The Physics of Digital Cameras

[Jonathan Wienke]

I appreciate your comments on the subject Jonathon, but I must say that for the time being I am not convinced. When I use my f/2.8 lens I don't see a noticeable improvement in brightness over my f/5.6 even though according to you there should be a difference of 2 STOPS.

The human eye quickly adapts to fluctuating light conditions, so you're not going to readily perceive a 2-stop change between lens changes. When you pull your eye from the viewfinder, it is adjusting to ambient conditions while you're changing lenses. What you need to do is a side-by-side comparison. Fortunately, there is an easy way to do this, even if you only have one camera and lens.

Simply attach an f/2.8 lens to your camera, set the aperture to 2.8, and adjust ISO and shutter speed manually to achieve proper exposure Shoot a few frames to make sure you have it right. Then set aperture to f/5.6, leaving all other settings the same. Shoot a few more frames, and compare the frames shot at 2.8 and 5.6. Be sure and let us know what you find.

[Jonathan Wienke]

If you use a magnifying glass to examine a small object, you are looking at a virtual image, but if you use it as a burning glass, you are focusing an image of the sun at the focal point of the lens. For a given focal length, the size of the imaged disc will be the same, but the intensity will vary with the square of the diameter of the lens.

This last point (intensity varying with the square of the diameter of the lens) is what the OP has completely confused, and why it is possible to concentrate light (at least light coming from quasi-point light sources) using only optics.

[dmerger]

Jonathan, wouldn't it be easier to see the difference between f2.8 and f5.6 if you just attached an f/2.8 lens to your camera, set the aperture to f5.6 and used the DOF preview button to toggle between f2.8 and f5.6?  Assuming, of course, your camera has a DOF preview button.

[Jonathan Wienke]

Assuming, of course, your camera has a DOF preview button.

I don't think his D60 has one of those...

[ErikKaffehr]

Hi,

I'd suggest that the construction of the viewfinder matters a lot here. Both Fresnel lenses and view finder screens are optimized for a certain aperture: For that reason the view in the viewfinder may not utilize all the light passing trough the lens when fast lenses are used. For the same reason, viewfinders may be a poor tool for visualizing depth of field at large apertures.

Best regards
Erik

The human eye quickly adapts to fluctuating light conditions, so you're not going to readily perceive a 2-stop change between lens changes. When you pull your eye from the viewfinder, it is adjusting to ambient conditions while you're changing lenses. What you need to do is a side-by-side comparison. Fortunately, there is an easy way to do this, even if you only have one camera and lens.

Simply attach an f/2.8 lens to your camera, set the aperture to 2.8, and adjust ISO and shutter speed manually to achieve proper exposure Shoot a few frames to make sure you have it right. Then set aperture to f/5.6, leaving all other settings the same. Shoot a few more frames, and compare the frames shot at 2.8 and 5.6. Be sure and let us know what you find.

[WarrenMars]

The human eye quickly adapts to fluctuating light conditions, so you're not going to readily perceive a 2-stop change between lens changes. When you pull your eye from the viewfinder, it is adjusting to ambient conditions while you're changing lenses. What you need to do is a side-by-side comparison. Fortunately, there is an easy way to do this, even if you only have one camera and lens.

What I DID Jonathon, was to look at a scene with my left eye closed first through the viewfinder and then direct. I shuttled back and forth between these two views not giving my eye a chance to adjust for brightness. There was a slight attenuation, certainly less than 1 stop. I then changed lenses and performed the same test. I noted a slight attenuation, certainly less than 1 stop. Thus it was that I was able to compare the brightness through the viewfinder of 2 lenses with a maximum aperture difference of 2 stops. I encourage all readers of this thread to try it for themselves.

Simply attach an f/2.8 lens to your camera, set the aperture to 2.8, and adjust ISO and shutter speed manually to achieve proper exposure Shoot a few frames to make sure you have it right. Then set aperture to f/5.6, leaving all other settings the same. Shoot a few more frames, and compare the frames shot at 2.8 and 5.6. Be sure and let us know what you find.

Really Jonathon, just because I made an error with the sun's focus, doesn't mean that I know zero about cameras. Surely you realise that I am talking about the brightness apparent THROUGH THE VIEWFINDER. Of course I know a faster lens gives you a faster exposure. Why else would I spend money on buying an f/2.8 lens?

The question here is whether optics can give light amplification to the eye! Whether it be through a magnifying glass, a telescope or an SLR viewfinder.

[col]

You may be suprised to find that current image quality is already within 2 stops of its theoretical maximum and is unlikely to improve by more than 1 stop. You may also be surpised to discover that current technology has already pushed photography 6 stops beyond what can be achieved theoretically and that the difference has been covered up with a combination of human tolererance, noise reduction and sharpening.

There may be some other results that may surprise you also. Go ahead and read my exposé on this fascinating and complex subject.
http://warrenmars.com/photography/technica...ion/photons.htm

I have read Warren's essay, and would like to make a few suggestions.

The first part of the essay is concerned with estimating how many (or more significantly, how few) photons are detected by each pixel. Basically, Warren starts out with known illumination levels in a variety of environments, and uses this to estimate in a very, very rough manner how many of these available photons might actually be detected by the camera CCD. This is an extremely bad way to estimate the number of detected photons. Problems include the unknown spectral distribution of the quoted illumination levels, all sorts of arguments about the efficiency of the camera optics, the detector quantum efficiency, the active area of the detector as a proportion of the toal area, the effectiveness of the detector microlenses, and who knows what else. The result is that the estimates obtained for the number of detected photons are very rough indeed, almost uselessly so, and I think Warren would acknowledge this.

Fortunately there is direct data available for the number of detected photons per pixel, elegantly sidestepping all the the guesswork and unknowns in Warren's approach. The total number of photons that can be detected per pixel is known as the "full well capacity", and this is shown for a large number of cameras and sensors at www.clarkvision.com Sure, it varies somewhat from camera to camera, but broadly speaking the full well capacity per pixel of a 12MP APS-C camera is about 25,000 electrons. Each electron corresponds to a detected photon, so we are talking about a MAXIMUM of 25,000 photons being detected per pixel for your typical APS-C SLR camera. This full well capacity corresponds to the maximum brightness level that can be recorded at the base ISO, which is usually pretty close to ISO 100. At higher ISOs, the number of detected photons is proportionally less. The full well capacity is roughly proportional to the pixel area, so full-frame cameras score better with typically 70,000 electrons, and point-n-shoot camera score much worse, in fact best not to talk about them.

Let's talk APS-C, where a maximum of around 25,000 photons are detected at base ISO, usually ISO 100. Warren can now more accurately re-calculate the horrific effects of his "poisson aliasing", which I personally don't believe is a problem, so I will instead talk very briefly about what I personally believe are the implications of detecting "only" 25,000 photons.

For simplicity I will consider only shot noise (photon counting noise), as this type of noise dominates in regions of medium to high intensity. The uncertainty in the measured photon count is SQRT(25000) or 158 photons. Therefore the signal-to-noise ratio of pixels in the brightest part of the image is 25000/158 or about one part in 158. That might not sound wonderful, but it is generally accepted (for example, by DXOmark) that an SNR of better than 30dB, which is one part in 31, is "good" image quality, so I am confident in stating that one part in 158 is very good indeed, and you won't see any noise at all. That is in accordance with general observation. Take an image with an APS-C camera at 100 ISO, and you won't see noise at the brightest parts of the image.

However, that is absolute best case. The average intensity level in most images is around 18% of the peak, corresponding to 0.18x25000 = 4500 photons, leading to an uncertainty of one part in 67. That is still pretty good, and again in accordance with the observation that the noise in the "average" parts of the image is still pretty low at 100 ISO.

At 400 ISO, those 4500 photons will be only 1125 photons, and the signal/noise wil be 1 part in 33.5, just a whisker above the 30dB accepted as "good" IQ. Yes, that is pretty much consistent with experience when using an APS-C camera.

At 1600 ISO the signal/noise has dropped to 1 part in 17, which is certainly visible, though not catastrophic. Keep in mind that all of these examples are in the worst case where the image is viewed at 100%. When the image is downsized, as it usually is, the effective noise is less.

In summary, the numbers obtained from conventional theory "add up" pretty well and match everyday experience, so I am yet to be convinced about the supposedly horrific effects of "poisson aliasing" though my mind is always open. The number of bits used for digitizing of course has nothing to do with it, just provided the number of bits is chosen conservatively so that quantization error is negligible compared to other noise, which is the case.

Warren's predictions of only modest improvements in the future seems likely. Shot noise is fundamentally related to the number of detected photons, and there really are only a limited number of ways to detect more photons.

On the detector front, expect gradual improvement in QE, pixel fill factor, microlenses, electronic noise and signal processing.

On the camera front, fundamentally the amount of light (= total number of photons) collected and imaged onto the sensor is set by the absolute aperture (not fnumber) of the lens, and a larger absolute aperture essentially means a physically larger lens. Therefore, if you want to collect more light, the lens will need to be larger and heavier, and this is true independently of sensor size. In practice, increasing the absolute aperture also requires scaling up the sensor size, to avoid impractically small Fnumbers. Experience strongly suggests that the cost of making larger sensors will continue to fall, but keep in mind that a larger aperture (= bigger, heavier) lens is required to collect more light onto the bigger sensor, so there are no free lunches here. How many kg of lens are you prepared to pay for and carry? The absolute aperture also sets the depth of field, so gathering more light onto the sensor fundamentally leads to a decrease in DOF, ultimately placing a practical limit on collecting more light from the lens. Warren made this point also.    

Cheers, Colin

[Jonathan Wienke]

What I DID Jonathon, was to look at a scene with my left eye closed first through the viewfinder and then direct. I shuttled back and forth between these two views not giving my eye a chance to adjust for brightness. There was a slight attenuation, certainly less than 1 stop. I then changed lenses and performed the same test. I noted a slight attenuation, certainly less than 1 stop. Thus it was that I was able to compare the brightness through the viewfinder of 2 lenses with a maximum aperture difference of 2 stops. I encourage all readers of this thread to try it for themselves.

That still isn't a valid test, because your eye is adjusting while you are transitioning from looking through the viewfinder to looking directly at the subject. And it's adjusting even more while you are changing lenses. The human eye is completely unsuited to measuring that sort of thing. Taking exposures with a sensor that does not constantly and automatically adjust to changing light levels is the only way to gather any meaningful data and draw any meaningful conclusions. If your camera has a DOF preview button, you can see the viewfinder brightness change when you use it to switch between f/2.8 and f/5.6 simply by mounting an f/2.8 lens, setting aperture to f/5.6, and pressing the button. But accurately judging the degree of brightness by eye alone is still preposterous.

You're also ignoring the light losses inherent in the focusing screen of the camera's viewfinder and the pentamirror assembly (which has much higher light loss than the pentaprism found in higher-end cameras such as the Canon 1-series). There is a considerable difference between the brightness of a 1Ds viewfinder and a 10D viewfinder, even with the same lens mounted on both.

Really Jonathon, just because I made an error with the sun's focus, doesn't mean that I know zero about cameras. Surely you realise that I am talking about the brightness apparent THROUGH THE VIEWFINDER. Of course I know a faster lens gives you a faster exposure. Why else would I spend money on buying an f/2.8 lens?

That error is only one of several blatant and fundamental errors in your essay, as I've already pointed out. You're struggling to grasp some of the kindergarten-level fundamentals of optical physics, and yet you expect us to recognize you as an authority in more advanced areas of optic physics where your grasp of the principles involved is demonstrably more flawed and erroneous.

The question here is whether optics can give light amplification to the eye! Whether it be through a magnifying glass, a telescope or an SLR viewfinder.

It's not a question, it's an easily demonstrable fact, as proven by a simple experiment of burning things with a magnifying lens outdoors on a sunny day. Photons striking the relatively large surface area of the lens are refracted and concentrated into a much smaller area at the plane of focus. The concentration of photons is increased so far above the initial concentration that their energy is sufficient to start fires, which isn't possible without the lens. If you compare the brightness of the focused sun image to the ambient light level, it is obvious that the focused sun image is orders of magnitude brighter.

A faster lens with a larger aperture (or diameter) concentrates more photons per second to each unit of area at the focal plane than a slower lens with a smaller aperture (or physical diameter). That is why opening up the aperture (or increasing lens diameter relative to focal length) allows you to increase shutter speed. More photons/second = greater brightness and more exposure, regardless of whether the lens is directing those photons to an ant crawling in your yard, a piece of film stock, a digital sensor, a viewfinder screen, or your retina. The larger the diameter of the magnifying glass, the brighter the sun image will be, as long as focal length is not changed.

Consider this: the human eye is approximately 25mm in diameter, and normal pupil diameter (the effective aperture of the lens) varies from approximately 2-5mm. That works out to a maximum f-number of approximately 4. Any lens faster than f/4 is capable of creating a greater concentration of photons at the focal plane than the lens found in the human eye.

[Jonathan Wienke]

The total number of photons that can be detected per pixel is known as the "full well capacity", and this is shown for a large number of cameras and sensors at www.clarkvision.com Sure, it varies somewhat from camera to camera, but broadly speaking the full well capacity per pixel of a 12MP APS-C camera is about 25,000 electrons. Each electron corresponds to a detected photon, so we are talking about a MAXIMUM of 25,000 photons being detected per pixel for your typical APS-C SLR camera. This full well capacity corresponds to the maximum brightness level that can be recorded at the base ISO, which is usually pretty close to ISO 100. At higher ISOs, the number of detected photons is proportionally less. The full well capacity is roughly proportional to the pixel area, so full-frame cameras score better with typically 70,000 electrons, and point-n-shoot camera score much worse, in fact best not to talk about them.

Well capacity refers to the number of electrons knocked across the junction of a photodetector by incoming photons, not the actual number of photons striking the detector. Not every photon adds an electron to the detector well, and in some circumstances a single photon can add more than one electron to the well. Google "multiple exciton generation" for more details on this. There is still room for improvement in the conversion rate from photons to electrons, especially considering that shorter-wavelength photons (blue) have more energy that longer-wavelength photons (red) and therefore are capable of transferring more electrons to the well.

This somewhat fuzzy relation between photon count and electron count is one source of sensor noise.

[col]

Well capacity refers to the number of electrons knocked across the junction of a photodetector by incoming photons, not the actual number of photons striking the detector. Not every photon adds an electron to the detector well, and in some circumstances a single photon can add more than one electron to the well. Google "multiple exciton generation" for more details on this. There is still room for improvement in the conversion rate from photons to electrons, especially considering that shorter-wavelength photons (blue) have more energy that longer-wavelength photons (red) and therefore are capable of transferring more electrons to the well.

This somewhat fuzzy relation between photon count and electron count is one source of sensor noise.

I maintain that what I said is just fine for the intended purpose of estimating the number of detected photons, which in turn sets the shot noise, which is what Warren's original article was essentially all about. I am not an expert re CCD and CMOS sensors, but my impression is that Roger Clarke (ClarkeVision.com) has researched the topic extensively and has a fair idea what he is talking about. Quoting from his website :-

The trapped electrons correspond to absorbed photons, and in the sensor industry, photons and electrons are interchanged in describing sensor performance.
Thus, when a digital camera reads 10,000 electrons, it corresponds to absorbing 10,000 photons. So the graphs shown in this article that are in units of electrons, like Sensor Full Capacity, also indicate how many photons the sensor pixel captured.

Unless you can find evidence to the contrary, I believe that "multiple exciton generation" is a second or third order effect, and that the noise from that effect is small compared to shot noise. I make an effort to cross check everything I write by independent means, and I'm sure it is not just coincidence that my back-of-envelope analysis gives predictions that closely match real world observations, which was my cross-check.  

I fully agree with you though that in a small number of cases, a single photon will lead to more than one electron.  

Well capacity refers to the number of electrons knocked across the junction of a photodetector by incoming photons, not the actual number of photons striking the detector. Not every photon adds an electron to the detector well ...

Unless I misunderstand you here, you seem to be implying that what matters is the number of photons striking the detector. Not so. What matters is the number of detected photons, and I maintain that the number of captured electrons is a pretty good estimate of this, and I do not know of any other estimate that would be better.

There is still room for improvement in the conversion rate from photons to electrons, especially considering that shorter-wavelength photons (blue) have more energy that longer-wavelength photons (red) and therefore are capable of transferring more electrons to the well.

Agreed, and I said the same thing when writing that a gradual increases in quantum efficiency(QE) could be expected, though the QE of the best detectors is already extraordinarily high in my view (60%), so we may confidently predict that there will be no "quantum leaps" in this area of performance, as at best it is only possible to approach a QE of 100%. However, if your implication is that shorter wavelength photons have the potential to transfer more than one electron to the well per photon, then of course that would not improve shot noise.
 

[Jonathan Wienke]

Unless you can find evidence to the contrary, I believe that "multiple exciton generation" is a second or third order effect, and that the noise from that effect is small compared to shot noise. I make an effort to cross check everything I write by independent means, and I'm sure it is not just coincidence that my back-of-envelope analysis gives predictions that closely match real world observations, which was my cross-check.  

I fully agree with you though that in a small number of cases, a single photon will lead to more than one electron.

I don't have a source to quantify how often multiple excitons are generated; I would imagine it depends heavily in the design of the photodetector and the materials it's made of. But if you google the term, there's plenty of evidence to support the existence of the phenomenon.

What matters is the number of detected photons, and I maintain that the number of captured electrons is a pretty good estimate of this, and I do not know of any other estimate that would be better.

I agree with this, to a point. My point was that electrons in the well != total photons entering the detector, due to both less-than-perfect quantum efficiency and multiple exciton generation. So while there is a good correlation between photons and electrons, it isn't precisely 1:1.

Agreed, and I said the same thing when writing that a gradual increases in quantum efficiency(QE) could be expected, though the QE of the best detectors is already extraordinarily high in my view (60%), so we may confidently predict that there will be no "quantum leaps" in this area of performance, as at best it is only possible to approach a QE of 100%.

Quantum efficiency of the detector itself is only part of the puzzle. Fill factor (the sensor area actually occupied by photodetectors) and possible alternatives to the Bayer matrix (which intrinsically limits overall QE to ~33%) leave room for a few stops of additional improvement. Then there's alternative sensor designs that increase maximum well capacity for a given detector size, further improvements in read noise performance (better amplifiers and A/D converters), improved microlens design, and probably things that haven't been figured out yet...

[col]

I don't have a source to quantify how often multiple excitons are generated; I would imagine it depends heavily in the design of the photodetector and the materials it's made of. But if you google the term, there's plenty of evidence to support the existence of the phenomenon.

I agree with this, to a point. My point was that electrons in the well != total photons entering the detector, due to both less-than-perfect quantum efficiency and multiple exciton generation. So while there is a good correlation between photons and electrons, it isn't precisely 1:1.

Quantum efficiency of the detector itself is only part of the puzzle. Fill factor (the sensor area actually occupied by photodetectors) and possible alternatives to the Bayer matrix (which intrinsically limits overall QE to ~33%) leave room for a few stops of additional improvement. Then there's alternative sensor designs that increase maximum well capacity for a given detector size, further improvements in read noise performance (better amplifiers and A/D converters), improved microlens design, and probably things that haven't been figured out yet...

Hi Jonathon,

We appear to be in pretty good agreement.

Your point about the number of collected electrons being potentially slightly greater than number of detected photons is interesting but, as it turns out, ultimately irrelevant to the discussion at hand.

I read Roger Clarke's experimental procedure carefully, and strictly speaking, he does not actually measure the number of collected electrons, and nor is there any point in doing so, as fundamentally it is the number of detected photons that dictates the shot noise. He measures the image noise at a variety of ISOs, and also measures the readout noise and dark noise. He then inputs this experimental data into a model incorporating shot noise, readout noise and dark noise, and finds that the data fits the model extremely well. Thus confident that his model is valid, and using the well known properties of shot noise, it is then a simple matter to infer how many photons must have been detected to obtain the measured signal-to-noise ratios. The result is that the “full well photon count” for a typical APS-C sensor is around 25,000 photons, which (fortunately) is exactly the same as I said previously. This is the maximum effective number of photons that can be faithfully detected by a single pixel, which is exactly what we wanted to know in the first place. Whether or not this is exactly the same as the number of collected electrons is irrelevant to our discussion about image noise.  

Basically, what all this means is that all of my previous back-of envelope calculations and conclusions are correct and valid, except that strictly speaking I should have exclusively used the term "full well photon count", and avoided the term "full well electron capacity" altogether.

Cheers,  Colin

[Graeme Nattress]

What I DID Jonathon, was to look at a scene with my left eye closed first through the viewfinder and then direct. I shuttled back and forth between these two views not giving my eye a chance to adjust for brightness. There was a slight attenuation, certainly less than 1 stop. I then changed lenses and performed the same test. I noted a slight attenuation, certainly less than 1 stop. Thus it was that I was able to compare the brightness through the viewfinder of 2 lenses with a maximum aperture difference of 2 stops. I encourage all readers of this thread to try it for themselves.



Really Jonathon, just because I made an error with the sun's focus, doesn't mean that I know zero about cameras. Surely you realise that I am talking about the brightness apparent THROUGH THE VIEWFINDER. Of course I know a faster lens gives you a faster exposure. Why else would I spend money on buying an f/2.8 lens?

The question here is whether optics can give light amplification to the eye! Whether it be through a magnifying glass, a telescope or an SLR viewfinder.

If you take a fast prime, hold down the "DOF Preview" button and scroll through from F1.2 upwards, viewing through the viewfinder, you'd expect to see the image get a stop darker for each stop of aperture you've closed down, right? But you don't. On my camera (Canon 1d Mk III) I see a very small amount of darkening from F1.4 through about f2.8, and it only really seems to get a stop darker per stop of aperture from about F4 onwards. This is because the viewfinder screen has an effective aperture of around F4 ish, and hence sets the effective widest aperture for the viewfinder system, and that does indeed explain what we see. It also effects the appearance of DOF at wide apertures where the viewfinder effectively "lies" to you showing much more to be in focus than really is.

http://www.dphotoexpert.com/2007/09/21/liv...slr-viewfinder/ shows quite clearly the DOF Lie effect.

Graeme

[WarrenMars]

You may be wondering why I have not responded to the later issues raised here... Well I have been away performing real world quantifiable experiments to verify my theorising and answering many of the points raised above. I believe that my results will be off interest to many of you. However, before I upload my findings I have one question to ask:

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?

[Bart_van_der_Wolf]

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)

Have a look at formula's 1, 2, 3, 4 and 11 on this page. The magnification factor at the focal plane, including specifics of focus distance and pupil magnification to account for optical design, will give you one element of the equation you are looking for. The size of the entrance pupil is another important factor.

Here is another consideration.

You take it from there, I rely on my exposure meter/histogram when I'm out shooting. There are too many variables for a practical use of the theory you're after, including transmission characteristics due to coatings and lens element thickness and groupings which could easily account for a few percent of loss.

Cheers,
Bart

[bjanes]

Have a look at formula's 1, 2, 3, 4 and 11 on this page. The magnification factor at the focal plane, including specifics of focus distance and pupil magnification to account for optical design, will give you one element of the equation you are looking for. The size of the entrance pupil is another important factor.

You take it from there, I rely on my exposure meter/histogram when I'm out shooting. There are too many variables for a practical use of the theory you're after, including transmission characteristics due to coatings and lens element thickness and groupings which could easily account for a few percent of loss.

Cheers,
Bart

Bart,

Thanks for the excellent link. I am not familiar with Paul van Walree, but from his web site I see that he is an underwater acoustician, apparently with a PhD, and an avid amateur photographer.  He knows what he is talking about and brings scientific rigor to his presentation.

Another source that might be helpful is Equation 7 on Doug Kerr's web site.
Bill

[Jonathan Wienke]

Another source that might be helpful is Equation 7 on Doug Kerr's web site.

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.

[BJL]

At what f number is the brightness of the object equal to the brightness of the image?

There is no specific f number that does it in all cases.

For example, with an SLR's viewfinder, brightness depends on factors like how efficient the VF screen is at passing light on to the viewer. Companies like www.brightscreen.com make replacement VF screens that give a brighter image than the standard issue ones that come in the camera. Conversely, the pentamirror VFs' used in cheaper SLRs usually give a dimmer image than good pentaprism VFs.

Other factors are the magnification of the VF: using lenses of the same focal length and aperture with a VF of lower magnification typically gives a smaller, brighter image, while adding a clip-on accessory VF magnifier gives a larger, dimmer image.

As a further complication, the f number of the human eye adjusts with the overall brightness of the scene in front of it, which changes the perceived brightness of parts of the scene.


No need to believe me; these are experiments that you can try.

[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.

That's right Jonathan. And for those who are too lazy or disinterested to do calculations this table shows the ratio of the scene luminance to image illuminance for various f/stops:

[attachment=19449:LuminanceRatios.gif]

And as BJL pointed out, the brightness in the viewfinder will be appreciably less bright than what you would obtain in the sensor plane.

Also, the eye has a log response to luminance and the perceived brightness can be predicted by the Stevens Power Law.

Note: Post edited 1/17/2010 to indicate that Ef is image illuminance, not luminance. Scene luminance is expressed in Cd/m^2 (lumens per steradian per meter^2) whereas illuminance is expressed in lumens/m^2. Luminance is invariant in geometric optics. Readers might also want to look up étendue . Photographic light meters measure lux (lumens/meter^2) and luminance is measured by a luminance photometer. See here for details.

[WarrenMars]

Thanks for the link to Doug Kerr's page. Note the following:

If we then follow a long trail of photometric algebra (which I will spare the reader here!)...

I think before I accept his formula I should like to see his derivation.
My own derivation has the pi canceled out, thus implying a theoretical "ideal" aperture of f/0.5.

I would like to see another source for this figure though. One would think that such an important quantity would be found all over the net with a variety of derivations. Strange that Mr Kerr's site is the only one found so far.