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horacekenneth
18-Aug-2012, 02:22
So I read in an old thread here that basically the two biggest factors to picture resolution (after format size) come from the lens: aberrations and diffraction. Don't stop me yet. I went on to read that the smallest point resolution possible is roughly the numerical equivalent of microns to whatever f/stop your using, so if I'm stopped down to f/32, I can't have a light point smaller than 32 microns. How accurate is this? Based on this reasoning, an f/1000 pinhole should have no light points smaller than 1mm, but prints from one show that it does have more resolution than that!
What am I missing?

Thanks,
Horace

ic-racer
18-Aug-2012, 06:59
The size of the disk-of-confusion is easily estimated. At half the distance from the camera to the point of exact focus, the disk is half the working diameter of our lens. At twice the distance to the point of focus, the disk is equal to the lens diameter.

When a lens is focused at infinity, the disk-of-confusion will be of constant diameter, regardless of the distance to the object. The diameter of the disk-of-confusion, S, will be equal to d, the working diameter of the lens, at all object distances.

E. von Hoegh
18-Aug-2012, 08:03
So I read in an old thread here that basically the two biggest factors to picture resolution (after format size) come from the lens: aberrations and diffraction. Don't stop me yet. I went on to read that the smallest point resolution possible is roughly the numerical equivalent of microns to whatever f/stop your using, so if I'm stopped down to f/32, I can't have a light point smaller than 32 microns. How accurate is this? Based on this reasoning, an f/1000 pinhole should have no light points smaller than 1mm, but prints from one show that it does have more resolution than that!
What am I missing?

Thanks,
Horace

The diffraction limit in LP/MM (line pairs per millimeter) can be estimated by dividing the F number into 1500.

Emmanuel BIGLER
18-Aug-2012, 09:03
The diffraction limit ..
... exressed as a period in metres of milimetres or microns is EXACTLY equal to N. lambda in monochromatic light of wavelength lambda, without any other numerical factor (no 1.22 there ;).
If we consider the worst case of what the human eye can actually see at lambda = 0.7 micron (red light) we end-up with something like 0.7 N microns for the cut-off diffraction period and 1400/N in cycles/mm. Hence 1500/N does make sense if we consider an average wavelength of 0.66 microns ;)

f/1000 pinhole should have no light points smaller than 1mm, but prints from one show that it does have more resolution than that!
What am I missing?

Difficult to argue without actually looking at the print, but most pinholes have a f/number closer to f/500, not f/1000.
As seen from 30 cm (1 foot) distance, what the human eye can resolve is about 5 to 7 cy/mm on the print.
Hence a print with only 1 cy/mm should look poor.
The diffraction limit is an absolute resolution limit that cannot be corrected by any digital post-processing, unlike geometrical aberrations and defocusing which can be corrected to some extent (remember the famous example of the first "wrong" mirror of the Hubble space telescope)

horacekenneth
18-Aug-2012, 21:22
Emmanuel,

I am using a .15mm pinhole plate with a focal length of 6" or a 152mm. I believe that's just about f/1000.

I don't understand the circle of confusion in terms of the way it actually plays out on paper but I can do 1500/n math. Based on 1500/n I should have 1.5 line pairs per mm in a 5x7 contact print from the said f/1000 pinhole. Maybe I don't know what I should be looking for, the resolution isn't great, but it does look a lot better than 1.5 line pairs sounds. Here's a link: http://www.flickr.com/photos/horacekenneth/7787384304/

Ben Syverson
18-Aug-2012, 21:30
Remember that a diffraction pattern has a bright center and concentric rings. So even when diffraction is overwhelming, you're still getting some sharpness along with the blur.

Emmanuel BIGLER
19-Aug-2012, 01:25
I am using a .15mm pinhole plate with a focal length of 6" or a 152mm

Well, according to well-know rules for the optimum pinhole diameter, your pinhole is too small for the said projection distance of 152 mm. And yes, 0.15 mm @ 150 mm makes f/1000.
"Too small" w/respect to the optimum diameter means that you loose both on exposure time and on resolution, hence there is really no gain ;)

As crazy that it may sound, you should try to record the image of a test target like the famous 1951 USAF targert, or a copy
http://www.takinami.com/yoshihiko/photo/lens_test/USAF.pdf
More details on how to use this target : http://www.takinami.com/yoshihiko/photo/lens_test/pdml-procedure_c.html
Print this target on a letter-size paper, In the bottom-right corner of the target you have the biggest test pattern, a grid whith a period of 4 mm, i.e. each bar has 2 mm width separated by a 2 mm whitespace. This makes a 4 mm period. place the target at 1.5 metre (5 feet) in front of your pinhole camera, your magnification ratio will be 1:10 if your projection distance is 150 mm, and see what happens. If you can resolve the period of 4 mm on the target it means that you can resolve 4:10 = 0.4 mm in your camera.

The best pinhole diameter @152 mm according to Stroebel in his book '"view camera technique" is
(1/28) (f in mm)^(1/2) (result in mm)
Stroebel shows example images shot in 8"x10" demonstrating that this value is actually the reasonable one, and not a theorician's dream ;)
Leslie Stroebel, "View Camera Technique" Focal Press, 7th ed. (1999) ISBN 0240803450, pp.46-48

Hence @ 152 mm, your optimum diameter shoud be (1/28).(12.3) = 0.44 mm ; f/number @150 = 150/.44 = 345.
So if you consider that your images are good @f/1000, by simply using the optimum pinhole, your exposure time will be reduced by about a factor 10 and your resolution wil improve by a factor 3 !

Remember that a diffraction pattern has a bright center and concentric rings.
Here we are entering theory ;)
This is valid for a diffraction-limited optical instrument where you record the sharpest possible picture of a point object, in monochromatic light.
In an optimum pinhole where you have a 50%-50% mixture of geometrical projection of rays and diffraction effects, even in monochromatic light, I challenge anybody to show me any Airy disk recorded on film !

And even in a diffraction-limited, perfect optical instrument, even if you can see the Airy disk, it has been shown since the 1940's that the cut-off diffraction period is N.lambda, and the cut-off frequency 1/(N.lambda) (in cy/mm) and absolutely nothing beyond ... this is not rocket science, and the formula is so simple ... but those ideas have hard times to make their way into the photographic world where all textbooks dealing with diffraction always refer to the Airy disk and its concentric rings.