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Author Topic: Canon 50D @ 15MP  (Read 116073 times)

ejmartin

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« Reply #120 on: August 31, 2008, 08:56:21 am »

sorry, double post.
« Last Edit: August 31, 2008, 08:57:25 am by ejmartin »
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emil

Ray

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« Reply #121 on: August 31, 2008, 10:57:26 am »

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That depends on what you mean by "the same amount of diffraction".  The range of angles over which the direction of the light spreads after passing through the aperture depends only on the physical size of the aperture; since aperture is (focal length/f-number), then indeed the range of angles will be the same for 50mm @ f8 and for 100mm @ f11. 

However, the relevant quantity for photography is the amount of blur (the Airy disk diameter) at the focal plane; for that, one should multiply the spread in angles by the distance from the aperture to the focal plane.  The spread in angles is proportional to 1/aperture, that times the distance gives the spot size at the focal plane proportional to (focal length/aperture)=f-number.

So the relevant quantity for photography (the diffraction spot size) depends only on the f-number and not the aperture or focal length separately.  And so that 50mm @f8 will have sqrt[2]~1.4 times less diffraction in the image than the 100mm @ f11.
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This is getting a bit confusing, Emil   . If we are talking about the same scene, the same composition, the same field of view from the same perspective, then the 100mm lens at F11 will produce less diffraction than the 50mm lens at F8, but of course the sensor with the 100mm lens would have to be bigger.

If we are talking about a fixed size sensor, then the diffraction spot size will vary only with F stop, but the FoV will vary in inversely with the focal length. Is this correct?
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Fine_Art

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« Reply #122 on: August 31, 2008, 01:03:19 pm »

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Yes. But as I said above this is not 'handling diffraction well'. It is just that other aberrations are handled well so diffraction is the limiting factor. BTW Diffraction is not caused by the 'edge'. It is caused by the hole.
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Diffraction is the limiting factor in lens design. Making it irrelevant is handling it well.

What about the hole causes it if not the edge? Diffraction happens when light hits an object. Any object. Its very small unless the amount of light getting through is small relative to the amount of diffraction. A small hole has a small area relative the the edges (circumference) where diffraction happens. Thats a result not a cause.
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ejmartin

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« Reply #123 on: August 31, 2008, 03:22:46 pm »

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This is getting a bit confusing, Emil   . If we are talking about the same scene, the same composition, the same field of view from the same perspective, then the 100mm lens at F11 will produce less diffraction than the 50mm lens at F8, but of course the sensor with the 100mm lens would have to be bigger.

If we are talking about a fixed size sensor, then the diffraction spot size will vary only with F stop, but the FoV will vary in inversely with the focal length. Is this correct?
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Sorry, I was referring to the physical size of the diffraction spot (Airy disk) on the sensor, at a fixed distance from the subject.  The size of that physical spot depends only on the f-number.   So for instance,  one could "zoom with the feet" and move twice as close with the 50mm compared to the 100mm to get a comparable field of view (ignoring perspective effects), then the diffraction spot size depends only on the f-number.

Alternatively, one can stay in the same place and use two different cameras.  For f11 to have the same diffraction spot size relative to frame size as f8 on another camera, the ratio of the sensor sizes would have to be sqrt[2] in linear dimension (f11 would need a larger sensor).  For comparable field of view, a 100mm lens must be on a sensor twice larger in linear dimension than a 50mm lens; so 100mm at f11 on FF is equivalent to 50mm @ f5.6 on a half-size sensor (eg 4/3 system) in terms of diffraction effects, or a bit under f8 @ ~66mm on APS-C.
« Last Edit: August 31, 2008, 09:07:47 pm by ejmartin »
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emil

ejmartin

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« Reply #124 on: August 31, 2008, 03:29:24 pm »

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Diffraction is the limiting factor in lens design. Making it irrelevant is handling it well.

What about the hole causes it if not the edge? Diffraction happens when light hits an object. Any object. Its very small unless the amount of light getting through is small relative to the amount of diffraction. A small hole has a small area relative the the edges (circumference) where diffraction happens. Thats a result not a cause.
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I would have said that optical aberrations are the limiting factor in lens design.  Diffraction is a function of f-number, and no change in the lens design is going to alter that.

From your choice of wording, I suspect that the mental construct you have in mind is that light is getting bent by some sort of collision or scattering of light as it bounces off an object.  That is not diffraction.  Diffraction is a purely wave phenomenon that results from the mutual interference of different parts of the light wave as it passes through the aperture.  It comes from no particular part of the aperture, rather it comes from all over the aperture since it is the interference of all the wave components that results in the diffraction pattern.
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emil

NikosR

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« Reply #125 on: August 31, 2008, 03:47:34 pm »

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Diffraction is the limiting factor in lens design. Making it irrelevant is handling it well.


You cannot make the diffraction irrelevant. You can minimise other optical aberrations so diffraction becomes the main limiting factor. Thus, in a good optical design you make diffraction highly relevant.

Diffraction is not an edge phenomenon. As it is a kind of interference it is a wave phenomenon that occurs all over the wavefront passing through the aperture. In that sense it is caused by the hole and not the edges. The other way of approaching it is probabilistically (quantum mechanics). Again it is the hole that is relevant. You cannot explain diffraction by a geometric (light ray) approach.
« Last Edit: August 31, 2008, 04:13:43 pm by NikosR »
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Nikos

Tony Beach

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« Reply #126 on: August 31, 2008, 04:55:57 pm »

Seems the point of this thread got lost in a technical discussion about diffraction, S/N ratios, etc.

Canon has released some sample images here:  http://web.canon.jp/imaging/eosd/eos50d/eos50d_sample-e.html

Make of them what you will, I still expect little or no difference at f/11 between 50D and a 40D.  Inerestingly, Canon has chosen a somewhat stand-offish scene shot at f/8 to illustrate the camera's landscape capabilities, which does nothing to put the DOF and Airy disc debate to rest.
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Panopeeper

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« Reply #127 on: August 31, 2008, 05:25:22 pm »

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It comes from no particular part of the aperture, rather it comes from all over the aperture since it is the interference of all the wave components that results in the diffraction pattern
I don't feel comfortable to dispute a physicist's statement regarding physics, but I looked it up in the McGraw Enciylopedia of Science, which cites from "Fundamentals of Optics" by White, "Introduction to Classical and Model Optics" by Meyer-Arendt and "Lectures on Theoretical Physics" by Sommerfeld. I don't know from which source following paragraph comes:

The bending of light, or other waves, into the region of the geometrical shadow of an obstacle. More exactly, diffraction refers to any redistribution in space of the intensity of waves that results from the presence of an object that causes variation of either the amplitude or phase of the waves.

Specifically related to the Fraunhofer diffraction, it says

... At the instant that the incident plane wave occupies the plane of diffracting screen, it may be regarded as sending out, from each element of its surface, a multitude of secondary waves, the joint effect of which is to be evaluated in the focal plane of the lens.

My questions are: if I accept, that not the edge of the opening causes the diffraction but the light waves cause it on their own, then

a. how does this depend on the size of opening?

b. why does the opening exactly at the location of the aperture count, not for example the diameter of the first lens element?

c. this is expanding on b: at which point of the lens (at which point of the travel of the light rays) does the diffraction start?

I do understand, that this is far away from everyday's photography, but perhaps I am not the only one intrigued by the notion, that the presence and shape of the aperture is not the cause of diffraction.
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Gabor

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« Reply #128 on: August 31, 2008, 07:54:14 pm »

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For comparable field of view, a 100mm lens must be on a sensor twice smaller in linear dimension than a 50mm lens; so 100mm at f11 on FF is equivalent to 50mm @ f5.6 on a half-size sensor (eg 4/3 system) in terms of diffraction effects, or a bit under f8 @ ~66mm on APS-C.
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Emil,
I think you meant to write twice bigger, didn't you?  
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Ray

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« Reply #129 on: August 31, 2008, 08:18:29 pm »

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Canon has released some sample images here:  http://web.canon.jp/imaging/eosd/eos50d/eos50d_sample-e.html

Make of them what you will, I still expect little or no difference at f/11 between 50D and a 40D. 
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Using the 50D in place of the 40D at F11 (in the F8 landscape shot), just might make the difference between being able to tell the time on that tower clock and not being able to discern the position of the hands.
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Fine_Art

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« Reply #130 on: August 31, 2008, 08:38:58 pm »

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I would have said that optical aberrations are the limiting factor in lens design.  Diffraction is a function of f-number, and no change in the lens design is going to alter that.

From your choice of wording, I suspect that the mental construct you have in mind is that light is getting bent by some sort of collision or scattering of light as it bounces off an object.  That is not diffraction.  Diffraction is a purely wave phenomenon that results from the mutual interference of different parts of the light wave as it passes through the aperture.  It comes from no particular part of the aperture, rather it comes from all over the aperture since it is the interference of all the wave components that results in the diffraction pattern.
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I was thinking of the 2 slit experiment that shows the waves as they propagate from small points. I understand light doesnt operate like particles. You are correct in that I was thinking the wave pattern was mainly influenced my the object (edge) that it forms around.

Your statement that its from all components sounds better that what I said which I see is wrong after thinking about your statement.
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ejmartin

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« Reply #131 on: August 31, 2008, 09:07:10 pm »

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Emil,
I think you meant to write twice bigger, didn't you? 
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Yep, my bad.
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emil

ejmartin

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« Reply #132 on: August 31, 2008, 09:41:48 pm »

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I don't feel comfortable to dispute a physicist's statement regarding physics, but I looked it up in the McGraw Enciylopedia of Science, which cites from "Fundamentals of Optics" by White, "Introduction to Classical and Model Optics" by Meyer-Arendt and "Lectures on Theoretical Physics" by Sommerfeld. I don't know from which source following paragraph comes:

The bending of light, or other waves, into the region of the geometrical shadow of an obstacle. More exactly, diffraction refers to any redistribution in space of the intensity of waves that results from the presence of an object that causes variation of either the amplitude or phase of the waves.

Specifically related to the Fraunhofer diffraction, it says

... At the instant that the incident plane wave occupies the plane of diffracting screen, it may be regarded as sending out, from each element of its surface, a multitude of secondary waves, the joint effect of which is to be evaluated in the focal plane of the lens.

This last statement is to me the important intuition, known as Huyghens' principle -- the idea that each point along a wave's oscillation is a source for the further propagation of the wave.  Consider dropping a stone into a pond.  The stone oscillates the surface of the water up and down at the point of impact, resulting in a spherical wave travelling out on the surface of the water.  Consider as well a point on the surface away from the point of impact; as the wave passes by, it oscillates the surface of the water up and down -- so each point along the wave is itself a source for further propagation of the wave, as each point oscillates up and down it sends out spherical waves of its own which combine to make the wave which is seen a little later.

In diffraction, what happens is that only the wave at the aperture are present, all the other points along the wavefront having been blocked by the obstacle/aperture.  Each point along the opening sends out its little spherical wave, and so in particular sends light into the region which one might naively think is in the shadow of the obstacle.  The wider the opening, the more of the incident wavefront is present, and the more these little spherical waves from all along the wavefront cohere to make a forward propagating wave (with a slight spread from the diffraction).  The narrower the opening, the more the further propagation comes from almost a single point, and so the more the further propagation emanates out spherically like the pebble in the pond rather than in the forward direction which one would naively think is the direction the light should be going.

Note that, according to Huyghens, the further wave pattern comes from the pattern created by propagation from all along the wavefront, not just at the edges of the obstacle.

There are different levels of approximation in the calculation of diffraction effects.  The leading order approximation is called Fraunhofer diffraction, and is the sort that goes on when light passes through a small aperture.  The next order is known as Fresnel diffraction, and is more relevant to diffraction effects in the shadow of an obstacle.  The next order beyond that is relevant to the formation of supernumerary rainbows, a rare occurance but beautiful to see when it happens.

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My questions are: if I accept, that not the edge of the opening causes the diffraction but the light waves cause it on their own, then

a. how does this depend on the size of opening?

b. why does the opening exactly at the location of the aperture count, not for example the diameter of the first lens element?

c. this is expanding on b: at which point of the lens (at which point of the travel of the light rays) does the diffraction start?

I do understand, that this is far away from everyday's photography, but perhaps I am not the only one intrigued by the notion, that the presence and shape of the aperture is not the cause of diffraction.
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a.  The angle over which the beam spreads is inversely proportional to the diameter of the aperture.

b.  All elements in the optical path will generate some diffraction, however given the relation of the spreading angle to the aperture size, it is the smallest constriction in the light path which has the dominant effect.

c.  I suppose you could say that it starts at the lens hood, but really the amount of diffraction there is negligible; it is the smallest opening, the aperture itself, which has the most important effect.
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emil

Panopeeper

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« Reply #133 on: August 31, 2008, 10:58:59 pm »

Emil,

thanks for the detailed explanation. I will chew on that later; however, now I have a question of practical relevance.

The claim, that the diffraction depends only on the f-number is based on a woodoo calculation: it starts out with a thin lens and puts the aperture directly to the lens, ignoring the distance between the aperture and the center plane of the lens (from where the focal length is measured in case of the thin lens). There is no such design in today's photography.

If the location of the aperture is the main "starting point" of the diffraction, then the aperture-focal plane distance should be used in the calculation, which is usually far from the focal length. This is true re both short and long focal lengths; the location of the aperture is at the second nodal plane only in particular retrofocal configurations.

Thus the statement, that the diffraction depends on the f-number is not correct. In fact, it does not depend on the focal length at all.

Another issue is, that the light does not travel in a straight line between the aperture and the focal plane. It changes the direction several times, there is even a convergence point in-between.

I have the feeling that the drawing (and associated calculation) illustrating the correlation between aperture, focal length and focal plane is meant for the kindergarten. It stank already, when I saw sin(x) substituted by x (for example in Sidney Ray's Applied Photographic Optics); x must be several orders smaller for this substitution than it is in the cases we are talking about. This has nothing to do with the dependance/independance of the diffraction on/from the focal length, but it does have to do with the diameter of the airy disk.

What do you think of that?
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Gabor

Tony Beach

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« Reply #134 on: September 01, 2008, 01:12:34 am »

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Using the 50D in place of the 40D at F11 (in the F8 landscape shot), just might make the difference between being able to tell the time on that tower clock and not being able to discern the position of the hands.
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You just don't get it, or choose not to believe what I'm asserting.  f/11 is going to be more blurry than f/8, and you will probably no longer be able to make out the hands on the clock, and that will be the same as what 40D resolves.
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NikosR

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« Reply #135 on: September 01, 2008, 01:25:48 am »

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I understand light doesnt operate like particles.
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You can approach this phenomena from the particle point of view (quantum mechanics, quantum interference). But then you have to think in terms of probabilities not what happens to a single particle.
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Nikos

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« Reply #136 on: September 01, 2008, 01:29:50 am »

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  f/11 is going to be more blurry than f/8, and you will probably no longer be able to make out the hands on the clock, and that will be the same as what 40D resolves.
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The first phrase in your sentence is probably correct (with most of today's lenses). But how do you infer the correctness of the third phrase?
« Last Edit: September 01, 2008, 04:02:11 am by NikosR »
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Ray

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« Reply #137 on: September 01, 2008, 02:30:29 am »

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You just don't get it, or choose not to believe what I'm asserting.  f/11 is going to be more blurry than f/8, and you will probably no longer be able to make out the hands on the clock, and that will be the same as what 40D resolves.
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No. You haven't interpreted correctly what I wrote. I'll try again. In circumstances where at F11 the time on that tower clock (the tower clock as identified in the F8 shot) is just legible in a 50D shot, it probably wouldn't be legible in a 40D shot at the same aperture of F11.

Of course I know that all good 35mm lenses will be marginally sharper at F8 than at F11. Why on earth would you think I wasn't aware of that?

However, some budget priced zooms are sharpest at F16, at certain focal lengths, which is another issue.
« Last Edit: September 01, 2008, 02:38:03 am by Ray »
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ejmartin

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« Reply #138 on: September 01, 2008, 09:04:27 am »

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Emil,

thanks for the detailed explanation. I will chew on that later; however, now I have a question of practical relevance.

The claim, that the diffraction depends only on the f-number is based on a woodoo calculation: it starts out with a thin lens and puts the aperture directly to the lens, ignoring the distance between the aperture and the center plane of the lens (from where the focal length is measured in case of the thin lens). There is no such design in today's photography.

If the location of the aperture is the main "starting point" of the diffraction, then the aperture-focal plane distance should be used in the calculation, which is usually far from the focal length. This is true re both short and long focal lengths; the location of the aperture is at the second nodal plane only in particular retrofocal configurations.

Thus the statement, that the diffraction depends on the f-number is not correct. In fact, it does not depend on the focal length at all.

Another issue is, that the light does not travel in a straight line between the aperture and the focal plane. It changes the direction several times, there is even a convergence point in-between.

I have the feeling that the drawing (and associated calculation) illustrating the correlation between aperture, focal length and focal plane is meant for the kindergarten. It stank already, when I saw sin(x) substituted by x (for example in Sidney Ray's Applied Photographic Optics); x must be several orders smaller for this substitution than it is in the cases we are talking about. This has nothing to do with the dependance/independance of the diffraction on/from the focal length, but it does have to do with the diameter of the airy disk.

What do you think of that?
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I wouldn't characterize it as a voodoo calculation, rather it is the sort of reasonable simplifying approximation that is made in physical models.   It is however important to understand what those simplifying assumptions are and how they affect the result.

For instance, in the oft-quoted formula for the location of the first minimum of the diffraction pattern (the size of the Airy disk), at 1.22*wavelength/f-number, the assumption is not only that the aperture is located at the focal length from the focal plane, also the 1.22 comes from assuming that the aperture is circular rather than hexagonal or octagonal (as for a real lens aperture); and as you are pointing out, there is a sequence of obstructions along the light path through the lens, each of which will contribute to diffraction effects.

All that a more refined calculation, using the actual configuration of a specific camera lens, will accomplish is to refine that numerical factor 1.22 into something slightly different.

How different is the aperture-to-focal plane distance from the focal length in a typical lens?  Does the ratio of these two quantities vary substantially from lens to lens?  I'm not knowledgeable about the details of current lens designs.
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emil

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« Reply #139 on: September 01, 2008, 04:04:57 pm »

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I wouldn't characterize it as a voodoo calculation, rather it is the sort of reasonable simplifying approximation that is made in physical models
I characterize it that way, for I don't find anything reasonable on it. There is nothing there related to real photographic lenses; though you may use that method when calculating the diffraction of a loupe.

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How different is the aperture-to-focal plane distance from the focal length in a typical lens?  Does the ratio of these two quantities vary substantially from lens to lens?  I'm not knowledgeable about the details of current lens designs.
The block diagrams of almost all Canon lenses are shown in Canon's Camera Museum; one can roughly estimate the position of aperture from the flange.

The flange-focal plane distance is about 36mm (rough measurement, for I was careful with the vernier caliper inside the chamber). I made a search for it and found 44mm; no way (anyway, this is relevant only for the calculation with specific lenses, not in general).

My estimation is, that the aperture/focal plane distance is close to the focal length in case of the 50mm lenses: about 55mm with the 50mm f/1.4. In case of the ultrawides, like the 14mm f/2.8L the aperture may be more than four times the focal length away. The proportion turns around with the long lenses; for example the 200mm f/2.8L: the aperture appears to be around 90mm away from the focal plane, with the 300mm f/2.8L IS this appears to be 120mm, with the 400mm it is about 130mm, roughly 1/3 of the focal length.

Thus there is a factor of at least twelve between the lenses, even more between extreme short  and extreme long ones.

Good by airy disk calculation based on the f-number.

(And we have no idea yet, how the multiple bending of the rays affects the outcome.)
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