Scattering and non-imaging optics

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ZeroFunGame
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I have seen the term "conventional scattering lens" being used (see https://arxiv.org/pdf/2112.08144) and I was wondering whether the term scattering lens is accurate? I always envisioned a lens as a refractive element, and wondering if the reference means refracting rather than scattering?

Separately, and this is due to my lack of understanding of the nomenclature, would a non-imaging lenses be also considered a refractive optical element? For example a Fresnel lens?
I am confused because https://en.wikipedia.org/wiki/Nonimaging_optics states radiative transfer as the technical term used to describe non-imaging optics, which linked to https://en.wikipedia.org/wiki/Radiative_transfer but did not include refraction (only, absorption, emission, and scattering)
 
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ZeroFunGame said:
I have seen the term "conventional scattering lens" being used (see https://arxiv.org/pdf/2112.08144) and I was wondering whether the term scattering lens is accurate? I always envisioned a lens as a refractive element, and wondering if the reference means refracting rather than scattering?

Separately, and this is due to my lack of understanding of the nomenclature, would a non-imaging lenses be also considered a refractive optical element? For example a Fresnel lens?
I am confused because https://en.wikipedia.org/wiki/Nonimaging_optics states radiative transfer as the technical term used to describe non-imaging optics, which linked to https://en.wikipedia.org/wiki/Radiative_transfer but did not include refraction (only, absorption, emission, and scattering)
There are lenses that are based on diffraction rather than refraction. That's especially useful if there is no material with a refractive index that is sufficiently different from 1. This is the case for X-rays, for example.

A Fresnel lens is a refractive lens that's simply split into zones in order to keep it thinner. It uses the same principles as a normal refractive lens.

In the first paper, however, they talk about lens-less imaging, using diffraction patterns from overlapping areas (ptychography). That's a non-imaging technique.
 
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ZeroFunGame said:
I have seen the term "conventional scattering lens" being used (see https://arxiv.org/pdf/2112.08144) and I was wondering whether the term scattering lens is accurate? I always envisioned a lens as a refractive element, and wondering if the reference means refracting rather than scattering?

Separately, and this is due to my lack of understanding of the nomenclature, would a non-imaging lenses be also considered a refractive optical element? For example a Fresnel lens?
I am confused because https://en.wikipedia.org/wiki/Nonimaging_optics states radiative transfer as the technical term used to describe non-imaging optics, which linked to https://en.wikipedia.org/wiki/Radiative_transfer but did not include refraction (only, absorption, emission, and scattering)
And: There are no non-imaging lenses.
 
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Philip Koeck said:
And: There are no non-imaging lenses.
A lens with a focal length greater than the Rayleigh Distance cannot bring rays to a focus and hence cannot form an image.
 
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tech99 said:
A lens with a focal length greater than the Rayleigh Distance cannot bring rays to a focus and hence cannot form an image.
Could you explain a bit?
I just looked up Rayleigh distance on Wikipedia: https://en.wikipedia.org/wiki/Rayleigh_distance

If D is much larger than λ in the expression they give then Z is quite a large distance compared to λ.

Did you mean "a focal length smaller than the Rayleigh distance"?
 
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ZeroFunGame said:
I have seen the term "conventional scattering lens" being used (see https://arxiv.org/pdf/2112.08144) and I was wondering whether the term scattering lens is accurate?

Looking at the way it's used in that preprint, I think they may mean an optical diffuser.
 
Philip Koeck said:
In the first paper, however, they talk about lens-less imaging, using diffraction patterns from overlapping areas (ptychography). That's a non-imaging technique.
I have to correct my terminology.
Ptychography is of course an imaging technique, but it doesn't use images produced optically.
It uses multiple diffraction patterns. The image is then calculated from these.
This means that you don't need a lens.

The same is possible if you have a single diffraction pattern of a periodic or aperiodic object.
You "just" need to find the phases of the Fourier transform and you can get an image by inverse FT.
Usually this requires some additional information about the object, though.

This is the field of "lens less imaging" and I'm sure there's more to it than I know about.
 
ZeroFunGame said:
I have seen the term "conventional scattering lens" being used (see https://arxiv.org/pdf/2112.08144) and I was wondering whether the term scattering lens is accurate? I always envisioned a lens as a refractive element, and wondering if the reference means refracting rather than scattering?
I would say a conventional lens for visible light is refractive.
The term you quote sounds rather odd.
 
Philip Koeck said:
There are lenses that are based on diffraction rather than refraction.
I remember (many years ago) being introduced to The Zone Plate, which involves a diffraction pattern (in this case, Newton's Rings) on a photographic film. This can be used to form an image, like that from a convex lens. That's a trivial example but could be useful when, as mentioned above, there is no material with suitable refractive index available for the wavelength involved.

I wouldn't lose sleep about the terminology used. Different words are often used for a particular process / object in different disciplines.
 
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sophiecentaur said:
I remember (many years ago) being introduced to The Zone Plate, which involves a diffraction pattern (in this case, Newton's Rings) on a photographic film. This can be used to form an image, like that from a convex lens. That's a trivial example but could be useful when, as mentioned above, there is no material with suitable refractive index available for the wavelength involved.

I wouldn't lose sleep about the terminology used. Different words are often used for a particular process / object in different disciplines.
The terminology is slightly confusing, I've realized after googling a bit.

A (Fresnel) zone plate is a diffraction lens, just as you say, whereas a Fresnel lens is really a refractive lens split up into concentric rings (zones?) in order to make it flatter and lighter.

Both are named after Fresnel just to make life difficult.

Here are two Wikipedia entries:
https://en.wikipedia.org/wiki/Zone_plate
https://en.wikipedia.org/wiki/Fresnel_lens
 
Philip Koeck said:
The terminology is slightly confusing, I've realized after googling a bit.

A (Fresnel) zone plate is a diffraction lens, just as you say, whereas a Fresnel lens is really a refractive lens split up into concentric rings (zones?) in order to make it flatter and lighter.

Both are named after Fresnel just to make life difficult.

Here are two Wikipedia entries:
https://en.wikipedia.org/wiki/Zone_plate
https://en.wikipedia.org/wiki/Fresnel_lens
Is there any other way to focus photons other than reflection, refraction and diffraction?

For charged particles there are electric and magnetic fields and also standing light waves
(see for example: https://www.researchgate.net/publication/363866622_Transverse_Electron-Beam_Shaping_with_Light).

Would the opposite work: Standing electron waves that focus x-rays for example?
 
Philip Koeck said:
Would the opposite work: Standing electron waves that focus x-rays for example?
I think there would be an Energy problem here. The density of a standing wave of electrons would be limited by the Coulomb force so could you produce conditions with electrons contained in enough quantity. But that would still constitute a Diffraction device.
 
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vanhees71 said:
The lenses, however, are a bit unhandy. It's good to have an entire galaxy or at least a supermassive black hole for that ;-).
It's just a matter of intergalactic collaboration.
 
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hutchphd said:
Gravitation?
Of course! I was thinking in terms of 'in the lab' but galaxies and black holes do it measurably. But, in that case you have to take what you're given.
 
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