How Many Photons in a Photograph

Robin Harrison

Robin Harrison

aka Harrison Cronbi
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There's no RFF sub-forum for the Science of Photography, but this seems to be the most scientifically-oriented one. It does relate to optics in one sense, though.

Anyway, the question is: How Many Photons in a Photograph?

And my wild guess is: 2.4 Trillion

I tried to tackle this from one direction, starting at a typical ambient light level in lux, but could not find out a way of converting that to light intensity on the focussed plain for a given focal length and aperture (this is the optics question, I guess).

Instead I went the other way, and found the definition of ISO sensor speed ratings based on the light levels they respond to, and then worked out the equivalent number of photons.

Has anyone done, or seen, such analysis before? Or at least come across a number? 2.4 trillion seems two low, if you ask me. But then again it does translate as approximately 100,000 per pixel of a 24MP full-frame image...maybe that's enough to stimulate those litted CCD elements. Actually, perhaps that's another angle I could take, trying to find on a low level the photons needed to create a certain mid-tone signal at a certain ISO for a single pixel.

My stream-of-conciousness ramblings:
http://www.cronbi.com/2011/09/20/how-many-photons-in-a-photograph/
 
The glib and flippant answer would be ... it depends how bright it is.

My second thought is one would need a known value for the light source, and the reflectivity of the scene, focal length of the objective, lens speed, size and sensitivity of the sensor, oh and a calculator ... at that point I get a headache
 
Sorry - yes trillion as in 10 ^ 12.

Single photon? Yowsers. Will have to read that.

Stuart - sorry, I did specify these in my blog ramblings. I assumed:
- 36x24mm capture area
- EV13
- ISO400
- 1/1000s @ f5.6
- A photo of an 18% grey squirrel in an 18% grey suit against an 18% grey background
- I wasn't sure if focal length mattered, because if I take a photo of a blank infinitely large evenly lit wall with a 28mm or 50mm lens, I get the same picture for the same exposure settings, and therefore I concluded I would capture the approximately same number of photons. Any hole in that logic?
 
It's impossible to be accurate theoretically. Film and sensors react differently with temperature, light colour, humidity and other factors. No medium is anything near 100% sensitive - it's probably close to an order of magnitude photon loss in sensitivity for most consumer sensors and more for film. You can work it out for each shot if you measure the sensor/film active radiation, then assess the true sensitivity, signal loss, signal to noise ratio, and image/area conversion fidelity.

Marty
 
Robin Harrison said:
Sorry - yes trillion as in 10 ^ 12.

Single photon? Yowsers. Will have to read that.

Stuart - sorry, I did specify these in my blog ramblings. I assumed:
- 36x24mm capture area
- EV13
- ISO400
- 1/1000s @ f5.6
- A photo of an 18% grey squirrel in an 18% grey suit against an 18% grey background
- I wasn't sure if focal length mattered, because if I take a photo of a blank infinitely large evenly lit wall with a 28mm or 50mm lens, I get the same picture for the same exposure settings, and therefore I concluded I would capture the approximately same number of photons. Any hole in that logic?
Click to expand...

I don't think focal length would matter then ... iirc assuming sunny f16 then 1 square metre (in full sunlight) is equivalent to 1kw ... anybody how many photons are needed to carry 1kw?
 
Sparrow said:
I don't think focal length would matter then ... iirc assuming sunny f16 then 1 square metre (in full sunlight) is equivalent to 1kw ... anybody how many photons are needed to carry 1kw?
Click to expand...

Ah, that's an interesting stat.

Although, that's per square meter? In which case I guess distance to subject also comes into play. BUT, going back to the 'blank wall' analogy, if I take a photo of that hypothetical wall at 1m, 5m, or 50m, it will produce the exact same photograph, so somehow I feel I should be able to cancel out the area and subject distance terms in this most simple of cases.

Also, thinking about that 1kw, I imagine that energy is being emitted distributed amongst all directions, not just the direction that points at my camera lens aperture?
 
I think it is possible to figure this out, armed with the appropriate data ahead of time. For instance, if you assume silver gelatin B/W emulsion, the crystals of which have about a 5-electron sensitization threshold (the newer Kodak emulsions have a 2-electron sensitization threshold), each dislodged electron representing a photon striking a crystal, and the 5 electron threshold representing the darkest shadow detail that can be recorded, and each subsequent zone higher in tone is a logarithmic progression.

You'd need other missing pieces of data, like the density of a fully exposed and developed emulsion (how many grains of silver per square mm, etc.), and the statistical variation in grain sizes within the emulsion (graphic arts films have crystals uniform in size, for instance, resulting in very little middle-gray tones, while continuous-tone films have a wide variation in crystal sizes, representing various probabilities of intermediate light levels interacting with the free electrons in the crystal halide lattice).

Some of the required data is fairly straight-forward, others (like statistics in crystal sizes) are proprietary to the film's emulsion design and would be difficult to determine ahead of time.

~Joe
 
Double Negative said:
None.

By the time there's a "photograph" there are no more photons. They've done their job and moved onto a better place.
Click to expand...

Well...true. A better way to phrase the question would have been "How many photons are needed to form a photograph".

More of a philosophical question, that - without photons, can a photograph be said to exist? A photograph is a visual thing, and unless viewed (which takes another set of photons to illuminate the various parts of the surface), we have ourselves a tree-falling-in-forest situation!
 
That sounds like the right answer to me. By the time you've got an image like a photograph, the photons are flying through the infinity of space. On the other hand, the light reflected by the print is now in the form of photons, isn't it? Or electromagnetic waves, if you prefer. Okay, my brain has just overheated. I need to lie down for a rest.

Double Negative said:
None.

By the time there's a "photograph" there are no more photons. They've done their job and moved onto a better place.
Click to expand...
 
Robin Harrison said:
Ah, that's an interesting stat.

Although, that's per square meter? In which case I guess distance to subject also comes into play. BUT, going back to the 'blank wall' analogy, if I take a photo of that hypothetical wall at 1m, 5m, or 50m, it will produce the exact same photograph, so somehow I feel I should be able to cancel out the area and subject distance terms in this most simple of cases.

Also, thinking about that 1kw, I imagine that energy is being emitted distributed amongst all directions, not just the direction that points at my camera lens aperture?
Click to expand...

Yes indeed meter, dyslexic sorry they always catch me out

Clearly if the wall is is 18% grey then it is absorbing some of the photons as, I'm unsure if it's absorbing 18% or 82% but either way the fact that some are reflected and some are absorbed as heat would suggest those reflected were of a shorter wavelength ... and therefor light is a wave, not a particle at all, and thereby rendering the question invalid ... 🙂
 
For a typical Kodak CCD, the pixels saturate at about 60,000 photons. So 10,000,000*60,000 would be the upper limit. 6.0E4 * 1.0E7 is 6.0E11, (600Billion) ballpark for a Leica M8. That puts the M9 at over 1 Trillion assuming the same saturation per pixel, but I do not have a spec sheet on the KAF18500.
 
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So an 18% grey card would be 0.18* 6.0E11.

Keep this up and I'll post FORTRAN.

(Actually a Bluff. I do not have the source code of the FORTRAN program that I used to convert Density on film to number of Photons that struck it. I last used it in 1980)
 
Brian Sweeney said:
For a typical Kodak CCD, the pixels saturate at about 60,000 photons. So 10,000,000*60,000 would be the upper limit. 6.0E4 * 1.0E7 is 6.0E11, (600Billion) ballpark for a Leica M8. That puts the M9 at over 1 Trillion assuming the same saturation per pixel, but I do not have a spec sheet on the KAF18500.
Click to expand...

Ok, that makes me feel a lot better, Brian. My 18% grey figure of 100,000 per pixel assumed:
a) all light that reached the sensor area actually reached a pixel, whereas even with the best microlens arrays, that won't be the case
b) a filter-less array

I'm guessing the RGB colour filters lose about two stops of light? And that the effective capture area is at best about 1/4 of the theoretical pixel pitch? In which case, we're talking about 100,000/16 = 6,250 photons per pixel, or 3.26 stops under the typical saturation levels you refer to. Not too far away!
 
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