When stacking tubes to make a normal lens a macro lens, you have significant light lost due to the increased distance. Do normal macro lenses suffer the same fate? Why or why not?


2 Answers 2


Yes, but it may be hidden. A unit-focus macro (the old school, where the lens elements are in a more-or-less fixed relationship to one another and the whole shebang is moved further away from the sensor/film plane) will usually work in exactly the same way as extension tubes. The lens is essentially a well-corrected ordinary lens with a built-in adjustable helicoid extension tube, and you need to be aware of the lens draw when metering externally. (When metering in-camera, the electronics take care of the hassle for you.)

Newer internal-focus (or rear focus) macros (and non-macros for that matter) actually focus by changing the focal length of the lens. The front element stays anchored in space, and the focal length of the lens is reduced. That does two very noticable things. First, there is no "focus breathing" -- the subject will stay the same size in the frame as you focus. Secondly, and most pertinently for your question, the size of the physical aperture stays the same as the focal length is decreased, so the effects of lens draw are masked (or moderated, depending on the individual lens) by an ever-increasing relative aperture as you focus closer. So with a "pure internal focus" design, at infinity you may have a lens that is 100mm set at, say, f/8, but when you focus much closer, you may actually have a 75mm lens at f/6 (using the same 12.5mm apparent aperture), but the lens draw due to focus reduces the light to the same level as it would have been at f/8. The physical length of the assembly is unchanged, so the relative aperture of the whole remains the same if the physical aperture is unchanged.

And just to complicate matters a bit further, some lenses seem to be of a hybrid design -- they use both internal focus and ordinary extension to get where they're going. That's not a problem with TTL metering, but it would mean creating a draw chart for use with an external meter. A couple of test shots with a grey card and manual exposure settings will tell you whether or not you need to go to the trouble of creating one.

  • \$\begingroup\$ Very interesting! I was really wondering about the whole internal focus part. That makes sense. \$\endgroup\$
    – rfusca
    Commented Jan 3, 2012 at 20:19

Yes - the closer you focus the longer the extension, and the inverse square law gets you all the time!

At 1:1, you will form an image two stops darker than when focussed at infinity, at 1:2 your image will be formed a stop darker than infinity focus.

However, if you're using TTL metering then your camera meters with this drop-off and your exposures will be correct.

  • \$\begingroup\$ Really 2 f-stops down at 1:1? I thought the light loss ratio relative to that considering f-stop alone was 1/(1+M) where M is the magnification. Normally M is very small (a mountain down to 24mm), so you ignore this effect. At 1:1, M is 1 and the overall light is 1/2 what you'd expect, or one f-stop down. This is all assuming a extension-focussing lens, not internal focussing like what Stan is talking about. \$\endgroup\$ Commented Jan 3, 2012 at 20:44
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    \$\begingroup\$ @Olin The formula I believe is correct is: compensation=(extension/focallength)**2, and as such this states that for example a 100mm lens focusing at 1:1 (200mm extension) would have a 4x exposure factor - i.e 4 stops. Take a look at [The QuickDisc] (salzgeber.at/disc/index.html) to see a practical example of how some LF photographers deal with this issue!! \$\endgroup\$
    – Tim Myers
    Commented Jan 5, 2012 at 20:43
  • \$\begingroup\$ Something doesn't seem right. Are you sure "extension" means the total distance from the lens to the film plane as you are using it here, or extension beyond what would be required to focus at infinity? \$\endgroup\$ Commented Jan 5, 2012 at 21:57
  • \$\begingroup\$ I just looked it up, and my formula above is wrong. I forgot to square the result. That brings it in line with what you are saying that you're down 2 f-stops (a factor of 4) at 1:1. Sorry for the confusion. \$\endgroup\$ Commented Jan 5, 2012 at 22:09

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