Lets say we have two image sensors, each one with 1MPixel resolution but differing in size. The first sensor has the size 1/3", the other one 1". Each sensor is equipped with a suitable lens (for the image sensor format, so no vignetting or cropping takes place). Both lenses have the same characteristics: F-stop f/2.0 and Ultrawide view angle circa 100°. Both sensors are running at 100fps, so that the image shown by the 1/3"-sensor is rather dark at normal illumination in a house.

Now how much does the image shown by the 1"-sensor differ (at the same illumination) in terms of brightness? I'd guess that the 1"-image would be much brighter because of the greater entrance pupil, is that correct, and if yes, how can I calculate which f-stop the bigger lens (which sits on the 1"-sensor) need to have to resemble the image brightness of the small sensor?

2 Answers 2


The image brightness will be the same. The entrance pupil size does not matter for image brightness - what matters is only the F number (or more accurately, T number which is equal to the true f number divided by the square root of the transmission.)

In a bit more detail:

If you fix your field of view at 100deg and vary the sensor size, the focal length scales between different image sizes, as does the entrance pupil diameter. Since the focal length directly affects magnification and shorter focal lengths provide greater magnification in image space (that is, behind the lens) the sensor sees a much bigger pupil for a 10mm f/2.8 lens than it does a 20mm f/2.8 lens.

  • Perhaps the OP made an unnecessary assertion and has his assumptions backwards? Presumably, given the same objective lens size, and holding field of view constant, a smaller sensor will get a brighter image because exit pupil can be smaller. I.e., the incident light is the same but it can be concentrated more on a smaller sensor area, right? (This assumes transmission is the same through both lenses.)
    – feetwet
    Jul 8, 2015 at 18:06
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    Not quite. Imagine an ultra-wide angle lens and a supertele. The supertele spreads maybe 4 degrees across 12mm, right? An ultra-wide spreads 40 degrees or more across that same 12mm. Thus, the magnification at the sensor is very very high. If you used a positive lens behind the same base objective to concentrate the previous image size onto a smaller area, you will get a faster f/number and brighter image. If you don't and let the excess be excess, it is identical to cropping from the bigger image. Jul 8, 2015 at 19:32
  • OK: It sounds like your statement is holding the lens constant. But if we assume as in the OP that the lens was designed for the sensor in each case (so the lens for the 30mm sensor has a 30mm spread while the lens for the 12mm sensor has a 12mm spread), then are we talking about different f numbers? (Again, assume same FoV, objective size, and overall lens transmission.) Or do we have the same f number but more illumination per mm^2 on the smaller sensor?
    – feetwet
    Jul 8, 2015 at 19:56
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    If a lens is f/2.8, it provides the same light per area regardless of focal length and image size. I.e it will provide the same amount of light to the centermost 1mm*1mm area as it will to a corner 1mm*1mm area, ignoring vignetting and natural cos^4 light falloff. Jul 8, 2015 at 21:18

Relative to your question about sensor size, sensor size does not matter. Illuminance is the light per unit area. For example, a hand held light meter only sees a very tiny area, but that reading is good for very tiny sensors or much larger sensors, like 8x10 inch sheet film.

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