0
$\begingroup$

In Afshar's paper, there is an image of the difference in the interferences with both as well as single slits open. enter image description here Did he measure the pattern on the same detector with both the slits open individually?

So my question is: In Afshar's double slit experiment, is the interference on 1 detector with both the slits open the superposition of the interference with each slit open one by one? Please tell me if my question is not clear.

$\endgroup$
  • 1
    $\begingroup$ Can you attach the paper you are referring to? $\endgroup$ – user35122 Oct 9 '16 at 10:54
  • $\begingroup$ @user35122 I haved added the link to the paper. $\endgroup$ – Prem kumar Oct 11 '16 at 17:58
2
$\begingroup$

Yes, you get the total wave amplitudes for both slits being open by adding the wave amplitudes you get for each slit being open, as long as there is nothing in the apparatus that can determine which slit the particle went through. This is just what you would do with sound waves or water waves also, by the way. (To clarify, wave amplitude is what each individual source of that wave produces, and you must add all the amplitudes from all the sources together to get the total amplitude, which then has both a magnitude and a phase. The magnitude is what the energy in the wave depends on, and the phase is what you think of as where the wave is locally in its oscillation. Since the contributing amplitudes at any given point can be 180 degrees out of phase with each other, they can cancel rather than add, and that is what produces interference fringes in the two-slit experiment.)

| cite | 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.