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I am photographing a scene with white and black elements in it. Starting at the f/22 stop, I widen the aperture one stop and decrease the exposure time by a factor of 2, take a picture, and keep doing this for all the f-stops on the lens. My expectation is that raw counts should stay the same inside a white region or a black region since halving the exposure time compensates for widening the aperture. But when I select a white region and average its pixel raw counts for each image, there is variability between the images (the standard deviation of the raw counts is ~5% of the mean). Same thing if I select and average a black region. I am not knowingly changing anything else (illumination, camera position), and the camera is a scientific CMOS. What could be causing this variation: noise, or something more systematic?

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    \$\begingroup\$ What is the lens? Is it also designed for scientific purposes? If so, does it come with a datasheet listing tolerances, or can you get that from the manufacturer? \$\endgroup\$
    – mattdm
    Mar 11, 2019 at 18:25
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    \$\begingroup\$ Nothing is "perfect" in the cameras used for artistic/documentary photography. As with any technical device, there are always tolerances to be considered. For typical cameras, historically that tolerance has been about 1/3 stop, though in more recent times we seem to have settled on 1/6 stop. If you need scientific quality measurements from a camera, you should consider lab grade instruments rather than consumer grade cameras (even the top "pro" models are consumer grade in this context). But be prepared to pay tens of thousands of dollars instead of hundreds or a few thousand dollars. \$\endgroup\$
    – Michael C
    Mar 11, 2019 at 23:02
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    \$\begingroup\$ Much of this ground has already been covered in great detail in the answers and comments to Is there a sane reason why ¹⁄₁₂₅ is not, instead, exactly half of ¹⁄₆₀? \$\endgroup\$
    – Michael C
    Mar 11, 2019 at 23:03
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    \$\begingroup\$ @mattdm It is an industrial lens. Good idea to contact the mfr for tolerance info. \$\endgroup\$
    – KAE
    Mar 12, 2019 at 11:21
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    \$\begingroup\$ If you really care about this (calibrated transmission intervals), what you probably want is a cinematic lens calibrated in T-Stops instead of F-Stops. An F-Stop is purely a measure of the aperture diameter with respect to the focal length. T-Stops are a measure of 50% light transmission levels. The former gives you a more precise measure of the depth of field, the latter of exposure. \$\endgroup\$
    – J...
    Mar 12, 2019 at 11:31

5 Answers 5

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This is normal behavior, caused by:

  1. Imperfections of aperture. Usually there are variations from technology process which cause not to have exact size of the hole. On 50mm lens f4 you should have 12.5mm opening, but it can be 12.4mm or 12.6mm
  2. Imperfections in shutter speed. The shutter is also mechanical unit and based on some factors as temperature, how precise are the blades and other elements inside, speed will be not 1/100s but can be 1/110s or 1/90s.
  3. The same is true about the sensor itself (from electronic point of view)

At the end even two consecutive photos can have different (slightly) exposure.

And add fluctuation of your illumination source...

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  • \$\begingroup\$ So 'noise' and not something systematic. That makes sense. \$\endgroup\$
    – KAE
    Mar 11, 2019 at 17:14
  • \$\begingroup\$ @KAE, you can name it noise :) There are so many variables so at the end it is random \$\endgroup\$
    – Romeo Ninov
    Mar 11, 2019 at 17:18
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    \$\begingroup\$ Funfact: some cameras store the real measured aperture and shutter speed in the metadatas. And I saw some really interesting values there. But I have also admit that some of these values are a little bit odd, maybe there is missing a scaling factor. \$\endgroup\$
    – Horitsu
    Mar 12, 2019 at 5:22
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    \$\begingroup\$ I would also add that f/22 also blocks much more diffuse ambient light than f/4, which results in more contrast. Dark is darker when the aperture is smaller. \$\endgroup\$
    – CamilB
    Mar 12, 2019 at 9:00
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    \$\begingroup\$ @CamiB I would think that this is a very significant factor. Anybody who has taken an against-the-light picture with and without shielding from the sun (which was not in the picture!) can tell the huge difference in contrast. The same effect, just smaller, is present with the normal ambient lit objects outside the picture frame. A smaller aperture greatly reduces scattered light, so it will darken the picture overall. \$\endgroup\$
    – Peter - Reinstate Monica
    Mar 12, 2019 at 14:37
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In theory, yes — stops are interchangeable. In practice, they do not perfectly cancel to complete precision.

the standard deviation of the raw counts is ~5% of the mean

In photographic terms, this is basically nothing. It is far below human perception, and even when the difference is noticeable, the generally-expected workflow involves working with each image individually, so the photographer can compensate either in the field or in post-production.

Cameras meant for photography are not measuring devices; using them as such is setting yourself up for disappointment. Making the devices much more precise would be a lot more expensive and provide no benefit for the target market. Even if you have a camera made for scientific purposes, these particular tolerances might not be within the relevant area of concern.

If you're trying to get perfection for something like a time-lapse or another series of photos, post-processing to even out the fluctuations is your best bet.

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    \$\begingroup\$ Good idea to even out the fluctuations - maybe I will take a lot of frames and average the pixel values. \$\endgroup\$
    – KAE
    Mar 11, 2019 at 17:17
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    \$\begingroup\$ Ideally you would analyze the exposures by a piece of software, let it measure the average exposure and then adjust each picture individually to match that target. However the math gets quite comples very fast (how do you define the "brightness", do you know how to transform your brightness into a linear colorspace instead of a gamma-corrected one etc.) Averaging multiple images is much simpler, hopefully the variation is random and not systematic. Or you can calculate the f/8 image as the average of f/5.6 and f/11, a kind of a "runnin average". \$\endgroup\$
    – NikoNyrh
    Mar 13, 2019 at 18:40
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The short answer is yes... they cancel. But there are some nuances.

Each time the diameter of a circle increases (or decreases) by a factor equal to the square root of 2 (approximately 1.4) the area of that circle is exactly doubled (or halved if decreased). The f-stop numbers are all based on powers of the square root of 2 (e.g. f/1 = √2^0; f/1.4 = √2^1; f/2 = √2^2; f/2.8 = √2^3; etc.)

Shutter exposures are more intuitive. 1/500th sec is obviously half as long as 1/250th sec, etc.

The nuances:

Cameras do a bit of rounding. E.g. if you have a 100mm lens it's probably not precisely 100mm (but it's probably not far off) and as you refocus, the lens may do a bit of focus breathing (for a good lens that stays with 5% of the stated focal length ... but some lenses have rather strong focus-breathing issues ... e.g. 30%. When this happens, it means the f-stop isn't strictly accurate.

F-stops aren't strictly accurate as it is. But they are "close enough" that the margin of error wont impact the exposure in a noticeable way.

There are other issues. When you shoot heavily stopped down (e.g. f/22), all light comes from a very small area near the center of the lens axis and is distributed across the sensor more evenly. When you shoot wide-open, light comes from a wide range of angles. Areas of the sensor near the center can collect light from many angles, but areas of the sensor near an edge or corner are more limited on the number of paths light can take through the lens to reach that particular spot. This results in vignetting. So while you can take two photos using "equivalent exposures" (trading a stop of aperture for a stop of shutter duration), changes in vignetting patterns can cause pixels to have a different amount of collected light depending on the pixel you choose to inspect.

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  • \$\begingroup\$ While this is a good answer, where are you getting the numbers for focus breathing from? \$\endgroup\$
    – Hairy Dresden
    Mar 11, 2019 at 17:53
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    \$\begingroup\$ @Hairy Dresdon, just reviews of various lenses. But these are tests you can also do yourself. E.g. while looking through a viewfinder (or even in live-view) run the focus through the full range and watch the field of view change. You can can also use calculators (like this one: tawbaware.com/maxlyons/calc.htm ) to work out the true focal length (relative to what it should be). There are a few lenses that have reputations for being "heavy breathers". \$\endgroup\$
    – Tim Campbell
    Mar 11, 2019 at 18:26
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    \$\begingroup\$ The AF-S Nikkor 70-200mm f/2.8G VR II is notorious for "breathing" to about 140mm when zoomed to 200mm and focused at MFD. Nikon corrected this with the more recent AF-S Nikkor 70-200mm f/2.8E FL VR (at a price of about $2,900!). \$\endgroup\$
    – Michael C
    Mar 11, 2019 at 23:12
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I think it wasn't mentioned: with increase in exposure time comes increase in thermal Dark Shot noise. You can read more here, for example

enter image description here

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With regard to systematic problems: you are taking into account that with opening up the aperture depth of focus decreases and thus the borders of out-of-focus scene parts blur? Also with small apertures you might get some blurring due to diffraction.

If you have a mechanical shutter, you actually can get diffraction with large apertures from the resulting short shutter times when a significant amount of the exposure time is spent near at least one of the shutter curtains moving across.

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    \$\begingroup\$ I check the raw counts well inside an all-black or all-white region of my scene, away from the borders, to avoid blurring due to the change in focus that you mention. \$\endgroup\$
    – KAE
    Mar 12, 2019 at 11:23

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