GRED Answer: Double slit with early divider

[RJN]

Guess the Result of the Experiment of the Day (GRED)

A classic double slit experiment is done with light (photons) that creates an interference pattern on the image screen. The experiment is repeated, except now a thin but opaque divider is placed from the source to the slit screen that divides the two slits. Here is an ASCII schematic:

Code: Select all

                                                            
                               |                               |
     s ------------------------|                               |
           opaque divider      |                               |

    Photon                   Double                          Image
    Source                    Slit                          Screen
                             Screen

What pattern appears on the image screen?

The initial poll, where spoilers were not allowed, can be found here: http://asterisk.apod.com/vie ... 30&t=19557 . If you are new to this GRED and want to ponder this question without seeing spoilers, please go there now instead of scrolling down.

The answer is now given below here: http://asterisk.apod.com/vie ... 66#p122606

This GRED and poll will remain open for comments however.

[Philip]

What is the width (and smoothness) of the divider? What is the width of the photon emission? My guess is that with the right thickness, no photons will hit the screen. Also what is the material of the divider? With some of those new metamaterials... it is as if there would be no divider and hence an interference would show. So the key questions are, in my opinion, divider constituency, its width and width/breath of the photonic source. If I am complicating things too much... then since this is QM and we have to expect the illogical, then the interference pattern will show.

[RJN]

The divider is thinner than the distance between the slits. It could be made of anything opaque. I have tried not to name specific length scales since theoretically scales are only relevant in comparison.

Still, in one realization of this, were the source a standard red laser pointer, I would think that a standard piece of printer paper could be used as the divider. The paper would be thin enough so that it would not block a signifcant amount of laser light from the slits. The paper would be thick enough, however, so that little laser light would go through it.

I hope this is a help.

[mscullin]

Since the divider does not prevent the source from seeing both slits (or both slits from seeing the source), there should be no effect. The divider doesn't really divide anything except the space. It creates only a thin shadow on the center of the "slit wall" which would have no effect on the light striking either slit.
Either that or global warming's effect on the apparatus makes the outcome unpredictable.

Mr.Mike

[rcharlet]

Howdy,
my guess = interference pattern. The divider dose not stop the source waves from reaching the slits. But the divider may cause reflections and additional waves. These additional waves may hit the slits and cause even further interference patterns. But the reflected set of wave will be much weaker than the direct waves.
I have pondered this out by thinking in terms of waves and I may or may not be correctly guessing. If there turns out to be some quantum/packet effect here, those things tend to be above my ability to ponder (but I still may or may not be correctly guessing).

[gbottrell]

The interference comes from the wave nature of the indivudual photon, i.e. without the divider, the photon goes
through both slits and causes interference. By putting the divider in, it insures that the photon only goes through
one slit. I think this would cause there to be no interference pattern.
GB

[makc]

without the divider, the photon goes
through both slits and causes interference. By putting the divider in, it insures that the photon only goes through
one slit.

why then opaque material between two slits is not enough to ensure that the photon only goes through one slit?

[Guest]

Assume for the sake of argument that one of the slits is blocked. The other slit would certainly produce a diffraction pattern on the final screen. The constructive/destructive interference of the diffraction patterns from two such slits is what produces the classical interference pattern from Young's double slit. So long as the source remains coherent, there's no way to get around this, so when both slits are illuminated, there MUST be an interference pattern.

The problem is that both slits might not be illuminated. No matter how thin any practical divider might be, it is considerably larger than the wavelength of light, and it must itself produce a diffraction pattern on the screen with the slits. (Think about the effect of a hair on a projector lens, or to choose a slightly more well-known example, the Fresnel spot.) Whether both slits are, in fact, illuminated depends on their separation. As a result, it's impossible to give a definitive answer in the absence of dimensions. Given the relatively small physical extent of the minima in the diffraction pattern projected onto the slit screen, there will PROBABLY be a visible interference pattern, but it's not guaranteed. The structure of that pattern, however, is likely to be quite different from the "normal" interference pattern produced by Young's double slit due to the (probably) different illumination of the two slits.

The response from gbottrell sounds plausible, but it doesn't hold up to scrutiny. A double-slit experiment can be (and has been) done using electrons, and it also produces an interference pattern. One might argue that a large collection of electrons must behave like a wave, but even when the emissivity of the electron source is so drastically reduced that only a "single electron" passes the slit plane at any given time, an interference pattern is slowly built up over many successive electron detections. This means that each individual photon needing to "go through" either one slit or the other doesn't necessarily preclude the development of an interference pattern. The the most important effect of the divider really is to change the illumination pattern at the slit plane, and the result can't be determined without more detailed information.

[hstarbuck]

If it were water waves in this scenario there would be an interference pattern. If the divider was infinitesimally thin, there would not be diffraction or reflection with orthogonal waves (I think) therefore I believe the easy answer here is interference. If photons only interfere with probabilities of themselves and not other photons, and since either path is equally probable, I believe any position on the screen is possible within the normal interference pattern for each photon. Over time the pattern will develop--with constant light or 1 by 1 photons. I am thinking the photon stays entangled until it is recorded (measured/observed) on the screen. Interference.

[Guest]

The response from gbottrell sounds plausible, but it doesn't hold up to scrutiny. A double-slit experiment can be (and has been) done using electrons, and it also produces an interference pattern. One might argue that a large collection of electrons must behave like a wave, but even when the emissivity of the electron source is so drastically reduced that only a "single electron" passes the slit plane at any given time, an interference pattern is slowly built up over many successive electron detections. This means that each individual photon needing to "go through" either one slit or the other doesn't necessarily preclude the development of an interference pattern. The the most important effect of the divider really is to change the illumination pattern at the slit plane, and the result can't be determined without more detailed information.

(please use [quote]tag[/quote] next time, thanks - makc)

The comparison to the electron interference was exactly what I was thinking about. When the electron source is reduced
so that only one electron at a time is emitted, an interference pattern is produced. The pattern is produced precisely
because the electron treated as a wave can not be said to have gone through one slit or the other. It has to be thought
of as going through both (see "The Feynman Lectures on Physics" vol.3 pg. 1-6). Whenever any experiment is done that
determines which slit the electron actually went through (by having some detector watch the slit), the interference pattern
is destroyed, and the electrons act like particles. That is the point of what is called the wave-particle duality. When you
do an experiment that looks for particle like behavior of objects, they exhibit particle like behavior. And when you examine
their wave like behavior, you see the wave like behavior.

Putting the opaque divider between the two slits does not allow the photon to have amplitude at both slits. In
effect, you are requiring the photon to go through one slit or the other. Thus you are measuring the particle like
behavior of the photon. True, each slit will produce a single slit diffraction pattern, but the two diffraction patterns
will merely overlap, they will not interfere.
GB

[makc]

Putting the opaque divider between the two slits does not allow the photon to have amplitude at both slits. In
effect, you are requiring the photon to go through one slit or the other.

why then opaque material between two slits is not enough to ensure that the photon only goes through one slit?

[RJN]

I believe that the answer is "An interference pattern". I have not done this experiment, but it seems to me that "which path" information does not exist. It appears to me that it is not possible to establish, even in principle, which slit a photon went through on its way to the image screen. The "early divider" cannot provide this information.

In retrospect, I should have included that the divider is not only opaque and so does not allow transmission, but black and so does not allow reflection either. Even so, added reflection will give only a more complicated interference pattern, but an interference pattern never-the-less.

This is a very easy experiment to do, and I would be grateful if someone would actually do it and report back here. In particular, a simple set up involving on a laser pointer and a pin is described in Scientific American here: http://www.scientificamerican.com/slide ... tum-eraser

- RJN

[poodlepeople]

I am a bit confused by references to both interference and diffraction patterns. A single slit will produce a diffraction pattern. The classic double slit will produce an interference pattern.

If you cover one slit of the two slits in the setup with one edge of a very long, opaque sheet, you will see a diffraction pattern from the other slit. Now "swing" the sheet toward the screen such that the one edge remains anchored at the point between the two slits. (This will be a "very" long sheet to reach the screen at the this initial position.) You will continue to see a diffraction pattern from the open slit and also a diffraction pattern from the second slit (mostly on the sheet if it is at all reflective and distorted by the angle of the sheet to the slit material).

If you continue to swing the opaque sheet toward the screen (it gets "shorter" as you do this, of course), you will continue to see diffraction patterns on each side - one from each slit. Once the sheet is perpendicular to the screen this will continue to be the case but at no time with there be an interference pattern from a photon going through both slits.

I am not sure just what additional assumptions are necessary for this argument to be valid but it seems to be a way to reason that there should be no interference pattern for the case that was originally provided.

[bystander]

@poodlepeople: That would be Double slit with late divider.

See: http://asterisk.apod.com/vie ... 30&t=19609