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This question already has an answer here:

I have a Canon 60D (crop factor 1.6x) and an 85mm f1.8 prime lens. Now I want to add a 2x teleconverter and attach the 85mm lens to it.

My question is: What will the effective aperture be?

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marked as duplicate by mattdm, Dan Wolfgang, drfrogsplat, AJ Henderson, Paul Cezanne Mar 27 '14 at 16:43

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Be sure the convertor you want to use is compatible with that lens. The Canon EF 2X series are not compatible with the EF 85mm f/1.8. – Michael Clark Mar 27 '14 at 11:12

With a ideal 2x teleconverter, you will be 2 f-stops down from what the lens is set to.

Think about the basic physics and this should be clear. A 2x teleconverter makes the dimension of anything in the image 2x larger. Something that would result in a 1x1 mm square with the bare lens results in a 2x2 mm square with the teleconverter. That 2x2 mm square has 4 times the area of the original 1x1 mm square. That means the same light coming out of the lens is now spread over 1/4 the area, which means it will be 1/4 as bright. A factor of 4 less light is 2 f-stops, since each f-stop represents a factor of 2.

Any real teleconverter will loose a little light due to it being reflected off the front surfaces of the lens elements and the like. Therefore a 2x teleconverter will loose a little more than 2 f-stops of light. However, this extra loss is generally small enough to ignore. In any case, the automatic metering system in the camera should compensate for whatever the teleconverter does. You will see this as slower shutter speed, or requiring you to open the lens more for the same shutter speed.

Therefore, to answer your question, if the lens is set to f/1.8, then the effective aperture for the purpose of light level will be 2 f-stops higher. f/1.8 is not one of the standard f-stop values, but computing 2 f-stops higher is still easy. Each f-stop is sqrt(2) higher from the previous, so 2 f-stops is 2x higher, so 2 f-stops past f/1.8 is f/3.6. If the lens was set to f/4.0, then you'd get f/8.0, if set to f/11, you'd get f/22, etc.

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Not worth another answer since this is a great explanation of what is happening, but I also find it sometimes helps to point out it is kind of like a magnifying glass. It just looks at the center 1/4 of the image and throws out the rest. – AJ Henderson Mar 27 '14 at 13:47

The effective maximum aperture will be F/3.6. You can stop down from there to F/22 which will be equivalent to F/44. The camera will get the same amount of light as F/3.6 to autofocus since that is done wide-open.

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