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The question Will a macro extension tube eliminate the cropping of an APS-C lens on a full-frame sensor? is certainly related, though I'm not sure if it's a duplicate.

As a follow up to my recent question about extension tubes I realized I had another question. I currently have a Canon 40mm f/2.8 pancake lens that's designed for a full frame camera. I have that on my Canon T5, which is an APS-C camera. Of course, technically this means that I'm not getting the full picture - some of the light that comes from the lens is hitting somewhere not-my-sensor.

If I add an extension tube or two to my lens, will more of the light make it from the lens to my sensor, or less?

  • Another way to ask basically the same question: Will there be less vignetting on my APS-C camera if I use a full frame lens with my extension tubes? – mattdm Aug 22 '16 at 20:37
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    How is this different from your previous question? An extension tube has a crop like effect, which means less then the full image circle of the lens hits the sensor. So the answer to this question is: yes, with an extension tube of negative length. – null Aug 22 '16 at 21:13
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    Possible duplicate of Why doesn't an extension tube "crop" the picture? – null Aug 22 '16 at 21:13
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If I add an extension tube or two to my lens, will more of the light make it from the lens to my sensor, or less?

The answer here is much the same as for your other question. As you move the lens away from the sensor, the image circle spreads out over a larger area. The sensor is obvious the same size, so it will see a smaller portion of the image. The field of view as seen by the sensor will therefore decrease, the brightness of the image falling on the sensor will decrease, and the part of the image that does fall on the sensor will be magnified more. To get more of the light projected by the lens to fall on the smaller sensor in your camera, you'd have to move the lens closer to the sensor, but extension rings move the lens farther away, not closer.

Just reading an answer doesn't always help you to really see how something works, so let me propose the following experiment...

Try this: Tape a 5"x7" index card in the center of an 8.5"x11" sheet of paper, and then tape that paper to a wall. Shine a flashlight at the paper and adjust the distance so that the flashlight's spot just covers the paper. This is a model of the inside of a camera: the index card represents the APS-C sensor, the larger paper represents a full-frame sensor, and the lighted end of the flashlight is the rear surface of the camera lens. Now move the flashlight farther from the wall while continuing to shine it at the paper. Does more of the light from the flashlight fall on the index card, or less?

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The standard camera lens/body combination stops the forward movement of the lens when 0.1 magnification is obtained. At this close up positon, a 10 cm height is imaged 1 cm on film/sensor. The f/numbers engraved on the lens barrel are only valid when the typical lens is imaging objects at infinity. As we move closer and closer by elongating the lens to film/sensor, the magnification obtained increases. Magnification means that the image forming light rays are being spread thinner. That translates to less light energy in any given unit area.

We can easily calculate the magnitude of the loss in image brightness due to magnification. We are talking about the Bellows Factor formula.
BF = (M +1) ^2 That’s magnification plus 1 squared. For the 0.1 magnification positon BF = (0.1+1) ^2 = 1.21 about 1/3 of an f/stop (we must compensate by aperture or shutter speed).

We handle the BF just like a filter factor, it is a multiplier. Suppose the exposure for this subject is is f/16 @ 1 second as determined by a hand-held light meter. We apply the BF = 1 X 1.21 = 1.21 seconds. In other words, at the 0.1 magnification position we are required to increase the exposure from 1 second to 1.21 seconds to compensate for light loss as the image forming rays transvers the lens. If we increase the lens to sensor distance an obtain magnification 0.5 the BF = 2.25. In other words, if we obtain magnification ½ life size we compensated by increasing the exposure from 1 second to 2.25 seconds. If we obtain life size, magnification 1, the BF is( 1 + 1) ^2 = 4. Now the exposure time must be increased from 1 second to 4 seconds. That’s two f/stops. If we obtain magnification 2 (twice life-size) the BF is (2 +1)^2 = 9. We increase the shutter speed from 1 second to 9 seconds.

All this is independent of the lens unless it is a macro. A macro is optimized to image at magnification 1 (unity). The design of the lens compensates for BF thus the f/numbers stay true.

  • What has any of that got to do directly with whether the crop factor is mitigated or multiplied by extension tubes? Your answer has true information, but it doesn't answer this question very clearly at all. – Michael C Aug 22 '16 at 23:12
  • @ Michael Clark -- Please help me understand this part of the OP's question - "If I add an extension tube or two to my lens, will more of the light make it from the lens to my sensor, or less"? I think I hit the nail directly on the head. – Alan Marcus Aug 22 '16 at 23:33
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    It seems to me, based on the rest of the question, that he is referring to the size of the projected image circle, not the field strength or quantity of light. It seems to be more concerned with field of view than anything else. – Michael C Aug 22 '16 at 23:57
  • As the extension tubes are added the lens to sensor/film distance increases. The more the lens is extended the more the image is magnified. As this occurs, the circle of good definition expands. The APS-C frame is 66% smaller that the FX frame. In other words, the magnification factor (crop factor) is 1.5. Stated differently the FX frame is 150% larger than the ACS-C frame. The 40mm with tubes on an FX body will surely cover the larger frame. The net result, the image will be enlarged 150% as compared to the APS-C frame. – Alan Marcus Aug 23 '16 at 1:14
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    That info needs to be more explicitly stated in the answer instead of only implied in a way that the only reader who would catch it is the person who already knows it. – Michael C Aug 23 '16 at 2:07

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