diffraction limiting?

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How does one determine what is the smallest aperture/(largest f#) one can use without 'problems'?

Is it dependent upon factors like format and degree of enlargement like choosing a circle of confusion number for a given format?

I want to insert a stop between cells in a large barrel lens to allow practical exposure times with no shutter, and don't want to go so far as to impact the quality (at least not drastically).

Thanks
 

Loose Gravel

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Not that simple. The smaller the aperature, the more diffraction. The trade is diffraction vs depth of focus. If you are using an older lens, then you trade resolution vs f/number, too. If the wind is blowing, you have to add in the blur due to movement vs all this other stuff. When all is said and done, you might just pick something about f/22 to f/45, depending on your format and enlargement ratio.

Try posting the question: "What's your favorite f/number?"
 

gma

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I asked the same question a while back and there was not any definitive answer. Opinion was that there is no absolute diameter, but that it varies with the focal length, type of lens, shape and location of diaphragm, etc.,etc.
I suggest that you will start to lose image quality when the diameter of the diapragm gets to be less than about 3 or 4 mm. Since you are into experimenting, maybe you can make some test negatives using standard line charts with different diameter diaphragms and report the results in the experimental gallery.
 

Loose Gravel

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Maybe I should rephrase...

It is hard to determine this from a scientific point unless you answer many questions, but from a practical point, set it for f/45 and take pictures. Diffraction does not all-of-a-sudden ruin your photograph. Diffraction is a gradual worsening of the image that slowly overtakes all the other 'problems'.
 

clay

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Loose Gravel said:
Maybe I should rephrase...

It is hard to determine this from a scientific point unless you answer many questions, but from a practical point, set it for f/45 and take pictures. Diffraction does not all-of-a-sudden ruin your photograph. Diffraction is a gradual worsening of the image that slowly overtakes all the other 'problems'.

I shoot the 355 G-claron regularly at f/64 and f/90 for contact printed 12x20's and 7x17's. Frankly, if there is any diffraction problem, it is not evident on the prints. I agree with Loose G, just relax and go.

But if you are really interested, just remember that diffraction is related to the absolute physical aperture size, NOT the relative aperture (ie f/numbers) For example f/45 on a 300mm lens is the same size opening as f/22 on a 150mm lens, and f/11 on a 75mm lens, and in each case would have the same theoretical diffraction effects with all other factors (lens type, coating, etc) being held the same. That's why ULF people can get away with shooting at such small f stops. (That and contact printing! :^))
 
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You guys with these funny little cameras... :smile:

Jorge O
 
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Great - I needed to hear 'real world answers'. I tend to be an 'armchair' photographer. I have so many projects going at once nothing gets finished so I rely on reading and planning, etc.

Thank you.

It will be some time before I get results but this gives me the perspective to inch forward with a couple ideas.

Murray
 

gma

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Consider this to complicate the answer even more. The pinhole camera has no glass lens at all. Some "precision pinholes" are amazingly sharp (in a contact print at least). The pinhole relies solely on diffraction to make the image. Without the effects of diffraction there would be no image in that type of camera. Glass lenses work by refraction.
 

sanking

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clay said:
I shoot the 355 G-claron regularly at f/64 and f/90 for contact printed 12x20's and 7x17's. Frankly, if there is any diffraction problem, it is not evident on the prints. I agree with Loose G, just relax and go.

For contact printing diffraction should never be a problem.

Diffraction limited resolution varies according to color of light but can be approximated by the following formula.

R = 1500-1800/f-N, with R resolution in lppm, f = aperture of lens. The figure varies from 1500 - 1800 because of different resolution with color of light.

For example, a 355 G-Claron used at an aperture of f/90 would have a diffraction limited resolution of between 17 - 20 lppm, depending on color of light. This is considered more than adequate for contact printing since the human eye is said to have maximum resolution of about 15 lppm at a viewing disance of 10 inches.

For projection printing the limits are more stringent since the negative will most likely be enlarged. For example, a 2X magnification of a 5X7 negative made with a lens aperture of f/90 would result in an effective resolution of only 8-10 lppm on the print, less than the maximum resolution of *some* people at a viewing distance of 10 inches.

Sandy
 

chrisg

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clay said:
But if you are really interested, just remember that diffraction is related to the absolute physical aperture size, NOT the relative aperture (ie f/numbers)

Actually, that's not correct. F-number is the relevant figure. Oversimplifying a little here, but a diffraction limited lens can focus parallel rays of light down to a spot with diameter = 2.4 x (wavelength of light) x (f-number). For the sake of coming up with a useful number, say the wavelength of visible light = 500 nm (0.5 um), so the minimum spot size will be about 1.2 um x f-number. Suppose you wanted at least 50 line pairs per mm on your neg? That's 20 um per line pair or 10 um per line, so you could aperture down to about f/8 before diffraction would prevent you from achieving your desired resolution. That's a 'factor of two' kind of estimate. In practice, the resolution of most lenses is limited by something other than diffraction at f/8. (I believe silver particles on the order of 10 um to a few 10's of um, so 50 line pairs/mm is about as high a resolution as you'd get.)

I think the human eye can perceive 5-10 line pairs/mm. (I suspect that's under ideal conditions when you're standing extremely close to what you're looking at. Someone may correct me on that.) If you're exposing negatives for contact printing, then I might expect to perceive a diffraction effect somewhere in the range of f/45 to f/90 if I was looking real hard for it. If you expect people to view the print from at least a few feet off, you can probably go considerably slower than that.

Chris
 

Ed Sukach

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Diffraction..

I've been searching ever since this topic surfaced.

I remember that there WAS a "practical" formula: that the parameters involved were the physical diameter of the aperture; the aperture-to-film distance; of course, the wavelength of the light (usually, 550 nM - nanometers - I think... I still think in terms of Angstoms...) and the number "1500". Now, If I could only remember how the @$# these went together...

It is really not a concern with factory lenses - the manufacturers are keenly aware of diffraction - and rarely will they produce a lens with an aperture small enough to make a material difference. I do recognize the problem in the installation of a custom diaphram to a large format lens.

I'm plying through my old notebooks - sooner or later - I'll find it..
 

JohnArs

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Hi

I never have ruined a picture true diffraction bud many times in the past with to short DOF! As a very easy and good rule I try if possible stop a lens with the smallest f stop of f64 only down to f 45 on 4x5 inch with 8x10 I stop fully down if needed with no problems!
If I remember correctly the pepper picture of Edward Weston was done with f 128 do you think it is a bad picture?
 

Roger Krueger

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JohnArs said:
If I remember correctly the pepper picture of Edward Weston was done with f 128 do you think it is a bad picture?

It's a good contact print. Even AA commented that much of EW's work wasn't sharp enough to enlarge.

Clay said:
But if you are really interested, just remember that diffraction is related to the absolute physical aperture size, NOT the relative aperture (ie f/numbers)

ChrisG is right, Clay's wrong. Absolute aperture determines how much the light is bent by diffraction, but distance traveled determines how much it spreads--which neatly reduces to f-stop.

Circle-of-confusion on film is one of my pet peeves--it's an absolutely meaningless number. CoC on your final intended print is what matters. f128 for a contact print looks great; f128 on 4x5 enlarged to 40x50 looks pretty sad.

I firmly believe that our ability see print resolution is at or below the low end of the oft-quoted 5-10 lp/mm range, BUT there is a visible difference between a 5 lp/mm print and a 10 lp/mm print--the MTF of 2-3 lp/mm detail.
 

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Most (if not all!) of what you need to know can be found either here or in one of the articles referenced.
Degree of enlargement and viewing distance of the final print are two major factors you need to determine/decide for yourself and then follow either the equations or tables...

Really helped me a lot in understanding the principles. For now, I'm happy to follow recommendations I understand. Perhaps one day I'll take a series of shots using smaller and smaller apertures and see for myself to what extent diffraction comes into play in perceived sharpness of the prints and how much blur I can tolerate. (BTW, once you've read the link this will make more sense : I decided to use a CoC of .05 - .075 mm for my 4x5 since I like to blow them up and look closely, but a more common one is 0.1mm)

Good luck and good light

Andy
 

felipemorgan

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Just a quick note to add to this discussion: some late 1800's and early 1900's lenses used the Uniform System of aperture measurement. See: <http://www.sizes.com/tools/uniform_system.htm>

U.S. 128 in this system corresponds to the f45 in the current system. Edward Weston in his earlier days used a lens marked in the U.S. style and thus there are references to him using an aperture of 128 or smaller (even a pinhole aperture he made himself out of paper, IIRC). I am basing my comments on a descriptive note added by the editor in the very back of his "Daybooks"; it describes some of his working methods. Although EW's "Pepper" photograph may have been exposed at f128, beware the ambiguity in the apocryphal information surround EW's photographs.

--Philip.
 
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Helen B

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gma said:
Consider this to complicate the answer even more. The pinhole camera has no glass lens at all. Some "precision pinholes" are amazingly sharp (in a contact print at least). The pinhole relies solely on diffraction to make the image. Without the effects of diffraction there would be no image in that type of camera. Glass lenses work by refraction.

Are you sure about that? The image is formed by the rays that pass straight through the pinhole. Diffraction lowers the resolution of pinhole cameras, just as it does with glass lenses.

Best,
Helen
 

Ed Sukach

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Helen B said:
Are you sure about that? The image is formed by the rays that pass straight through the pinhole. Diffraction lowers the resolution of pinhole cameras, just as it does with glass lenses.
Best,
Helen

Not ALL of the rays "go straight."

Some are deflected by passing near to an edge ... a phenomenon known as "diffraction." Without that "bending", no image would form.
There is a "best" diameter where resolution will be greatest, and theoretically the image will degrade after a certain decrease in pinhole diameter, but, by then, exposure time will be so long as not to be practical.
 

Helen B

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Am I the only one here to say that a pinhole creates an image from the straight rays (geometric optics), and that the diffracted rays (diffraction effects) do nothing other than degrade the image? That is why there is an optimum pinhole size: the balance between geometric sharpness against diffraction unsharpness - both of which increase with decreasing hole size. That optimum point is usually considered to be when the diameter of the hole equals about 0.035 times the square root of the hole-to-film distance.

There are diffractive optical elements (DOE - ie optical elements that form an image by diffraction, usually in combination with refractive elements) but they are rather more complex than a simple pinhole, and they are usually formed on glass or crystalline surfaces.

Best, Helen
 

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OT "The sun in the church" is a fascinating book about using cathedrals as solar observatories - by making a small hole in the top of the tower and following the disc of light on the floor, they could track the calendar quite precisely (to ensure Easter and Christmas were celebrated on the right day, etc.) and it was during these efforts that diffraction's effects were first observed, around 1650! If the hole was too small, the edge of the disk was fuzzy, so they enlarged it until it was "acceptably" sharp. Interesting piece of trivia ;-)
Andy
 

Ed Sukach

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Helen B said:
Am I the only one here to say that a pinhole creates an image from the straight rays (geometric optics), and that the diffracted rays (diffraction effects) do nothing other than degrade the image?

I wish there was a way to describe what happens in the confines of APUG .... I tried - I returned to ""What is Light", by A.C.S. van Heel and C.H.F. Velzel -- the text used in *one* of my optics classes.
It was not an easy read - We had wonderful - and necessary - instructors.

It should be remembered that light can be cosideed as a wave form, and that the laws of "propagation of waves" are signicantly applicable ... and ... and ...

Oh, that I had a magic wand....

I'll copy some of the text...

"18 Airy's disc

One can generally limit the abberations of a lens by using only the central portion of its surface. The smaller piece of the wavefront that then reaches the image side will then more closely approach a spherical shape. One can achieve that objective by grinding away the outer edge of the lens, but also simply by putting in front of it a screen with a hole, a diaphram. Unfortunately, the image now becomes less bright; the lens element transmits a smaller fraction of the spherical wavefront emitted by the object.
But if we take greater care, it becomes apparent that this solution is also not complete. If we make the diaphram smaller and smaller, then to our amazement, we see that the image spot becomes larger again. Careful study shows that there is a disc surrounded by weaker concentric rings, called Airy's disc, after its discoverer. This image is similar to the already familiar image with spherical abberations, but is more regular...."

I won't continue... this is only the very beginning to the introduction of the sections on diffraction - and considering the material these guys are dealing with, they are *masters* at concise presentation.

One last try -- If the image was formed *only* by the "straight rays" with no bending to different destinations, the image would be nearly the size of the aperture. There is no question that it is not, at "pinhole aperture sizes, it will cover a considerably larger area.
 

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Ed Sukach said:
One last try -- If the image was formed *only* by the "straight rays" with no bending to different destinations, the image would be nearly the size of the aperture. There is no question that it is not, at "pinhole aperture sizes, it will cover a considerably larger area.

Sorry Ed, but you'll have to continue. I've always thought that if it weren't for diffraction, the resolution on film would be equal to the size of the pinhole - all other things being ideal (perfect film, perfect hole, infinity distance an all that). So that the CoC=hole size.

And that diffraction (in a simple hole) would only serve to reduce this resolution, making the perfect pinhole size a compromise between the decreased hole size and the increased resolution.
Since there is no glass in a pinhole, spherical resolution doesn't enter into the equation at all. Chromatic aberration is obvious, as the effect of diffraction increases with decreasing wavelength. But not quite the same as in lens systems...
 

Helen B

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Ed,

I'm not sure how far to get into this - up to now I admit that I have not been particularly careful and complete in my postings because I didn't expect that a refutation of the concept that there would be no image if there was no diffraction (my main reason for wading in) would cause this depth of reply, nor did I expect that anybody would be all that interested!

Briefly then, using the text you quote:

Ed Sukach said:
...If we make the diaphram smaller and smaller, then to our amazement, we see that the image spot becomes larger again. Careful study shows that there is a disc surrounded by weaker concentric rings, called Airy's disc, after its discoverer. This image is similar to the already familiar image with spherical abberations, but is more regular...."

"The image spot becomes larger again." Exactly! The circle formed by a point of light becomes larger because of diffraction. Airy's disc refers to the image (actually an interference pattern) made by a point source, so diffraction turns a point source into an enlarged point surrounded by rings. That's why it reduces resolution. In fact the Airy pattern is also known as the Point Spread Function (PSF).

The optimum pinhole size pretty much coincides with the diameter at which the geometric image of a point source is nearly the same diameter as the first dark ring of the Airy disc. This is not a surprise. It does mean that the image formed by an optimum pinhole can be adequately described in terms of Fraunhofer diffraction only. This is not the same as saying that the image would not exist if it weren't for diffraction. The 'diffraction only' explanation would not work if the pinhole was significantly larger. Experience shows that large pinholes do work - they just produce low resolution images. I guess that that is the main weakness I see in the 'image formed by diffraction only' argument.

Ed Sukach said:
...One last try -- If the image was formed *only* by the "straight rays" with no bending to different destinations, the image would be nearly the size of the aperture. There is no question that it is not, at "pinhole aperture sizes, it will cover a considerably larger area.


I don't agree that the image circle would be the size of the pinhole if it weren't for diffraction. Oblique rays can come though the pinhole, and anyone who has made a pinhole camera will know that the size of the image circle is highly dependent on the thickness of the material with the hole in it - which, along with the hole diameter, determines the cut-off angle. However, I would be very interested to learn of image circles that exceed the theoretical diameter calculated by geometric optics, and I admit to never having looked into that aspect of the pinhole cameras that I have made.

Anyway Ed, thanks for an interesting and most intelligent explanation of a viewpoint that I had not come across before.

Best,
Helen
 

glbeas

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Helen is right about the pinhole imaging. It has nothing to do with diffraction. Visualise this...You are in a dark room with a very small window centered on the opposite wall. From the position you are at you can see a blob of light through it that is part of something. Move over some and you see another part. As you move around you can eventually see all of what is out there but in pieces. The classic description of how a pinhole camera works is similar. A ray of light bounces of the surface of an object and is partially absorbed. The remainder happens to enter the pinhole and its trajectory lands it on a certain spot on the film. Another ray hit adjacent to it and is altered and its trajectory lands it next to the previous rays landing site. And so on. Each ray has its own brightness and color and direction which when entering the pinhole detemines where on the film that spot of light hits and how bright and what color it is. They all merge like pixels on a dye transfer print into a recognizeable image in the camera and the film records it..
 

Tom Hoskinson

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I have been lurking here, watching this thread. Here is some pertinent material I abstracted from a paper by Kazuo Sayanagi.

This information is from a paper titled: Pinhole Imagery, by Kazuo Sayanagi, published in September 1967 by the Journal Of The Optical Society Of America, Vol. 57, No. 9 1091-1099. In this paper, Sayanagi presents a modulation transfer function approach to calculating optimum pinhole diameters for various different conditions.

From Sayangi’s paper:

“The pinhole has been used as an imaging device for centuries. The oldest description of pinhole imagery I could find was written by the Arabian scholar Ibn Al-Haithan (A.D. 965-1038). He described the use of a pinhole to observe the projected image of a solar eclipse. He pointed out that the finest imaging with the pinhole can be obtained by using a very small hole and discussed the image quality when the size and shape of the hole are changed, as follows:”

“The image of the sun only shows this (crescent shape) property when the hole is very small. If the hole is larger the image changes, and the change is more marked with increasing size of the hole. If the hole is very large, the crescent shape of the image disappears altogether, and the light becomes round if the hole is round, quadrangular if it is quadrangular, and with any shaped opening you like, the image takes the same shape, always provided the hole is large and the receiving surface parallel to it.”

The first physical consideration of pinhole imagery, based on a mixed treatment of geometrical and physical optics, was written by Petzval (1857, 1859). He expressed the diameter D of the image point made by the pinhole as the sum of the geometrical diameter of the aperture d and the diameter of the diffraction pattern caused by the aperture d,

D = d + k l lambda / d

Where k is an optimal condition constant (Petzval chose k=2), l is the distance between the pinhole and the receiving plane and lambda is the wavelength of the incident light. The optimum diameter of the hole is defined so as to give the minimum diameter of the point image.

By solving the resulting partial differential equations, Petzval obtained:
d squared = k l lambda, with k = 2, where d is the optimum diameter of the pinhole.

Based on Fraunhofer diffraction in the presence of defocus, Lord Rayleigh’s analysis (1891) arrived at a k value of 3.6

Sayanagi (1967) used a modulation transfer function approach to calculate a general-purpose k value of 3.8

These approaches all assume an infinitely thin pinhole.

Equations don't work well in this text editor, Sean.
 

Ed Sukach

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I was trying to say that the study of light - and diffraction was *far* more complex than the scope of APUG would allow.
It is *not* as simple as "The image is only formed by the straight rays..."

I'll suggest the book ... that I quoted verbatim et literatim:

A.C.S. van Heel
C.H.F. Velzel

What is Light?

Translated from the Dutch
by J.L.J. Rosenfeld

World University Library
McGraw-Hill Book Company
New York Toronto

(c) mrs. H.G. van Heel and C.H.F. Velzel 1968
Translation (c) George Weidenfeld and Nicolson Limited 1968
Library of Congress Catalog Card Number 67-24448
Phototyped by BAS Printers Limited, Wallop, Hampshire, England
Printed by Officine Grafiche Arnoldo Mondavori, Verona, Italy.

Another sample (struck with a sudden masochistic urge fro typing:

"Until now we have used as a model for the propagation of light the principles of Huygens, which we define as follows. One can imagine a wave front to originate out of the previous one by supporting each point in the latter to be a secondary source of spherical waves. The envelope of these spherical waves forms the new wave front. Notice that here, in contrast to our previous formulation of Huygens' principle, we are talking in the language of waves (italics mine -E.S.). But we have in the meantime discovered that waves show interference. Fresnel's addition to Huygens' principle consists, as we shall see in the following, in this, that he takes account of these interference phenomena. We should add a word about the range of validity of Fresnel's theory. The electromagnetic theory of light, which we shall describe later on, explains diffraction effects in an unforced manner and is, on deeper examination more precise. But Fresnel's theory of diffraction does, in most instances give the correct explanation even of details, and for this reason is considered by opticians to be a good working tool."

OK, class ... close your books. Time for a pop quiz...

Hmmm... another passage:
"This can be described as follows. It is unfortunatey a somewhat long and complicated tale, but it cannot be shortened and is most easily understood with the aid of a recipe, Think of a sphere about P which just touches V at Q and call the distance QP, l. Imagine further other spheres with P as centre whose radii are respectively ..."

No more ... Is Jim beam a good remedy for typer's (and brain) cramps?
 
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