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 Author Topic: Mark Dubovoy's essay  (Read 37939 times)
Antisthenes
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Mark's article is good because it's a relevant question to ask (regarding false ISO setting), but it's not excellent because his ideas of a cause and his whole "depth of field" idea is totally irrelevant here.

Depth of field appears "irrelevant" to you because you don't understand the notion of "Circle of Confusion" used in DoF calculations.

The intersection of the lens' light cone and the image plane is a disc.  The light rays near the periphery of the cone are more tilted than the central light ray (the chief ray).

If the tilt exceeds a critical angle, the rays cannot reach, and therefore are not recorded by the photodiodes.  Ergo, the recorded diameter of that disc is affected by the sensel's acceptance angle, and that recorded diameter can therefore be smaller than the geometric diameter of the light cone at the imaging plane intersection.

The CoC is also a disc, and is conceptually and physically identical to the disc formed by the intersection — discussed above — of the light cone and the imaging plane.

Ergo, a sensel with a limited acceptance angle will record a disc that is smaller than the disc recorded by a medium like film that hasn't any acceptance angle limits.

A smaller disc, in CoC terms, is a sharper, less blurred disc.  If the constitutive points of an image are sharper, it means that the sharpness zone in the object field, i.e. the DoF, must be deeper.  Ergo, a sensel with a limited acceptance angle necessarily results — as Mark Dubovoy correctly points out — in a deeper DoF than you'd get with film.

The photozone.de OOF disc samples, while not ideal as they haven't been taken at the center of the imaging field, provide a good measure of the plateauing of the disc's dimensions at a dimension corresponding to about f/1.5

Considering the typical price, size and weight differentials between a f/1.2 and a f/1.4 lens, the fact that — when used on a DSLR like the Canon 5D2 — the effective speed of an f/1.2 lens, as well as its DoF and bokeh might be equivalent only to that of a f/1.5 lens is a quite relevant observation on a photo forum.

The apparent deformation affecting the exit pupil at larger image heights — that is, at points far from the image center — is essentially an elliptical one on photozone.de's pictures.  Elliptical deformations affect the dimensions of a circular exit pupil only along one axis, and quite valid dimensional measurements can thus be made along the axis unaffected by the elliptical squeezing.

These photozone.de measurements are tangible and totally consistent with the physics of light cone propagation and microlens acceptance angle limitation models.

All your assertions, OTOH, are unsubstantiated, including the one that a test with the NEX-5 and a 40mm lens of unknown exit pupil distance is somehow relevant to judge a DSLR's sensel's response to a f/1.2 cone of light.

The diagonal of an APS-C sensor like the one used in the NEX-5 is 28mm.  If we assume, as you do, that the lens' exit pupil distance from the imaging plane is 40mm, the chief ray's angle in the corner has a tilt of arctan(14mm/40mm) — i.e. corresponding to the tilt of the marginal rays of a light cone of a 40mm/28mm = f/1.4 lens.

So, even if the NEX-5 had no offset microlenses — something that is impossible for me and you to know, — your assertion that your f/1.4 NEX-5 thought experiment "proves" that ray tilt cannot be a vignetting factor with a f/1.2 lens is totally irrelevant.
Obviously, if the NEX-5 had offset microlenses, your thought experiment would be even more irrelevant, as the corner sensels' sensitivity to the chief ray's tilt angle would be even less observable.
 « Last Edit: November 08, 2010, 07:39:44 AM by Antisthenes » Logged
BartvanderWolf
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The marginal rays of the light cone projected by a fast — e.g. f/1.2 — lens are quite tilted.

I wouldn't characterize the maximum possible angle as "quite tilted" (whatever that may mean):

That doesn't seem like a very problematic angle for a microlens to me, especially when we consider the modest light fall-off in the corners of the image when stopped down where the central rays strike at an even more oblique angle of 29.65 degrees ...

Cheers,
Bart
 « Last Edit: November 08, 2010, 07:52:48 AM by BartvanderWolf » Logged
Antisthenes
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I wouldn't characterize the maximum possible angle as "quite tilted" (whatever that may mean):

An incidence angle that is larger than the acceptance angle of common DSLR sensors is, in my mind, quite tilted.

I also invite you to review what "exit pupil" means, optically speaking.
Considering that most SLR lenses with a 50mm focal length have an exit pupil to image plane distance of about 50mm, the notion that a short tele 85mm Canon lens designed to be mounted on an SLR could possibly have an exit pupil distance of 38mm is, well, quite amusing.

Furthermore, any assertion that the sensel's acceptance angles are not a factor should also be accompanied with a physically coherent explanation as to why the OOF discs recorded with a DSLR happen to be clipped at a radius corresponding to around f/1.5 even with f/1.2 lenses while, with film cameras, the recorded OOF disc diameter maintains a strict proportionality with the lens' selected aperture.
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pegelli
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@ Antithenes, interesting discussion but in my mind still inconclusive.

One thing that might clear up some doubt for me is if the light fall off proportionally (or gradually) drops down as the inclination angle increases, or if there is a very steep fall off from near 100% to near 0% once you exceed a certain threshold inclination angle. What's your understanding how that works?
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pieter, aka pegelli
b_z
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Depth of field appears "irrelevant" to you because you don't understand the notion of "Circle of Confusion" used in DoF calculations.

You make me laugh... Seriously. I perfectly understand  circle of confusion and depth of field calculation. They are irrelevant here because you will never EVER see any modification of DoF because of ray inclination.

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The intersection of the lens' light cone and the image plane is a disc.  The light rays near the periphery of the cone are more tilted than the central light ray (the chief ray).

Which cone ? In fact it's a lot more complicated than that. A point in object field send a light cone which intersect sensor=image plane in a disc (provided there is no mechanical vignetting, thing you seem unable to understand), yes.

But, as I already said, the chief ray is the less inclined of the disc ONLY in the middle of the sensor. At the border, the chief ray is MORE tilted than some of the border rays.

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If the tilt exceeds a critical angle, the rays cannot reach, and therefore are not recorded by the photodiodes.

That's utterly wrong. The transition is PROGRESSIVE (I said that earlier, you did not understand ?), there's no "critical angle after which NO light is recorded".

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Ergo, the recorded diameter of that disc is affected by the sensel's acceptance angle, and that recorded diameter can therefore be smaller than the geometric diameter of the light cone at the imaging plane intersection.

Again, this would only be true right at the middle of the sensor, if there was such an "acceptance angle" (which is NOT the case at any angle we're speaking of).

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The CoC is also a disc, and is conceptually and physically identical to the disc formed by the intersection — discussed above — of the light cone and the imaging plane.

Not true. CoC is the diameter a disc can have on the sensor and yet be considered ponctual by an observer placed at the right distance of the printing. When the light cone of an object point intersect the sensor in a circle which is smaller than CoC, then it is in focus.

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Ergo, a sensel with a limited acceptance angle will record a disc that is smaller than the disc recorded by a medium like film that hasn't any acceptance angle limits.

Not true. At most you'll have more vignetting, but appearently this "pixel/sensor vignetting" is always weaker than the cos4 law.

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A smaller disc, in CoC terms, is a sharper, less blurred disc.  If the constitutive points of an image are sharper, it means that the sharpness zone in the object field, i.e. the DoF, must be deeper.  Ergo, a sensel with a limited acceptance angle necessarily results — as Mark Dubovoy correctly points out — in a deeper DoF than you'd get with film.

Not true. You're totally confused here, it makes absolutely no sense.

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The photozone.de OOF disc samples, while not ideal as they haven't been taken at the center of the imaging field, provide a good measure of the plateauing of the disc's dimensions at a dimension corresponding to about f/1.5

Wrong. Their vertical size is approximately the same than at f/1.5, BECAUSE OF LENS MECHANICAL VIGNETTING AND NOTHING ELSE, but if you were honest, you would point out that their horizontal size is really what's expected for f/1.2 lens. And it proves you're wrong.
And had this crop been taken at center of the frame, you'd see a perfezct circle, the size you except to have for f/1.2 (unless ther is mechanical vignetting due to the chamber, it sometimes happens as I said earlier).

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Considering the typical price, size and weight differentials between a f/1.2 and a f/1.4 lens, the fact that — when used on a DSLR like the Canon 5D2 — the effective speed of an f/1.2 lens, as well as its DoF and bokeh might be equivalent only to that of a f/1.5 lens is a quite relevant observation on a photo forum.

Except this "might be" is in fact a "is not".

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The apparent deformation affecting the exit pupil at larger image heights — that is, at points far from the image center — is essentially an elliptical one on photozone.de's pictures.  Elliptical deformations affect the dimensions of a circular exit pupil only along one axis, and quite valid dimensional measurements can thus be made along the axis unaffected by the elliptical squeezing.

Yes, and had you been honest you would have seen only vertical size is smaller than expected. Horizontal size is fine.

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These photozone.de measurements are tangible and totally consistent with the physics of light cone propagation and microlens acceptance angle limitation models.

Yes, and measuring these disks along horizontal axis prove that it's really a f/1.2 bokeh.

I'm laughing at you saying things about "physics of light cone propagation" when in fact you fail to understand even the simplier arguments. "microlens acceptance angle limitation models" is just b***s*** when you talk of 30° angles. It's an issue only for a lot higher angles, and even so, cos4 law is a waaaay bigger issue.

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All your assertions, OTOH, are unsubstantiated, including the one that a test with the NEX-5 and a 40mm lens of unknown exit pupil distance is somehow relevant to judge a DSLR's sensel's response to a f/1.2 cone of light.

I'm laughing. Really. "unsubstantiated" describes very well your own posts. Mine consist of facts I've checked, or when it's not I say it.
The exit pupil is not unkown, you failed to read it was between 35 and 45mm ? Yes, I measured it, because I know how to do...

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The diagonal of an APS-C sensor like the one used in the NEX-5 is 28mm.  If we assume, as you do, that the lens' exit pupil distance from the imaging plane is 40mm, the chief ray's angle in the corner has a tilt of arctan(14mm/40mm) — i.e. corresponding to the tilt of the marginal rays of a light cone of a 40mm/28mm = f/1.4 lens.

Yes, you're right, for once ! But I'm not "assuming", I don't "assume" things or I say I do. I measured it.

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So, even if the NEX-5 had no offset microlenses — something that is impossible for me and you to know, — your assertion that your f/1.4 NEX-5 thought experiment "proves" that ray tilt cannot be a vignetting factor with a f/1.2 lens is totally irrelevant.

The fact you don't know that doesn't mean nobody knoes. I do know Nex has no offset microlenses. And I proved that tilted rays are a totally irrelevant issue for f/1.4 lenses with that lens/sensor combination.
And read one of my above post again : I also measured a 50~55mm exit pupil for m 50/2.8 macro Sony lens, and on FF sensor (Alpha 850/900) it has no vignetting. Meaning for f/1.2 lenses (the angle is almost the same), this issue is irrelevant too.

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Obviously, if the NEX-5 had offset microlenses, your thought experiment would be even more irrelevant, as the corner sensels' sensitivity to the chief ray's tilt angle would be even less observable.

Except it doesn't have. And I'm not guessing, it's a fact.
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b_z
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An incidence angle that is larger than the acceptance angle of common DSLR sensors is, in my mind, quite tilted.

Define acceptance angle.

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I also invite you to review what "exit pupil" means, optically speaking.
Considering that most SLR lenses with a 50mm focal length have an exit pupil to image plane distance of about 50mm, the notion that a short tele 85mm Canon lens designed to be mounted on an SLR could possibly have an exit pupil distance of 38mm is, well, quite amusing.

"Exist pupil" on this scheme is in fact "rear lens element".

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Furthermore, any assertion that the sensel's acceptance angles are not a factor should also be accompanied with a physically coherent explanation as to why the OOF discs recorded with a DSLR happen to be clipped at a radius corresponding to around f/1.5 even with f/1.2 lenses while, with film cameras, the recorded OOF disc diameter maintains a strict proportionality with the lens' selected aperture.

Define acceptance angle before even thinking of going on with this.

The bold sentence is completely false. There is no such OOS disc clipping APART FROM MECHANICAL VIGNETTING. But you seem unable to understand that.

Try with film and sensor, you'll get the exact same OOF discs shape. Exactly. Not even a 1% difference in size. Maybe a 1~10% luminosity decrease at borders with sensor compared to film, but no shape difference. At all.
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BartvanderWolf
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An incidence angle that is larger than the acceptance angle of common DSLR sensors is, in my mind, quite tilted.

That is quite a bit of circular reasoning. What is the acceptance angle of common DSLRs? You seem to conclude that it's exceeded, so you suggest you know.

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I also invite you to review what "exit pupil" means, optically speaking.

And I invite you to review the difference between exit pupil (especially in tele lenses) and (secondary) principle plane ...

In fact, the diameter of the exit pupil of the EF 85mm f/1.2 at 44mm distance from the flange is almost exactly the same at the rear lens element and the virtual image of the front lens aperture. True, the exact distance of the exit pupil may be at a different position, but it's visible diameter isn't different (as viewed from the fixed distance at the center of the sensor). When moving off optical axis, one can see (from the position of the sensor through the rear lens element) the front of the lens barrel, which explains the vignetting at full aperture, the aperture is no longer exactly  circular. All of the above is when focused at infinity.

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Considering that most SLR lenses with a 50mm focal length have an exit pupil to image plane distance of about 50mm, the notion that a short tele 85mm Canon lens designed to be mounted on an SLR could possibly have an exit pupil distance of 38mm is, well, quite amusing.

Unless you are confusing exit pupil of a telelens (where the rear element acts as an aperture) with principle plane, I fail to grasp your misplaced amusement.

Cheers,
Bart
 « Last Edit: November 08, 2010, 10:14:44 AM by BartvanderWolf » Logged
b_z
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@ Antithenes, interesting discussion but in my mind still inconclusive.

One thing that might clear up some doubt for me is if the light fall off proportionally (or gradually) drops down as the inclination angle increases, or if there is a very steep fall off from near 100% to near 0% once you exceed a certain threshold inclination angle. What's your understanding how that works?

http://en.wikipedia.org/wiki/Vignetting very good article on vignetting.

Two kinds of light fall-off gradually increase depending on the inclination angle.

The biggest fall-off in regard of inclination angle is the cos4 law, which is the same on film and sensor. It's a physical law, you can only correct that by making your lenses telecentric (meaning rays are less inclined in the borders of the frame).

The second is the pixel/sensor vignetting, which is Mark's concern. And in fact it's known to appear in some bizarre kind of way (purple or magenta zones because of selective refraction on microlenses, I read somewhere, and added vignetting) only at very tilted angles you can get with a 15~20mm non telecentric lens with APSC or bigger sensor, EVEN with offset microlenses (Leica strongly advise not to use old design (non telecentric) ultra wide angles on M9 because of that).
The angle concerned is approximately what you'd get with a f/0.7 lens. With smaller angles, no noticeable sign of pixel/sensor vignetting.
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pegelli
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@b_z, BartvanderWolf, also thanks for all the insights and links you're providing, should have included you in my earlier posted question At least I can follow your logic since it's clear concepts and data (I'm a ChemE by training, so have some appreciation for that  ). I'll continue to follow, but might not be able to contribute a lot to the insights other than asking a stupid question now and then
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pieter, aka pegelli
PierreVandevenne
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As far as mechanical vignetting is concerned for fast optics, I can confirm it seems to exist in Canon cameras

http://dl.dropbox.com/u/2445259/boxeffect.jpg

I have consistently observed the above vignetting pattern (with slight variations) from my Canon cameras on a few of my instruments and lenses (on 10D, 20D, 40D, 1DMKII, 5D, 5D MKII) - it is usually not perceptible when looking at short exposures full size, barely noticeable on long exposures full size, but quite noticeable on processed images that haven't been properly calibrated with a flat field. It's orientation is always the same relative to the mirror box: there is always a darker linear area at the bottom of the frame.

The only other explanation I haven't been able to rule out with absolute certainty is that there would be a sensitivity issue at the bottom of all Canon sensors I have and does not show up with the astro CCDs I own. I haven't looked at the issue in depth with canon lenses as they are often not ideally suited for astro-photography (not to say that you can't get very good results, but they don't compare with the 2.7 FFC).

Last but not least, the whole subject is extremely complex, involves advanced lens design, advanced sensor design, some murky (at least in our minds) definitions, the inability to test components individually, etc. The consequences are relatively marginal in both senses of the term, and there are glaring holes in most of the complex arguments developed here (as they would be in my arguments if I went into them).

Let's be realistic here, we can't really second guess entire teams of competent engineers, each with their own domains of expertise, who have at their disposal the full resources of large manufacturers who happen to be locked in a fierce competitive war.

No point in getting into complex point by point rebukes or flame wars imho.
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Antisthenes
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@ Antithenes, interesting discussion but in my mind still inconclusive.

One thing that might clear up some doubt for me is if the light fall off proportionally (or gradually) drops down as the inclination angle increases, or if there is a very steep fall off from near 100% to near 0% once you exceed a certain threshold inclination angle. What's your understanding how that works?

The photodiode at the bottom of the CMOS pixel stack has a finite size.  Provided the light ray lands somewhere, anywhere in the photodiode's active area, it will presumably generate electrical charges, indicating that light has been detected.

As the ray's tilt increases, at some point, the ray won't fall anymore within the photodiode's active area.  Beyond that critical angle, the light ray can be considered, for imaging purposes, as lost.

The photodiode's active area is not "fuzzy".  It is created with IC manufacturing-class precision — perimeter variations of a few tens of nanometers at most against total sensel longitudinal dimensions of several thousand nanometers.
The angle transition between "detectable" and "not detectable" will thus presumably be quite sharp.

And indeed, the clipped f/1.5-equivalent OOF disc visible in photozone.de's picture has an clearly recognizable boundary — i.e. its perimeter is emphatically not progressively dissolving with, say, a Gaussian kind of blur where no definite boundary exists.
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Antisthenes
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Unless you are confusing exit pupil of a telelens (where the rear element acts as an aperture) with principle plane, I fail to grasp your misplaced amusement.

I found it amusing because you're obviously conflating the rear element with the "exit pupil".
The exit pupil is the virtual image of the lens aperture — or diaphragm blades, if you prefer — as seen from the back of the lens.

Your drawing showing an SLR tele lens with an exit pupil distance of 38mm is therefore utterly unrealistic.
The chief ray tilt calculations based on that unrealistic placement are thus, of course, totally meaningless.

Let F be the f-number of a lens. The half-angle of the light cone emitted by that lens will necessarily have as value arctan(1/(2*F))

For an f/1.2 lens, the half-angle must thus be arctan(1/(2*1.2)) ~ 22.62 degrees.

The fact that you indicated 27.76-degree half-angle for a 85mm f/1.2 lens is also evidence that your calculations of chief ray tilt angles at various image heights are, I regret to say, absolutely unrealistic.
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Antisthenes
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I perfectly understand  circle of confusion and depth of field calculation. They are irrelevant here because you will never EVER see any modification of DoF because of ray inclination.

Yawn.  Will you ever be able to understand that it is not sufficient to make assertions, but that you must also provide a modicum of evidence or physics- or math-based reasoning supporting your assertions ?

So, do please explain, preferably using sound optics principles, why "one will never EVER see any modification of DoF because of ray inclination"

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But, as I already said, the chief ray is the less inclined of the disc ONLY in the middle of the sensor. At the border, the chief ray is MORE tilted than some of the border rays.

Yawn.
In your thought experiment, your lens has an exit pupil at a distance of 40mm from the imaging plane.
This happens to correspond to the half-angle of the light cone of a f/1.4 lens.

Problem is, we're talking about the light loss affecting the light cone of a f/1.2 lens, not f/1.4.

Last time I checked, f/1.2 lenses had wider light cones than f/1.4 lenses, meaning that the f/1.2 cone's marginal rays are MORE, not LESS tilted than the f/1.4 light cone's marginal rays.

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If the tilt exceeds a critical angle, the rays cannot reach, and therefore are not recorded by the photodiodes.

That's utterly wrong. The transition is PROGRESSIVE (I said that earlier, you did not understand ?), there's no "critical angle after which NO light is recorded".

Yawn.  Where is the substantiation of your gratuitous assertion that the transition is PROGRESSIVE ?  Are the depletion zones of the photodiodes at the bottom of a CMOS pixel stack drawn using some kind of fuzzy semiconductor technology ?

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Ergo, the recorded diameter of that disc is affected by the sensel's acceptance angle, and that recorded diameter can therefore be smaller than the geometric diameter of the light cone at the imaging plane intersection.
Again, this would only be true right at the middle of the sensor, if there was such an "acceptance angle" (which is NOT the case at any angle we're speaking of).

Yawn.  Where is the substantiation of your gratuitous assertion that there can't possibly be acceptance angle effects with large aperture lenses ?

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The CoC is also a disc, and is conceptually and physically identical to the disc formed by the intersection — discussed above — of the light cone and the imaging plane.
Not true. CoC is the diameter a disc can have on the sensor and yet be considered ponctual by an observer placed at the right distance of the printing. When the light cone of an object point intersect the sensor in a circle which is smaller than CoC, then it is in focus.

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Ergo, a sensel with a limited acceptance angle will record a disc that is smaller than the disc recorded by a medium like film that hasn't any acceptance angle limits.
Not true. At most you'll have more vignetting, but appearently this "pixel/sensor vignetting" is always weaker than the cos4 law.

Yawn.  Where is the substantiation of your gratuitous assertion that "pixel/sensor vignetting" is always weaker than the cos4 law ?

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A smaller disc, in CoC terms, is a sharper, less blurred disc.  If the constitutive points of an image are sharper, it means that the sharpness zone in the object field, i.e. the DoF, must be deeper.  Ergo, a sensel with a limited acceptance angle necessarily results — as Mark Dubovoy correctly points out — in a deeper DoF than you'd get with film.
Not true. You're totally confused here, it makes absolutely no sense.

Yawn.  Has it occurred to you that it might not make sense to you because:
and
2) your understanding of geometric optics concepts like CoC is basically non-existent ?

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The photozone.de OOF disc samples, while not ideal as they haven't been taken at the center of the imaging field, provide a good measure of the plateauing of the disc's dimensions at a dimension corresponding to about f/1.5
Wrong. Their vertical size is approximately the same than at f/1.5, BECAUSE OF LENS MECHANICAL VIGNETTING AND NOTHING ELSE, but if you were honest, you would point out that their horizontal size is really what's expected for f/1.2 lens. And it proves you're wrong.
And had this crop been taken at center of the frame, you'd see a perfezct circle, the size you except to have for f/1.2 (unless ther is mechanical vignetting due to the chamber, it sometimes happens as I said earlier).

Yawn.  Where is the substantiation of your gratuitous assertion that only mechanical vignetting is to blame, and that the horizontal size is what's expected for a f/1.2 lens ?
How about, well, actually calculating yourself what that horizontal size should be, using sound optical principles, and then providing us with your calculated value, and the measured value ?

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Considering the typical price, size and weight differentials between a f/1.2 and a f/1.4 lens, the fact that — when used on a DSLR like the Canon 5D2 — the effective speed of an f/1.2 lens, as well as its DoF and bokeh might be equivalent only to that of a f/1.5 lens is a quite relevant observation on a photo forum.
Except this "might be" is in fact a "is not".

Yawn.  Where is the substantiation of your gratuitous assertion that it "is not" the case that a f/1.2 lens can behave as a f/1.5 lens with a DSLR like the Canon 5D2 ?

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The apparent deformation affecting the exit pupil at larger image heights — that is, at points far from the image center — is essentially an elliptical one on photozone.de's pictures.  Elliptical deformations affect the dimensions of a circular exit pupil only along one axis, and quite valid dimensional measurements can thus be made along the axis unaffected by the elliptical squeezing.
Yes, and had you been honest you would have seen only vertical size is smaller than expected. Horizontal size is fine.

Yawn.  Where is the substantiation of your gratuitous assertion that the horizontal size is fine ?  Are you even able to calculate what that horizontal size should be ?

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I'm laughing. Really. "unsubstantiated" describes very well your own posts. Mine consist of facts I've checked, or when it's not I say it.
The exit pupil is not unkown, you failed to read it was between 35 and 45mm ? Yes, I measured it, because I know how to do...

Yawn.  You "know how to do" an exit pupil distance measurement of a rangefinder lens with a 40mm focal length, yet you can't even provide that value with an incertitude better than 10mm ?

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So, even if the NEX-5 had no offset microlenses — something that is impossible for me and you to know, — your assertion that your f/1.4 NEX-5 thought experiment "proves" that ray tilt cannot be a vignetting factor with a f/1.2 lens is totally irrelevant.
The fact you don't know that doesn't mean nobody knoes. I do know Nex has no offset microlenses.

Yawn.  Where is the substantiation of your gratuitous assertion that the NEX has no offset microlenses ?

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And I proved that tilted rays are a totally irrelevant issue for f/1.4 lenses with that lens/sensor combination.

Yawn.  Where is that "proof" [sic] ?

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And read one of my above post again : I also measured a 50~55mm exit pupil for m 50/2.8 macro Sony lens, and on FF sensor (Alpha 850/900) it has no vignetting. Meaning for f/1.2 lenses (the angle is almost the same), this issue is irrelevant too.

Yawn.  I suppose you don't even realize that your measurements are meaningless if you cannot provide solid evidence that:

• the Sony 50/2.8 macro lens has an exit pupil distance relative to the imaging plane of 50-55mm

and

• the Sony A900 sensor does not have offset microlenses.
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BartvanderWolf
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For an f/1.2 lens, the half-angle must thus be arctan(1/(2*1.2)) ~ 22.62 degrees.

On a symmetrical lens design. However, we are not dealing with a symmetrical lens design, were're dealing with a telelens. Ever hear about a concept called pupil magnification?

Cheers,
Bart
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pegelli
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Will you ever be able to understand that it is not sufficient to make assertions, but that you must also provide a modicum of evidence or physics- or math-based reasoning supporting your assertions ?

Maybe this discussion might get somewhere if you provided some data and evidence versus your current attitude of "I know it all and you guys don't"
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pieter, aka pegelli
Antisthenes
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For an f/1.2 lens, the half-angle must thus be arctan(1/(2*1.2)) ~ 22.62 degrees.
On a symmetrical lens design. However, we are not dealing with a symmetrical lens design, were're dealing with a telelens. Ever hear about a concept called pupil magnification?

Do you actually read the information on the pages you're linking to ?
I've highlighted an, um, relevant section on that quite interesting web page.
What part of "paraxial approximation" and "sin(theta) ~= tan(theta)", or the relation "theta=arctan(tan(theta))" don't you understand ?

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Antisthenes
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Will you ever be able to understand that it is not sufficient to make assertions, but that you must also provide a modicum of evidence or physics- or math-based reasoning supporting your assertions ?
Maybe this discussion might get somewhere if you provided some data and evidence versus your current attitude of "I know it all and you guys don't"

Maybe this discussion might get somewhere if some people actually had experience with film-based photography and knew that with film, the diameter of an OOF disc is directly proportional to the lens' aperture.

Maybe this discussion might get somewhere if some people had a sufficient intellect to understand that when a digital sensor delivers an OOF disc whose diameter is less than what you'd get on film, that this necessarily implies that light rays are being lost — i.e. not recorded by the photosites.
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b_z
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Antisthenes, you're the one providing no evidence at all.

I won't even reply to you, you just proved your uselessness : your only argument is "b_z might have done some error here, so all he said is wrong".
Besides, the fact you need to criticize my English to feel better about your inability to counter my arguments with facts and not assumptions that I'm wrong is just lame.

I strongly advise everyone here not to pay attention to you or answer any more.
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Alan Goldhammer
Sr. Member

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Posts: 1739

Antisthenes, you're the one providing no evidence at all.

I won't even reply to you, you just proved your uselessness : your only argument is "b_z might have done some error here, so all he said is wrong".
Besides, the fact you need to criticize my English to feel better about your inability to counter my arguments with facts and not assumptions that I'm wrong is just lame.

I strongly advise everyone here not to pay attention to you or answer any more.
And it would be of great help if the two of you would act in a civil manner, use your own names rather than some pseudonym (as most of us do because of a general request last year), and provide some evidence of your scientific background.  There are enough of us with advanced degrees in engineering and the physical sciences that we don't engage in these shouting matches but do, in fact, understand the physics of all this.

Alan
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BartvanderWolf
Sr. Member

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Posts: 3909

Quote
On a symmetrical lens design. However, we are not dealing with a symmetrical lens design, were're dealing with a telelens. Ever hear about a concept called pupil magnification?

Do you actually read the information on the pages you're linking to ?

Yes. However, the part you seem to have difficulty with is that the rear element of the 85mm f/1.2 restricts (or is of equal size from the perspective of the sensor) the exiting cone of light. While not at the actual exit pupil distance, it would virtually act as a rear aperture if it wasn't dimensioned exactly right (that's why it's so large).

Photo courtesy of www.photozone.de

The rear lens element is situated 6mm behind the lens flange, as indicated in my earlier diagram.

Cheers,
Bart
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