What really causes "diffraction blur" ?

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Harold Gough
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Post by Harold Gough »

Rik,

Yes, the mouseover effects I regularly see are transient and it would be a nuisance if they were not.

Harold
My images are a medium for sharing some of my experiences: they are not me.

Lou Jost
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Post by Lou Jost »

I'm struggling to understand this old thread.

Rik wrote "In addition, all of that light has to be coherent enough to form a real image by interference."

I am puzzling over that. We can make an image one photon at a time. The wave function for that photon will interfere with itself, of course, so even building up an image one photon at a time, we will still see interference phenomena. But I can't see where the coherence is involved; what does that even mean for single photons?

phil m
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Post by phil m »

Which brings me to pin holes. What happens when the aperture and the lens are the same thing and extremely small, possibly one photon in size?

Lou Jost
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Post by Lou Jost »

"....one photon in size"

Well, that's a tricky concept. A photon doesn't really have a size. It has a wavelength, and a wave function which can be used to calculate the probability of it's detection at any given point. But you can't really say it has a certain diameter.

The pinhole causes diffraction of the single photon's wave function. If the hole is small compared to the wavelength of the photon, the wave will spread out and you'll have no idea where that photon will hit the sensor.

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Post by rjlittlefield »

Lou Jost wrote:The pinhole causes diffraction of the single photon's wave function. If the hole is small compared to the wavelength of the photon, the wave will spread out and you'll have no idea where that photon will hit the sensor.
Adding to this point...

In my own thinking, I have come to prefer some phrasing like "where the energy will be deposited", rather than "where that photon will hit the sensor". That's because it's simpler for me to keep in mind that the photon acts like a distributed wave until it suddenly collapses and deposits its energy in some identifiable location.

The probability of depositing at any particular point depends on the wavefunction at that point.

If you sequentially squeeze a bunch of photons through a moderately sized aperture, one at a time, and accumulate the results, you'll end up with a discrete approximation to a classic Airy disk.

The width of the disk depends inversely on the diameter of the aperture. If you're serious about the aperture being around the same size as the photon's wavelength, then the disk will get blown out to being just a single central peak with no rings. The classic ring structure is seen with larger apertures, many wavelengths across.

--Rik

rjlittlefield
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Post by rjlittlefield »

Lou Jost wrote:I'm struggling to understand this old thread.

Rik wrote "In addition, all of that light has to be coherent enough to form a real image by interference."

I am puzzling over that. We can make an image one photon at a time. The wave function for that photon will interfere with itself, of course, so even building up an image one photon at a time, we will still see interference phenomena. But I can't see where the coherence is involved; what does that even mean for single photons?
The word "coherent" means different things to different people. I used it loosely in that explanation as a way of trying to help people who are new to the problem to think about some sort of wave that can interfere with itself to form a useful real image. If you're already comfortable with the concept of a single photon interfering with itself, then you don't need that, and in that case the word "coherent" becomes an un-helpful distraction because of all its associations with lasers and other multi-photon sources. So for you, the answer is that coherence, as you understand the word, is not involved.

--Rik

Lou Jost
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Post by Lou Jost »

OK, thanks for the clarification. That helps a lot.

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