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Phone manufacturers have recently started advertising the size of the photosites on smartphone camera sensors. They argue that larger photosites lead to better low-light performance. I think a good analogy would be car manufacturers claiming that larger wheels lead to faster cars. True, given the same f-number, a larger photosite captures more light and given the same axle RPM, larger wheels make a car go faster. However, larger wheels means that less force is applied to the road given the same engine torque, cancelling the effect.

I thought it would be the same for cameras. A larger sensor requires a larger focal length to deliver the same field of view. Given the same entrance pupil size, this increases the f-number, cancelling the effect of capturing more light. I was under the impression that the f-number of phone camera lenses was limited by the largest entrance pupil size that can fit into a smartphone. Surprisingly, phone manufacturers have been able to enlarge the entrance pupil enough to keep the f-number constant while increasing the sensor size. In fact, they were even able to lower the f-number significantly in some cases.

A similar phenomenon can be observed with regard to full-frame vs. crop cameras. The f-number of a crop lens is usually at least as high as that of its full-frame counterpart and there are hardly any fast prime lenses for crop cameras at all. It seems like it is much easier to get large entrance pupils on lenses built for large sensors. Take a 35mm f/1.4 full-frame lens with an entrance pupil diameter of 25mm. Is it not possible to build a similar lens that concentrates the light captured by this entrance pupil to a smaller image circle fit for a crop sensor, yielding a 22mm f/0.88 lens? Why does it seem like large sensors are necessary for good low-light performance?

Note: I know that sensor size also influences electrical characteristics, but I am only interested in optical considerations here. Let us pretend that sensors are ideal photon detectors with infinite dynamic range, leaving only shot noise to determine low-light performance. Let us also assume that all these sensors have the same number of photosites.

The question Why are larger sensors better at low light? does not answer my question as all the answers assume constant f-number, which is exactly the proposition I would like to challenge.

EDIT: I should mention that my question is not about what happens when you mount full-frame lenses on crop bodies. Obviously, this wastes a lot of the light captured by the lens so it's an apples to oranges comparison in my book. Instead, I'm wondering why it seems to be so difficult to concentrate all of this light to the crop sensor image circle in order to get a lens that makes the same total amount of light available to the smaller sensor.

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    \$\begingroup\$ The reason there are hardly any fast primes (or primes of any type for the most part) made specifically for crop body cameras is that primes made for full frame/35mm film cameras work just fine with cameras using smaller sensors. The reverse, however, is not the case. \$\endgroup\$
    – Michael C
    Commented Feb 9, 2016 at 2:29
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    \$\begingroup\$ There have recently been a few crop camera zoom lenses introduced that take advantage of the smaller image circle needed on such a camera to provide a lower f-number. For example the Sigma 18-35mm f/1.8 lens made for crop sensored cameras. The fastest zoom lenses made for FF cameras are generally f/2.8. The front element of a 35mm f/1.8 lens needs to be roughly the same diameter as a 50mm f/2.8 lens. Essentially what such a lens does is take a 28-50mm f/2.8 FF lens design and use a focal reducer to concentrate all of that light in an APS-C sized image circle. \$\endgroup\$
    – Michael C
    Commented Feb 9, 2016 at 2:48
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    \$\begingroup\$ They can do so more cheaply because the smaller image circle allows them to no longer worry about correcting aberrations to the same degree that such a lens casting a larger image circle would require. \$\endgroup\$
    – Michael C
    Commented Feb 9, 2016 at 2:52

5 Answers 5

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One thing to consider is that the size of the image that the lens projects is completely independent of its f-number and focal length, but a factor of the lens design. F-number is a product of the width of the aperture, and the focal length. The focal length is the distance between the point of convergence, and the sensor, completely ignoring factors beyond that further into the lens.

You can produce an image of the exact same quality as an FF sensor, on a phone sensor as long as you design an 'equivalent' lens. But to create an equivalent field of view, as you alluded, the focal length needs to be reduced, as it needs to bend the light more to compress it into that image circle. With the size of phone sensors, these figures are increasingly small, often less than 5mm.

When you reduce both the sensor size and the focal length, the f-number needs to be increased to compensate. While decreasing the focal length will collect more light, you're only using the light in the same field of view. The lens design can take this into account, and simply not collect the wasted light. The actual aperture width (or entrance pupil) will need to stay the same, which increases the f-number.

When the focal length is so small, and the aperture so comparatively large, this represents an increasingly difficult technical challenge, as glass takes up space, and as the aperture increases, so will the size of these elements, but with the focal length so small, they still need to be either absurdly close together where they're occupying the same space, or so absurdly large and heavy that you'd need impractical and massive contraptions on the back of the device.

It's also worth noting the effect of chromatic abberation, caused by the fact that different wavelengths of light passing through a lens will bend differently, slightly more or less, depending on the wavelength. Lens makers have gotten pretty good at correcting for this effect to some degree, but it becomes increasingly difficult when taken to extremes.

Total light gathered is the important factor in image quality, sensor efficiency is basically the same across all modern sensors. Larger sensors are not necessarily better, but they reduce the challenge of lens design significantly.

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    \$\begingroup\$ You understood my question even though it was apparently misleading to many other people and answered it perfectly. Thank you! \$\endgroup\$
    – Jules
    Commented Feb 10, 2016 at 1:56
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    \$\begingroup\$ The second sentence should read: The F-number is the focal length divided by the diameter of the aperture. (Wikipedia) \$\endgroup\$
    – Nayuki
    Commented Feb 12, 2016 at 7:06
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Why does it seem like large sensors are necessary for good low-light performance?

Because for the same amount of light passing through a lens a larger sensor will collect more of it.

Your tire size analogy is seriously flawed. A better analogy would be increasing the diameter of the engine's cylinders. The size of the individual molecules of the fuel/air mixture entering the cylinder stay the same in the larger cylinders, but you can put more fuel/air molecules into the cylinder and maintain the same amount of fuel/air per cubic centimeter of cylinder displacement. Thus the larger cylinder will yield more power for the same fuel/air mixture density. With a camera lens, the entrance pupil is the cylinder diameter, the f-number is the density of the fuel/air mixture. To compute the total amount of fuel/air (and thus energy) in the cylinder, the density and the displacement must both be factored in. The gear ratios and tire sizes are equivalent to the amount of amplification needed to give a higher or lower ISO sensitivity.

Surprisingly, phone manufacturers have been able to enlarge the entrance pupil enough to keep the f-number constant while increasing the sensor size.

The reason f-number is so useful is because it is a measure of field density. It is not a measure of the total amount of light passing through the lens. Rather, it is a measure of the field density of light cast by the lens on a unit area of the surface of a sensor or film at the lens' focal length. The lens projects the same amount of light per mm² of sensor area regardless of the size of the sensor. If a lens projects an image circle 44mm in diameter it is just large enough to cover a 36x24 mm FF sensor. If that same lens is placed on a camera with an APS-C sized sensor it is still projecting the same 44mm wide image circle, but the 24x16 mm sensor, with a diagonal measurement of only 29mm and a surface area only 44% the size of the larger sensor is not collecting as much of the light in the image circle as the larger sensor does.

I'm wondering why it seems to be so difficult to concentrate all of this light to the crop sensor image circle in order to get a lens that makes the same total amount of light available to the smaller sensor.

If you modify the same lens and reduce the size of the image circle cast so that all of the light collected is now projected in an image circle only 29mm in diameter (instead of 44mm), you have changed the focal length of the lens by a factor of 1/1.5X. Thus you have also changed the f-number by the same factor. You've concentrated the same total amount of light in a smaller area, thus increasing the field density. This is true whether applying it to FF vs. APS-C sized light circles or 7.5mm sensor chips vs. 5mm sensor chips. You've also required the lens to be either 50% shorter in focal length (ironically increasing the thickness of the phone that contains such a rectilinear retro-focal lens) or for the lens materials used to be significantly denser while maintaining the same amount of light transmission (increasing the cost substantially).

Why aren't there any prime lenses specifically designed for crop sensor cameras?

The f-number of a crop lens is usually at least as high as that of its full-frame counterpart and there are hardly any fast prime lenses for crop cameras at all.

There have recently been a few crop camera zoom lenses introduced that take advantage of the smaller image circle needed on such a camera to provide a lower f-number. For example the Sigma 18-35mm f/1.8 lens made for crop sensored cameras. The fastest zoom lenses made for FF cameras are generally f/2.8. The front element of a 35mm f/1.8 lens needs to be roughly the same diameter as a 50mm f/2.8 lens. Essentially what a lens such as the Sigma 18-35mm f/1.8 does is take a 28-50mm f/2.8 FF lens design and use a focal reducer to concentrate all of that light in an APS-C sized image circle. They can do so more cheaply because the smaller image circle allows them to no longer worry about correcting aberrations to the same degree that a FF 28-50mm lens with an f/1.8 sized front and casting a larger image circle would require.

There are also more affordable zoom lenses, such as the Canon EF-S 17-55mm f/2.8 or the Tamron 17-50mm f/2.8 Di II for crop sensored cameras than their 24-70mm f/2.8 FF counterparts.

The reason there are hardly any fast primes (or primes of any type for the most part) made specifically for crop body cameras is that primes made for full frame/35mm film cameras work just fine with cameras using smaller sensors. The reverse, however, is not the case. Consumer grade primes designed for FF cameras do very well on cropped cameras because the weaknesses of such lenses are almost always on the edges of the image circle that falls outside the area of the cropped sensors. And just like cameras with cropped sensors, they're a lot more affordable than premium grade FF lenses. There isn't a lot of market demand for premium grade prime lenses for cropped cameras because presumably anyone willing and able to pay for premium lenses is also willing and able, at least eventually, to pay for the benefit of a full frame camera.

What is the benefit of a larger sensor?

Assuming the same number of photosites, a larger sensor has larger photosites. This means for the same field density of light, each photosite on a larger sensor collects more total light. It also means the differences introduced by the randomness of the distribution of photons (shot noise) are averaged more evenly, thus reducing the overall impact. If the read noise per photosite is constant regardless of the size of the photosite, then a larger photosite will yield a better signal-to-noise ratio both in terms of read noise and in terms of shot noise.

And although your question attempts to exclude it, the difference in efficiency (what percentage of the photons striking the surface of the sensor actually make it to the bottom of pixel wells and are actually detected and counted by the sensor) between a larger and smaller sensor with the same number of photosites is significant. As the ratio of surface area to linear circumference increases with larger photosites, so does the efficiency. (Consider a 2x2 square has a linear circumference of 8 linear units and an area of 4 areal units. A 4x4 square doubles the circumference to 16 linear units but quadruples the area to 16 areal units.) Because light demonstrates properties of both wave energy and particle energy most of the photons that are lost, even on so called "gapless sensors", are lost at the edges of the individual photosites.

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    \$\begingroup\$ Excellent answer. Would upvote twice if I could. \$\endgroup\$
    – JohannesD
    Commented Feb 9, 2016 at 10:30
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    \$\begingroup\$ I should have mentioned that my question is not about what happens when you mount full-frame lenses on crop bodies. Obviously, this wastes a lot of the light captured by the lens so it's an apples to oranges comparison in my book. Instead, I'm wondering why it seems to be so difficult to concentrate all of this light to the crop sensor image circle in order to get a lens that makes the same total amount of light available to the smaller sensor. \$\endgroup\$
    – Jules
    Commented Feb 9, 2016 at 13:13
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    \$\begingroup\$ Probably because anyone worried that much about the difference in noise performance between a FF and cropped sensor is also probably going to desire higher quality lenses than are typically available in cropped sensor only designs. And again, although you can make the equivalent of a speed boosted lens that concentrates the light into a smaller image circle, in increases the cost of producing a lens significantly enough that camera phone manufacturers are going to try more cost effective measures instead. \$\endgroup\$
    – Michael C
    Commented Feb 9, 2016 at 13:23
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    \$\begingroup\$ And there is no way around the fact that, even apart from noise performance, larger sensors are capable of creating images with many qualities deemed more desirable than smaller sensors can create. They allow shallower depth of field with lens designs that remain in the practical realm. They allow higher resolution with lenses of the same optical limits in terms of line pairs per inch. And they are more efficient with the light that does fall upon them. \$\endgroup\$
    – Michael C
    Commented Feb 9, 2016 at 13:31
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    \$\begingroup\$ Or compare the EF-S 24mm f/2.8 ($150) to the EF 40mm f/2.8 (also $150). Even though the housings are pretty much the same size, observe how much further forward in the housing the front element for the 28mm is compared to the 40mm. Also consider that, due to projecting a FF image circle, the 40mm either includes more aberration correction than the crop body 24mm lens or the advantages of a focal length nearly the same as the diameter of the image circle projected (44mm) doesn't require such additional correction. \$\endgroup\$
    – Michael C
    Commented Feb 11, 2016 at 19:18
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A larger sensor requires a larger focal length to deliver the same field of view. Given the same entrance pupil size, this increases the f-number, cancelling the effect of capturing more light.

This is at least part of your misunderstanding. f-number is not affected by sensor size. Field of view is, of course, because a smaller sensor sees only a part of the image circle. But the lens gathers the same amount of light and delivers it to the same area regardless of the sensor size.

The focal length is determined by the distance from the optical center of the lens to the image plane when the lens is focused at infinity. That number is the same for a given lens regardless of the size of the sensor. Likewise, the apparent size of the entrance pupil isn't affected by the sensor size. So, f/8 is f/8 no matter how big the sensor is.

Consider what would happen if you were using a 35mm film SLR and shooting at a fixed shutter speed, say 1/100s. You adjust the aperture so that the meter indicates a reasonable exposure, and then you snap a photo and advance the film to the next frame. Next, you put the camera into a changing bag so that you can work with it in complete darkness, pop open the back, and insert a thin, opaque card with a APS-C-sized hole in the middle between the film and the shutter. Close it all up, pull it out of the bag, and snap another photo. Finally, develop the film. Do the exposed parts of the two frames have the same exposure, or are they different?

I know that sensor size also influences electrical characteristics, but I am only interested in optical considerations here. Let us pretend that sensors are ideal photon detectors with infinite dynamic range, leaving only shot noise to determine low-light performance.

The problem here is that the improved electrical characteristics are the main reason that larger sensors give better results. When you boost the sensitivity (ISO), you're essentially increasing the multiplier that's applied to the data for each pixel. With a weak signal (low light) and high gain (big multiplier), the tiny variations due to electrical noise get amplified into a noisier image. Due to their increased area, larger photosites collect more photons and generate a stronger signal in the same way that larger solar panels provide more power. With a stronger signal, less amplification is required and you get more image and less noise.

I guess at some point you get to a situation where the sensor really is counting fairly small numbers of photons. At that point, even with ideal photosites, a large sensor will record a better (smoother, less "noisy") image simply due to the law of large numbers: small photosites will record fewer photons than large ones due to their smaller area, so the relative scale of random differences between sites will be larger. Small random differences in the relatively large numbers of photons collected by larger photosites will have less impact. I honestly don't know if the sensors in mobile phones are working at this level -- I strongly doubt it, but that's just a guess.

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  • \$\begingroup\$ I'm not sure I understand. The f-number determines how much light hits each unit area of the image circle, right? Furthermore, it is defined as focal length divided by entrance pupil diameter. So when you increase the focal length to adjust for the field of view of the larger sensor without also changing the entrance pupil, you get less light per unit area on the image circle. If my math works out, this exactly cancels the effect of the larger photosites on the larger sensor capturing more light since they take up more area. I don't think I've said anything that would contradict your example. \$\endgroup\$
    – Jules
    Commented Feb 8, 2016 at 20:52
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    \$\begingroup\$ Didn't you already specify manufacturers have been able to enlarge the entrance pupil enough to keep the f-number constant while increasing the sensor size? \$\endgroup\$
    – Caleb
    Commented Feb 8, 2016 at 21:40
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    \$\begingroup\$ "Decreasing sensor size while keeping the entrance pupil the same to increase the f-number." Concentrating the total light of the larger image circle of a lens designed for one sensor size into a smaller image circle for a smaller sensor area requires you to change the focal length of the lens. And though it might seem counterintuitive, as focal lengths get shorter and shorter lens designs need to get longer due to the retro-focal design needed to maintain rectilinear (instead of fisheye) projection. And how thick can you make a phone that will sell? \$\endgroup\$
    – Michael C
    Commented Feb 9, 2016 at 3:41
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    \$\begingroup\$ Those designs also get expensive in a hurry. Just look at the cost difference between the Canon 11-24mm f/4 rectilinear FF lens ($3K) compared to the 8-14mm f/4 fisheye ($1.25K). \$\endgroup\$
    – Michael C
    Commented Feb 9, 2016 at 3:44
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    \$\begingroup\$ See the following question for why there's a sweet spot for cost/performance of lenses with particular focal lengths in the middle of the wide angle and telephoto extremes. photo.stackexchange.com/a/54435/15871 \$\endgroup\$
    – Michael C
    Commented Feb 9, 2016 at 3:52
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Let us pretend that sensors are ideal photon detectors with infinite dynamic range [...]. Let us also assume that all these sensors have the same number of photosites.

The question "Why are larger sensors better at low light?" does not answer my question as all the answers assume constant f-number, which is exactly the proposition I would like to challenge.

Your premise is rigging the rules of the game, begging the question. Fixing the number of photosites to be the same is fine; increasing the focal length is fine. But by not allowing f-number to stay the same, you are rigging the comparison against any size increase.

Let's use a bit of absurdity to draw a clear distinction. The iPhone 6S has a 12 MP, 7.21 crop factor, 4.15 mm focal length, ƒ/2.2 max aperture. Therefore, the apparent aperture is 1.9 mm across. So let's "embiggen" the iPhone 6S up to full frame sensor size and mount the 1.9 mm diameter lens to the camera at 30 mm. That gives us a ƒ/15.8 aperture, hardly a fair or reasonable comparison.

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  • \$\begingroup\$ I think it is a fair comparison. The lens on the iPhone camera is a tiny thing, it should come as no surprise that it has a tiny aperture. \$\endgroup\$
    – Jules
    Commented Feb 9, 2016 at 13:17
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    \$\begingroup\$ But the aperture is not tiny relative to the camera's geometry. Keeping the absolute lens diameter constant is akin to giving 1 m^2 of cloth each to a 5 year old and to a above-average height and weight adult and asking them to cover themselves modestly. The results will not be comparable! =) \$\endgroup\$
    – scottbb
    Commented Feb 9, 2016 at 13:45
  • \$\begingroup\$ In understand what you're trying to to do. You're conducting a thought experiment, trying to control variables. I applaud that. But in this scenario, holding apparent aperture constant has several real world side effects: (1) larger depth of field, no ability to control bokeh; (2) pro bodies stop autofocusing after ƒ/8, the rest cannot AF above ƒ/5.6; (3) increased diffraction limitation, negating one of the prime benefits of having larger photosites. \$\endgroup\$
    – scottbb
    Commented Feb 9, 2016 at 13:56
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    \$\begingroup\$ You analogy is funny but I don't think it works out ;). If I'm not mistaken, making the sensor larger but giving it the same total amount of light to work with would not change anything. The photons will simply spread out over the larger sensor area, but in the end, the same number of photons hit each photosite. \$\endgroup\$
    – Jules
    Commented Feb 9, 2016 at 13:57
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    \$\begingroup\$ Sure, there are a lot of other benefits to having a larger sensor, I'm well aware of that. But just for this question, I'm interested in how (or rather why) they affect the amount of light gathered on the image circle in total. \$\endgroup\$
    – Jules
    Commented Feb 9, 2016 at 14:00
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The light that falls on a unit area is governed by the size of the entrance pupil. A larger sensor requires a longer focal length to achieve the same field of view. That larger sensor requires a larger entrance pupil to maintain a given f-number. The larger pupil gathers more light.

You must maintain the f-number as that, together with the shutter speed and ISO sensitivity, is directly related to the amount of light emanating from the scene.

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