A lens has a focal length, and its entrance pupil has a diameter. The ratio of these is an f-number. The size of the entrance pupil is limited by the size of the front element, and we describe lenses by the f-number at this limit. So, a 50 mm lens with a 35 mm diameter front element is an f/1.4 lens. So far so good.

One of the reasons we care about f-number is that it describes the light-gathering ability of the lens. At any given focal length, a lens with a faster f-number (closer to f/0) will gather more light, and two lenses with the same f-number will gather the same amount of light.

But what happens to this light?

The thing is, lenses also have an image circle diameter. The image circle is where the cone of light radiating from the rear element reaches the plane at the lens's flange focal distance (aka register distance). That's where the sensor (or film!) sits. That circle has a diameter. For a traditional full-frame 35 mm lens, the image circle is 43.3 mm in diameter; for a Micro Four Thirds lens, it's 21.6 mm in diameter (numbers taken from this remarkable page). The µFT image circle is almost exactly half the diameter of the FF one.

The light that is gathered is then spread out over the image circle. It's not spread out homogenously, obviously - light from a given point in the scene goes to a corresponding point in the image circle. That's why it's called an image circle, because there's an image on it. But the fact is that every photon gathered by the lens, forgetting the weak ones that get lost in the camera, ends up somewhere on the image circle.

The camera's sensor (or film!) then sits inside the image circle. Photons that fall on the sensor go towards making a picture; those that don't, don't.

When you use a camera with a native lens, the sensor is sized so that it occupies as much of the image circle as it can. For 35 mm, that's 59% of it; for µFT, that's 61% of it. But when you adapt a lens from a different system (using a glassless, purely mechanical, adaptor), the sensor may not be the ideal size; if you (somehow) but a µFT lens on a 35 mm camera, the image circle would be too small, and you would get severe vignetting. If you put a 35 mm lens on a µFT camera (which may of us can and do), it works, but the image circle is far bigger than the sensor - the sensor covers just 15% of the image circle.

So, if i make two lenses, both 50 mm and f/1.4, but i make one for a 35 mm mount, and one for a µFT mount, but i put both of those on a µFT camera, i will get very different performance. My understanding of optics is shaky, but i believe that the magnification of the images will be the same, as that's purely related to focal length. Because the lenses have the same f-number, the amount of light gathered will be the same. But because the one made for a 35 mm mount has a larger image circle, of four times the area, the amount of light actually falling on the little µFT sensor will be smaller - four times less. That's the equivalent of two stops!

This contrast is not just theoretical. I have a µFT camera. I have a Sigma 60 mm f/2.8 native lens. I also have a Canon 50 mm f/1.4 FD-mount lens on an adaptor. These are not of the same focal length, but they're similar. You would think from the f-numbers that the Canon is substantially faster than the Sigma. But if i'm right, then the effect of the Canon's larger image circle means that it behaves more like a f/2.4. That's barely any faster at all!

Is my analysis correct? If not, what am i missing?

As i said, my understanding of optics is shaky, and i haven't been able to find any discussions of this subject from this angle (these questions are unrelated). I abase myself in gratitude for any insight offered. The Master replied: Do you have a question to ask, or do you want to make a speech?. Apologies for the length and turgidity of this question.


1 Answer 1


You're close, but not right. The light falling on any given area of the image circle is constant regardless of the format the lens is designed for. Otherwise, cropping a photo would change the exposure, which is obviously nonsensical.

To put it another way, the f/stop is representative of the light at each point regardless of sensor size — not of the total amount spread across the whole covered area of the image circle. (I think everything else you have written is correct, except this point which is causing your confusion.) If you take a printed photograph and tear it in half, each part will have received half of the light of the whole — but that measurement is not relevant to exposure.

On the other hand, bigger sensors do inherently receive more light overall. That's why cameras with bigger sensors have an advantage in having less noise at the same ISO (given roughly equal sensor technology).

But, the lens doesn't matter, because the image circle is irrelevant to your photographs — only the part of it you actually record. (Unless of course you are trying to use a lens with an image circle that doesn't cover your sensor, but that's a different issue.) An f/1.4 lens is always two stops (4x) faster than an f/2.8 lens.

  • \$\begingroup\$ Just to confuse things, though: assuming you print at the same size, the effect of depth of field is altered by sensor size. (Smaller sensors give less blur in proportion to the difference in sensor size. So your 4/3rds f/2.8 has depth of field similar to an f/5.6 lens on a full-frame camera. \$\endgroup\$
    – mattdm
    Oct 12, 2013 at 14:20
  • \$\begingroup\$ Are you sure you didn't get that last part backwards? When you magnify the viewing size of an image don't you reduce the depth of field because you are enlarging the blur at the edge of the DoF from imperceptibly small to barely large enough to be perceptible? Isn't the reason smaller sensors generally yield less DoF is because of the need for shorter focal lengths to produce the same FoV. \$\endgroup\$
    – Michael C
    Oct 12, 2013 at 19:03
  • \$\begingroup\$ @MichaelClark yeah that came out wrong. Deleted -- will rephrase and repost. \$\endgroup\$
    – mattdm
    Oct 12, 2013 at 22:51
  • 1
    \$\begingroup\$ It's exactly as you say: "the f/stop is representative of the light at each point" on the image plane. To put it yet another way, what the f-number is about is how much light is gathered from any given point in the scene. That's a matter of simple geometry (which this margin is too small to contain) to show. Because this is a lens, the light from any given point in the scene ends up at a given point on the image plane. A lens with a bigger image circle is simply gathering light from more points in the scene - a 50 mm is a normal prime on a 35 mm camera, but a telephoto on a µFT camera. \$\endgroup\$ Oct 13, 2013 at 22:32

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