GRED Answer: Double slit with sharp eyed intruder

[RJN]

Guess the Result of the Experiment of the Day (GRED): The Double Slit Experiment with a Sharp Eyed Intruder

A classic double slit experiment is done with light (photons) that creates an interference pattern on a distant image screen. The experiment is repeated, except now a sharp-eyed intruder (person) places their head in front of the image screen and looks toward the slits. The person's eyesight is good enough to discern each slit individually.

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    Photon                   Double                          Sharp      Image
    Source                    Slit                            Eyed      Screen
                             Screen                          Person

Will the sharp-eyed person see a net flux from the slits that corresponds to the interference pattern behind them? In other words, is the net flux the same at any position regardless of whether the detector is an eye or a screen?

Please post answers, comments, and discussion below. In a few days I will post what I believe to be the correct answer. OK, it has now been a few days, and the answer is posted below and also visible here: http://asterisk.apod.com/vie ... 52#p123952 . Comments, polite disagreements, and discussion are still welcome.

The initial poll, where spoilers were not allowed, can be found here: http://asterisk.apod.com/vie ... 30&t=19720 . If you are new to this GRED and want to ponder this question without seeing spoilers, please go there now instead of scrolling down.

- RJN

[hstarbuck]

I initially chose no pattern, but I wish to change to pattern. After all, the eye is simply a detector as the screen is.

[Chris Peterson]

I initially chose no pattern, but I wish to change to pattern. After all, the eye is simply a detector as the screen is.

It is, but unlike the screen it is an imaging detector. So it won't see an interference pattern (that is, facing the slits, there will be no evidence of interference structure). But as the observer moves back and forth, he will see the apparent brightness of the light coming from the slits change- which I suspect is why RJN was careful to express the question as he did, in terms of net flux. The net flux is the same at any position regardless of whether the detector is an eye or a screen. But the observer will only "see" an interference pattern by physically moving so his eye scans across it from side to side.

[RJN]

The net flux is the same at any position regardless of whether the detector is an eye or a screen. But the observer will only "see" an interference pattern by physically moving so his eye scans across it from side to side.

This is essentially what I am trying to ask: Is the net flux the same at any position regardless of whether the detector is an eye or a screen? I am not asking if the observer will see, standing still, bands of interference in their field of view. They will not -- they will see two slits and possibly the source through one or both of the slits. I have now edited the initial GRED text to make this point more clear.

- RJN

[jabm67]

I guess I don't see the difference between this set-up and the college freshman physics lab examining two-slit diffraction patterns, when you ignore safety rules and look back toward the slits from in front of the screen. The eye is lousy at discerning intensities, and looking with both eyes at once will definitely confuse the neuro-visual engine if the interference fringe width is comparable to or smaller than the distance between the eyes.

I realize I'm making a couple of assumptions here, the first being that the slits are too close together for the eye to separate them, which is the usual way the freshman lab is set up. Are you asking about the opposite situation, where the slits are far enough apart that the eye could resolve them (when illuminated from behind with non-coherent light), but the eye subtends an angle comparable to the angular width of a single interference fringe? I'm assuming you are using a single eye here; I'd expect that looking with two eyes (when the eye separation is larger than a fringe width) would be simply confusing and you'd get an impression similar to that of looking though leaves at the Sun

[ergophobic]

The observer will see two slits of light of equal brightness at any one position. As the observer moves from side to side, the light level from the two slits will brighten or dim equally, in synchrony with the interference pattern...

[Beta]

...unlike the screen [the eye] is an imaging detector. So it won't see an interference pattern (that is, facing the slits, there will be no evidence of interference structure). But as the observer moves back and forth, he will see the apparent brightness of the light coming from the slits change- which I suspect is why RJN was careful to express the question as he did, in terms of net flux. The net flux is the same at any position regardless of whether the detector is an eye or a screen. But the observer will only "see" an interference pattern by physically moving so his eye scans across it from side to side.

I must disagree with Chris Peterson, hstarbuck and ergophobic. As the observer moves across the screen, the brightness of the two slits will not change. The observer will not see the bright-dim-bright-dim interference pattern. In particular, there will be places where the eye sees light but the non-resolving light sensor detects no light (in a dark band), and places where the sensor detects light twice as bright as the total flux reported by the eye (in a bright band).

The eye detects light rays from the two slits separately (with different parts of the retina if you're human, but, hey, whatever) and does not allow them to combine and interfere.

EDIT: It occurs to me that something is wrong in my analysis: if we cover the bright rings with non-resolving sensors, and the dark rings with resolving sensors, the energy doesn't add up. I think I'm basically right, but there's something I'm not seeing...

[bystander]

The act of observing the photon will change behavior from wave (uncertain) (interference pattern) to particle (certain) (diffraction pattern).

[rboatright]

one of the posters had the critical question I think.

There are two issues here. If the size of the entry pupil of the eye is LARGE enough to resolve the two slits, then it is probably also WIDER than any single interference fringe.

If the eye is NARROWER than the fringe, then it is unlikely to be able to resolve the slits.

Assume that the eye CAN resolve the slits. that means that there will MANY interference fringes hitting the front surface of the lens of the eye, and these will be focused into an IMAGE of the slits... not an interference pattern on the retina.

Assume that the eye can NOT resolve the slits. Then, the eye is just a photo detector which is either in a crest or a trough of the interference pattern, so it will see either nothing, or a single bright line.

You can't have it both ways. We could do the math, the size of the fringes vs the angular resolution of a lens of diameter x, but it seems clear what the result is. For an eye narrower than a fringe, the theoretical resolution of the eye is lower than the slit spacing. For an eye wider than a single fringe, you have the data input from MANY fringes hitting the lens and those will be combined BY the lens into an image of the two slits.

There no such thing as a "sharp eyed observer" without a discussion of the size of the observer. The picture as drawn in the question shows an observer who's eyes are proportionately SMALLER than the individual fringes. That observer, AS DRAWN will see a single bright (or dark) line as he moves across the field. But then, he isn't really "sharp eyed" enough to resolve the slits as the question stated. If he was, his eyes would have to be MUCH BIGGER in which case, we're back to the eye intersecting multiple fringes and the result being an image.

So, the correct answer is "None of the above."

[Emiel]

Actually, if I remember well, in highschool we did this experiment. Both with light and sound.

I believe the pattern is best discernable at some distances, so if the sharp eyed intruder is able to come in the same focus as the screen is, he will be able to see the differences in the light...

[Ken Winters]

Let's say the slits are resolvable by the human observer and you put a half-silvered mirror behind the slits so that the person can observe the slits without blocking the screen (and that the laser power is low enough to make this safe). Based on results from other experiments I have read about, I think that when the person looks at the screen they will see an interference pattern and when they turn to look at the slits (that is, at the image of the slits in the mirror) that they will not. In addition if A and B are looking at the screen, observing the fringe pattern, and A turns to look at the slits, B will observe that the interference pattern disappears. If A blinks, the interference pattern will reappear for a fraction of a second. My one doubt here is whether the human eye is really detecting single photons. If not, then the fringes stay. (Based on the information the last from post, I have to conclude that we aren't detecting individual photons.)

This also answers the point about using resolving detectors located in the dark fringes and non-resolving detectors in the bright ones, or vice versa. The total flux does add up correctly in this case, since the existence of a single resolving detector destroys the fringe pattern. It's actually even stranger than that: If you use a resolving detector (let's say you get an electrical signal from the detector) the fringes disappear; but if you short out the output, destroying the already gathered information, the fringes reappear. (A different, but similar, experiment has been done, and that's really how it worked.) A measurement which causes "the collapse of the wave function" (as this is called) collapses the wave function everywhere, even in cases where this seems to require transmission of information faster than the speed of light. The EPR experiment, which was deliberately designed to show that the randomness inherent in quantum mechanics leads to impossible conclusions, has been done, and the "impossible conclusions" were observed. So we know that something that seems like information travels instantaneously (in the frame of reference of the particles involved).

One really esoteric complication that I'm really NOT sure about With a resolving detector which only measures some fraction of the photons, do the undetected photons still form a fringe pattern? I think that we have to go back to the basic low-level description and consider each photon separately. Before the photon is detected, it is described by a wave function which consists of two components - one describing it going through one slit, the other describing it going through the other. It's these components of the wave function which interfere with each other and cause the interference pattern. Since that wave function has a non-zero probability of interacting with (and being detected by) the eye or the resolving detector it "collapses" (one of the two components disappears and the other becomes the entire wave function) and the interference pattern disappears completely. I think that's what would happen. I'd like to see this done. I'm trying to think of ways to set up the situation so that the screen would show some of the light spread evenly and some of it in fringes - some of the photons acting as waves and some acting as particles. It seems that that should be possible, but I haven't come up with anything that I think would work. It could be faked using a physical resolving detector and turning it on and off fast enough that we couldn't see the flickering between the two patterns.

[alter-ego]

Adding a "sharp eye" to the problem does change the result => no interference pattern is seen as long as the viewer stays focused!
The "eye" is equivalent to using a lens and the same viewing screen. The most simple and definitive way for me to think about it is based in fundamental optics, so I will describe the possible outcomes with a lens and the existing viewing screen.

1. IF the lens is positioned to exactly image the slits on the screen (the sharp eye resolving the slits), then by optical theory, the intensity distribution of object plane is duplicated (with magnification) at the image plane (the viewing screen). Since there is no interference at the plane of the slits, there will be no interference at the image plane. What you will see is the image of the two slits, back-lit. (Beta, you may be the closest)

2. IF the lens is positioned so the viewing screen is at its focus (ol' Sharp Eye is fatigued and relaxes to focus at infinity), the interference pattern will then be visible. First of all, normal creation and analysis of interference is typically done at the focal plane a lens. Why? The lens focus, or transform plane, "images" the far-field of the incident light on the lens - at the lens focal plane, the "image" is the intensity distribution at infintiy. Pure interference is ideally characterized in "angle space", or far field, which is explicitly, and ALWAYS obtained at the lens focal plane. I.e. at the lens focal plane, the intensity distribution is identical to the intensity distribution at infinity (in angle space) without the lens. Therefore, in a dark room, the relaxed eye will see (if not already blinded by a laser ) an interference pattern.

3. IF the lens is position somewhere in between ('ol Sharp Eye is struggling to regain focus of the slits), I speculate that there will be a mix of interference and image information. So maybe like Chris's explanation, the slightly blurred, partially resolved slits will show some intensity modulation (ol' Sharp Eye moves his head side to side and sees intensity changes overlaid on the slit images).

Let me just say, my profession is working / building / designing laser resonators, simple imaging systems and interferometers. Although I haven't done this experiment, I am using acquired knowledge and experience, and these descriptions are based on fundamental properties of lenses and imaging.

[Chris Peterson]

I must disagree with Chris Peterson, hstarbuck and ergophobic. As the observer moves across the screen, the brightness of the two slits will not change.

It will. I've done this in the lab, and I've also used a mechanical scanner to move across diffraction patterns and record their intensity. If you move your head so that your eye is in an interference minimum, you'll see little or no light from either slit. Move to a maximum and you'll see the slits brightly illuminated.

[Beta]

I must disagree with Chris Peterson... the brightness of the two slits will not change.

It will. I've done this in the lab... If you move your head so that your eye is in an interference minimum, you'll see little or no light from either slit. Move to a maximum and you'll see the slits brightly illuminated.

Experiment beats theory, and I'm not in a position to try the experiment (because I don't have an imaging sensor I'm not afraid of damaging with the laser), and this would solve my energy problem, but... really? You can distinguish the two slits and they get dimmer and brighter together? If so then there's something really wrong with my ideas about wave mechanics.

[alter-ego]

Experiment beats theory, ...

True, but only if the experiment is done correctly (said as a general comment, no critique intended towards anyone)

Regarding this puzzle question, I think a theoretical explanation is the intended response. No doubt I stand by the expanations I've stated - the fundamentals I've mentioned are accepted without question (ideally). However, there are real-world contributors that affect observations even in the simplest of experiments.

Real World meets Theory
1. A lens, eye or whatever is not perfect. The above explanations assume aberration-free and paraxial performance, focal plane is precisely locatable and singularly observable, and no un-wanted secondary reflections. Also, I'm assuming the lens collects light from both slits.
2. Interference occurs over a relatively long distance away from the slits, and not in an narrow region near the viewing screen. This behavior is enhanced using a laser because they are so bright AND typically have a long coherence length.

I think there are two key factors that can "contaminate" an observation:
A laser is used that has a very long coherence length which results in a long path length having high fringe visibility, and that imaging has a non-zero depth of field. These two factors could conspire to mix properly imaged slits with an overlaid interference pattern. Common are HeNe lasers with a coherence length of at least 20cm. HeNe interference might be visible beginning centimeters in front of the slits which may be in the depth of field of the imaging lens, albeit somewhat fuzzy fringes.

I am speculating on the real-world complication(s). Clearly, to un-complicate the results, I might try to use a source having a shorter coherence length (coherence length goes as 1/Spectral Bandwidth) and use a ccd camera in place of the eye. A diode or laser pointer might be better. In any case, there are viable reasons why an observed result won't fully agree with the simplified theory / explanations.

[alter-ego]

1. IF the lens is positioned to exactly image the slits on the screen (the sharp eye resolving the slits), then by optical theory, the intensity distribution of object plane is duplicated (with magnification) at the image plane (the viewing screen). Since there is no interference at the plane of the slits, there will be no interference at the image plane.

I've been confidently blabbin' away without an supporting links . In order to be a little accountable, and realizing Robert takes time to post supporting information for his stance for the answer, I took a few minutes to locate something myself. There's a ton of information out there, but the crux of the puzzle is really centered on item 1. A fundamental part of optical theory involves the Modular Transfer Function (MTF) for intensity distributions which is what we're talking about here. I think the reference below is all that is needed... and I feel better

http://www.schneiderkreuznach.com/foto/mtf.pdf

[RJN]

I believe the best answer is 'No, resolving the slits gives "which-way" information that negates interference.' Now this is a tough one and I may well be wrong in whole or in part of my brief explanation, but here goes.

Once "which-path" information is obtainable for a specific photon between two paths, that photon does not show interference with itself over those two paths. This should be true for every photon going through this setup. So when a sharp-eyed intruder resolves the two slits with her eye, she can tell which photons went through which slit, and those photons should not show interference with themselves, anymore, between those two paths. And, as the famous physicist Dirac once said, "Each photon interferes only with itself."

Does this mean that if the sharp-eyed observer stands in front of a dark interference band, she will see dark only when her eyes are closed, but light when they are open? Actually, this is what I thought when I first posed this question, and it really bothered me. Sadly, I think about these things way more than I should, and this type of situation has bothered me before. They seem relevant, for example, to the Afshar experiment.

I then learned a key point when rboatright gave his answer here: http://asterisk.apod.com/vie ... 21#p123748 . I learned that in order to resolve the two slits, the sharp eyed observer must have an eye larger than a single interference band! Therefore, it seems to me, the sharp-eyed observer sees an average of at least two interference bands, and the blink/dark, open/light situation I worried about does not occur.

What I now think is that as the sharp-eyed observer moves her head in front of the image screen that shows an interference pattern, she truly sees flux levels that correspond to there being no-interference pattern. Also, however, her eye records the same flux than can be found on the interference pattern directly behind her, when averaged over the large size of her eye.

An odd facet is that this situation seems to illuminate a case where photons do interfere with themselves even given which-path information. Oddly, though, the path-known photons do so only in a way where this interference is averaged out and cannot be detected. We live in one strange universe!

[bystander]

I think it is only necessary to observe one slit, that is enough to make the position of the photon certain (either it came through the slit we are watching, or it came through the other one). Once the position of the photon is certain, no interference pattern can be observed, only the diffraction pattern.

From Wikipedia: Double-slit experiment: Quantum version

There is a variation of the double-slit experiment in which detectors are placed in either or both of the two slits in an attempt to determine which slit the photon passes through on its way to the screen. Placing a detector even in just one of the slits will result in the disappearance of the interference pattern. The detection of a photon involves a physical interaction between the photon and the detector of the sort that physically changes the detector. (If nothing changed in the detector, it would not detect anything.) If two photons of the same frequency were emitted at the same time they would be coherent. If they went through two unobstructed slits then they would remain coherent and arriving at the screen at the same time but laterally displaced from each other they would exhibit interference. However, if one or both of them were to encounter a detector, time could be required for each to interact with its detector and they would most likely fall out of step with each other—that is, they would decohere. They would then arrive at the screen at slightly different times and could not interfere because the first to arrive would have already interacted with the screen before the second got there. If only one photon is involved, it must be detected at one or the other detector, and its continued path goes forward only from the slit where it was detected.

[alter-ego]

I think it is only necessary to observe one slit, that is enough to make the position of the photon certain (either it came through the slit we are watching, or it came through the other one). Once the position of the photon is certain, no interference pattern can be observed, only the diffraction pattern.

You are correct, but one slit or two, the key requirement is the same - the slits have to be resolved. Whether you monitor one or both is irrelevant, but you need to resolve them.

[Chris Peterson]

You are correct, but one slit or two, the key requirement is the same - the slits have to be resolved. Whether you monitor one or both is irrelevant, but you need to resolve them.

No, you don't. The slits could be infinitely narrow, for instance. You need only be able to detect the energy coming from them. The only "resolution" that is required here is the ability to distinguish two slits as two slits. That is, they need to be far enough apart to see them as two energy sources from the viewing distance with the viewing optics.

[alter-ego]

You are correct, but one slit or two, the key requirement is the same - the slits have to be resolved. Whether you monitor one or both is irrelevant, but you need to resolve them.

No, you don't. The slits could be infinitely narrow, for instance. You need only be able to detect the energy coming from them. The only "resolution" that is required here is the ability to distinguish two slits as two slits. That is, they need to be far enough apart to see them as two energy sources from the viewing distance with the viewing optics.

Yes, that is exactly what I meant - resolving the two slits is saying the same thing. Given the discussion so far, spatially resolving an individual slit has not been the primary point of interest and discussion, and, in that vein, does not make sense. In his post, Robert used "resolve" in the same context as I did, and it is common to refer to "resolving binaries" pertaining to separations, not stellar size (that's how I was thinking). It's interesting you picked up on that at this point. That semantic confusion was not even on my radar.

I appreciate your sharp eye in this matter, but I think you may be a little too picky in this case. In my first sentence, if I said, ".. a slit has to be resolved" I think that is more confusing. However, to give you the benefit of the doubt, your precise comment, "ability to distinguish two slits as two slits" is eminently clear.

[Chris Peterson]

I appreciate your sharp eye in this matter, but I think you may be a little too picky in this case.

Not trying to be picky. I was actually responding to your statement

but one slit or two, the key requirement is the same - the slits have to be resolved

which makes perfect sense with two slits, but by referencing a single slit was unclear.

We're obviously on the same wavelength here <g>.

[Henning Makholm]

And, as the famous physicist Dirac once said, "Each photon interferes only with itself."

I do hope he said so in the particular context of double-slit experiments, rather than as a general proposition.

I don't know whether interference between clearly distinct photons can be observed directly in the visual spectrum, but at radio frequencies it is a simple matter to measure interference patterns between photons that are definitely different because they come from different transmitters (the practical point being that it is easy to keep RF emitters coherent for long enough to observe the effect).