0
$\begingroup$

I don't understand why the probability density in the double slit experiment in the case of both slits opened, has a minimum corresponding to the maximum of intensity. Shouldn't $P_{12}$ have the same trend as intensity?

enter image description here

Intensity patterns in the double-slit experiment (a)Photon intensity $I_1$ on the screen with slit-1 only opened (b) Photon intensity I_2 with slit2 only opened (c) Interference pattern when both slits were opened d) probability distribution $P_{12}$ of photons when both the slits were opened and when the film was replaced by an array of photon detectors.

The manual says Probability distribution P_12 of photons when both the slits were opened and when the film was replaced by an array of photon detectors

Improve this question
$\endgroup$
4
  • $\begingroup$ How do you define "intensity"? How is it different from the probability density? (I would say that they are the same, expect that one term is used for electrons and the other for electromagnetic waves.) $\endgroup$ – Roger Vadim Sep 30 '20 at 8:29
  • $\begingroup$ In the case of em waves is defined like this $I = |\mathbf{E}(\mathbf{x},t|^2 = I_1 + I_2 + 2Re(\mathbf{E_1}^{*} \cdot \mathbf{E_2})$ $\endgroup$ – Giuliano Malatesta Sep 30 '20 at 8:32
  • $\begingroup$ Precisely. And in the case of electrons we have probability density: $\rho(\mathbf{x},t) = |\psi(\mathbf{x},t)|^2 = |\psi_1|^2 + |\psi_2|^2 + 2\Re(\psi_1^*\psi_2)$ $\endgroup$ – Roger Vadim Sep 30 '20 at 8:34
  • $\begingroup$ It is unclear what these curves represent. a) and b) do not show any diffraction. c) Looks asymmetric and the zero order light is quite weak. I have no idea what d) can be. Perhaps a relative 180 degrees phase was applied between the two slits. In the present form I propose to close the question. $\endgroup$ – my2cts Oct 29 '20 at 9:45
0
$\begingroup$

The hands on way to understand this is by assuming that a wave parting from each of the slits is a plane wave, i.e. $$ \psi_1(\mathbf{x}) = e^{i\mathbf{k}_1(\mathbf{x}-\mathbf{x}_1)}, \psi_2(\mathbf{x}) = e^{i\mathbf{k}_2(\mathbf{x}-\mathbf{x}_2)} $$ Then one can look at the probability density in the plane $x=x_0$ and see that it will exhibit oscillatory behavior: $$ |\psi_1(\mathbf{x}) + \psi_2(\mathbf{x})|^2 = 1 + 1 + 2\Re\left[e^{i\mathbf{k}_1(\mathbf{x}-\mathbf{x}_1)}e^{i\mathbf{k}_2(\mathbf{x}-\mathbf{x}_2)}\right] $$

Improve this answer
$\endgroup$
0
$\begingroup$

You didn't say what the manual claims that (a) through (d) are supposed to represent, but as far as I can tell the diagram is simply wrong.

If (a) and (b) are supposed to be the intensities on the screen when only one slit or the other is open, then the intensity when they're both open will have to look essentially like the sum of those two curves. Interference can only happen at points on the screen that get a significant amount of light from both slits, and in their setup that's nowhere.

I also don't know what they mean by "the film was replaced by an array of photon detectors". A film is an array of photon detectors. Nor can I see why (c) would be different from (d).

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.