The Physics of Digital Cameras

[bjanes]

Actually, equation 8 is the one that puts it in terms of the f-number as opposed to diameter. Rearranging it to solve Warren's question reduces to:

sqrt(pi/4) = f/0.886226925

The important thing to keep in mind is that this formula is for calculating the ratio of subject surface luminance to image plane luminance; i.e. a subject surface emitting 10 lumens/m^2 creates an image receiving 10 lumens/m^2 from the lens. This is not the same as the flux density in front of (or to the side of) the lens. The focused image of the subject will be brighter than the unfocused light from the subject not passing through the lens well before f/0.886226925.

Jonathan,

Each time I look at your post, it seems to have an additional qualification. Unfortunately, LuLa does not keep track of or show the editing history. Definitions are important, and we need to specify exactly what is being calculated. I'm not certain what you are trying to calculate.  As Mr. Kerr states in a footnote:

"4 Note here that since luminance and illuminance are measures of different physical
properties, the statements we sometimes hear about “what fraction of the scene
luminance ends up on the focal plane” are misguided and meaningless."

[col]

At what f number is the brightness of the object equal to the brightness of the image? Or to put it more precisely:
At what f-ratio is the Luminous Emittance of an object equal to the Luminosity of its corresponding image at the sensor? (assuming no lens losses)

I had assumed that the magic ratio was f/1.0 but my own derivation from 1st principles yielded a different figure.

Come on folks: Anyone think they know the the answer?

The best answer I have seen so far is :-

Note here that since luminance and illuminance are measures of different physical
properties, the statements we sometimes hear about “what fraction of the scene
luminance ends up on the focal plane” are misguided and meaningless.

Can Warren (or anyone) precisely explain what they intend the question to mean, with examples of how you might experimentally verify the answer? To put this another way, can Warren explain why he is asking this question in the first place? What physical/experimental situation do people have in mind when posing the question? What does the question really mean?

Cheers,  Colin

[Jonathan Wienke]

Jonathan,

Each time I look at your post, it seems to have an additional qualification.

I edited the post to add a few additional thoughts. The formula calculates (for an ideal lens focused at infinity) what aperture is required for outgoing photon flux density measured at a point on the surface of the subject to equal incoming photon flux density of the corresponding point on the image plane. In other words, if the lens is imaging a light source emitting 1 W/m^2 of light in the visible spectrum, if the lens has an aperture of f/0.886226925, the image plane will receive 1 W/m^2 of visible light energy in the area of the image plane occupied by the image of the light source.

The distinction I was trying to make is that in this situation, the photon flux density at the front element of the lens is going to be significantly less than 1 W/m^2. How much less will depend on the distance between the subject and the lens and various other factors. So while an aperture of f/0.886226925 is needed to make the flux density of photons leaving the subject equal to the flux density of photons intersecting the image plane, one can get away with a much smaller aperture if the objective is to make the flux density of photons intersecting the image plane greater than the flux density of photons intersecting the front element of the lens.

The latter condition is all that is necessary to amplify image brightness by optics alone. Therefore, if we use a magnifying glass with an aperture of f/2 to burn paper, it will work just fine. The visible-spectrum energy density (W/m^2) at the image plane is not as great as it is at the surface of the sun, but it is significantly greater than what it would be if the lens was not present. As a result, the lens can be used to ignite the paper.

[col]

Note here that since luminance and illuminance are measures of different physical
properties, the statements we sometimes hear about “what fraction of the scene
luminance ends up on the focal plane” are misguided and meaningless.

This statement is correct, but requires explanation. Firstly though, I will from here on refer only to the equivalent radiometric quantity power density, in units of W/m^2.  This is easier to think about, and also the correct and relevant quantity when talking about burning holes in paper using a simple magnifying glass to form a focused image of the sun on a sheet of paper, which is the example that I will later use.

In general, it is impossible to calculate the power density at the sensor (focal plane) as a function only of the Fnumber and surface power density at the source. The reason is that power density pays no regard to the angular distribution of light leaving the source, and is concerned only with the TOTAL amount of power per unit area. However, the amount of light collected by a distant lens forming an image of the source most definitely DOES depend on the angular distribution of light from the source, so merely knowing the total emitted power per unit are is NOT enough – you also need to know, or make some assumption about, the angular distribution of light leaving the source, and in general you do not and cannot know that angular distribution. Therefore the question as to what Fnumber corresponds to the power density at the focal plane being equal to the surface power density of the source does not (in general) have an answer, as the answer depends on unknown factors other than the Fnumber, namely on the angular distribution of the light leaving the object.

However, with all that said, it still makes sense to ask the question under very specific conditions, such as when the source is known to emit light equally in all directions, AKA isotropically. An excellent example of such an object is the sun, which is a spherical, isotropic radiator. Therefore, I will derive the correct and meaningful equation that relates the surface power density of the sun, to the power density on a focused image of the sun, produced by a simple magnifying glass. For simplicity, I’ll quote the result first, so those not inclined to do so don’t need to wade through the derivation.

PDs = Power density at the surface of the sun (=5900W/cm^2)
PDi = Power density at the image of the sun (W/cm^2)
F = f-number of the lens (magnifying glass), = L/D
L = focal length of the lens
D = absolute aperture diameter of the lens

PDi = PDs / 4(F^2)

Therefore, an F0.5 lens (magnifying glass) would result in the power density on at the focused image on the paper being the same as at the surface of the sun, which I find really cute. A typical magnifying glass has a focal length of 200mm, and a diameter of 70mm, so in this case the achievable power density is 0.25(70/200)^2 = 0.03, which is “only” 3% of the power density at the sun’s surface. Still not bad, and quite enough to burn paper. I note that Warren also derived the value of F0.5, though I suspect he does not realize that it is not generally applicable.

By far the more significant point I wish to make is that, in general, there is no formula that describes image power density at the sensor (focal plane) as a function only of the Fnumber and surface power density at the source, so people are wasting their time trying to derive and use a general formula. Remember, the formula that I have given, and that Warren appears to have independently derived, applies only to an isotropically emitting source, which is not generally the case. Maybe this is why you don't see too much about it in the textbooks?

I will present the derivation in a future posting.

Cheers,  Colin

[Jonathan Wienke]

However, with all that said, it still makes sense to ask the question under very specific conditions, such as when the source is known to emit light equally in all directions, AKA isotropically.

The condition of isotropic emission is mentioned in Doug's web page.

I will present the derivation in a future posting.

I look forward to seeing it.

[WarrenMars]

Indeed I made the assumption that the object light source emitted isotropically. Yes, there are sources that emit mostly in one direction but the isotropic emitter is a worse case scenario, close enough for real-world photography and a good place to start.
I am gratified Colin that you derived the same result that I did. For the time being at least we are in a majority.

The figure f/0.5 is interesting as a limit since it is also the limit imposed by the Abbe Sine Condition. Together these limits mean that as far as air based photography goes we will never get a greater energy density at the image plane that there is at the object. I don't know about you but I find this symmetry profoundly soothing.

For those that are wondering what this has to do with real world photography the answer is that it establishes the theoretical limit to tolerable noise ISO levels.
Consider a spherical light source in the distance focused onto the sensor such that its image area is exactly that of one pixel. The photon flux density at that pixel can never be greater than the flux density of the object. Since image quality is dependent on per pixel photon flux per exposure, this set the upper limit.

The magnifying glass example is not a counter-example: as Colin pointed out, you don't need the full energy density of the surface of the sun to burn paper!
Yes, as you reduce the focal length you condense the radiant energy of a distant object, but there is a limit to this: the Abbe Sine Condition. When you reach the state where the focal length is half the aperture, then the only way to reduce the focal length further is to reduce the aperture and then, of course, you reduce the afferent flux at the lens so the energy density of the image remains the same. You cannot achieve light amplification through optics alone.

[Bart_van_der_Wolf]

By far the more significant point I wish to make is that, in general, there is no formula that describes image power density at the sensor (focal plane) as a function only of the Fnumber and surface power density at the source, so people are wasting their time trying to derive and use a general formula.

Hi Colin,

Wouldn't that be solved by measuring the luminous flux incident at the aperture (entrance pupil)?

Cheers,
Bart

[col]

Hi to all. As promised, the derivation for the power density at the image, as a function of the surface power density at the source, and the Fnumber, for the case of an isotropically emitting source, such as the sun.

 
PDs = Power density at the surface of the sun (=5900W/cm^2)
PDi = Power density at the image of the sun (W/cm^2)
PDlens = Power density, from the sun, at the lens (W/m^2)
Plens = total power incident upon the lens (W)
F = f-number of the lens (magnifying glass), = L/D
L = focal length of the lens = distance from lens to image
Dlens = absolute aperture diameter of the lens
Rlens = radius of the lens
r = radius of the sun
d = diameter of the sun
R = distance from the centre of the sun, to the lens

For a spherical isotropic radiator (such as the sun), the power density (W/m^2) is inversely proportional to the square of the distance from the centre of the sphere. Therefore, in our example of the sun, the power density at the lens is :-

PDlens = PDs x (r/R)^2

The total power (W) incident upon the lens is equal to the power density at the lens, multiplied by the lens area :-

Plens = PDs x (r/R)^2   x   pi x Rlens^2


The magnification of this simple lens with object at infinity is given by L/R. Therefore, the image radius is rL/R, and therefore the area of the image is :

Ai = pi x (rL/R)^2


In this example, as all of the power entering the lens appears in the image, the power density at the image plane is equal to the power entering the lens, divided by the area of the image.
 
PDi = PDs x (r/R)^2   x   pi x Rlens^2  / ( pi x (rL/R)^2)

PDi = PDs x Rlens^2 / L^2

PDi = PDs / 4(F^2)

Therefore, an F0.5 lens is required to make the power density at the image plane equal to the surface power density at the object, for this non-general example of imaging the sun.

In general, there is no exact formula that describes power density at the sensor (focal plane) as a function only of the Fnumber and surface power density at the source. All of the above algebra applies only to an isotropically emitting source, which is true for the sun, but will not be true in general.

Colin

[col]

Actually, equation 8 (Doug Kerr) is the one that puts it in terms of the f-number as opposed to diameter. Rearranging it to solve Warren's question reduces to:

sqrt(pi/4) = f/0.886226925

The important thing to keep in mind is that this formula is for calculating the ratio of subject surface luminance to image plane luminance; i.e. a subject surface emitting 10 lumens/m^2 creates an image receiving 10 lumens/m^2 from the lens.

Hi Jonathon,

Actually, you appear to have the photometric quantities and units slightly mixed up here.
Doug Kerr's equation 8 shows how the Illuminance (Lumens/m^2) on the image plane is a function of the Fnumber, and the Scene Luminance (Lumens/steradian/m^2).

Interesting though Doug's formula is, it does not answer Warrens question, which was how the Illuminance on the image plane is a function of Fnumber and the Luminous Emittance (Lumens/m^2) at the surface of the object.

That could explain why I got a result of F/0.5, while Dougs formula gives F/0.886

Colin

[Jonathan Wienke]

You cannot achieve light amplification through optics alone.

You may not be able to make a focused image with a greater flux density (W/m^2) than what is emitted by the surface of the subject, but you certainly can create a focused image that has a much higher flux density than what is present immediately in front of the front element of the lens, or than what would be present at the focal plane in the absence of the lens. In other words, a magnifying lens may not be able to create a focused image of the sun with a flux density greater than the surface of the sun, but the flux density of the focused image can be orders of magnitude greater than the flux density of the sunlight reaching the surface of the earth that is not being focused by the lens. And the greatest flux density will occur at the plane where the image of the sun is most precisely focused.

[Jonathan Wienke]

Hi Jonathon,

Actually, you appear to have the photometric quantities and units slightly mixed up here.
Doug Kerr's equation 8 shows how the Illuminance (Lumens/m^2) on the image plane is a function of the Fnumber, and the Scene Luminance (Lumens/steradian/m^2).

W/m^2 is the total energy emitted by the surface, and W/steradian/m^2 denotes the portion of that energy that reaches the aperture of the lens. I don't think confusing one with the other would account for the discrepancy between f/0.5 and f/0.886, especially when distances approach infinity. The energy passing through the lens aperture is much less than the total emitted/reflected energy emanating from the subject surface.

[col]

The figure f/0.5 is interesting as a limit since it is also the limit imposed by the Abbe Sine Condition. Together these limits mean that as far as air based photography goes we will never get a greater energy density at the image plane that there is at the object. I don't know about you but I find this symmetry profoundly soothing.

For those that are wondering what this has to do with real world photography the answer is that it establishes the theoretical limit to tolerable noise ISO levels.
Consider a spherical light source in the distance focused onto the sensor such that its image area is exactly that of one pixel. The photon flux density at that pixel can never be greater than the flux density of the object. Since image quality is dependent on per pixel photon flux per exposure, this set the upper limit.

Warren:
For those that are wondering what this has to do with real world photography the answer is that it establishes the theoretical limit to tolerable noise ISO levels.

No. This statement if wrong.
Warren has apparently assumed that the photon counting (shot) image noise is set and limited by the photon flux density (photons/area/second) falling on the sensor. This is not true, with the result that all of the recent discussions are irrelevant, albeit sometimes interesting in their own right. Here are the (very simple) facts of the matter.

Image shot noise is set by the total number of detected photons which, all else equal, is set by the total number of photons striking the sensor, NOT by the intensity of light striking the sensor per se.

For the same field of view and exposure time, the number of photons collected by the lens and delivered to the sensor is set by the absolute aperture diameter of the lens. Period. For those that have heard me say that many times before, I apologize.

One method of increasing absolute aperture is to built a "faster" lens, with a smaller F-number. Smaller F-numbers result in a more intense image, more photons/area/time. We all agree that there is a limit to the smallest theoretical and practical F-number and, therefore, a limit to attainable image intensity.

The other method of increasing absolute aperture (and thus improving shot noise by delivering more photons to the sensor) is to simultaneously scale up the absolute aperture and focal length, thus maintaining F-number, and then use a larger sensor to maintain the same field of view. In this case, the image intensity is not increased, but the all-important number of photons striking the sensor is increased due to the larger sensor area.

Therefore, the limit to how many photons can be collected is more practical than theoretical. One practical limit is the cost, weight and bulk of physically large lenses. More light means a larger absolute aperture means a larger lens. Real simple.

The other limit is tolerable depth of field. More light means a larger absolute aperture means a poorer DOF. Again, the limit here is more practical than in the nature of a theoretical limit.

That's it in a nutshell, Warren.

Cheers, Colin  
 

[WarrenMars]

I see Colin that your derivation followed the same basic idea as mine. Here is mine:



On reflection I believe that this derivation is more generally applicable than just to single spherical isotropic light sources such as the sun.
The critical point is that the luminosity ratio is relative to the surface of the ball, NOT the imaginary light source in the centre of it. Thus the emission need not be isotropic. For this reason a ball that emits ALL its luminous flux from the hemisphere facing toward the lens will still find the above analysis applies. Pack a whole bunch of these hemispheres as closely together as you like and you have surface that is not unlike the surface of things in the real world, such as skin, leaves, dirt etc.

[col]

Hi Warren,

I need to ask a favour of you. Read my previous post (post#92) carefully, and tell me if you disagree with anything.

Unless and until we agree on all I have said in post#92, all further discussion is pointless.

Cheers,  Colin

[joofa]

Warren has apparently assumed that the photon counting (shot) image noise is set and limited by the photon flux density (photons/area/second) falling on the sensor. This is not true, with the result that all of the recent discussions are irrelevant, albeit sometimes interesting in their own right. Here are the (very simple) facts of the matter.

Image shot noise is set by the total number of detected photons which, all else equal, is set by the total number of photons striking the sensor, NOT by the intensity of light striking the sensor per se.

The above is incorrect. Poisson noise is not determined by the total number of detected photons. It is determined by the average number of detected photons for the underlying distribution. Further more, for any given pixel in a particular image, even the sqrt of average number of photons does not represent the actual noise on that pixel. The actual noise is the deviation from the average value. The sqrt thingy just says that on average the noise is that far from the mean value, but, for a given pixel and a given image it could fluctuate, and not given by sqrt (average photon count).

In fact, the underlying Poisson variable has a photon intensity parameter, in this case measured as photons/seconds/area, which together with the integration time determine the average on which noise calculations are based.

I had some comments on how to calculate these numbers on dpreview and you are welcome to have a look at an example. Click here.

Joofa

[col]

I wrote:

Image shot noise is set by the total number of detected photons which, all else equal, is set by the total number of photons striking the sensor, NOT by the intensity of light striking the sensor per se.

The above is incorrect. Poisson noise is not determined by the total number of detected photons. It is determined by the average number of detected photons for the underlying distribution. Further more, for any given pixel in a particular image, even the sqrt of average number of photons does not represent the actual noise on that pixel. The actual noise is the deviation from the average value. The sqrt thingy just says that on average the noise is that far from the mean value, but, for a given pixel and a given image it could fluctuate, and not given by sqrt (average photon count).

In fact, the underlying Poisson variable has a photon intensity parameter, in this case measured as photons/seconds/area, which together with the integration time determine the average on which noise calculations are based.

I had some comments on how to calculate these numbers on dpreview and you are welcome to have a look at an example.

Hi Joofa,

What you say about the Poisson noise being determined by the average number of detected photons for the underlying distribution is technically correct, though I think that what I said was just fine within the context of the point I was trying to get across.

I think you may have read too much into my term "detected photons". My intention here was only to make it clear that it doesn't matter a rat's arse how many photons strike the detector - what actually matters  re image noise is how many you are able to detect, on average of course. If I was sloppy enough to say that what mattered was the number of photons striking the detector, rather than the number of detected photons, I would have many people saying "that's wrong because you haven't considered QE, fill factor, microlenses etc"

If you read the entire thread, you would realize that Warren appears to be hung up on the light intensity (photons/second/area) striking the sensor, and he appears to have concluded that there is a theoretical limit to how low the shot noise could be made, because F-number limitations put an upper limit on the intensity of light striking the sensor.

The intention of my quoted sentence above was only to point out that it is not the image intensity that matters per se, but the total number of detected photons (you now now the context in which I say this) and you can collect and detect more photons with larger pixels while keeping the F-number and image intensity the same, though you still need a larger lens of course.  

Basically, I don't believe that your posting in any way alters the substance or thrust of what I was saying. If you still think it does, then get back to me, and we'll thrash it out

Cheers,  Colin

This addition edited in later.
Hi again.
At first I thought you were just nitpicking, for example, making the distinction between the detected photon count, and the "true" photon count of the underlying distribution. However, I have now read your post on DPreview, and on the basis of the errors you have made there, I now tend to the view that you were not nitpicking at all, but just plain wrong!
It starts look like we will indeed need to thrash this out.

[joofa]

At first I thought you were just nitpicking, for example, making the distinction between the detected photon count, and the "true" photon count of the underlying distribution. However, I have now read your post on DPreview, and on the basis of the errors you have made there, I now tend to the view that you were not nitpicking at all, but just plain wrong!
It starts look like we will indeed need to thrash this out.

What is wrong? There is a typo there in a number that I calculated, which I have to correct, but that typo does not change the thrust of what I am saying.

[col]

What is wrong? There is a typo there in a number that I calculated, which I have to correct, but that typo does not change the thrust of what I am saying.

Hi Joofa,

Just for now, I would like to concentrate on whether you still feel I have made any significant error in anything I said in my post #92, as that is from where this discussion started. I certainly maintain that everything I said is just fine.

Exactly which of my statements or claims do you disagree with?

Firstly though, some terminology. There is a difference between "estimated uncertainty" and "error". The error is the difference between the measured value, and the "true" value. In terms of photon counting, the "true" value is the average value you would get if the count was repeated an infinite number of times. The estimated uncertainty is a prediction of the expected error. For example, if you measure 100 photon counts, then the estimated uncertainty is SQRT(100), which is 10. In general, you do not know the true value, so the estimated uncertainty is usually all you have.

I think (but am not sure) that you have a fundamental problem with my statement that the shot noise in an image is a function of the number of detected photons, rather than the intensity (photons/second/area) as such, so I will attempt to spell that out in detail with an example.

Let the "true" intensity (photons/second/area) be 100 photons/mm^2/second, and let this intensity be perfectly uniform over the entire detector. By "true", I mean the intensity you would measure if you had the luxury of measuring a very large number of photons.

Let the area of each pixel be 1mm^2, and let the exposure time be 1 second, so the "true" photon count is 100 photons/pixel.

By any meaningful measure, the uncertainty in the photon count for each pixel is 10, so the image SNR is 100/10, or 10. There is nothing to be gained (IMHO) by pointing out that the actual error in the count for some pixels will not be 10.

To improve the image SNR, we need to detect more photons, on average, of course. To take a specific example, if we wish to double the SNR, we need to detect 4 times as many photons. All my post#92 is saying, is that there are many ways by which we could increase the number of detected photons, and the improvement in image SNR would be the same in each case. For example :-

(1) Increase the detector QE by a factor of 4
(2) Increase the image intensity by a factor of 4, for example by fitting a lense with smaller Fnumber
(3) Double the sensor dimensions (4 times the area), keep Fnumber the same, and maintain FOV by doubling focal length

All of these will have an identical effect of improving image SNR by a factor of two. Agreed or not?

Cheers,  Colin

[WarrenMars]

Colin, you seem to have strangely misunderstood me.

If you read the entire thread, you would realize that Warren appears to be hung up on the light intensity (photons/second/area) striking the sensor, and he appears to have concluded that there is a theoretical limit to how low the shot noise could be made, because F-number limitations put an upper limit on the intensity of light striking the sensor.

The intention of my quoted sentence above was only to point out that it is not the image intensity that matters per se, but the total number of detected photons (you now now the context in which I say this) and you can collect and detect more photons with larger pixels while keeping the F-number and image intensity the same, though you still need a larger lens of course.

When you say "total number of detected photons" I will assume you mean per pixel. Surely you must see that the number of detected photons per pixel per exposure is proportional to the light intensity at that pixel.

Since it is "photons/second/area" you can increase your photon count by increasing the pixel area or the exposure. However exposure is constrained by other factors so it is out. Since the light intensity at the pixel is dependent on the f-stop and the source intensity you can also increase the photon count by increasing the aperture or by lighting the source, say with flash for example.

Any of these techniques will increase the photon count per pixel and hence decrease the Poisson noise.

Increasing the pixel pitch can be achieved by 1) reducing the number of pixels for a given sensor size, a good move, but you can't go too low else you have insufficient resolution for your final image 2) increasing the sensor size for a given number of pixels, fine except that it makes larger cameras with shallower DoF. Lighting the scene is fine but is not always possible or artistically desirable. Increasing aperture is 1) achieved at the expense of DoF and 2) eventually limited by the Abbe Sine Condition. Such are the limiting factors in minimising Poisson noise.

[joofa]

Just for now, I would like to concentrate on whether you still feel I have made any significant error in anything I said in my post #92, as that is from where this discussion started. I certainly maintain that everything I said is just fine.

Hi Col,

You seem to be going back and forth between the errors you thought you found in my dpreview post and your post # 92.

Exactly which of my statements or claims do you disagree with?

That the Poisson noise is the sqrt of total number of detected photons.  This is a misunderstanding.

I think (but am not sure) that you have a fundamental problem with my statement that the shot noise in an image is a function of the number of detected photons, rather than the intensity (photons/second/area) as such, so I will attempt to spell that out in detail with an example.

Let the "true" intensity (photons/second/area) be 100 photons/mm^2/second, and let this intensity be perfectly uniform over the entire detector.
Let the area of each pixel be 1mm^2, and let the exposure time be 1 second, so the "true" photon count is 100 photons/pixel.

It is interesting that you are now adopting the photon intensity in photons/area/seccond, where as you chided Warren for that and I repeat it here:

Warren has apparently assumed that the photon counting (shot) image noise is set and limited by the photon flux density (photons/area/second) falling on the sensor. This is not true, ...

By any meaningful measure, the uncertainty in the photon count for each pixel is 10, so the image SNR is 100/10, or 10. There is nothing to be gained (IMHO) by pointing out that the actual error in the count for some pixels will not be 10.

There is. First of all there is the correct interpretation of theory. Secondly, suppose you are denoising images. Then, the point is to develop appreciation of image correlation structure in SNR determination.

To improve the image SNR, we need to detect more photons, on average, of course. To take a specific example, if we wish to double the SNR, we need to detect 4 times as many photons. All my post#92 is saying, is that there are many ways by which we could increase the number of detected photons, and the improvement in image SNR would be the same in each case. For example :-

(1) Increase the detector QE by a factor of 4
(2) Increase the image intensity by a factor of 4, for example by fitting a lense with smaller Fnumber
(3) Double the sensor dimensions (4 times the area), keep Fnumber the same, and maintain FOV by doubling focal length

All of these will have an identical effect of improving image SNR by a factor of two. Agreed or not?

As long as you maintain a difference between actual and average values some of the above stuff can be sorted out. However, we are back to the original proposition here. Taking our example where the average count for a given pixel is 100. Suppose you are able to predict/measure with certainty that the average value is now 400. Again that does not mean that the SNR for the same pixel is now up by a factor of 2, in a given image. What Poisson statistics is saying is that if you acquire a large number of images (i.e., not a single image), under the same conditions, then the average SNR for that pixel is up by a factor of 2.