposure compensation or subject/object distances. Those (subject/object distances cancel out to become magnification in certain formulae) - but I've been googling for some time and can't find the correct formulae - though I know they exist since I've used them before but it must be 15 years since I have.

I wouldn't really call it 'ultra macro' or anything of the sort. It's just regular 'macro' photography. I figure a micro nikkor would probably work the best... maybe i'll just stick with one of those for everything... but I was just hoping there was a quick formula i could use to derive the extension from the F.L. and the magnification - but I guess it's not too well known...

8. Here's an example of a page where someone's comparing enlarger vs macro lenses... so I think it should work just fine... I've been doing LOTS of 1:1 work on 4x5 with old sironars etc and the results have been quite outstanding (40x50 prints with TONS of detail)... I would encourage others to try...

http://savazzi.freehostia.com/photog...erlensespm.htm

9. The Kodak Professional Photoguide says: Lens-to-film Distance = (Magnification + 1) X Focal Length and Effective Aperture = f-number X (Magnification + 1)

10. Originally Posted by Prof_Pixel
The Kodak Professional Photoguide says: Lens-to-film Distance = (Magnification + 1) X Focal Length and Effective Aperture = f-number X (Magnification + 1)
hmmm... okay that sounds more fruitful... maybe I should look it up - I think i even have one of those somewhere! only question - what do you mean by 'and' as in FL and eff. aperture?

I'm pretty sure you can cancel out the aperture somehow - i.e. - it really shouldn't come into play (?) except where exposure compensation is needed I think?

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