I have a Nikon 60mm f/2.8 Micro lens (DX sensor) and I haven't been able to exactly determine what magnification I can get with Kenko 12mm, 20mm, 36mm extension tubes. Suppose I would use all of them together.

Also how many stops of light would I lose using them?

Depending on the answer, is it worth it to obtain the tubes or the macro effect wouldn't be so different and I would lost a lot of light instead? I would buy them only if I could create images with magnification similar to this one


  • 1
    \$\begingroup\$ possible duplicate of How can I calculate what the effect of an extension tube will be? \$\endgroup\$
    – mattdm
    Commented Dec 22, 2012 at 14:29
  • \$\begingroup\$ @mattdm I still don't know if it helps on macro lenses (which can focus at close distance without help) or I have to use some zoom (70-300) with the tubes to get good magnification. \$\endgroup\$
    – user219882
    Commented Dec 22, 2012 at 14:44
  • \$\begingroup\$ A "macro lens" is just a (well-corrected) ordinary lens with a built-in, adjustable extension tube. Adding additional tubes will get you closer than 1:1 (by how much depends on the actual focal length of the lens at its 1:1 focusing distance), as with any other lens. And, as with any other lens, you may hit a point where the required subject distance from the optical centre of the lens for a given magnification lies inside of the lens, at which point the lens would need to be reversed. \$\endgroup\$
    – user2719
    Commented Dec 22, 2012 at 19:18

2 Answers 2


The previous post link talks about magnification.

As far as light loss, I would first suggest you just use the meter in your camera. It should show the relative setting no matter if you have neutral density filters or extension tubes attached.

If you really want to calculate this stuff manually, adding extension tubes simply alters the focal length of a lens. And the aperture of a lens is a ratio of its focal length.

Using the formula N = f/D (F-stop = focal length divided by diameter of lens opening), your 60mm f/2.8 lens has an pupil diamter of 21.43mm.

Now, you add 12mm of extension and your focal length becomes 72mm, but the physical aperture and diameter sizes of the lens did not change.

This means that your new f-stop value, with 12mm extension, becomes f/3.5 which is 72mm / 21.43 opening size, or, 2/3 of a stop less light.


I don't think the previous answer can be generalized. what if you use a 8mm f/2 the pupil diameter is not 4mm right ? I found another formula: effective aperture (EA), is the relation between f-stop value(f-stop), lens magnification (LM) and pupil magnification (PM), such that:

EA= f-stop*(1+LM/PM)

with PM= rear pupil diameter / front pupil diameter of the lens

My concern is that the value of PM is not given by the manufacturer as far as I know ?

And then, how do you calculate effective aperture for a stack of a 200mm f/4 with a reversed 10mm f/2 lens?

(sources: http://toothwalker.org/optics/dofderivation.html#eq1)


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