The Stack Overflow podcast is back! Listen to an interview with our new CEO.

New answers tagged

0

Bright spots in the pattern are where almost all photons hit, dark spots in the pattern are where almost no photons hit. The emission will occur in the bright areas. In double slit interference the classical explanations implied photons were hitting in the dark areas but cancelling each other out (which is a violation of conservation of energy). The ...


3

It will emit electrons according to where photons in the interference pattern hit.


0

It is difficult to give a definitive explanation of the MWI, since there seems to be so many variations of it. However, a logical common denominator is the assumption that the branching occurs at those points at which the Copenhagen interpretation would assume that the wave function had collapsed, which in the two slits experiment would be when the particle ...


0

Interference is the most poorly explained concept in physics, often it is based on older classical thinking which is OK for many problems. The modern method is the probability wave but it is complex to apply the correct boundary conditions for the double slit. One way to understand it is to see that the wave function requires that the photon or electron must ...


1

The many worlds interpretation (MWI) and the Copenhagen interpretation (CI) are not different physical theories, because in all real-world experiments devised so far, they make the same predictions. The theory is quantum mechanics. MWI and CI are interpretations. Physicists therefore do not have to pick one or the other to believe in. These days, attention ...


0

In this experiment a 1000 electrons land on a sensor that creates an image of their position distribution. The probability of an electron to land on a particular location is given by the square of the wave function. This squared wave function is the famous interference pattern. Assuming the detector has N pixels, it takes in principle $2^N$ worlds to ...


0

I think your idea is an ingenious one, but for it to make physical sense you would have to develop it to explain all quantum phenomena, and not just the two slits experiment with photons, and show how it was consistent with relativity. As to your suggested experiment, it would be good to suggest a rationale for why 30s was a sufficient time lag, given that ...


1

The traditional interpretation of quantum mechanics was that during measurements the wave function of a particle switched instantaneously to one of the allowed 'eigenfunctions' associated with the property being measured. If you subsequently measured some other property with a different set of allowed eigenfunctions then the wave function would switch again ...


0

The wave function for a free particle experience dispersion (this means that components with different energies, travels at slightly different speeds). This causes a free particle, that is initially well localised, to become more and more delocalized. The wave function spreads as it moves around in space (propagates).


1

Collapse is an unfortunate terminology, imo. The wavefunction is a mathematical function , not a balloon When you throw a dice and it comes up 6, has anything collapsed? It is but one event, that follows the probability distribution. Probabilities are the same in quantum and classical states. The confusion arises by the people who want to think of the ...


0

Good question but I believe the pattern is unaffected. Just the intensity would decrease with increased thickness as photons have to be on smaller path angles to make it through.


0

It's a great idea, but if you follow its implications it can't be quite right. The interference pattern is there whether or not you look at it. You can get a robot to perform the experiment and video the results so that millions of people can watch it on Youtube. Best wishes


0

It is entirely equivalent to tilting the projection screen relative the sources. The angular pattern of minima nd maxima remains the same relative the line joining the two light sources, but those rays are now projecting onto a different surface.


0

On page three of your paper is the sentence The double-slit consists of two 50-nm-wide slits with a center-to-center separation of 280 nm (see inset 1 in figure 1) So in this case the gap between the slits is about four times the width of each slit. That relative spacing helps the double-slit interference pattern to dominate over the single-slit ...


1

Your title question and your question in the body is different. You are not specifically saying where the detector (and what type of detector it is) is positioned. Now in the body your the question you are asking what would happen if you would turn on the detector after the photon has already hit the screen and interacted with it (left a dot). In this case, ...


1

To a reasonable approximation it is a cylindrical wave convolved with the single slit amplitude function in the angular direction. The axis of the cylinder is lies at the center of the slit. (Here I have assumed that the slit is much longer than it is wide and that we don't care about the behavior in the long direction.) It will keep it's frequency and ...


6

Which one is right? I think that neither of the two options that you presented is completely right, but I think that the true answer contains elements of both choices. Without getting into the real physics (i.e., the math): The geometry of the experiment (two slits, and the screen) defines a wave function. Any photon that gets past the slits will make at ...


11

The photon is a quantum mechanical entity. Number 1) cannot be right, because the experiments with single photons at a time give single dots as photon footprints not bright regions . Single-photon camera recording of photons from a double slit illuminated by very weak laser light. Left to right: single frame, superposition of 200, 1’000, and 500’000 ...


0

Any edge deflects photons into areas with higher and lower density. To measure such a result different measurement instruments can be used. Beside a screen one may use a photodiode which is moved and the outcoming current shows the same intensity distribution like the pattern on a screen. The detection of photons direct behind the slit destroys the ...


0

Whatever people see in your demo setup will not be the photons themselves, but something representing the photons. So, you might as well just use a TV screen, displaying the output of an imaging photomultiplier. If you want the setup to be visible to onlookers while it's operating, you might be able to put a single-wavelength filter over the ...


0

Here are 2 basic documents on the double slit from Harvard and MIT. The first having a nice demo based on single photons but could be easily converted to a cheaper camera and many photons (same results). And the second gives the "classical" interference math which gives a good approximation of the image geometries based on lamda and other variables that ...


0

If you don't demand that you are doing a "two slit" experiment, then there are inexpensive diffraction gratings that will give you first maximum angles of ten degrees or more with visible light. Which means projection distances less than a meter will work, and that in turn means that an inexpensive diode laser is more than good enough. Such grating can be ...


1

On the assumption that the screen is far away from the two slits and the angles involved are small the fringe separation is $\Delta x = \frac {\lambda D}{d}$ where $\lambda$ is the wavelength, $d$is the slit separation and $D$ the two slit to screen distance. So $\Delta x \propto D$. A diagram with the angles exaggerated is shown below with the radial ...


2

Your question is an excellent one, and puzzled me when I was studying for my PhD in quantum theory. The answer is that the intermediate screen in which the slits are formed can indeed collapse the wave functions of electrons projected towards it; but they are the electrons that make it through neither slit, being instead absorbed or scattered by the ...


1

The exact path difference is: $$ \Delta = \sqrt{D^2 + \big(\frac w 2\big)^2} -D $$ $$ \Delta = D\big[ \sqrt{1 + \big(\frac w {2D}\big)^2} - 1\big] $$ $$ \Delta = D\big[ \sqrt{1 + \epsilon^2} - 1\big]$$ with $$ \epsilon \equiv \frac w {2D}$$ as a small parameter under $w/D \ll 1$. Now Taylor expand around $\epsilon = 0$: $$\Delta = D\big[ 1+\frac 1 2 ...


-1

Sometimes yes, observation of very small particles influence those particles because of the extremely small size of the particle that is attempted to be observed. I guess it depends, this can be both right and wrong. Sometimes you can observe it and influence the photon to become like a particle or wave as you observe it. Interference can also happen in ...


Top 50 recent answers are included