Confusion on teleconverter


TPF Noob!
Oct 11, 2007
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I am confused on the two F stop of a 2X tele. I was explained today that it would take a F6.5 and go to 8(1 step) then 11 (2nd step) Can someone explain that to me. I understand F-stop, I understand why the tele does this, what I don't understand is how that is two f-stops?
Imagine your 70-300 at f/5.6, 300 mm. With a 2x telephoto that goes to f/11, 600 mm as you say. (small 6.5 - 5.6 typo in your post)

The f-number of the lens is the focal length of the lens divided by the diameter of the 'entrance pupil'. The entrance pupil is what the iris (the aperture blades) looks like when you look at the front of the lens. A 300 mm f/5.6 lens has an entrance pupil 300/5.6 = 53.6 mm in diameter. (300/53.6 = 5.6)

When you put the teleconverter on, the entrance pupil is unaffected: the teleconverter goes behind the lens. Now you have a 600 mm lens with the same entrance pupil. You have doubled the focal length, so the f-number also doubles. (600/53.6 = 11).

The f-stops between 1, 2, 4, 8, 16, 32, 64 etc are not exact, hence the small rounding error between 11 and 2 x 5.6

Another way of putting it is that the same amount of light gets into the lens, but it is spread out over four times the area (two times linear magnification = four times area magnification), and therefore it is a quarter as bright - ie two stops less bright.

I can never remember this myself, I don't suppose you have a lens that has an aperture ring on it do you? Those rings are usually marked at 1 stop intervals. 22 16 11 8 5.6 4 2.8

As Helen said the relationship is via the sqrt(2). so 5.6 * sqrt(2) * sqrt(2), or 5.6 * 2 = 11 give or take a rounding error.

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