There must be a mathematical description of the difference that an extension tube makes to a lens -- is it something that can be easily described?

(For example, with teleconverters you can say things like "a 2x teleconverter will turn a Y-mm lens into a 2Y-mm lens, and will lose you 2 stops." Is there something similar for extension tubes?)

If there's nothing much you can say about magnification, in general, what about the change in closest focal distance? Is that also lens-dependent?

What about if we factor out the lens: is there any general way to compare the effects of (say) a 12mm and a 24mm extension tube on the same lens?


5 Answers 5


I do believe there are some formulas you can use. To Matt Grum's point, I have not tested these with zoom lenses, and to my current knowledge, they apply only to prime (fixed focal length) lenses. You did not specifically specify zoom lenses, so...

The simplest way to calculate the magnification of a lens is via the following formula, with magnification \$M\$, total extension \$E_\textrm{total}\$, and focal length \$f\$:

$$ M = \frac{E_\textrm{total}}{f} $$

To calculate the magnification with an extension tube, you need to know the total extension \$E_\textrm{total}\$... that is, the extension provided by the lens itself (\$E_\textrm{intrinsic}\$), as well as that provided by the extension tube, \$E_\textrm{tube}\$. Most lens statistics these days include the intrinsic magnification. If we take Canon's 50mm f/1.8 lens, the intrinsic magnification is 0.15×. We can solve for the lens's built-in extension like so:

$$ \begin{align} 0.15 &= \frac{E_\textrm{intrinsic}}{50} \\ E_\textrm{intrinsic} &= 50\times 0.15 \\ E_\textrm{intrinsic} &= 7.5\textrm{ mm} \end{align} $$

The magnification with additional extension can now be computed as follows:

$$ M = \frac{E_\textrm{intrinsic} + E_\textrm{tube}}{f} $$

If we assume 25 mm of additional extension via an extension tube:

$$ \begin{align} M &= \frac{7.5\textrm{ mm} + 25\textrm{ mm}}{50 \textrm{ mm}} \\ &= \frac{32.5\textrm{ mm}}{50\textrm{ mm}} \\ &= 0.65\,\times \end{align} $$

A fairly simple formula that allows us to calculate magnification fairly easily, assuming you know the intrinsic magnification of the lens (or its intrinsic extension.) If we assume the wonderful 50 mm lens is the lens you are extending, to create a 1:1 macro magnification, you would need 50 mm worth of extension. The problem here is that if you add too much extension, the plane of the world that is in focus (the virtual image) might just end up inside the lens itself. Additionally, this assumes a "simple" lens, one with very well-defined and well-known characteristics (i.e. a simple single-element lens.)

In a real-world scenario, having a clear understanding of any particular lenses characteristics is unlikely. With lenses that focus internally, or zoom lenses, the simple formula above is insufficient to allow you to calculate exactly what your minimum focusing distance and magnification can be for any given lens, focal length, and extension. There are too many variables, most of which are likely to be unknown, to calculate a meaningful value.

Here are some resources that I have found that provide some useful information that might help in your endeavor:

  • 2
    \$\begingroup\$ Oddly enough, the 50mm f/1.8 is the lens (well, OK, one of the lenses) I'm extending -- and those links look really useful, too. Thanks! \$\endgroup\$ Commented Dec 11, 2010 at 12:29
  • 1
    \$\begingroup\$ Good Answer, excactly what I was looking for! It only bugs me that you are using TE to abbreviate "TotalExtension" and also "TubeExtension". \$\endgroup\$
    – smow
    Commented Dec 3, 2016 at 23:10

I think it can be described, in fact Wikipedia has the relevant formula:

$$ \frac{1}{S_1} + \frac{1}{S_2} = \frac{1}{f} $$

where \$S_1\$ is the distance from the subject to the front nodal point, \$S_2\$ is the distance of the rear nodal point to the sensor, and \$f\$ is the focal length. Since extension tubes increase \$S_2\$, it then allows you make \$S_1\$ smaller, thus you can focus much closer to the subject.

  • 1
    \$\begingroup\$ That formula assumes you know the front and rear modal points which in general aren't manufacturer specified, so you'll have to measure them for each lens. Plus the formula isn't valid for lenses which change focal length when focussing, so I don't think it's quite what the questioner as after. \$\endgroup\$
    – Matt Grum
    Commented Dec 10, 2010 at 15:11
  • \$\begingroup\$ With a simple (i.e. single-element) lens, the focal length never ever changes (unless you change the shape of the lens -- or unless you're being very specific and talking about different colors of light, or the like), and this is absolutely correct (in fact, moving the position of the lens is all you're doing to change the focus anyway, so an extension tube just lets you move it further). For complex (multi-element) lenses, I don't understand the optics principles well enough to be sure if the same holds true. But the film plane is always the "target" of the focus, right? So... I'd think so. \$\endgroup\$
    – lindes
    Commented Dec 10, 2010 at 17:51
  • \$\begingroup\$ Some of my sources for learning (which I'll hopefully later pull together in an answer of my own -- no time for that now, though): youtube.com/view_play_list?p=F703024381DE9004 -- and in particular, these two: youtube.com/watch?v=oKfqO4tBfPc&p=F703024381DE9004 and youtube.com/watch?v=mjIfdXnhyQI&p=F703024381DE9004 \$\endgroup\$
    – lindes
    Commented Dec 10, 2010 at 18:14
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    \$\begingroup\$ @Matt Grum - I think the equation illustrates the principle behind it which seems to be the crux of the question. At least it did to me. :) \$\endgroup\$
    – Joanne C
    Commented Dec 10, 2010 at 18:28
  • \$\begingroup\$ Joanne - the formula illustrates well why extension tubes decrease the minimum focussing distance, but I think the questioner was looking for a formula that he can use to judge what length extension tube you need to buy for a given lens in order to increase magnification x times, which unfortunately is not possible in the general case... \$\endgroup\$
    – Matt Grum
    Commented Dec 10, 2010 at 18:39

edit to respond to follow up questions given you know the effects of a tube of a certain length on a certain lens you can work out the missing values from John's equations you should be able to get an estimate of the effect of a different length tube. Again the values will be subject to the foibles of the lens focussing method, but should give you a good enough idea.

In general no. There is a formula, of course, but you need to know the internal configuration of the lens and usually some elements of the lens design.

Extension tubes usually change the effective focal length slightly (the actual focal length of the lens is a property of the bending power of the glass so doesn't change when you move it) but how much depends on the lens design. A lot of it is to do with the angle at which the light rays leave the back of the lens. If you take an object space telecentric lens (a special type of lens where the rays exit parallel to each other) then the distance to the film plane doesn't matter since the rays are parallel they wont converge or diverge any more.

If you look at the back of a wide angle lens the rear element is very close to the rear of the lens. Now look at a telephoto lens, there will be a gap between the last piece of glass and the mount, as if the lens already has an extension tube. An extension tube will behave quite differently on these two different lenses. The method of focus (internal vs. external) also affects the results of adding extension tubes.

So in short I'm afraid there is no formula that's as simple as the one for telecoverters.

  • \$\begingroup\$ Is it truly accurate to say that the focal length changes? My understanding of optics in any detail is in its infancy, but my understanding thus far is that by moving the lens (which is all an extension tube really does), the focal length won't change (though perhaps the magnification might? Or what we've collectively started calling the "effective focal length"), but rather, the distance changes for the focal plane, which causes the in-focus plane to change... I'll try to find some resources and post them in an answer. I -think-, though, that this answer is factually questionable. I think. \$\endgroup\$
    – lindes
    Commented Dec 10, 2010 at 16:44
  • \$\begingroup\$ Whether or not the focal length changes and to what degree depends on the lens, as I stated. For the simplest case of a pinhole lens it's easy to see that the focal length changes if you move the pinhole further from the camera, since the focal length is defined as the distance from the pinhole to the imaging plane! \$\endgroup\$
    – Matt Grum
    Commented Dec 10, 2010 at 17:21
  • \$\begingroup\$ Ahh, but a pinhole is not a lens, and as I understand it, for lenses (or optical systems in general??), the focal distance is defined not by the distance between the point and the imaging plane, but between the center of the lens and a point, given an input of parallel lines. Is that not correct? (Note: see video links on my comment to Joanne's answer -- How can I calculate what the effect of an extension tube will be? -- also, note that I'm genuinely asking; I'm relatively new to understanding optics at this level.) \$\endgroup\$
    – lindes
    Commented Dec 11, 2010 at 1:42
  • \$\begingroup\$ Yes you are correct a pinhole has no focal length as the focal length describes a lenses ability to bend light, what I meant to say is that the effective focal length of a pinehole system is the distance between the pinhole and screen, i.e. it gives the same field of view as a lens with the same fl. The point is you need to make assumptions about an imaging system in order to predict how it will behave if you change one of the parameters without knowing the others. \$\endgroup\$
    – Matt Grum
    Commented Dec 11, 2010 at 14:44

Cambridge in colour has an online magnification ratio calculator. And to quote the web site:

An extension tube increases lens magnification by an amount equal to the extension distance divided by the lens focal length.

Which translates into:

$$ M_\textrm{ExtendedLens} = M_\textrm{Lens} + \frac{\textrm{Extended length}}{f} $$


What about if we factor out the lens: is there any general way to compare the effects of (say) a 12 mm and a 24 mm extension tube on the same lens?

I decided to test the relationship mentioned in other answers here re: magnification= extension of lens and tubes / focal length of lens.

I have an OM50mm f1.8 (I don't know it's magnification or extension properties) and three extension tubes, 12 mm, 20 mm and 36 mm. I mounted the 12 mm and 36 mm through an adapter to an Olympus E-520 with a 4/3 sensor measuring 17.3 mm by 13 mm (according to Wikipedia). This 10 megapixel camera gives 3648 × 2736 pixel JPEG images so: 3720/17.3 = 215.03 pixels per millimetre (ppm) or 2800/13 = 215.39 ppm.

I then took an image of a ruler at the minimum focus distance and an aperture of f8 and opened it in the open source graphics editor Gimp. In the menu Image > Print Size I changed the dpi setting to 215.2 ppm giving a print size of 17.29 mm x 13.01 mm (if your View menu setting is not dot for dot your image will be displayed at this size on the screen at 100% view magnification). Pressing shift + m activates the measuring tool and with units selected as mm, on checking the ruler on the image the notches on the ruler measured off at 1.022 mm in Gimp.

screenshot: picture of steel ruler opened in Gimp

Not exact, hardly scientific, but an indication that 48 mm of extension tubes gives me approximateley 1:1 magnification on my 50 mm lens. Using the same method you could get a feel for the magnification gained by using different combinations of extension tube.


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