[BasilG]

Point taken regarding the use of incorrect terminology (I don't deal with photons much usually). However, as for the rest, I suggest you have a look here:

 

http://en.wikipedia.org/wiki/Optical_transfer_function

 

4th figure looks a lot like what we're looking at above.

 

"In fact, the contrast becomes zero on several occasions [...]. This explains the gray circular bands in the spoke image shown in the above figure. In between the gray bands, the spokes appear to invert from black to white and <i>vice versa</i>, this is referred to as contrast inversion, directly related to the sign reversal in the real part of the optical transfer function, and represents itself as a shift by half a period for some periodic patterns."

 

I realize that this (optical system with aberrations) does not apply exactly to the problem here (defocus), but the effect certainly does look close.

 

Further, I'd be interested in your explanation of the half (or quarter, by your account) period shift between the star shape inside and outside the ring of zero contrast/destructive interference if there is indeed no such thing as contrast inversion.

 

Lastly, I'd be interested whether this is the product used for the test shoot:

 

http://cvp.com/index.php?t=product/carl_zeiss_1849-755

 

And if so, why the center point in the shot shown in this thread is white, while it is black in the original chart (would be nice if the original poster could confirm this), assuming that, as you say, "no imaging function will invert anything".

 

I might be totally off with my interpretation, but I don't think so.

[Scythels]

Withdrawn

[BasilG]

For what it's worth: I've just reproduced the effect with my own camera and that test chart. My conclusions are: (i) the zero contrast rings move according to the amount of defocus, and (ii) there is contrast inversion across each ring.

 

http://forum.photozone.de/index.php?/gal...hart-shot/

 

Download the test chart here:

 

http://diglloyd.com/blog/2014/20140416_2...ation.html

 

and try it for yourself.

 

I don't quite understand why you say that phase doesn't contribute - you need amplitude and phase contribution for image formation, no?

[Guest]

That was in fact my exact test chart. Some other lenses I tested in the same way did not show the effect, but it is repeatable with the 40mm. (Which is a really good lens, by the way. Only real flaw is the strange effect in the bokeh sometimes.)

[Scythels]

Withdrawn

[BasilG]

I am open to learning something new here, but so far, I have not seen an explanation for the phenomenon at hand - zero contrast transfer rings and contrast inversion - that is better than mine (with support from Wikipedia, for what it's worth ). So, my conclusion is - the transfer function drops below zero due to defocus and in fact oscillates back and forth across zero. The zero crossings are seen as grey rings lacking stripes, and where the transfer function is below zero, contrast inversion occurs, giving rise to the alternating zones of inverted contrast in my image (and the image in the original post).

 

As I said, photons are not my favorite discipline, but if you scroll down to the Figure 8 in the following link, you will see that at least these guys agree with me (in addition to the guys at Wikipedia):

 

http://micro.magnet.fsu.edu/primer/anato...intro.html

[Scythels]

Withdrawn

[BasilG]

1) Phase matters. Without phase information, you don't have an image (well, an image of your object in real space representation - you can have a diffraction pattern).

 

2) Why do we observe contrast inversion, in your opinion?

[Scythels]

Withdrawn

[BasilG]

re: 1

 

Sin(0x) is 0.  Sin(0x + pi/2) is 0.  Sin(0x + pi) is 0.  Without phase information you still have an image.  It is most likely that all imaging devices you have ever used are "phase blind."  Unless you do holographic imaging there is no phase information recorded. 

 

re: 2

 

It's constructive/destructive interference patterns, as I have said previously.

 

 

Re 1 - While you may not directly record the phases when recording a 2D image, the phase information still contributes to image formation. Otherwise you could calculate the original image from the amplitudes (without phases) of the Fourier transform of that image.

 

Re 2 - I remember what you claimed. But I have not seen any reasonable explanation for why your hypothesis would be true, while you seem to be really quick to dismiss any evidence presented be me (well, I agree it's not great evidence, but it's what was available on the web). In particular, I don't understand why your statement that "microscope objectives are different than camera lenses" would in any way falsify the idea that the transfer function can cross zero when the lens is defocused, giving rise to both effects observed in the images.