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Suppose I have two different cameras, one with a sensor whose area is twice the other's. Further assume the sensors have the same aspect ratio.

Suppose I put 35 mm-equivalent lenses on both, thus giving the same FOV (field of view).

If they both have the same f-stop, would the luminous flux on the larger sensor be 2x that on the smaller sensor?

I.e., for the same FOV and f-stop, would the large-sensor camera only need 1/2 the exposure time to achieve the same EV as the small-sensor camera?*

And would that in turn mean that, for the same FOV and f-stop, the aperture diameter of the lens on the large-sensor camera would need to be sqrt(2) times that on the small-sensor camera?

Finally, how would their depth of fields compare?

*This answer said that, for the same f-stop and FOV, the images falling on the sensors would be equally bright (How does sensor size affect the brightness of an image given the same field of view and aperture?), but didn't precisely define brightness. I think by brightness they meant lumens per unit area (even though brightness is actually a perceptual term), not total luminous flux. If so, then I believe what I wrote in italics would be correct.

Edit: Here are some additional details to clarify the intent of my question:

Based on the comments, it seems it's agreed that, for the same FOV and f-stop (and with the same lens transmissivity), a camera with double the sensor area will have double the lens entrance pupil area, and double the light per unit time will hit its sensor.

Now, according to Wikipedia:

In low light, correct exposure requires the use of slow shutter speed (i.e. long exposure time) or an opened aperture (lower f-number), or both, to increase the amount of light (photons) captured which in turn reduces the impact of shot noise...

The size of the image sensor, or effective light collection area per pixel sensor, is the largest determinant of signal levels that determine signal-to-noise ratio and hence apparent noise levels, assuming the aperture area is proportional to sensor area, or that the f-number or focal-plane illuminance is held constant. That is, for a constant f-number, the sensitivity of an imager scales roughly with the sensor area, so larger sensors typically create lower noise images than smaller sensors. [emphasis mine]

So let's suppose you're in a low-light situation, and are deciding what shutter speed you need to choose to to acheive a certain picture quality—in particular, to keep S:N above a certain level. For simplicity, let's assume the two cameras have the same resolution. Thus the camera with double the sensor size will have double the pixel size, and thus gather double the light per pixel. Hence a camera with double the sensor size will have approximately double the sensitivity, and will only need approximately half the exposure time [or would it be 1/sqrt(2) the exposure time, since S:N varies as sqrt(signal)?] of the smaller-sensor camera to achieve the same low-light performance.

See also: https://www.dpreview.com/articles/5365920428/the-effect-of-pixel-and-sensor-sizes-on-noise/1

The total amount of light that goes to make up your image is the most important factor in determining image quality.

As I tried to explain (and demonstrate) in my article on equivalence, f-numbers dictate the light intensity of an exposure (light per unit area). However, this ignores the sensor size. To understand how much total light is available to make up your image, you need to multiply this light per unit area by the area of your sensor.

The assertion that a smaller sensor sees less total light and is, therefore, noisier (since it receives less light, and hence has a worse signal-to-noise ratio because of shot noise), is contentious. This is understandable, as it means thinking in terms of the whole image, not the individual pixels or the light intensity per unit area that underpins the f-number/ISO system of exposure (that is specifically designed to work independently of sensor size).

However, while it may seem heretical to anyone strongly attached to the conventional exposure model, it is, nonetheless, demonstrably true.

I.e., in determining shutter speed, what matters is the total luminous flux on the sensor, not the luminous flux per unit area. Luminous flux per unit area would only be the relevant metric if people printed out or viewed their photos at a size in proportion to the size of the sensor used to take them.

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For the same FOV and f-stop, will total luminous flux increase linearly with sensor area?

Yes, total luminous flux will increase. But photography doesn't measure exposure or brightness¹ using total luminous flux. It bases exposure on light per unit area.

I.e., for the same FOV and f-stop, would the large-sensor camera only need 1/2 the exposure time to achieve the same EV as the small-sensor camera?

No. This is because, by definition, EV, or Exposure Value, is a combination of equivalent f-numbers (Av for aperture value) and exposure times (Tv for time value) that produce the same amount of exposure per unit area

Again, the entire concept of Exposure Value is based on light per unit area.

It's usually assumed, unless explicitly indicated otherwise, that EV = EV100.

For instance, some of the combinations for EV9 (EV1009) are f/1 @ 1/500, f/1.4 @ 1/250, f/2 @ 1/125, f/2.8 @ 1/60, f/4 @ 1/30, and so on. Taking it into the 1/3 stop scale, f/2.2 @ 1/160 and f/2.5 @ 1/200 are also EV9.

Likewise, EV2009 are combinations of f-number and Tv that are one stop dimmer (i.e. half as much light is allowed to enter the camera): f/1 @ 1/1000, f/1.4 @ 1/500, f/2 @ 1/250, and so on.

The same combinations of Av and Tv for EV9 are the combinations for EV20010. Because ISO 200 is one stop more sensitive (film) or amplified one stop more (digital) than ISO 100, then EV200 = EV100 + 1.

Since ISO 800 is three stops more sensitive/amplified than ISO 100, then EV800 = EV100 + 3. That is, combinations of Av and Tv that are EV9 would be combinations of EV80012.

Notice that if the scene illuminance is constant, higher EVs allow less light into the camera, not more. What we tend to think of as "bright" EVs, such as EV16 that is approximately equal to the "sunny 16" rule of thumb actually allow less light into the camera with a scene of constant light than what we tend to think of as "dim" EVs, such as EV8 as is often used for night sports under artificial lighting. Lower EVs are brighter¹ Tv/Av combinations used to compensate for dimmer scenes. Higher EVs are dimmer Tv/Av combinations used to compensate for brighter scenes.

Strictly speaking, Exposure Value is not a measure of light intensity, though today it is often misused in that way. When someone says something like "the scene is EV9", what they are really saying is that the scene is, on average, a particular brightness level such that it will be "properly" exposed (whatever that is) using any combination of Tv, Av, and ISO that equals EV9.

This will be more true of a scene with the same overall general brightness, like a city park under overcast skies at mid-day, than a scene with stark differences in brightness levels from one part to another, like a cityscape at night that is mostly very dark with small areas of bright lights and small bright areas illuminated by lights. In the second case, if one were to use an incident light meter and expose based on the average overall light level of the scene, the image would be grossly overexposed with the very dark areas of the scene rendered as medium bright ("18% gray" if we were shooting in B&W) and the specular highlights and other small areas of the scene that are brightly lit would be completely blown out.

So let's suppose you're in a low-light situation, and are deciding what shutter speed you need to choose to to acheive (sic) a certain picture quality—in particular, to keep S:N above a certain level.

Earlier in your question you are concerned with maintaining the same exposure. Now you want the same SNR? The two are not the same thing. In fact, the difference between the two is why larger sensors perform better in low light than smaller sensors.

For simplicity, let's assume the two cameras have the same resolution. Thus the camera with double the sensor size will have double the pixel size, and thus gather double the light per pixel.

True. (If one acknowledges that sensors do not have pixels, they have photosites or sensels. Only recorded digital images and the mediums that display them have pixels.)

Hence a camera with double the sensor size will have approximately double the sensitivity, and will only need approximately half the exposure time [or would it be 1/sqrt(2) the exposure time, since S:N varies as sqrt(signal)?] of the smaller-sensor camera to achieve the same low-light performance.

What you miss here is that sensors with larger photosites (a/k/a pixels (sic) or sensels) don't use the same amount of analog amplification to be rated at the same ISO sensitivity. If the resolution of both sensors is the same, then the photosites of the larger sensor are twice as large as the photosites on the smaller sensor. Reducing the amplification allows higher well values for larger photosites before highlights are clipped, thus giving greater dynamic range. Increasing the size of photosites also reduces the variability from one photosite to the next due to the random distribution of photons within a light field. The larger the area of each photosite, the more this Poisson distribution noise is averaged.

"S:N varies as sqrt(signal)" is only true for the noise attributable to the random distribution of photons within a light field, sometimes called shot noise or Poisson distribution noise. Electronic noise and dark energy added by a camera do not vary by the square root of the signal, but rather remain fairly constant for any particular set of environmental conditions, primarily temperature of the sensor and circuitry, regardless of the signal level resulting from photons falling on the sensor. At lower ISO and brighter exposure the primary source of noise is the camera's electronics. At higher ISO and dimmer exposure the primary source of noise is shot noise.

A larger sensor isn't more sensitive at the same ISO, it's amplified less than the smaller sensor for the same ISO setting. If the total number of photons collected are the same, lower analog amplification means lower voltages are fed into the ADC and the numbers that come out the other side of the ADC are lower for the larger sensor. Due to the lower analog amplification, the larger sensor needs twice as much light energy falling on it to end up with the same brightness¹ level at the ADC.

Thus, for the same exposure (shutter speed), each pixel in the larger-sensor camera is picking up the same proportion of the image as in the small-sensor camera, but is getting twice as much light for that same sub-image. With the same resolution, each pixel is responsible for the same percentage of the image, regardless of the sensor size. I.e., if you want to compare across different format sizes, what matters isn't light per unit sensor area, it's total light per total image.

In a sense, you are correct, but that's not how it works out in practice because (digital) camera makers and the film chemists before them chose to create a system of f-numbers and film sensitivity in a way that accommodates using the same numbers for aperture, as a ratio between focal length and entrance pupil diameter, that holds the same amount of light per unit area regardless of format size. ISO in a digital camera isn't an absolute, it's a calibrated number designed to allow the same overall image brightness for the same combination of f-number, exposure time, and ISO, regardless of the format size. This follows directly the way that the EV system was created for films of various areal sizes and lesnes of various focal lengths.

To make the numbers, some of which are based on √2 and others which are based on the log2(√2) easier to deal with, let's assume for a moment that the larger sensor is twice the linear size and four times the areal size of the smaller one.

If your goal is to use the larger sensor to produce an image with the same SNR of the smaller sensor, you would halve the exposure time. But to make the image equally bright¹ at the ADC, you would also need to double the ISO. Thus, you are giving away the advantage of the larger sensor's ability to perform better than the smaller one in low light in exchange for a shorter exposure time. There are cases where this would be appropriate, such as when camera or subject motion come into play.

Luminous flux per unit area would only be the relevant metric if people printed out or viewed their photos at a size in proportion to the size of the sensor used to take them."

Have you ever used an enlarger in a wet darkroom? We "cheat" in the darkroom by using more light to print an 11x14 from the same negative than we use to print a 5x7. Likewise, assuming they are both calibrated to the same cd/m² (nits), a 32" monitor will allow more total energy to pass through the panel and be emitted from the face of the panel to display the same photo on it's full screen than a 24" monitor would. Everything in the chain between an actual scene and a viewed photograph is based upon light per unit area!

In determining shutter speed, what matters is the total luminous flux on the sensor..."

... multiplied by the analog amplification of the signal generated.

If a 4X larger sensor amplifies half as much as a smaller sensor, then the larger sensor needs to catch 4X as many photons to wind up with the same signal at the ADC, where analog voltages are converted to digitized numbers.

1/4 x 4/1 = 1

Thus, to get the same exposure, one must use the same Ev regardless of format size.

¹ Brightness, as the term is used in photography to indicate relative values between the storage (digital file or negative) or display (print or monitor) medium's maximum and minimum possible values and not as is defined when talking about direct human perception of a primary light source, for which a photograph is but a less than perfect representation.

² Technically, it should be noted as Ev to indicate that it is a logarithmic scale. See the EXIF section of CIPA 2016 and applicable ASA, ANSI, and ISO Standards. But EV has been commonly used for quite a while.

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  • I think, with your focus on the changing historical definitions of EV, you're missing the essential intent of my question. Perhaps I should have instead posted on this on Physics SE, but I've added an addendum to the end of my question in the hope that it will provide some clarification.
    – theorist
    Dec 5 '20 at 5:34
  • I did look at your answer at photo.stackexchange.com/questions/88083/…, where you write "The brightness, which is defined as the amount of light energy per unit area,...." Since you are concered with precision (e.g., distinguishing f-stop vs. t-stop, and aperture vs. entrance pupil), I thought you'd want to know that is not the defintion of brightness. As explained at en.wikipedia.org/wiki/Brightness "brightness is the perception elicited by the luminance of a visual target. It is not necessarily proportional to luminance."
    – theorist
    Dec 5 '20 at 7:43
  • @theorist If we were talking in the context of a physics classroom you would be correct about the use of brightness. But sometimes various disciplines use the same words differently. For instance, "focal plane" in an optics lab does not mean the same thing it means when a camera manufacturer marks a camera's "focal plane" as the depth in the camera where the sensor or film is located. Different disciplines, different nomenclature. In photography, "brightness" means exposure level. With prints, it has absolutely nothing to do with how much light is being emitted by a photo...
    – Michael C
    Dec 5 '20 at 12:50
  • ... (because there is none), nor even with how much light is being reflected by a printed photo. This is because that is dependent upon the level of illumination striking the photo at any particular time. Brightness in photography is about the relative amount of reflectance between the paper's maximum and minimum capabilities. This usage has also been carried over to digital photography. "Brightness" is an expression about what level each pixel has for each color channel between the minimum and maximum possible values (i.e. between 0 and 255 for an 8-bit per channel image format).
    – Michael C
    Dec 5 '20 at 12:56
  • Whatever you think of Ken Wheeler - he probably put it very succinctly in saying that a camera sensor is not a solar cell :) Dec 5 '20 at 19:17
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The part you are missing is the Inverse Square Law (ISL). The inverse square law states that if the distance of a *point light source is 2x, then the density/luminous flux of the light is 1/2^2 (1/4). The opposite of that is at 1/2 the distance it is 4x. So two stops difference, not one. I.e. not 1/2 the SS (which is incorrect anyway).

The inverse square law works exactly the same in regards to the size something is recorded... i.e. if a light bulb is recorded at 1/2 the distance it will also be 2x the size and 4x the luminous flux. But that doesn't change the exposure settings required. Because when it is recorded at 2x the size the light is also spread out 4x farther at the image plane (in area).

It does not matter what causes this change in size. I.e. the exposure of something does not change if you zoom in on it with a constant aperture zoom lens; nor if you reduce the distance to it. The size of the sensor also has no affect on it, only whether it still fits on the sensor.

However, the increase in luminance flux you noted is exactly why larger sensors are better in low light. Because for the same exposure (SS/f#), of the same scene/composition, a larger sensor actually receives/records more light. Many mistakenly equate this as more light from a larger photosite/pixel, because they can be larger on the larger sensor for the same resolution. But that is mostly irrelevant, it is the light/area that matters.

*technically, we are not generally dealing with point light sources. But the ISL still holds true (close enough) other than at extremely short distances.

Edit to add: The entrance pupil diameter increases at the SqRt of area for a stop of light, which is 1.4x diameter. E.g. 1, 1.4, 2, 2.8...

The DoF of the smaller sensor will be greater for an equivalent recorded FOV due to requiring either a shorter FL or longer subject distance; both of which affect DoF more than aperture does. The difference will be effectively the difference in crop factor... i.e. 2 stops greater for the 1/2 sized 2x crop factor sensor.

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  • I am familiar with the inverse square law; it's universal in the sciences, and thus finds its way into both the general and physical chemistry courses I teach. And I don't believe it applies here, at least not in the way you think it does. My hypothetical involved two different camera-lens systems with the same FOV. Thus, to take the same picture, they're the same distance from the subject. The difference in light entering the camera between the two is thus not b/c of the inverse square law...
    – theorist
    Dec 5 '20 at 5:01
  • ....Rather, it's due to what you wrote in your 4th para.: I'm getting twice the light, so I should be getting about twice the low-light performance.
    – theorist
    Dec 5 '20 at 5:01
  • @theorist, I did not say the ISL was affecting the light entering the camera. Light can pass through an optical system from either direction, so look at it from the opposite side as the light recorded being the source of what you are viewing. Because one sensor is 2x the size (LxW, 4x area), and the light is recorded 2x larger at the same exposure, the ISL states that it will contain 4x as much light (2 stops more). Dec 5 '20 at 13:26
  • What causes it to record 2x larger at an equal exposure doesn't matter. It could be caused by moving closer w/ the same lens (the ISL increasing the light entering); or it could be caused by using a longer lens with a larger entrance pupil (reduced etendue) as in your example. Dec 5 '20 at 13:26
  • An entrance pupil that is 2x in diameter equals 4x area and 4x the light able to pass, not 2x... it's the ISL again. The FF camera has a 2 stop advantage over the 1/2 sized m4/3 sensor in your example. You keep insisting on "double;" it's not, its (2x length) x (2x width) = 4x area... Dec 5 '20 at 13:43
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We can calculate a rectangle with 2x more surface area by multiplying both height and width by the square root of 2 = 1.4 (rounded). Such a rectangle has a diagonal measure that is 1.4 x longer.

This 1.4 delta is important as it is also the crop factor.

Therefor to mount a 35mm equivalent, its focal length must be 35mm x 1.4 = 50mm. If both lenses are set to the same f-number, the diameter of the iris for the 50mm must be 1.4x larger to maintain the same exposure. Therefor the two different focal lengths differ by a delta of 1.4 and the iris dimeters also differ by the same 1.4 delta.

The f-number is a ratio that intertwines focal length with iris diameter. Any lens set to the same f-number delivers the same exposure. Difference exist because of the transparencies of the glass may differ as may the number of glass elements. For the most part these differences are moot for pictorial photography. We resort to “T-stops” when transmission difference is important, mainly is cine photography.

DOF is complex: The span of depth of field will differ because we must apply more magnification to view the image made by the smaller sensor. The size of circle of confusion used to calculate DOF for the smaller sensor must be reduced to allow for increased magnification needed to make the displayed image.

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