Diffraction -- advanced confusion!

sarah_fox · July 27, 2012

Last night it came to me in a dream: The diffraction limit of a lens/sensor system might not actually be constant across focal lengths. (Yes, I dream about such things. I actually had a good discussion about it with one of my dissertation committee members, and when I recounted the discussion upon waking, it actually made sense.) Here's what I'm thinking:

If we consider simple apertures where theoretical meniscus lenses are located, then yes, the size of the airy disc increases in direct proportion to the distance from the lens to the focal plane (i.e. focal length, assuming infinite focus), and the ratio of that size to a constant aperture stays constant across focal length. I get that.

Here's the thing, though: In real world lens design, long lenses are generally of a telephoto design, which reduces the back focus distance. Wide angle lenses are generally of retrofocus design, which increases back focus distance. Considering that light diffracts around the edges of the aperture blades, it seems far more important how far the aperture is from the focal plane, in relation to its size.

Again, in a telephoto design, the aperture would be pushed back towards the focal plane, and in a retrofocus design, the aperture would be pulled forward, relative to a simple meniscus lens model. It seems, therefore, that at a given aperture, there will be less distance for diffraction to occur in a telephoto design than for a retrofocus design, relatively speaking. Should the diffraction limit therefore be more permissive for a telephoto design than for a retrofocus design?

By way of example, consider my Sigma 12-24mm lens at f/8 and 12mm. It may have a 1.5mm aperture, but that aperture is not placed at 12mm from the focal plane, as it would if the lens were a simple meniscus lens. Rather, it's spaced quite far from the focal plane, well in excess of the 44m lens register (Canon EOS). There is therefore a much longer distance that light can diffract through this aperture. Diffraction could only be worse than for a 1.5mm dia meniscus lens at 12mm distance from the focal plane. Would diffraction limitation for a retrofocus lens not occur at a larger aperture than for a simple meniscus lens? Would the opposite not be true for a telephoto lens?

Caveats: I do realize it's the apparent and not physical aperture diameter that's important, but I have no idea how/whether this would relate to telephoto vs. retrofocus design. I also have not a clue how an aperture is constructed/moved in relation to the fixed and moving groups of a zoom lens! It makes my head hurt to think about it.

Anyway, if I am right, this would suggest that there is no simple diffraction limiting aperture for a given pixel density. Rather, the diffraction limit would depend very much on the design of the lens and would favor somewhat larger-aperture diffraction limits for wider lenses. (I do realize the diffraction limit also relates to the resolving capabilities of the lens, but I'm ignoring that for the sake of this discussion.)

I know there are some optical gurus here on this site. Can someone clarify this issue for me? Is my thinking somewhat clear/valid on this issue? Any help in resolving this issue and thus ameliorating my headache would be greatly appreciated! :-)

bkpix · July 27, 2012

Perhaps the question is best settled experimentally: i.e., take some pictures.

scott_ferris · July 27, 2012

Surely it is irrelevant what the retro-focus or tele-focus of the lens is, you are always dealing with apparent aperture and apparent focal length. This would mean the airy disc stays constant for the various apertures as they would be increased, or reduced, in size as if they were simple lenses.

JDMvW · July 27, 2012

Everytime I try to figure out why a long lens on a large format camera can be stopped down so far (e.g., f/64) with so little diffraction, I get a headache.

As the neo-fascists say, "Act, not think!"*

_________ * hmm, you suppose Yoda was a neo-fascist?

scott_ferris · July 27, 2012

JDM,

Because diffraction calculations take reproduction size into account. Your bigger sensor needs less enlargement, hence the larger format airy disc is enlarged less so you don't see it as soon.

The diffraction on the sensor/film is the same whatever size that sensor or film is, the amount you enlarge that diffraction to make a same sized reproduction, either print or on screen, is not, so the bigger the format the larger the airy disc can be before you see it in a same sized print.

dan_fromm2 · July 27, 2012

Sarah, read Emmanuel Bigler's contributions to this discussion: http://www.largeformatphotography.info/forum/showthread.php?55118-Confused-about-diffraction&highlight=emmanuel+bigler+diffraction

Emmanuel Bigler is a professor in optics and microtechnology at ENSMM, Besançon, France, an engineering college (École Nationale Supérieure d'Ingénieurs) in mechanical engineering and microtechnology . He got his Ph.D. degree from Institut d'Optique, Orsay (France).

sarah_fox · July 27, 2012

But Scott, on the example I cite (my Sigma 12-24 at 12mm), the apparent aperture "appears" to be maybe 120 or 130mm from the focal plane. If it were so, then the 1.5mm'ish aperture would reflect an aperture ratio roughly in the neighborhood of f/80. The apparent aperture doesn't appear to be 15mm in diameter -- nowhere close. So the retrofocus design is reflected quite well in the apparent size and position of the aperture.

Besides that, it's really the physical distance from the aperture to the focal plane over which diffraction scatters the light. The longer the distance, the more the scatter.

JDM, as Scott says, a larger format means a larger acceptable airy disc. It's quite probable a large format 300mm lens is going to have much less resolving power than a 300mm lens designed for a 35mm camera, for example, so diffraction limits are going to occur at a smaller aperture anyway.

jcuknz · July 27, 2012

Sarah ... ignore the whole subject as there are other factors which will spoil your image before diffraction comes into it ...not my comment but a good attitude I read from a very skilled tog.

Yes I wondered at the lack of diff when working at f/64 with a halfplate camera and it is the relative magnifcation for a given image size

JDMvW · July 27, 2012

I vass choking. Jag vorstor it. I didn't imagine people would actually try to explain, but I forgot about all the engineers (real and putative) that haunt the site. Indeed,somewhere in the archive there is an explanation, sort of, by me.... But thanks all the same. :)

However, I was not joking about the headache, but that is another matter.

scott_ferris · July 27, 2012

Sarah,

Look at the apparent aperture from both sides of your lens, to do that with EF lenses put it on the camera, power it up, select an aperture and push the dof preview button, whilst pushing that take your lens off, it will maintain the selected f stop. When you look in from the front the apparent aperture is correct, when you look from the back it is dramatically different. For my ultrawide retrofocus lenses the apparent aperture from the sensor side is much too big, but it is further away, so it evens out.

The point is, to the sensor the apparent aperture and focal length are correct, therefore the airy discs are the right size.

Dan,

That Emmanuel certainly knows his stuff, thanks for the link, but the thread was only dealing with simple lenses, not retro-focus or tele-focus.

sarah_fox · July 27, 2012

JC and JDM, the topic would be interesting to me even if I weren't a photographer. That's just my nature. ;-)

Regarding the choking... You mean Yoda isn't a neofascist character?

Dan, thanks so much for the link. The whole thread looks like interesting reading... for tomorrow when my head hurts less. Unfortunately I think Dr. Bigler is citing other reasons why longer lenses may be stopped down more than wider ones. However, I won't be sure until I dive into that thread.

dan_fromm2 · July 27, 2012

Sarah, when it comes to thinking about how far one can stop down, it helps to remember how much the image as captured will have to be enlarged to make the final print.

To make this concrete, consider a 20" x 24" negative that will be contact printed. Since the print will look sharp at normal -- ~ 12" -- viewing distance if it resolves 8 lp/mm the lens/negative system has to record no less than 8 lp/mm. This can be done at around f/200.

Now consider a 1" x 1.2" negative to be enlarged to 20" x 24". It will have to have at least 160 lp/mm to make a print that will deliver as much fine detail as a 20" x 24" contact print.

And that's why LF shooters can use smaller relative apertures that smaller format shooters. Its simple arithmetic combined with the realization that what matters is what's in the final print. The more a negative is enlarged the better it has to be.

Bill C · July 28, 2012

Hi Sarah, I have a suggestion for your dreams tonight - try to go back to these ideas: think of a single image point as the tip of a cone-of-light originating from the virtual aperture. I haven't reviewed formulas, much less worked anything out (recently, anyway), but I think the blur circle (Airy disk) will end up the same no matter where in the cone the actual aperture is placed.

It looks like you've already been here, but ruled it out based on how things "seem":

But Scott, on the example I cite (my Sigma 12-24 at 12mm), the apparent aperture "appears" to be maybe 120 or 130mm from the focal plane. If it were so, then the 1.5mm'ish aperture would reflect an aperture ratio roughly in the neighborhood of f/80. The apparent aperture doesn't appear to be 15mm in diameter -- nowhere close.

</blockquote>

I think that your visual evaluation is just way off, while the virtual distance and size are actually correct. I don't know how you're estimating the distance, I know this is very difficult with the naked eye - you really need a reference or measuring tool.

How can you verify this? I can describe two ways: one is fairly direct - measuring the optical distance and size with homemade gear, the other is indirect - since the geometrical cone-of-light is the basis for depth-of-field, one can shoot depth of field comparisons. (Your example, estimated aperture ratio about f/80), would have unmistakably large depth-of-field, right?)

I'll write a separate post on "measuring," I'll hurry so you can hopefully dream about it tonight!

Bill C · July 28, 2012

Ok, here's a way to rig up a crude telescope to locate and (roughly) measure aperture size. I'm not going too deep into the mechanical details, just the principles. However, I think someone could do this on a tabletop, using various blocks of wood to hold the pieces.

First, rig up a close-range telescope, that has a narrow depth-of-field. A pair of lenses, one with a longish focal length of about 300-400mm, the other with short focal length of about 25mm (or a strong loupe) should work. Using a thin board as a base, mount the long lens (a piece of 2x4 wood should do, if you cut a v-shape on top for the lens to lay in). Many people know that a same-size image results if the lens is two focal lengths from a target, with the image plane two focal lengths behind. (These distances are not critical my scheme, it just gets you in the ballpark so you know where the image is.)

Just to be clear about this, if you tape a ruler on the wall, set your 300 mm lens with its rough center about 600mm from the wall, an image of the ruler will be formed about 600mm away on the other side of the lens. (These will not be exact because the two pupils of the lens probably don't coincide.) Try to find the image while looking through your loupe (or short focal length lens). Mount the loupe in position; resting on another v-notched 2x4 should work. With both lenses on top of the thin board, you now have a close-up telescope that you can move around on the tabletop. I think you'll find it goes out of focus quickly if the distance changes very much, say a half inch or so; this means you can locate other things at that focus distance, including a "virtual aperture" (aka a lens pupil).

To find the a pupil distance, close down the lens aperture small enough that you can see it all through your "telescope." Move the scope forward and back (it's all on a single board) until focused. Now, mark your tabletop versus some fixed point on the telescope-holding board. Next, we want a reference that we can acually touch - the rear base of the lens, the mounting flange, will work fine. So slide your telescope back until the mount flange comes into focus. Again mark the tabletop. The distance between your two marks is equal to the distance from lens-mount flange to the virtual aperture. You still need to add on the flange-to-imager distance, I leave it to you to get this distance on your own.

At this point, you should now have an approximate optical distance from the "apparent aperture," or virtual aperture, or pupil. (It's not an actual physical distance, it's an "apparent" distance, in air, buit is IS what you want to work with. All that remains is to measure the apparent aperture diameter.

To get the diameter, you would ideally view from the image position and find the angle from one side of the aperture to the other, then calculate the angular width. However, you ought to be plenty close, for our purposes, by actually measuring the "diameter". To do this, we need a reference mark in our telescope. If you used a loupe with built-in reticle, you're already there. Otherwise, find another...you guessed it... 2x4. Put a small nail, or pin, etc., sticking straight up, so it can be seen through the eyepiece. (Note: if you used a short focal length camera lens, and can't focus on the pin, turn the lens around.)

Now, while observing the lens aperture through the telescope, move the "reticle" (the nail or pin) to match focus of the aperture. Then, move the telescope assembly so the nail/pin lines up with one edge of the aperture. Mark the table, then slide the telescope sideways, lining up the other edge of the aperture. The distance moved sideways is equal to the aperture diameter. Note: if the telescope turns even a little sideways, your measurement will be wrong, so it's a good idea to set up a guide. For example, another 2x4 turned sideways; the telescope assembly can be pressed against this while moving side to side.

Disclaimer: I've only measured apertures through the front of a lens, where this should be a legitimate method. I'm guessing that it works well enough through the rear, but I'm not for certain. (I don't think it would work for a telecentric lens.)

So, if you don't mind wasting an afternoon, and IF you can get enough precision out of this, you can confirm whether or not the virtual aperture distance divided by virtual aperture diameter always matches the f-number setting. Or you can just take my word for it - I think it will.

charles_stobbs3 · July 28, 2012

Why would a lens manufacturer bother to put unusable f stops on a product? If you've got an f stop, use it. Or is this too simplistic?

scott_ferris · July 28, 2012

Charles,

This is a casual photo conversation, I can understand it not holding any interest for the majority of people, I can't

understand people taking the time and trouble to be so uninvolved, your input is a telling reflection of you, simplistic

enough for you?

sarah_fox · July 28, 2012

Why would a lens manufacturer bother to put unusable f stops on a product? If you've got an f stop, use it. Or is this too simplistic?

</blockquote>

Actually it's much too simplistic, even for practical photography. One can take a very nice photograph at f/32, for instance, but it will be a very different sort of photograph than one taken at f/8, even ignoring differences in depth of field. I do sometimes take photographs with extremely small apertures where heavy diffraction is afoot, but I always do it for a specific reason.

Bill, while I will concede that my estimate of apparent aperture position might not be accurate, my point would be that it is not even anywhere remotely close to 6mm from the focal plane (38mm recessed from the mounting flange), at least viewed from the front of the lens. It is also extremely tiny -- not anywhere close to anything I would characterize as an f/8 entrance pupil.

That said, Scott makes a good point, that it is the apparent aperture as viewed from the rear of the lens that might be important. Scott, I realize that I can remove the lens with the aperture stopped down, so that I can view the apparent aperture from the rear of the lens, but I confess I really hate to do this with equipment I care about. So I'm going to take a pass. That said, I think I can lay my hands on a 24mm M42 Tamron to compare apparent aperture from fore and aft. Very curious about this.

Even so, even if it is the apparent aperture position/size that matters with regard to exposure, it is still the physical position of the entrance pupil from which diffraction begins, and it is the physical distance from that position to the focal plane over which the light continues to scatter. The longer the distance, the more the scatter, thus my original point/query.

I feel like a poor host, because I'm going to have to return to a day full of chores and will have to neglect this issue for today. I very much appreciate all the input so far and look forward to other ideas. I have yet to read/digest that large format thread, which might hold some interesting clues. (However, to my knowledge, retrofocus/telephoto lens designs are utilized for large format photography, and that's what my issue concerns.) More later...

Alan Marcus · July 28, 2012

Diffraction:

Light rays passing through an opening, such as a lens aperture, display a disturbed pattern we call diffraction and interference. A study of light shows that it behaves in a similar manor water and sound waves. Now it is well known that a hill or a building will not completely shadow sound because we can hear shouts and the like coming from around the obstruction. Now obstruction do block light waves but if we look carefully we find the blockage is not complete. We cannot see around an obstructions however light can be observed to fill in some of the space behind sharp edges of an obstruction by spreading around them. We are talking about the phenomena called diffraction.

This is the stuff that the French physicist Augustin Jean Fresnel studied and in 1818 published that light will indeed travel around a barrier producing a pattern of light and dark bands. Measuring the spacing of the bands, Fresnel was able to work out what the wave length that the light must have. In other words, his math worked out the mechanics light wave motion.

Now it is well known that camera lenses image by reproducing tiny points on the subject and project them as tiny points on the surface of film or digital chip. It is the size and shape of these image circles that we are taking about. In perfect optical system would be so tiny as to be immeasurable. Sorry to report, the best we can do is make tiny circles, not points and we can measure there diameter. These are called circles of confusion because under the microscope show us as a fuzzy scalloped circle.

'

One man devoted lots of time on this subject, George B. Airy, British 1801 ~ 1892, Director Cambridge Observatory. Because he enlightened us, in his honor we call these circles "Airy disc".

Under magnification, the Airy disc has a central core of light surrounded by concentric circles of light with dark bands in between. About 84% of the brightness is in the central disk, 1.75% in the first ring, 0.42% in the second, 0.16% in the third. After that the rings are too dim to be of concern. Because the Airy disk is the least possible image element, that conveys intelligence, we are concerned about their size and spacing. If too big they overlap an it becomes impossible detect separation.

Now enters Lord John Rayleigh 1842 ~ 1919 British Nobel prize physics 1904. He worked out the principles of circle size as they relate to the working diameter of the optical system. This is the stuff of resolving power and acuity. We know from the work of Airy and Rayleigh that the resolving power of a lens decreases as the aperture is reduced as causes the diameter of the Airy disc to enlarge. This is because as the aperture is reduced a greater percentage of the rays must pass near the blades of the diaphragm. Those close to the blades are not completely shadowed and that light that otherwise would be blocked bends into the image path and bleed in with the image forming rays. This causes circles of confusion to enlarge. This is the stuff diffraction.

Rayleigh famous equation define lens performance and this is called the Rayleigh Criterion. This is the measure of a lens as to its theoretical resolving power. The Rayleigh Criterion continues in use today.

Resolving power = 1392 ÷ f/number (answer in lines per millimeter)

Below is a table as computed by the Rayleigh equation for wavelength 589 millimicrons, a standard used to let know the effect of diffraction on resolving power.

f/1 = 1392 lines per mm

f/2 = 696 lines per mm

f/2.8 = 487 lines per mm

f/4 = 320 lines per mm

f/5.6 = 249 lines per mm

f/8 = 184 lines per mm

f/11 = 127 line per mm

f/16 = 87 lines per mm

f/22 = 63 lines per mm

f/32 = 44 lines per mm

The resolving power of an optical system a function of the working diameter of the system. The back focus distance is incidental. The resolving power at f/8 is higher than pictorially useful film emulsions.

More gobbledygook from Alan Marcus

Bill C · July 29, 2012

Bill, while I will concede that my estimate of apparent aperture position might not be accurate, my point would be that it is not even anywhere remotely close to 6mm from the focal plane (38mm recessed from the mounting flange), at least viewed from the front of the lens. It is also extremely tiny -- not anywhere close to anything I would characterize as an f/8 entrance pupil.

That said, Scott makes a good point, that it is the apparent aperture as viewed from the rear of the lens that might be important.

</blockquote>

Sarah, I initially thought you were trying to locate the virtual aperture position from the REAR of the lens; on rereading your first post, I now see that you made no indication of this. I erroneously interpreted it this way.

That aside, I agree with Scott; he is giving you the straight info.

I'm now doubtful that anyone sees the point of my test method, so I estimated data on a real lens for demonstration. It's a modern lens, for a 4/3 digital camera, 18mm to something, f/3.5 to something. I checked it at the 18mm focal length, wide open. If you use the f-number formula, you'd figure the "working aperture" to be 18mm/3.5 = 5.1mm. (~0.2 inch).

Examining from the rear of the lens, I estimate the "virtual aperture" diameter to be much larger - about 18mm (~0.7 inch). So it seems too large, by a factor of 3 or 4 times. But when I estimate distance from the image to the virtual aperture the distance, ~58mm (~2.3 inch), turns out to be 3 or 4 times larger than the focal length. The net effect is that the distance divided by the diameter, 58mm/18mm, turns out to be 3.2, pretty close to the spec'd f-number of 3.5. (The discrepancy is well within the limits of my very crude measurements.)

It is exactly as Scott Ferris said in his 8:15pm post yesterday. The seemingly too-large aperture is cancelled out by the seemingly too-large distance. The cone of light, from apparent aperture to an image point, has the same angle as a simple lens with same f-number. Thus, the airy disc should be same size.

ps: I didn't use the "telescope" I described; I used a digital camera. For diameter, I photographed the aperture, then, with the same focus setting, a ruler. Since both are focused the same, image sizes are the same, thus the photo of the ruler can measure the diameter of the photo of the aperture. Etc.

charles_stobbs3 · July 29, 2012

Getting less simplistic, isn't the thickness of the diaphragm also important? Doen't diffraction occur at both the object and film side edges of the diaphragm? And because the diphragm is a stack of diaphragm blades the more blades, the thicker the total package. Maybe the old fashioned Waterhouse stop had an advantage.

scott_ferris · July 29, 2012

Waterhouse stops are much thicker than modern aperture blades, with modern blades no light travels past more than two thicknesses of it, and that is only at the very small intersection point, the vast majority of the light passes only one very thin blade.

sarah_fox · July 29, 2012

Hey guys, thanks!

Scott and Bill, I pulled out that Tamron Adaptall M42 24mm (which can only have a retrofocus design) and confirmed that indeed the apparent aperture is rather large from the REAR of the lens, and with respect to its apparent position does have the correct f/ ratio relationship. I had hoped to remove the rear cell to view the diaphragm directly, but I was unsuccessful.

That said, here's what I think might be happening: Perhaps the physical aperture of a retrofocus lens is also larger than the diameter of a meniscus lens at the same focal length and f/ ratio. If so, then the diffraction would be less, despite the greater physical distance to the focal plane. Unfortunately I do not have a lens I'm able to (easily/safely) disassemble to see this relationship.

@Alan: Thanks for the gobbledygook. While it didn't address the question direction, it did give me a couple of tiny leads/perspectives.

@Charles: I don't think the thickness of the diaphragm is too important, except with regards to light scattering from reflectance off of the "sides" of the aperture blades. I believe that's the big bugaboo with pinhole photography, for instance.

There's no diffraction on the object side of the diaphragm, unless you're talking about light reflecting from the film plane back out of the lens.

The advantage of Waterhouse stops is that they are (ordinarily) perfectly round, yielding a perfectly uniform OOF pattern, particularly with respect to small specular highlights. You can also make special Waterhouse stops with different shapes to create special patterns. I've experimented with it a bit here: <a href="http://www.graphic-fusion.com/beyondbokeh.htm">http://www.graphic-fusion.com/beyondbokeh.htm</a>. What a perfectly round Waterhouse stop denies you is the ability to create starry patterns in your specular highlights -- a much better approach to starry highlights than the "star filter."

So does anyone happen to know: Does a retrofocus or telephoto design of lens preserve a "correct" ratio between the physical aperture diameter and the physical distance from the focal plane to the aperture. If so, that would pretty much answer my question, confirming how a retrofocus lens would not be diffraction limited prior to a normal lens, prior to a telephoto lens, all other things being equal.

sarah_fox · July 30, 2012

Apples to apples:

This morning I realized something. My 24mm Tamron is also a very myopic telephoto lens when flipped around backwards. Thus I have a retrofocus and telephoto lens of exactly the same design which I can compare to itself!

Very roughly speaking, the apparent apertures have about a 2:1 difference in diameter when viewed from their respective sides. This corresponds to roughly a 2:1 difference in back focus distance from the apparent aperture positions, thus preserving apparent aperture ratio. Interesting!

Here's the thing, though: When focused at infinity (if I could even get the lens focused to infinity in the telephoto orientation), the physical aperture, which is exactly the same diameter in either orientation, is much closer to the focal plane in the telephoto orientation than in the retrofocus orientation. (Note: I can view the mechanism directly, sans optics, by removing the Adaptall mount, so I'm certain of its physical location.) Because of this, I must still conclude that a retrofocus design puts a lens at a larger-aperture diffraction limitation than a telephoto lens, all other things being equal -- for the express reason that the same physical aperture is located farther from the focal plane, so that light can take a greater distance to scatter from diffraction.

Looking at this in an even more refined manner, in the telephoto orientation, the image is dimmer than the image formed in the retrofocus orientation, which makes sense from the perspective of the apparent apertures. If the larger apparent aperture of the retrofocus orientation is stopped down roughly two-fold in the retrofocus orientation, the images are of roughly equivalent brightness, and the apparent aperture ratios (apparent dia to apparent distance from focal plane) are roughly equal. Comparing that way, the difference between retrofocus and telephoto orientations becomes even greater. In the telephoto orientation, the physical aperture is roughly twice the diameter AND roughly half the distance to the focal plane (to achieve the same apparent ratio). Thus there would be roughly a 4-fold advantage in the telephoto orientation with respect to diffraction limitations -- e.g. the difference between f/4 and f/16.

I do believe that telephoto lenses are, in general, more permissive with regard to diffraction limitations, but I don't think I would characterize this as a 4-fold difference! When I play with the interactive blur charts on <a href="http://www.dslr.com">www.dslr.com</a>, I notice that zoom lenses often see a slight advantage towards the telephoto end in diffraction limitation, but this difference is only maybe a stop. And again, I have very little understanding of the mysteries of a zoom lens design!

Bill C · July 30, 2012

I do believe that telephoto lenses are, in general, more permissive with regard to diffraction limitations

</blockquote>

Sarah, I'm doubtful about this. Actually, I don't believe it at all. Do you have any argument to support this? (I tried your link, it says, "Buy this domain name..." and has a bunch of external links.)

My arguments have mostly been said. I think the diffraction effects can be based on the cone-of-light from the virtual aperture at rear of lens (the exit pupil) to the image. I don't know of any reason to doubt this.

In essence, I think the lens, however complicated, can be viewed simply as a black box. Examine it from the outside, and however it "seems to be" is how it functions. I might be wrong about these things, but I'd be willing to bet quite a lot of beer on it (I'd give you good odds, too).

ps: I don't think the physical diameter/position of the aperture are important in these issues.

pps: It seems odd that reversing a lens would change its effective aperture, but if the pupils are different sizes, then it must be true! I've not thought this through, so I'm not sure one way or the other. Still, nothing should be gained diffraction-wise - if the entrance pupil is larger, this is a wider aperture and vice versa. Diffraction must be considered based on the effective aperture in use.

scott_ferris · July 30, 2012

The physical aperture is irrelevant, just as the initial focal length of the primary optic is. It is what the glass does to convert that to an apparent aperture, or focal length, that is important.

In a simple lens a long focus f4 and a wide angle f4 have the same diffraction limit because the larger opening of the tele f4 is further away than the wide angle one, hence the airy discs are the same size. In a retrofocus design all the designers are doing is emulating a longer focal length with a wider apparent aperture that also appears further away than it should.

A telephoto lens does the same thing in reverse, the tele lens is shorter than it should be because the focal length of the primary optic is not long enough, a group of lens elements divert the light path and the apparent aperture to give the effect of a simple lens with a longer focal length.

It is all about where the aperture appears to be, and its size, with relation to the sensor, the physical position and size are not relevant if there are lens elements between the aperture and the imaging plane that distort the apparent positions.

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Diffraction -- advanced confusion!

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