B Interference fringes, what if you sample them?

[David Byrden]

Imagine a standard Young's Slit experiment using photons. We obtain large, distinct interference fringes on a target screen.

Then, we embed into the target screen a telescope focused on and capable of resolving the slits. The width of the lens must be greater than about one fringe to obtain this resolution. So we use a nice big lens with two or three interference bands falling upon the lens glass.

The telescope's display gradually builds up an image of the two slits.

Now, the interesting part. We paint our lens surface with opaque paint, leaving only some narrow strips unpainted. These have the same spacing as the interference bands. They are too wide to cause significant diffraction in any light passing through them.

We place our telescope in two positions;
1. the unpainted strips on the lens coincide with dark interference fringes
2. the unpainted strips on the lens coincide with bright interference fringes

A naive person might expect the telescope to register a higher rate of photon detection in [2] than in [1]. Personally I expect the rates to be the same.

But I would appreciate if somebody who really knows the subject, would tell us what will result here?

Thank you.

David

 

[RPinPA]

My immediate response was going to be that in single-photon "interference" experiments, when you count the photons and plot the counts vs position, you get the classic double-slit pattern building up over time. So that's definitely "higher rate in [2] than [1]".

But hold on, things are a little more complicated than that.
https://sciencedemonstrations.fas.harvard.edu/presentations/single-photon-interference
"The experiment is designed to also make it possible (in principle) to know which of the two slits the photons are passing through. In this so-called which-path case, the Young's double-slit interference pattern does not manifest itself. This can be followed by a quantum eraser (to erase the which-path information) to recover interference. Thus, the act of measurement and the design of the experiment affect what is being measured. Even if not actually measured, the mere possibility that an observer could determine which slit the photon passed through causes the interference pattern to switch to non-interference."

So you can set it up so that you do not get the classic pattern.

Single-photon interference is an experiment where you have light intensity so low that photons arrive at your detector individually, and you can count the individual photons.

 

[David Byrden]

My immediate response was going to be that in single-photon "interference" experiments, when you count the photons and plot the counts vs position, you get the classic double-slit pattern building up over time.
...
So you can set it up so that you do not get the classic pattern.

In my scenario, there is an interference pattern on the target screen, and simultaneously I think there is not an interference pattern detected by the telescope which is embedded in the target screen, sampling photons functionally equivalent to those that hit the screen.

I don't think that you need QM to predict the outcome in my scenario, I expect that classical wave theory will tell us the correct answer. I hope that somebody knows the outcome for sure.

What I am hoping to highlight is that interference does not occur in the quantum particles while they fly from the slits to the target; it occurs in the environment after they hit the target.

David

 

[BvU]

Now, the interesting part.

What you describe here is called spatial filtering - which google.

Personally I expect the rates to be the same.

Then you will be severely disappointed.
Anyway, this is not the way it works: experiment decides. A simple laserpointer and a lens on an improvised optical bench can teach you a whole lot.

What I am hoping to highlight is that interference does not occur in the quantum particles while they fly from the slits to the target; it occurs in the environment after they hit the target.

Any basis for this speculation ? Any ideas for experiments to verify/falsify this ?

 

[Vanadium 50]

Then, we embed into the target screen a telescope focused on and capable of resolving the slits.

Such a thing does not exist.

 

[David Byrden]

Such a thing does not exist.

There is no such thing as a telescope capable of imaging Young's slits from, say, two meters away?

Is this because the slits would be very close together? If so, why could we not resolve them by using a sufficiently large lens? I'm not seeing the show-stopper here.

David

 

[BvU]

There is no such thing

It's called a lens

 

[David Byrden]

Any basis for this speculation ?

This is the basis; the particles, while crossing the gap to the target screen, still bear directional information. Therefore they can in principle be distinguished. Therefore interference (addition of wave functions) is not occurring in that gap.

Any ideas for experiments to verify/falsify this ?

Yes; the thought experiment that I just described.

David

 

[BvU]

Yes; the thought experiment that I just described.

I withdraw from this thread

 

[CWatters]

If it helps...

Many people have asked what happens if you set up a double slit experiment then make holes in the screen to see what appears on a second screen located behind the first. The hole being either on a fringe or in the dark area between fringes. Google it.

 

[DrChinese]

This is the basis; the particles, while crossing the gap to the target screen, still bear directional information. Therefore they can in principle be distinguished. Therefore interference (addition of wave functions) is not occurring in that gap.

Your speculation here is not correct. You could simply move the target screen up to see that there is interference occurring.

 

[Nugatory]

This is the basis; the particles, while crossing the gap to the target screen, still bear directional information. Therefore they can in principle be distinguished. Therefore interference (addition of wave functions) is not occurring in that gap.

That line of thinking doesn't work (unless you move the detector/telescope so close to one of the slits that there is no appreciable amplitude for the paths to it from the other slit, in which case the lack of interference is to be expected). The problem is that the interaction with the slits leaves the particle with an uncertain transverse momentum so there's no directional information in the sense that you're thinking.

(This is a heuristic argument from Ghirardi. You can arrive at the same conclusion by doing the sum of paths thing).

 

[David Byrden]

Your speculation here is not correct. You could simply move the target screen up to see that there is interference occurring.

I'm sorry if I did not make myself clear.

As I see it, there is no interference in the air gap, but when the particle's wave function interacts with that of the target screen, then momentum, charge and most other properties are preserved but directional information is erased.
The interference, as I see it, occurs only after that interaction, because only when directional info is removed do the wave functions pertaining to the two slits become indistinguishable. They add (as complex amplitudes) only from that point on. Their interactions with the environment then also interfere.

So, if you ask me why there are few photons in the "dark" band, I say it's because there are two indistinguishable versions of you who do see impacts there, but the two "yous" almost cancel out.

Moving the screen up closer to the slits will of course not affect this process.

Am I looking at this wrongly?

David

 

[David Byrden]

the interaction with the slits leaves the particle with an uncertain transverse momentum so there's no directional information in the sense that you're thinking.

Doesn't that imply that no lens, no optical intrument of any kind, placed at the target screen, could possibly resolve the two slits as such?
Which, at first glance, would make astronomy an impossible pursuit... ?

David

 

[Nugatory]

Doesn't that imply that no lens, no optical intrument of any kind, placed at the target screen, could possibly resolve the two slits as such?

No (and that confusion may be behind the talking-past-one-another exchange between you and Vanadium50 farther up in this thread). Forming an image of the slits, or of an object on the source side of the barrier, is a different thing than assigning a path to a particular photon detection in the region illuminated by both slits.

 

[DrChinese]

As I see it, there is no interference in the air gap...

Depending on your perspective: this is either wrong (there IS interference there) or meaningless (because you can't meaningfully discuss what happens when no measurement is made).

 

[David Byrden]

Forming an image of the slits, or of an object on the source side of the barrier, is a different thing than assigning a path to a particular photon detection in the region illuminated by both slits.

Well, now you have me totally flummoxed. I'll spell it out as I see it, please indicate my mistake;

1.An image of the slits looks like two bright lines.
2. It is obtained by summing many, many impacts of individual photons.
3. Every photon that arrives on the detector screen contributes to either one line or the other.
4 That photon therefore appears to have transited one slit or the other - the slit whose image it contributes to.

No?

David

 

[Vanadium 50]

An image of the slits looks like two bright lines.

If you have the conditions for interference, and you stand in the light zone and look back at the slits, you will not see two bright lines. I couldn't find a picture with lines, but with dots (figure from the University of Tennessee at Knoxville) it will look like the figure on the right, not the left:

 

[Charles Link]

On the experiment presented by the OP, suggestion is to use a lens as opposed to a telescope. $\\$ If the paint stripes on the lens are placed on the bright stripes of the interference pattern at 2 meters, the energy is basically wiped away, so very little energy will be observed past this region. Any observations done in the region beyond the lens will receive very little energy. $\\$ If the opaque paint stripes are placed in front of the dark regions of the two-slit pattern at 2 meters, they will have little effect, and the results would be similar as if there were no paint stripes at all. $\\$ It's all a completely classical description.

 

[David Byrden]

If you have the conditions for interference...it will look like the figure on the right, not the left:
View attachment 237655

Thank you. But my proposal was to substitute a larger lens/telescope/instrument until the image resolves. Is that not possible?

David

 

[David Byrden]

...the results would be similar as if there were no paint stripes at all. $\\$ It's all a completely classical description.

Thank you. That sounds reasonable. It's also completely contrary to my understanding of QM, so I need to think about it.
Can you refer me to any experiments demonstrating this?

David

 

[zonde]

But I would appreciate if somebody who really knows the subject, would tell us what will result here?

Does this experiment answer your question?
https://en.wikipedia.org/wiki/Afshar_experiment

 

[Vanadium 50]

But my proposal was to substitute a larger lens/telescope/instrument until the image resolves. Is that not possible?

No. That just makes the picture I posted bigger.

 

[Vanadium 50]

It's also completely contrary to my understanding of QM

I don't think your problem is with QM. I think it goes back to classical waves and optics. For example, if you have a diffraction limited system, adding a lens doesn't make it clearer. It just makes the fuzziness bigger. You might want to review that before going on.

 

[David Byrden]

if you have a diffraction limited system, adding a lens doesn't make it clearer. It just makes the fuzziness bigger.

Everything that I read about telescopes today seems to agree with this statement;
"The minimum angular separation of two sources that can be distinguished by a telescope depends on the wavelength of the light being observed and the diameter of the telescope. This angle is called the DIFFRACTION LIMIT."

Therefore, the diffraction limit (which you apparently referred to) can be arbitrarily reduced by increasing the diameter of the lens, as I did suggest. (I never mentioned "adding a lens")

No?

David

 

[David Byrden]

Does this experiment answer your question?
https://en.wikipedia.org/wiki/Afshar_experiment

Absolutely ! It's very much along the same lines as I was thinking, and the result is not what I expected - so I'm learning.
Thank you !

David

 

[Vanadium 50]

If you have an interference pattern stand in the light spot and look at the slits, the separation between the two slits is below the diffraction limit. I know you don't believe this, but nevertheless it is true.

 

[David Byrden]

Vanadium : this disagreement is no mere tangential issue. It is central to the point that I am making, to the demonstration that I proposed.

You said that an observer, standing among Young's fringes, could not possibly resolve the two slits. Why not?
I suspect that you deduce this from the principle that interference occurs only if which-way info is absent.
The ability to resolve the slits would imply the info is present, therefore interference could not occur.
And (I think) you assume that there's already interference in the photons while they cross the air gap.
Have I correctly summarised your thinking?

Anway, the point that I wanted to make, is that interference does not exist in the air gap. The photons (or particles) do carry which-way info.
Interference occurs only within the detector after this info has been erased.

Now, here are my supporting equations. Multiple, seemingly competent articles about telescopes tell us this;

"The diffraction limit is θ=1.22 λ/D, where θ is the angle you can resolve, λ is the wavelength of the light, and D is the diameter of your lens."

And it's easy to see that λ = θ F where F is the fringe spacing and θ is the angular distance between slits as seen from the central fringe.

So, a telescope embedded in the target screen, among the fringes, will be able to resolve the slits if its lens diameter is at least 1.22 fringes.

Needless to say, if you can resolve the slits, you can see which slit was used by each photon.
The which-way information is present and you have extracted it.

David

 

[DrChinese]

Needless to say, if you can resolve the slits, you can see which slit was used by each photon.

So you are planning to place a telescope at one of the interference fringes (a spot that wouldn't get any light if there were no interference), And you think you can resolve to "see" which slit it came through?

Logically, the interfering light paths are meeting at the same point where it enters the telescope. And that light has but a single path through the telescope to your eye. It's not as if the photon went through one slit or the other. There is nothing to resolve at this point.

 

[David Byrden]

So you are planning to place a telescope at one of the interference fringes

Sorry if I didn't make my notation clear; by "fringe spacing" I mean bright-to-bright, i.e. the "wavelength" of the pattern. A lens that's 1.22 fringes wide will have both a bright and a dark band impinging on it, regardless of how you shift it around in the illuminated area. It can't be placed in a dark band.

a spot that wouldn't get any light if there were no interference

I really don't understand this comment.
I intended to put the lens near the center of the illuminated area, but really, I think that the whole area illuminated by interference bands will get light when the interference is not present.

Where is the "spot that wouldn't get any light if there were no interference" ?

And you think you can resolve to "see" which slit it came through?

Yes.
Refer to the experiment that Zonde linked to, in post 22. That experimenter claims to have resolved the slits almost exactly as I proposed. And others have repeated the experiment.

Logically, the interfering light paths are meeting at the same point where it enters the telescope. And that light has but a single path through the telescope to your eye.

No, that's not how telescopes work.

David

 

[DrChinese]

1. It can't be placed in a dark band. ... I really don't understand this comment.
I intended to put the lens near the center of the illuminated area, but really, I think that the whole area illuminated by interference bands will get light when the interference is not present.

Where is the "spot that wouldn't get any light if there were no interference" ?

2. No, that's not how telescopes work.

1. When there is interference, there are bands that appear strongly that do not much appear otherwise (when there is no interference, you are not investigating the "no interference" case). So this is where you will place a telescope. You hope to "resolve" the one true path, as you believe there secretly exists which-slit information that your telescope can detect.

2. That's not how light travels. There will be only 1 path through the telescope, not 2. That's because it arrives at a point when it enters the telescope. That point being a place where there is interference from the 2 slits. One effective path to your eye. Keep in mind that if there is interference, by definition it did not go through one slit or the other. It went through "both"*.*Unless you are a Bohmian.

 

[David Byrden]

there are bands that appear strongly...So this is where you will place a telescope.

This is how the minimum-size lens sits on the interference bands. It's not possible to place it exclusively in a "strong" band. At least one strong band and one weak band will fall on it. And with this lens, the image of the slits is just about discernible. We would need a larger lens for good clarity.

Much more significant : it's not possible for the telescope to detect the interference pattern by shifting across it.
Because it's large enough to detect the which-way info, it's too large to sense the interference bands.

(All right, I agree, the lens can slightly sense the band pattern if its diameter is not an exact multiple of the band spacing. But by the same token, it forms an image of the slits that is somewhat blurred, so it can't deduce which-way info for all the incoming photons. There is a tradeoff.)

You hope to "resolve" the one true path, as you believe there secretly exists which-slit information that your telescope can detect.

There's nothing secret about it. It's called "momentum" and it can steer photons in incredibly straight lines across ten billion light years of space.

There will be only 1 path through the telescope, not 2. That's because it arrives at a point when it enters the telescope.

You really should read about how telescopes work. Light that falls on the entire lens surface gets steered to a single point on the detector / eye.
For the single-photon case, the portion of that photon's wave function that falls on the entire lens surface gets steered to a single point on the detector / eye.

Interference doesn't happen until the photon gets absorbed. As I have been saying.
David

 

[DrChinese]

1. It's called "momentum" and it can steer photons in incredibly straight lines across ten billion light years of space.

2. You really should read about how telescopes work. Light that falls on the entire lens surface gets steered to a single point on the detector / eye.
For the single-photon case, the portion of that photon's wave function that falls on the entire lens surface gets steered to a single point on the detector / eye.

3. Interference doesn't happen until the photon gets absorbed. As I have been saying.

1. It will go in a straight line. And it won't be from a specific slit, it will be from between the slits to the extent it is anything. You must be aware that momentum and position are non-commuting. Therefore when position is well resolved, momentum will not be, and vice versa.

2. Were that true, in this case, you couldn't resolve the photon as coming from one slit or the other. Which is your premise.

3. There are a lot of issues with trying to discuss quantum interference when it is not *directly* observed. What is true is that the point of photon absorption is when irreversible processes occur.

 

[DrChinese]

View attachment 237721

This is how the minimum-size lens sits on the interference bands. It's not possible to place it exclusively in a "strong" band. At least one strong band and one weak band will fall on it. And with this lens, the image of the slits is just about discernible. We would need a larger lens for good clarity.

To be clear: you will not see 2 slits with the telescope, but you will "see" the bands.

 

[David Byrden]

A telescope performs a Fourier transform on the incoming photon pattern. Photon angle is transformed into position at the detector. Photon position at the lens does not appear at the detector.

The bands impinging on the lens do not appear in the image formed. The only way for the telescope to detect interference bands at the lens is through the overall brightness of the light gathered.

There is no reason at all why a telescope can't form an image of slits two meters away, if the Hubble can form an image of a black hole at the center of the galaxy - much smaller angles are involved. Quantum interference doesn't magically disable the function of telescopes.

David

 

[DrChinese]

Quantum interference doesn't magically disable the function of telescopes.

Optical telescopes are not going to be able to see many quantum events. In this case: you cannot see which slit a photon passes through because, by definition (or rule or whatever), path interference cannot occur when you can. The only imaging related to the 2 slits you will see is the portion of detections for which interference did NOT occur. That will be some, and you could possibly resolve that at some level. But not the ones that interfere with themselves. That information simply is not present because it does not exist.

 

[David Byrden]

Let me spell out what happens to a single photon in the telescope:

1. Two lobes of its wave function pass through the slits (call them A and B)
2. A portion of lobe A impacts the entire lens glass
3. Simultaneously a portion of lobe B impacts the lens
4. If the lens were opaque they would be absorbed, and interference would begin here and we would see fringes on the lens
5. But no, it's not opaque, so both lobes pass through the glass
6. Within the telescope, both lobes are focused onto separate points on the imaging plane
7. They are no longer interfering because they no longer overlap
8. The imaging plane can tell which slit the photon used, by the position where it is absorbed

Does anybody disagree?

David

 

[Vanadium 50]

You really should read about how telescopes work.

You really don't want to take this tone with people who are trying to explain things. I am beginning to think your intent was never to ask a question but rather to push your own iconoclastic view. If that's not your intent, you might be less aggressive.

Consider a case where you have slits separated by a distance d, with a screen at a distance D away and at a point x on that screen. If you are at the first bright fringe, the condition is x \approx \lambda D/d. If you then position a telescope at x and look back at the source, you cannot resolve an angle \lambda/d or smaller. These are the exact same conditions.

It doesn't matter how cleverly you make or position the telescope, since this is a property of the light you are using.

 

[DrChinese]

The wave interference does not begin at the point it enters the lens. It occurs at many spots prior, and in fact it would be incorrect to limit the paths taken to one or two particular ones. You are treating quantum objects like classical ones, and that will always get you in trouble - at least in the quantum forum.

If you study quantum reflection, how index of refraction changes the direction of light, Mach-Zehnder Interferometers, etc. you will see that all possible paths contribute to the final results. You are attempting to oversimplify something that is a complex subject. Before you speculate, I would recommend you beef up your understanding.

 

[Charles Link]

The wave interference does not begin at the point it enters the lens. It occurs at many spots prior, and in fact it would be incorrect to limit the paths taken to one or two particular ones. You are treating quantum objects like classical ones, and that will always get you in trouble - at least in the quantum forum.

If you study quantum reflection, how index of refraction changes the direction of light, Mach-Zehnder Interferometers, etc. you will see that all possible paths contribute to the final results. You are attempting to oversimplify something that is a complex subject. Before you speculate, I would recommend you beef up your understanding.

To the OP: @David Byrden If you study interferometers, even at the classical level, I think you will find the results are quite phenomenal: It is simply amazing that the presence of the second beam can cause all of the energy to wind up at one receiver, while with a single beam, there is a 50-50 energy split. (This phenomenal interference and wave behavior can be explained completely using the Fresnel coefficients for the amount of amplitude reflection that occurs from the beamsplitter, along with a little algebra to sum the waves).

 

[David Byrden]

You really don't want to take this tone with people who are trying to explain things.

Tell me, do you think that his explanation was correct?
The notion that all photons passing through a chosen point on the telescope lens will all meet up at a single point on its image plane - do you agree with it?
It would make my "thought experiment" look like rubbish - but is it true?

I am beginning to think your intent was never to ask a question but rather to push your own iconoclastic view.

I did ask a question about QM, and zonde effectively answered it (post 22).
Than, to my astonishment, my high school understanding of optics and telescopes was questioned by two people.
What kind of tone are you using with me? Quote: "You might want to review that"

Consider a case where you have slits separated by a distance d, with a screen at a distance D away and at a point x on that screen. If you are at the first bright fringe, the condition is x \approx \lambda D/d.

That's equivalent to the second equation that I wrote in post 28.
But OK, let's use your notation.
D is the screen distance, x is the fringe spacing, and so on.

If you then position a telescope at x and look back at the source, you cannot resolve an angle \lambda/d or smaller.

That's odd - where are D and x ?
Why did you introduce them, if you're not using them now?

But let's assume you are correct here. A telescope can resolve angles greater than λ / d
Let's take a slit spacing of 0.1mm as an example.
And we'll use a red laser as the source.

You're telling me that a telescope can resolve the slits if the screen is closer than about 1.5cm from the lens.
And you're telling me that the lens diameter has no effect at all.
So, I might as well use a tiny lens, such as we find in a phone. Put the screen 12mm away, and there we are!
You're promising me that it will work !

I must ask you, where did you get that limit, λ / d

David

 

[Charles Link]

Tell me, do you think that his explanation was correct?
The notion that all photons passing through a chosen point on the telescope lens will all meet up at a single point on its image plane - do you agree with it?

@David Byrden The QM description of summing wave amplitudes over all possible paths, if I understand it correctly, is very much like diffraction theory. I have a better understanding of the classical (diffraction) description, but I do believe @DrChinese has considerable expertise in this (QM) area. I strongly recommend you study his explanations in detail before coming to the conclusion that they are anything but completely accurate.

 

[berkeman]

Thread is closed for Moderation...

Thread is re-opened.

 

[bhobba]

A telescope performs a Fourier transform on the incoming photon pattern

It does not. I do not know a lot about optics, but do know a bit about QED and a fair bit about Fourier Transforms and that is not what a telescope does. In fact what it does is quite complicated and requires a solid state physicist to fully explain but here is an overview. When a "photon" gets to a material, it is absorbed by the material. The material then sets up an internal electromagnetic vibration called a phonon (its what is known as a quasi-particle). The phonon has a less-than-light velocity that depends on the properties of the material. When the phonon reaches the "other side" of the material, it may create a new photon that then goes out and travels off at the speed of light.

We have a rule here - when you make a statement you need to be able back it up. Now precisely what do you have to back your statement up?

A bit of friendly advice from a mentor - around here, unless you are 100% sure about what you write, best to listen to what others say.

Thanks
Bill

 

[David Byrden]

When a "photon" gets to a material, it is absorbed... phonon...may create a new photon

Ah, but look at the bigger picture, Bill. That description covers only a single photon passing through a transparent material of undefined shape. But I'm talking about the pattern of photons processed by a telescope.
Let me explain where the Fourier transform comes into it:

Here's a very simple telescope, with one lens. Real telescopes have more lenses and mirrors, but they're essentially doing the same thing.
Incoming photons are focussed by the lens (on the left) to an image sensor (on the right).

The coloured lines represent the paths of individual photons.
All of the photons from a source are near parallel, so e.g. the photon paths "a1" and "a2" are from a "red source".

The telescope focusses light such that we get distinct images of the "red source" and "green source" on the image plane.
Notice that all of the photons collected from a source, from the whole surface of the lens, are brought to a single point.

Now: how is that a Fourier transform?

The incoming photons can be described as a function of position and momentum.
Photons from different sources have different momentum vectors. But they may have the same position - for example, a2 and b2 pass through the same point on the lens.

At the image plane, the momentum of the photons has been transformed into position (light from different sources goes to different points)
The position has been transformed into momentum (light from different parts of the lens arrives at different angles).
I'm sure you will recognise that's the essence of what a Fourier transform does!

The field of Fourier Optics, where we manipulate light using concepts from Communications Theory, is a fascinating one. There's an introduction here;
https://en.wikipedia.org/wiki/Fourier_optics

And that's what I referred to earlier.

Finally; my example telescope is handling an ensemble of photons, a "pattern" as I called it, but we're talking about quantum events here. What will happen with a single isolated photon?
To answer that, I call upon a wonderful property of photons. They don't affect each other.
The behaviour of a huge ensemble of photons arriving together, is the same as if they arrive one by one and we slowly accumulate the image. Every description of the Young's experiment will remind you of that.
And it's important to understand why that is; it's because the Wave Function of the single photon behaves as an ensemble of photons would. The Wave Function impacts the whole of the lens, it gets focussed as I described, and it defines the single photon's probability distribution to match exactly the image that a large ensemble of photons would form.

Has anybody any questions?

David

 

[bhobba]

Has anybody any questions?

The above is full of so much misinformation I do not know where to start. For example for photons position is not an observable:
https://www.physicsforums.com/threads/why-no-position-operator-for-photon.906932/

Really I should delete it , but I will give you one more chance to start listening instead of lecturing about what you do not know. This is a formal warning in my capacity as a mentor. Ignoring it will mean stronger action.

Thanks
Bill

 

[Vanadium 50]

Let's sweep aside some brush. This has nothing to do with QM; it's pure optics. This doesn't even really have anything to do with interference. Fundamentally, the claim is that one can beat the diffraction limitation for optics via clever construction and/or placement of a telescope. This is what we should be focusing (no pun intended, mostly) on.

This is impossible because the limitation is not a property of telescopes; it's a property of light.

Furthermore, there are zillions of pictures out there that show diffraction limitation. It's a real thing.

 

[David Byrden]

...the claim is that one can beat the diffraction limitation...
...This is impossible because the limitation is not a property of telescopes; it's a property of light.

Yes, you're right, there is a limit to attainable resolution for telescopes.
But the question is, will that limit prevent me doing what I proposed to do?
What is its actual size?

You posted an equation :

... slits separated by a distance d, with a screen at a distance D away and at a point x ...
...you cannot resolve an angle λ/d or smaller.

I'm sorry but I can't make sense of that.
I calculated that it's not a show-stopper at certain scales, but you insist that it is.

I have to ask you again, could we have a source for that equation?

David

 

[David Byrden]

I will give you one more chance to start listening instead of lecturing about what you do not know.

I'm not sure what you are saying. Is it;

[1] classical approximations of QM phenomena are not wanted in this forum,
or
[2] lens don't create the results you described
or
[3] other ?

David

 

[zonde]

This is impossible because the limitation is not a property of telescopes; it's a property of light.

Aperture of a telescope certainly is a factor in determining diffraction limit. https://en.wikipedia.org/wiki/Airy_disk
Here is reference for experiment described by David Byrden: https://arxiv.org/abs/quant-ph/0701027
In this experiment two pinholes produce two well resolved peaks in the image behind the lens.