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How can I calculate what the effect of an extension tube will be?
My understanding is that extension tubes increase the magnification of lenses by reducing their minimum focusing distance.
But I can't figure out how to calculate the change an extension tube makes on to the maximum magnification of a lens. Is there a formula of some sort I can use?
This is a question that would have been simple to answer in the '70s, and there's still a relatively simple answer for unit focus lenses (lenses where the entire optical group moves as a unit when you focus — with, perhaps, the exception of one or two lens elements that float within the group to correct aberrations for close focusing). By far the vast majority of currently-available lenses, though, are internal focus.
Internal focus lenses actually change their focal length when you focus — the lens gets shorter as you focus closer, so the front element(s) don't have to move. The closer the lens focuses, the shorter the minimum focal length of the lens becomes. You need to find out what the actual focal length of the lens you are using is going to be — and unless it's pegged at infinity, it probably won't have much to do with the number printed on the lens barrel or the number you see on the zoom ring.
At a 1:1 reproduction ratio, the distance from the optical center of your lens to the sensor plane will be exactly one-half of the distance between your point of focus (your subject) and the sensor plane.
At magnifications other than 1:1, the distances can be calculated using the formula:
$$ {1\over s_\mathrm{O}} + {1\over s_\mathrm{I}} = {1\over f} $$
where \$s_\mathrm{O}\$ (displacement of the object) is the distance from the subject to the optical center of the lens, \$s_\mathrm{I}\$ (displacement of the image) is the distance between the film/sensor plane and the optical center of the lens, and \$f\$ is the effective focal length of the lens at the set focal distance, and:
$$ {h_\mathrm{I}\over h_\mathrm{O}} = \frac{s_\mathrm{I}-f}{s_\mathrm{O}-f} $$
(the the ratio of the heights of the image and the object is the same as the ratio between the distances of the image and the object from "infinity focus" on either side of the lens)
They're simple enough formulas if you know the focal length of the lens. With an internal focus lens, you either have to know it (that is, you can get the data from an outside source) or you can deduce it if you focus to the close focus limit on a well-defined subject (something with hard edges), measure carefully from the sensor plane marker on your camera body to the subject, then take a picture and count the pixels.
The size of your sensor and the number of pixels horizontally and vertically are known quantities (they should be in your manual), and your measurement will give you the value of \$s_\mathrm{I} + s_\mathrm{O}\$. That should be enough to figure out \$f\$ — and once you have \$f\$, you know what's really going on between the lens and the sensor, so you'll know how much extra magnification a given thickness of extension tube will give you at the closest focus point. At the furthest focus point (when the lens is set to infinity), the calculation is simpler because you can use the \$f\$ value that's printed on the lens. It's probably the close focus distance/magnification you're interested in, though, and if you can't find the focal length data for your lens on the intarwebs, you've got to experiment. (And $_DEITY help you if your lens is both an internal focus and a zoom.)
Or you could just buy a set of tubes and try them out.
