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Let's say I take a photo using an APS-C camera and using an FF camera, from the same position, with the same ISO and shutter speed.

Let's say I use a 50mm focal length for the FF camera. To match the field of view, I use a 50/1.5 = 33mm focal length for the APS-C camera.

Now, let's say that both lenses have the same aperture of f/2.8. But since the focal length is different (50 vs 33), the actual value of f/2.8 will be higher for the FF lens, which means it's letting in more light [1].

Am I correct in understanding that the FF system performs better not because of the sensor but actually because of the wider opening of lens? If we measured aperture in sq mm, and if I take a photo using the same actual aperture, like 20 square mm, with an FF and an APS-C camera, won't I get the same amount of noise?

In other words, the supposed advantage of the FF system is merely an artefact of the way we measure aperture, as f-numbers. If we chose to measure aperture in sq mm, the advantage of the FF sensor goes away.

I'm assuming we are comparing lenses of FF-equivalent focal length, and we don't have a mismatched system like mounting an APS-C lens on an FF body or vice-versa (unless you use something like the MetaBones Speed Booster).

There are two ways of measuring aperture:

  1. As an f-number, which is the conventional way of measuring aperture.

  2. As a total area, in square mm.

(1) measures the light per unit area and (2) measures the total light incident on the sensor.

(1) is useful to determine aperture in relation to shutter speed and ISO (the so-called exposure triangle).

(2) is useful when you're interested in noise under low-light. Since noise is caused by lack of light, so you want to measure the total light coming in to the sensor, not the amount of light coming in per unit area of the sensor. (I'm assuming you're viewing both photos at the same size on your monitor.)

Since (2) assumes that you don't have a mismatched system, where you mount an FF lens on an APS-C camera or vice-versa (without something like the MetaBones Speed Booster adapter), and also assumes you're not comparing lenses of different FF-equivalent focal lengths, it may not be a very useful measurement of aperture in photography in general. But it helps to determine noise, taking sensor size out of the question.

Let's say you're trying to buy a camera + lens for low-light work, and you've identified two options within your budget:

  1. An NEX body with a Sony f/2.8 lens.

  2. An Olympus µ 4/3 body with an Olympus f/2 or an f/1.4 lens.

The f-number measurement makes it hard for you to see which system performs better under low-light. Whereas, if we measured aperture in square mm, you'd instantly see which system worked better under low-light -- the one with the bigger aperture in square mm. (Again, let's assume both lenses have the same FF-equivalent focal length.)

A comment below says that it's an academic question whether we attribute the advantage of the FF system to the bigger sensor or to the wider entrance pupil of the lens. The existence of the MetaBones Speed Booster disproves that statement, since you can now mount an FF lens on a crop-factor sensor without losing light or, more formally, without having an image circle larger than the sensor. When you can mix and match FF and APS-C lenses and bodies, it's no longer academic whether you attribute the advantage of the FF system to the body or to the lens.

  • Folks, I've closed this question because it has fired off a ton of moderation flags as a consequence of the massive quantity of comments (92 total) and edits as a result of being less than clear. Each answer has then led to subsequent, large, comment-based discussions and this is not a discussion forum. This could, however, be a really good chat event that becomes a blog post. – John Cavan Jan 6 '14 at 18:42
  • That's fine with me, John. After all, the discussion was going nowhere. I do agree that I was not able to communicate this in a form you guys could understand. I did make a couple of offers to rephrase this, but in any case, thanks for trying to help me. – Vaddadi Kartick Jan 7 '14 at 6:42
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(Edited for Compliance:)

The short answer is yes, the way you're looking at the issue of attributing low-light capability to the total light-gathering power of the lens instead of to the area of the sensor does also work to explain the observed differences between different systems. However it is more complicated conceptually and doesn't really provide any advantage over the simpler and more commonly used model of 'bigger sensor = better'.

Everything in photography involves compromises and every system is the result of the combination of its parts. Combining different parts in different ways can achieve the same results, and there can simultaneously be other compromises involved that some people will accept and others will reject. There is never a single correct answer.

The idea that you're on the edges of here has been called "equivalence". More than you'll ever want to read about it is here: josephjamesphotography.com/equivalence/

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    This doesn't attempt to answer the question correctly so much as it tells the asker what he already wanted to hear when he asked it. – Michael C Jan 6 '14 at 8:32
  • I don't think you've even understood the question, Michael. – Vaddadi Kartick Jan 6 '14 at 10:28
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    @KartickVaddadi - Answering the question with a link to "go there and read" is not answering the question on this site. It is very likely to get deleted as a consequence. – John Cavan Jan 6 '14 at 11:33
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    @mpr I think there are still too many slightly and entirely incorrect statements in the question for "yes, you're right" to be the direct answer, even with an "even if". – mattdm Jan 6 '14 at 14:06
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    For example "total [aperture] area, in square mm ... measures the total light incident on the sensor". This is just wrong, and it's precisely because focal length isn't taken into account, which is, in turn, really the basis of the question. – mattdm Jan 6 '14 at 14:08
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Edit - One More Try (original below the rule)

The noise level for a given photograph is specifically a consequence of one or more of:

  • the ISO, with higher ISOs having more noise
  • the size of the sensor wells, bigger wells tend to be less noisy for the same generation of sensors because they get more light for the same pixel
  • the heat level in the camera, as long exposures can have more noise at a given ISO due to heating of the sensor
  • the generation of the sensor as more modern sensors are often better at noise management

You'll note that at no point did I mention lenses. The effect of a lens on the above is with respect to the exposure triangle. The aperture (f/#) dictates the volume of light transmitted to the sensor and is not specifically the diameter of the entrance pupil (this is what you keep measuring). If you fix the size of the entrance pupil, then for different focal lengths you will have different apertures. In fact, for longer focal lengths you will actually let in less light for the same diameter of the entrance pupil resulting in a need to change the shutter speed or ISO to achieve the same exposure.

Again, you'll note that I haven't talked about the size of the sensor. A larger sensor receives more light only the sense that it will capture more of the image circle, but it will get the same amount of light at any point in the image circle as the smaller sensor. The size of the entrance pupil that you list above isn't changing the size of the image circle (usually, skip the concept vignetting for a moment, we're going to assume perfect optics) and so the only effect of the size of the entrance pupil is the relationship to focal length and thus what aperture is in play. If the aperture is the same, then the amount of light at any point in the image circle is the same for either sized sensor.

If the two cameras are of the same generation, have the same ISO setting, same heat levels, aperture, and shutter speed then the potential difference here is sensor well size (point 2 above) and as typically these are larger on FF cameras they may have better noise performance because more light is gathered for the same physical pixel in the image because the surface area for the pixel is bigger.

So... the size of entrance pupil is not related to the performance of the sensor. It is related to the aperture and that controls how much light comes in, but the same amount of light will come in for the same aperture regardless of focal length.


For the purpose of discussion we'll keep this based on simple lens models. Complex optics actually result in physical openings not actually being the size you would think they would be.

At any rate, what I think what you were trying to say is that to get the same field of view on a FF camera, you would need a longer lens. That longer lens, for the same aperture setting, would need to actually be physically wider at the opening to expose the scene the same.

Yes, this is true, but if you think about the physical construction of a simple lens, the longer lens will also need to be wider to transmit the same amount of light and that's the point of using f numbers for aperture: it's a dimensionless way to measure the same amount of light coming through a lens. Wikipedia probably explains this better than I:

A 100 mm focal length f/4 lens has an entrance pupil diameter of 25 mm. A 200 mm focal length f/4 lens has an entrance pupil diameter of 50 mm. The 200 mm lens's entrance pupil has four times the area of the 100 mm lens's entrance pupil, and thus collects four times as much light from each object in the lens's field of view. But compared to the 100 mm lens, the 200 mm lens projects an image of each object twice as high and twice as wide, covering four times the area, and so both lenses produce the same illuminance at the focal plane when imaging a scene of a given luminance.

Basically, the designs of the lenses are used to focus the light at the same plane where the sensor happens to be located. So the fact that the sensor surface may only "see" some portion of that image circle at that plane is independent of the size of the circle. So, if you understand the exposure triangle then you would notice that it does not include focal length for calculating light volume.

All of that means is:

The amount of light entering the camera is the same for the same aperture regardless of the focal length.

If we take it from a different angle and assume a fixed physical diameter of the aperture of, say 25mm. Then what you actually have is a different aperture for focal lengths. So, on a 50mm lens, the aperture is f/2 and on a 75mm lens, the aperture if f/3. So, then the longer lens is transmitting less light.

Knowing that, it's pretty clear that the light gathering performance of a larger sensor is unrelated to the lens focal length. The "noise" levels of the sensor are specifically impacted by the ISO chosen (hence the reason you need to understand the exposure triangle) and the quality of the sensor itself. There are modern APS-C sensors, for example, that have better high ISO performance than older FF cameras. Technology improves as time goes by, but it's still not related to the physical size of the aperture opening.

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    @Kartick: If you think about it, your question is actually tautological. That is, you're right that more light is gathered overall on the full frame sensor in this situation, but it doesn't really matter what part of the system is responsible. Furthermore, the actual larger sensor area is the most direct, obvious, and uncomplicated part to talk about. – mattdm Jan 5 '14 at 16:57
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    That is, if I put a full-frame lens on a APS-C camera, nothing has changed in the lens, so it makes more sense to talk about the sensor. – mattdm Jan 5 '14 at 16:59
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    @Kartick You say "The reason I find it counterintuitive to say that the FF sensor has less noise is that, it's receiving more light overall." I think the whole... kerfluffle... here is because this isn't counterintuitive, it is why we say that. – mattdm Jan 5 '14 at 17:14
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    @KartickVaddadi - Every time you try to clarify, the semantic meaning of your question seems to change, I can't read your mind. In any event, you don't measure the physical diameter of the aperture that way for a reason. The point of f/# is to be dimensionless with respect to the lens length so that we can know for the same f/# that the amount of light will be the same, allowing is to select appropriate ISO and shutter speeds. The ISO selection will absolutely impact your noise amount, but if all things are equal, then the sensor quality will determine noise. – John Cavan Jan 5 '14 at 17:14
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    @KartickVaddadi - Aperture is only measured one way, it's a ratio of the size of the entrance pupil to the focal length. If you measure the entrance pupil, then you get different apertures. The aperture is what matters for the amount of light being transmitted. – John Cavan Jan 5 '14 at 18:05
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Okay, so, I think it comes down to this: a bigger sensor doesn't gather more light by magic, but because... bigger things are bigger. It is true that in order to get equivalent framing and exposure when using a larger sensor, one uses a larger focal length and therefore a larger actual aperture (or entrance pupil). But that doesn't mean that this larger physical aperture gives any advantage per area, so if you use a crop sensor or crop from the middle of the full frame sensor the result is the same.

You only actually realize the benefit of more actual light when you record that light, and that's why we talk about the sensor size being the important thing here: without that, the other is irrelevant (you would use a shorter focal length to get the same framing).

The larger area is what allows you to use a lens set in that way in order to get the agreed-on image. If you use the same lens in the same configuration with a smaller area, you won't get the same picture. You would need to reduce focal length and absolute aperture size equivalently. So, for getting the same picture (same framing and exposure) with more overall light gathered, it is the sensor size which matters.

So, to the second part of your question: isn't this just half a dozen of one and six of the other? Well, let's look at effective absolute aperture diameter without regard to sensor size. A 400mm f/2.8 lens has a an effective aperture area of a whopping 16,000mm². A 100mm lens has an effective aperture area of ¹⁄₁₆th that — 1000mm². Does the 400mm lens have better low light performance? Of course not — it's the same.

In fact, these lenses both literally gather the same amount of light for a given frame. That's because zooming and cropping are equivalent operations. That is, when you zoom in from 100mm to 400mm, you're cutting the size of the frame by 4 in every dimension — which means that the overall area captured is ¹⁄₁₆th the size. So, you're gathering ¹⁄₁₆th of the light — and it's obvious you need a 16× larger aperture to compensate for that. Raw aperture size without knowing the focal length is not helpful for talking about overall light gathering. And, conveniently, we already have a system for abstracting that out, the f/numbers.

I think what you really want to do here is compare lenses and bodies between systems. To a rough approximation, you can do this simply by multiplying the "crop factor" of the sensor by the aperture. See Why don't comparisons of aperture take sensor size into account? for more. However, don't get too hung up on this, because (although we are getting there) our sensors aren't at their theoretical perfect maximum yet, and it's likely that the sensors from one manufacturer are actually somewhat better than the other, making any abstract comparison just a bunch of talk with no real bearing on the world.

And speaking of that: although they aren't at their maximum, all of today's larger-sensor cameras are very good in low light and have amazing high ISO performance. It doesn't hurt to better understand things, but it's also not the most useful to get all hung up on contrasting differing degrees of "very good" vs. "really good". Better to have either camera and be taking pictures.

On your very last paragraph about the MetaBones Speed Booster: actually, no, this doesn't change anything. All camera lenses (Lensbaby aside) are made up of multiple glass elements arranged in groups. The Speed Booster simply adds an additional group which works as a reverse teleconverter, literally decreasing the focal length of the lens. This comes at the expense of a shrunken image circle, but that's actually irrelevant. (It's only important in that there happens to be a ready source of lenses designed for a larger image circle than used by other readily-available digital cameras.) What's important is that since the absolute aperture doesn't change, the relative aperture does — the lens gets faster. But a "converted" lens at that new relative maximum aperture has no advantage over a "native" lens with the same maximum aperture. In fact, if the native lens has the same focal length as the converted focal length of the lens using the Speed Booster, the effective aperture area will be identical. In short, this is entirely a red herring.

  • I don't think we're actually disagreeing here, @mattdm. Yes, the larger physical aperture doesn't give any advantage per area, but it gives an advantage overall. It's just two ways of looking at the same thing. I really see no disagreement here. – Vaddadi Kartick Jan 6 '14 at 10:31
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    @KartickVaddadi Well, yes and no. You say "In other words, the supposed advantage of the FF system is merely an artefact of the way we measure aperture, as f-numbers." But that's not the case. It is a real advantage, and it really is fundamentally because the area captured is larger. The larger absolute aperture happens to follow as a logical consequence of getting the same framing and exposure with that larger area. – mattdm Jan 6 '14 at 12:14
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    Here's another way to think about it: take a photo (from any camera) and cut it in half. Now blow that half up to the size of the original. Put it next to another copy of the original. In which one is noise more apparent? – mattdm Jan 6 '14 at 14:14
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Lots of very overcomplicated answers here...

Basically the same fstop means same light per unit area. More area thus means more light. More light means less noise. That's all there is to it.

Whether the advantage is down to lens or sensor is a philosophical argument as you can't get anyany advantage without both.

As to why manufactures don't specify the absolute size of the aperture instead of fstop, it would confuse beginners into thinking longer focal lenses produce brighter images, which is untrue due the fact the angle view is narrower.

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    If you read the comments, you'll see why they got complicated... – John Cavan Jan 6 '14 at 1:44
  • I updated the question to clarify -- doesn't the existence of the MetaBones Speed Booster mean that it's relevant to discuss whether the advantage of FF is due to the sensor or due to the lens? – Vaddadi Kartick Jan 6 '14 at 5:16
  • No, because the MetaBones Speed Booster swaps higher field intensity for smaller pixel size. – Michael C Jan 6 '14 at 8:17
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    @KartickVaddadi Yes, the Speed Booster absolutely changes the field of view. In fact, that's how it works. – mattdm Jan 6 '14 at 13:44
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Put simply, it is the fact that the larger sensor covers more area and thus takes more of the light that the lens outputs. On a crop body, the entire image is taken from only a portion of the light that goes through the lens because the image circle projected within the camera is bigger than the sensor on a crop sensor.

This means that more pixels come from less light on a cropped sensor camera. On the full frame camera however, the entire image circle is used and thus more light is available for each (larger) pixel. If there is a large resolution difference however, this would be counteracted.

What matters is the pixel pitch of the sensor. This is the actual physical size of each pixel. The lens puts a certain amount of light per area and the larger area that each pixel of the sensor covers, the more of that light it can bring in.

The only difference that the lens makes is that a lens made for a crop body won't make as big of an image circle as a lens designed for full frame. Even at the same aperture, the full frame lens will capture more light, however that light goes to waste on a crop sensor since it isn't projected on to the sensor.

  • I actually know all that, but I can see that I'm having trouble communicating my question in an understandable manner. I revised my question. Please take a look in case that helps. – Vaddadi Kartick Jan 6 '14 at 5:36
  • I agree with your last comment. In fact, I think we're saying the same thing. Thanks for your help. – Vaddadi Kartick Jan 6 '14 at 6:14
  • Sorry for the wrong terminology. Would it help if I go through my question and replace "aperture" with "entrance pupil" when I'm referring to the area? – Vaddadi Kartick Jan 6 '14 at 6:17
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Look at it this way: The effective aperture (more properly called the entrance pupil) is the diameter, not area, of the aperture as viewed through the front of the lens. This means:

  • If you double the diameter of the aperture, you increase the amount of light allowed to pass four-fold. But all of that light is still falling on the same size image circle. This means that each point on the image circle is receiving four times the illuminance.
  • As the focal length of a lens increases, the minimum diameter of the front element must increase to maintain the same f-number. An f/2 aperture is 50mm wide for a 100mm lens, so the front element of a 100mm f/2 lens must be at least 50mm wide. Otherwise you could not measure the aperture as 50mm wide as viewed through the front of the lens. A 200mm f/2 lens must have a front element at least 100mm wide.
  • If we referred to apertures by total area, rather than the f-number, we would need different combinations of Tv/Av/ISO for the same amount of light at every different focal length! By using the f-ratio, it allows the correct exposure values for a subject of a specific luminosity to remain the same, regardless of the focal length.

For more on why exposure is determined by the amount of light per unit area, and not the total amount of light gathered, see Lens f-number and speed on adapted lenses

I opened that link and will read it, but I wanted to quickly say that I understand and agree with the reasons why we measure aperture in f-numbers rather than the total area. However, in this case, I find that it obscures rather than helps me understand what's going on. Namely that the FF sensor is performing better only because it's receiving more light, not because it's inherently more sensitive.

It is only receiving more light because the sensor is larger. The amount of light per cm² from a 50mm f/2 lens is exactly the same as the amount of light per cm² from a 100mm f/2 lens (assuming both are viewing the same scene).

Doesn't noise depend on the total light coming in to the sensor, and not the light per unit area? If the latter were the case, you'd find that an FF sensor with an f/2.8 lens performs no better than an APS-C sensor with an f/2.8 lens, right?

NO. Noise in the image depends on the signal to noise ratio. Since the noise is fairly constant at each pixel, the stronger the signal is at each pixel, the lower percentage the noise floor will be of the total readout of that pixel. That is why larger pixels are inherently less noisy: each pixel can collect more light/photons/signal while generating no more read noise than a smaller pixel would.

Larger sensors allow either: larger pixels for the same resolution/number of pixels or higher resolution/number of pixels for the same pixel size or a combination of both (moderately larger pixels and moderately more of them).

If the pixels are the same size (and identical in terms of other technological questions) on both the FF and APS-C sensors then you are correct at the pixel-peeping 100% large display level that the noise level in the FF and APS-C cameras would be the same. BUT: if you are then displaying the images from the different sized sensors at the same display size (i.e 8x10, or 16x20, even 36x24, or even larger), the higher magnification needed to display the 10MP APS-C sensor compared to the lower magnification needed to display the 22MP FF image would also make the perceivable noise higher.

Remember that if both sensors have the same pixel size, then at 100% viewing size on your monitor the APS-C image would be less than half the total area of the FF image.

On the other hand, if both the APS-C and FF sensors have the same number of pixels, then each pixel on the FF sensor is 2.25 times the surface area of each pixel on the APS-C sensor. That means for the same scene through the same aperture lens the FF camera collects 2.25 times more light/photons/signal per pixel than the APS-C camera does, which means the SNR at each pixel is over twice (one stop) that of the APS-C camera.

I agree fully with your last five comments, Michael. I don't think we actually have a disagreement here. To be clear, I'm viewing the photos on a fixed size computer monitor, as I wrote in my updated question, and not 100%. For this setup, since the noise depends on the total light falling on the sensor (not per unit area), I don't think it's inaccurate to say that it's not the sensor per se, but the larger entrance pupil of the lens that's responsible for the better image quality. Correct?

If you are viewing the output from two differently sized sensors on the same size monitor the the difference in magnification (between the size of each sensor and the size of your screen) will make a difference relative to noise.

"For this setup, since the noise depends on the total light falling on the sensor (not per unit area)...

Just because you keep repeating this does not make it any more correct than it was the first time you said it.

You're also ignoring the elephant in the room since you clarified that this is not a comparison of a FF and APS-C that utilize the exact same technology but rather an APS-C made by Sony and a µ4/3 made by Olympus. The difference in the way each manufacturer chose to design the sensor and how each chose to process the output from that sensor may likely have more to do with the relative performance of each.

No matter what sensor size I have, a lens with an entrance pupil of 20 sq mm is going to give me photos with lower noise under low light than a lens with entrance pupil 10 sq mm. Even if these two lenses are mounted on different sized sensors. As long as the FF-equivalent focal length of these two lenses is the same, and we don't have a mismatched system (like mounting an FF lens on APS-C without the Speed Booster, thereby wasting light, or an APS-C lens on FF body).

You're still ignoring the effect of the different magnification factors on noise. Depending on the difference between the two sensors it can be quite significant when comparing photos taken in low light.

All other things being equal, a full frame sensor will outperform a 1.5X APS-C sensor in terms of SNR by a factor of 2.25 (roughly 1.15 stops). A 50mm lens on the APS-C camera provides the same FoV as a 75mm lens on a FF camera. The 26.8mm wide e.p. on the 75mm f/2.8 translates to an f/1.9 aperture on the 50mm lens. That is roughly 1.15 stops. But here's where the "all other things being equal ends." The smaller sensor needs just as much circuitry per pixel on the sensor die as the FF sensor does, which means almost certainly the smaller sensor is using less of its total surface area to actually collect light.

Let's also assume the same sensel size. I'm clarifying since that issue came up. For that matter, one can have two APS-C sensors with a different megapixel count and hence different sensel sizes. Let's not bring that other factor in.

That "other factor" is the real world vs. the purely theoretical.

  • I opened that link and will read it, but I wanted to quickly say that I understand and agree with the reasons why we measure aperture in f-numbers rather than the total area. However, in this case, I find that it obscures rather than helps me understand what's going on. Namely that the FF sensor is performing better only because it's receiving more light, not because it's inherently more sensitive. – Vaddadi Kartick Jan 5 '14 at 17:42
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    It is only receiving more light because the sensor is larger. The amount of light per cm² from a 50mm f/2 lens is exactly the same as the amount of light per cm² from a 100mm f/2 lens (assuming both are viewing the same scene). – Michael C Jan 5 '14 at 17:47
  • Doesn't noise depend on the total light coming in to the sensor, and not the light per unit area? If the latter were the case, you'd find that an FF sensor with an f/2.8 lens performs no better than an APS-C sensor with an f/2.8 lens, right? – Vaddadi Kartick Jan 6 '14 at 5:15
  • NO. Noise in the image depends on the signal to noise ratio. Since the noise is fairly constant at each pixel, the stronger the signal is at each pixel, the lower percentage the noise floor will be of the total readout of that pixel. That is why larger pixels are inherently less noisy: each pixel can collect more light/photons/*signal* while generating no more read noise than a smaller pixel would. – Michael C Jan 6 '14 at 7:57
  • Larger sensors allow either: larger pixels for the same resolution/number of pixels or higher resolution/number of pixels for the same pixel size or a combination of both (moderately larger pixels and moderately more of them). – Michael C Jan 6 '14 at 8:01

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