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I just bought a Canon 2X extender and I always believed that it would only result in 2 stops' loss of light and not change the aperture.

However, when I tested this extender on a Canon 135mm f/2 lens, the maximum aperture I could set went down to F/4. So it kind of confused me, because if I step down the aperture to f/5.6, is it actually setting the aperture to f/5.6 or f/2.8 (i.e., 1 stop below f/2)?

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4 Answers 4

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This is really simple when you think about it. The additional element changes the focal length of the lens, without changing the apparent size of the aperture. That means that the relative size of the aperture decreases, so the f number does in fact actually change. (If this is unclear to you, see the bit about f numbers in this other answer.)

This is also why rear wide-angle converters can go the other way, effectively increasing the aperture. (See How can a speedbooster improve the light performance of a lens? for more.)

Some converters communicate intelligently with the camera body, so the aperture displayed will be correct. This is the case with the Canon extender you have, but might not be with third-party ones. This explains the part you were confused about: the camera is aware of the change already and the numbers it is showing you are what you will actually get. When you set the aperture on the camera to f/5.6, the aperture on the lens is set to the same position that would be f/2.8 without the extender (but which genuinely is f/5.6 with it).

Note that teleside converters and wide-angle converters which go on the front of the lens do change the effective aperture (see What's the difference between real and effective aperture?), so they don't change the f number. (They are usually lower quality, however, and can introduce vignetting and other artifacts.)

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Yes: Remember that the f-stop is a relation between your aperture and the focal length.

In theory if you have a f/1.0 50mm lens, that means that you have an aperture of 50mm.

If you double the focal length (2x) the relationship reduces to half, so it will be a f/2.0 aperture.

There can be some other factors as the quality of the lens elements, the internal reflections, so probably you can lose some more light in the conversion process.

In short: Just multiply the teleconverter value by your current aperture.

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    \$\begingroup\$ Not just in theory. It actually affects the exposure in practice too. \$\endgroup\$
    – H H
    Feb 28, 2015 at 19:12
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Another way to think about this is that you are now fitting 1/4 of the light through the same size hole. You extended the focal length by using a teleconverter (basically a magnifying glass). You are now only looking at 1/4 of the original image on the sensor. This means you lose two stops worth of light that you are now discarding.

The opening for the light didn't change any, but the focal length was lengthened which means less light makes it to the sensor. Technically at the point of the aperture, it's still the unconverted focal length, but since the effective focal length is doubled, this drops the f number since the f number is based on the focal length divided by the size of the entrance pupil.

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  • \$\begingroup\$ Actually the light does make it to the lens. It even makes it to the light box. But it is spread out in the light box beyond the edges of the sensor. And spreading out light reduces its field density. \$\endgroup\$
    – Michael C
    Dec 27, 2015 at 7:00
  • \$\begingroup\$ Yeah, lens was the wrong word. I'm not sure why I used it when I'm pretty sure I meant sensor. Didn't even notice what you were pointing out until I looked closer. Thanks for pointing it out. \$\endgroup\$
    – AJ Henderson
    Dec 27, 2015 at 8:42
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I think this is happening because let’s say you had a 70-200mm f2.8 lens and you were at 70mm, f2.8. If you put a 2x extender on that, you would multiply everything by 2, so you would be at 140mm f5.6. I think that is what is happening.

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