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05-09-2012, 08:05 AM   #16
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I wanted to see the effect being discussed on this thread.
So I drilled a black plate 1.9 mm hole and put a 630 nanometre (red) led behind
https://www.box.com/s/0b1eb759d2def45fcb50

Using M4/3 (sorry no Q here) and the SMC Pentax -M 1:4 100mm macro and the Pentax Bellows M at 127 mm
M4/3 width is 17.3 mm and 4032 pixels for 4.29 micrometre per, that is about 7 wavelengths per pixel at 1:1.
The 1.9 mm hole is about 630 pixels wide corresponding to 1.5 mag which is about right per Herbert Keppler.
Stopped the lens down, spot metered, shutter to 0.0 in each case and focussed as sharply as possible.
f/8
https://www.box.com/s/8167d7cdc1e57a11d186
f/32
https://www.box.com/s/2410420baee5f88c4322

05-09-2012, 08:10 AM   #17
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Thanks for another interesting thread dudes, it's a question, I've thought about in the field, but never at my computer, it's great to get some of your thoughts on the question.
05-09-2012, 05:56 PM   #18
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"I wanted to see the effect being discussed on this thread.
So I . . . .
Using . . . . the SMC Pentax -M 1:4 100mm macro and the Pentax Bellows M at 127 mm
M4/3 width is 17.3 mm and 4032 pixels for 4.29 micrometre per, that is about 7 wavelengths per pixel at 1:1 . . . . "

Thanks wombat2go for an interesting and illuminating post with not too much technical mumbo-jumbo for a tyro like me. The difference between the f/8 and f/22 shots is spectacular!

I accept your calculations on trust but just two questions, if I may be allowed these:-

1. Is the lens a "bellows lens" i.e. having no focussing travel provided for in its own mount?

2. Does the degree of magnification actually affect the relativity of wavelength to pixel size? I notice you say ". . . 7 wavelengths per pixel at 1:1." [my stress added]

Surely the pixel is a physical thing and its size stays the same regardless. If the wavelength of the light alters wouldn't that alter the colour of the light?
05-09-2012, 07:08 PM   #19
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Good thread, Baron, and posts by others; it has prompted me to revisit the old 1968 engineering physics book about limits of resolution in analytical diffraction terms.
I had been vaguely aware that a too small diaphragm was no good and today i re-learned why.
Kendrick, for your questions:
1.The lens was the SMC Pentax 1:4 100mm macro (actually dental macro) which is a conventional lens with a focussing ring.
The lens focus adjustment is not relevant for bellows use (I am learning how to use the bellows)
First set the magnification by moving the camera and lens racks apart.
Then focus by using the lower rack which moves the camera and lens racks in unison with respect to the subject.

2. Yes, at the magnification i used (1.5) there were less than 5 wavelengths per pixel. I just wanted to check the pixel resolution compared to our typical lenses.
I suppose that does not matter, because as pixel density increases, we will eventually see the minima/maxima fringes, per pixel, of the lenses, which will be averaged for the viewer.

If we increase the magnification (extension tube length) or reduce the diaphragm, the pixel count does not matter because the limitations are all in the lens at f/32 in this test.

I was particularly interested to see the rough drilling swarf on the hole. Fairly clearly resolved at f/8, but at f/32 the lens was "blind" to the small details which is why astromomers need big apertures.

Your last comment is interesting, I was going to do the test with a tungsten filament light as was done in past ages, (they used 555 nanometre) but woke up remembering I had some red leds on hand. In such a test, it is easier to see the effects with a light of a limited frequency range.
Your comment about color and waveform is correct. I suspect the blurring at f/32 would be somewhat worse if we had sunlight shining through the hole.

05-09-2012, 09:23 PM   #20
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It is a good thread for me because of all the knowledge it holds. I know now that using a teleconerter with the Q and the DA*300 is doomed to failure. The only one that has any chance is the 1.5X. Even it is very doubtful.

My Olympus Om 350 f2.8 seems a better choice. With the matched Olympus converter it is a 500 f4. Alas, a problem different than diffraction appears. The lens shows ca and fringing. The Q is like a microscope on this. The teleconverter makes it worse. On a m4/3 body the Om out performs the Pentax 300. Hard to beat the resolving power of that big front optic. On the Q it reverses and the Pentax 300 wins. It has the least color fringing of any tele so far. Need a good program to reduce fringing in editing. More investigation required.

Thanks to Falconeye for the answers and Supplying the equations asked for in the original post. I now have a better understanding of the principles involved. This helps formulate a game plan.

Diffraction is important. Have a friend that went from 35mm film to aps-c digital to FF. He was amazed at how much sharper FF was than aps-c. I saw the prints and there is no way there should be that big of a difference. Found out he was a landscape photographer used to shooting film at f 16-22. Doesn't work with aps-c. Worked better with FF.
thanks
barondla
05-10-2012, 08:49 AM   #21
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QuoteOriginally posted by Kendrick Pereira Quote
So what are you disagreeing with?
I think Adam wanted to express that a TC deteriotes image quality enough even without taking diffraction effects into account.

And to a point, he's right because image quality at the usual apertures like effective f/5.6 or f/8 is not high indeed.

Nevertheless, I disagreed. Because w/o diffraction, you could stop down to an effective aperture like f/16 (f/9.4 + 1.7xTC) and get a good image quality because as you may know, stopping down minimizes most classical aberration issues.

But it is because of diffraction you can't. So, diffraction and classical deteriotion can't be separated, the sweet spot is where both effects are about equal and therefore you can't consider one effect while ignoring the other.

I hoped I made that clear.
05-10-2012, 10:27 AM   #22
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QuoteOriginally posted by barondla Quote
If we are aiming for f4 or 5.6 and use a lens that is f4 with a 1.4 teleconverter do we get the diffraction of an f4 or f5.6 lens? How about extension tubes in macro?
thanks
barondla
the amount of loss of sharpness from diffraction is a function of the percentage of aperture length to aperture open area. I say this because diffraction impacts a small defined width area at the edge of the aperture blade, and therefore the length is the only determining variable.

Since length of a circle is proportional to radius, but area of the opening is proportional to the square of the radius, the shorter the focal length at a given F Stop the more diffraction impacts the image.

Now, when we add a TC, we add this behind the entire lens and aperture, therefore a TC does not impact diffraction, because the diffraction is in the main lens not the TC, However, the TC amplifies the distortion because it amplifies the entire image from the main lens.

but for any lens, the physical aperture let's say 50mmF4 is the same diameter than the aperture of a 100mmF8 therefore the diffraction distortion of the two would be the same, in percentage, but the 50 F4 plus a 2x TC would be worse, because the TC amplifies the distortion as well as the image.
05-10-2012, 11:43 AM   #23
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Hi Lowell,
Diffraction on our camera lenses is not an edge effect, it is just a function of diameter.
Angular resolution - Wikipedia, the free encyclopedia
The loss of resolution occurs right across the area not just at the edges.

The angular limit of resolution of any round aperture is approximately alpha= 0.61 *lambda/diameter
where alpha = size/distance and Lambda is wavelength say 700nm
So a 50mm at f/11should be able to just resolve a 10mm leaf at 100 metres distance any further stopdown being blurred.

I made a rough little function from the alpha for my lenses for general use
fstop = FocalLenght*Size/(427e-9*Distance)
(all dimensions in metre eg FocalLength is 50e-3 m)

so for my 28mm to resolve 10mm leaf at 100 metre:
fstop = 28e-3 * 10e-3 /( 427e-9 * 100)
fstop = 6.5 max.
Any fstop higher would be blurred

Here is the tree at 100 metres away using the SMC Pentax 1:2.8 28mm
f/6.4
https://www.box.com/s/22849dae83b6db39bd10
f/22
https://www.box.com/s/d129451953668988ca04

It is worth mentioning that this is physics, the same loss of sharpness applies to the most humble or the most costly 28 mm lenses!

05-10-2012, 12:02 PM   #24
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QuoteOriginally posted by wombat2go Quote
Hi Lowell,
Diffraction on our camera lenses is not an edge effect, it is just a function of diameter.
Angular resolution - Wikipedia, the free encyclopedia
The loss of resolution occurs right across the area not just at the edges.

The angular limit of resolution of any round aperture is approximately alpha= 0.61 *lambda/diameter
where alpha = size/distance and Lambda is wavelength say 700nm
So a 50mm at f/11should be able to just resolve a 10mm leaf at 100 metres distance any further stopdown being blurred.

I made a rough little function from the alpha for my lenses for general use
fstop = FocalLenght*Size/(427e-9*Distance)
(all dimensions in metre eg FocalLength is 50e-3 m)

so for my 28mm to resolve 10mm leaf at 100 metre:
fstop = 28e-3 * 10e-3 /( 427e-9 * 100)
fstop = 6.5 max.
Any fstop higher would be blurred

Here is the tree at 100 metres away using the SMC Pentax 1:2.8 28mm
f/6.4
https://www.box.com/s/22849dae83b6db39bd10
f/22
https://www.box.com/s/d129451953668988ca04

It is worth mentioning that this is physics, the same loss of sharpness applies to the most humble or the most costly 28 mm lenses!
by an edge effect, what I mean is that when you consider the opening made by the lens and aperture, the diffraction takes place only at the edge of the aperture, not in the middle of the lens far away from the aperture. of course, since light from every part of the subject interacts with any point on the lens, including any and all points on the edge of the aperture blade, the impact is spread across the whole image, BUT this impact is proportional to the ratio of the light hitting the edge area of the aperture to the entire un-impacted area of the open lens.

that is why i consider it related to the ratio of impacted to non impacted area. a little blurr in 1% of the light passing through will not impact sharpness greatly, because it has little impact on exposure in general, but as the percentage increases, the area of the lens opening subject to difraction becomes more and more dominant in the overall image, leading to reduced sharpness
05-11-2012, 01:01 AM   #25
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everybody may think diffraction is an edge effect.

In reality, it is an interference effect in wave-optical terms or an uncertainty effect in particle terms, whatever you prefer. Both are caused by constraining the photon's spatial extent when passing a lens, lens elements as well as aperture blades. In wave-optical terms, it is an aberration caused by removing (blocking) parts of an incoming perfect wavefront, making it imperfect. You could make the block's border arbitrarily smooth (like some SFT lenses do) and still end up with the same diffraction magnitude. The more you remove from the incoming perfect wave front, the poorer the remaining part of it will behave when it comes to make a good light spot on the focus plane.

The following image from wikipedia which I cite in my image sharpness white paper may help you understand better



You may be able to see in an intuitive manner from this image that the entire lens, incl. all lens surfaces contribute to the image of a point source. There is no blur coming from the edges ...

Moreover, it isn't intuitive to consider a TC to be a part magnifying the diffraction coming from the "lens". Rather, in wave-optical terms, a TC is just altering a lens optical formula, i.e., its focal length and thereby altering it's f-stop number which determines the magnitude of diffraction effects.

Last edited by falconeye; 05-11-2012 at 01:23 AM.
05-11-2012, 05:33 AM   #26
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Falconeye, my schmidt-cassegrain telescope has a telecompressor that screws into the back of the scope and optically converts it from an f10 to an f6.3. The scope then has a wider field of view and projects a smaller image circle ( barely covers 35mm ). So this would reduce diffraction- correct? Wish there was one of these for the Q ( and 4/3). Camera people borrowed the teleconverter from the telescope makers years ago.
thanks
barondla -enjoying the technical discussion
05-11-2012, 07:10 AM   #27
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Falconeye I am not sure your explanation really helps, which is why I tried to make it simple. Yes, diffraction is an interference pattern caused by the light passing the edge blocking the perfect passage, What I was trying to stress, is that when you consider the impact, relative to the size of the hole and total area of the hole, the smaller the portion of perfect passage, relative to the portion creating the disturbance,

Since the perfect passage is the area with no obstruction, and the imperfect passage is related to an area surrounding the parimeter of the lens as the hole gets smaller the disturbance of the edge impacts the overall image to a larger extent. that is all
05-11-2012, 08:21 AM   #28
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Lowell, My reading is it is
Light off all points arrives as a diffraction pattern minima/maxima at wavelength spacing, regardless of aperture edge obstructions or not.
It is not an edge effect.
The image is an addition of all the patterns
Ability to resolve is just related to the the angle and the opening: in the limit:angle=0.61*lambda/dia
The 0.61 comes from the bessel function of the roundness, being 1.0 for a plane wave on an wide slit.

We probably deserve to get hosed for such a deep discussion of it here!
05-13-2012, 04:03 PM   #29
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QuoteOriginally posted by Lowell Goudge Quote
Falconeye I am not sure your explanation really helps
After reading your post, I'm sure it does. But I'll leave it where it is as it isn't that important.
05-14-2012, 05:36 AM   #30
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QuoteOriginally posted by falconeye Quote
After reading your post, I'm sure it does. But I'll leave it where it is as it isn't that important.
I have long gotten past this discussion, but in re-reading your post, there is one significant point you make, but i think it needs refinement Specifically "You may be able to see in an intuitive manner from this image that the entire lens, incl. all lens surfaces contribute to the image of a point source. There is no blur coming from the edges ..."

the point i am trying to get across is that as you indicate, all of the lens contributes to the image you get, but as you also state "The more you remove from the incoming perfect wave front, the poorer the remaining part of it will behave when it comes to make a good light spot on the focus plane."

the smaller the aperture you remove much more of the "perfect front" in proportion to the distorted front from the blades,

it is the ratio of perfect to imperfect that is important, and this ratio changes as you stop down.

That is the point I was trying to make.
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