I've been delving into this issue and also reading all the posts in other threads here. Lots of good info to chew over.

But, here's the problem...

Consulting a diffraction limit chart F32 shows to produce an airy disc of about 41 microns across, that's pretty big compared to the pixel size of the Canon 400D at 5.7 microns. It covers 7 pixels.

In theory the image should be extremely soft but doing some tests with newspaper prints (sad I know) the loss of quality between F5.6 and F32 is nowhere near as much as it should be. There's a distinct loss of contrast but the real resolution is not too bad.

Can anyone tell me why this might be?

I'm using a Canon 400D with a 135F2L lens on a tripod with mirror up etc. First frame F5.6, second frame F32

Click to view attachment

I only have a very basic understanding of this, so I might be complete off here, but...

A 7 pixel wide airy disk would cover 3.5 pixels worth of "blur" on the edges of the image. Looking at the "C" in the top left corner the white dots are very faint and hard to distinguish compared to the f5.6 image. I could believe the black is blurred by 3.5 pixels inward making the white dots almost disappear.

Even applying a 2px Gaussian Blur is enough to degrade the left image more than the right. A 3.5px blur makes it almost unreadable. I suspect this is an unrealistic measure but still...

The center of the airy disk is much brighter. That's why you notice a large loss of contrast (light spilling over to the surrounding pixels), while there is still a fair amount of sharpness.

For a nice illustration, look at:
http://www.olympusmicro.com/primer/java/di...tion/index.html

-- Tilo

True, but the figures I have for disc diameter are based on the width of the first dark ring which pretty much defines the central spike. If this is so big, 40 microns at F22, how come the image is not more degraded with respect to sharpness?

I do not know Canon equipment, but is that the effective or real aperture? If the former, that would explain why the image is not as soft as you expect.

Not sure what you mean, an aperture is an aperture as in F/diaphragm hole size. The theory says that the diffraction disk size is independent of focal length, merely depending on aperture.

When you take closeups, you have to extend the lens to achieve focus, and the effective aperture decreases (f/number gets larger). Ne = N*(1+M), where Ne is the effective aperture, M is the magnification, and N is the marked f/number. If you are at a magnification of 1, you lose 2 f/stops.

See the Lens Faq. Many newer lenses are equipped with microchip so that the actual f/stop is indicated on the camera. TTL metering takes the needed increase in exposure into account.

Bill

Won't that simply make the diffraction worse? You're saying that the effective f-number is larger than 32.

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It looks as though at f32 you are close to losing resolution of the white dots in the big "C" at upper left; how big across are those in pixels?

The photozone tests of the Canon 100mm/2.8 macro give about 1000 lines/picture height on APS-C at f32. That works out to about 35 line pairs/mm resolved, or a resolved line spacing of about 30µ. That's to be compared with the "size" of the diffraction spot which is ~40µ. So, not far off.

A further issue to consider is whether the size of the Airy disk is the limit of resolution. The disk radius is the location of the first zero of the diffraction intensity pattern. I plotted the intensity pattern of two diffraction spots of increasing separation -- 4, 5, 6, and 8 reading from left to right and top to bottom:



In these somewhat arbitrary units, the first minimum of the diffraction pattern is at a radius of about 3.8. The black circles are at the radius of the Airy disk, centered on each spot.

So clearly at a separation of the imaged spots equal to the diameter of the Airy disk (about 7.8 in the above plots, essentially the lower right figure) the objects are completely resolved. But I also imagine I'd be able to make them out at a separation of 6, the lower left figure; 5 (upper right) is starting to be merged, and 4 (upper left) shows basically one spot.

This is consistent with the Photozone tests, which show a resolution down to about 3/4 the Airy disk diameter.

Here is the C's at 1200% magnification, the white dots appear to be about 3 pixels wide.

I'm not sure that the "radius" amount for the Gaussian blur in Photoshop actually corresponds to the standard deviation of the gaussian used to convolve the image. In fact, IIRC it is not, but I don't remember in which direction. Photoshop is almost useless for precision numerical work; better to use ImageJ.

With pure two-tone text like this (dark, light, no intermediate shades) I would expect that the "centre-weighted" nature of diffraction lightens the dark parts, darkens the light parts, but without eliminating the difference: the dark parts will still be darker than the light parts, with some gradation at the edges.

That sounds a lot like a major reduction in contrast, some loss of acutance (edge sharpness) but not so much a loss of actual resolution.

Also, as Emil's illustration shows, the brightness at the nominal Airy disk edge is distinctly less than at the center, with the bulk of blurring over a somewhat smaller distance.

Agreed. Part of the problem here is that there is nothing really testing resolution directly, as in how closely spaced can a pair of lines get and still be resolved. Instead one has the indirect notion of how much a sharp light/dark transition in the source is broadened, the sharp edge becomes a transition region from light to dark whose width is governed by the resolution. One can recover the resolution from that analytically but it's much harder to just eyeball it.

Yes, the refraction would be worse with the smaller aperture. I don't know what the magnification in the sensor plane was, but it most likely is 1:1 or so. I don't know how the 400D and lens used for the test report aperture, effective or nominal. The newer Nikon lenses and cameras report the effective f/number and some don't even have the aperture ring. My old Nikkor macro 35 years ago actually automatically opened up the diaphragm to compensate for the exposure increase needed with closeups. If you used TTL metering, you had to undo the correction and it was a bit of a hassle.



A very nice demonstration of the Airy discs. My own testing and the results on Photozone.de show that MTF starts to decline when the size of the Airy disc exceeds the pixel spacing of the camera. Some authors use 2 times the pixel spacing.

If you can just make out line pairs, that corresponds to the Rayleigh criterion at a MTF of about 9%. For f/22 the Airy disc is about 25 microns for green light and MTF at 50% contrast is 35 lp/mm and 75% at Rayleigh.

This 2D graphic illustrates the Rayleigh resolution limit.

Thanks for the discussion, it helps me understand what I'm seeing.

FWIW the camera lens distance was about 1.5 and the lens was a Canon 135F2L. There should therefore be little adjustment needed for effective aperture. The example shot is a 100% crop of the unsharpened image converted at default from RAW. Exposure was +1 in camera.

It's a very sharp lens, I figured it would be good for this purpose.

I presume the distance is in meters (we still use feet in the USA). One can calculate the magnification and effective f/stop from the equations in the lens faq.

1/So + 1/Si = 1/f, where So is object distance, Si image distance, and f the focal length.

m = Si/So, where m = magnification

Si = 0.1484, m = 0.22008, Neff = f/22.48.

So you are correct, the correction factor is negligible.

If you like a more objective test, the Koren Lens Test Chart is nice.

Bill

Hi,

Just to clear the issue. Photozone does not calculate resolution figures but gives LP/height values for MTF of 50%. This corresponds to perceived sharpness according to litterature.

Resolution is probably mostly a function of sensor and antialiasing filter as much as of lens MTF.


Erik

Here, in The Luminous Landscape:

http://luminous-landscape.com/tutorials/resolution.shtml

R.

It is easy to convert to and from lp/ph and lp/mm: LW/PH is equal to 2 * lp/mm * (picture height in mm). See Norman Koren. Of course, Photozone determines the system MTF, not merely the lens MTF.

Bill