I understand the way double-slit is shown and ridges pattern calculated (e.g. https://personal.math.ubc.ca/~cass/courses/m309-03a/m309-projects/fun/Slits.html#topic1) is by assuming two identical waves coming out of the slits. But wave length of visible light is about 1/1000 of slits separation distance.

I recall reading photon is huge in volume even for small wave length, only that wave density is tiny at large distances from "center" so it can go to two slits, no problem. But does it go in a symmetrical way? Or in fact two waves from slits are not same in "intensity" (if such way of thinking even makes sense for photon), but by averaging out many photon the pattern is still the same as if each photon were doing through two slits in a symmetrical way? TIA

  • $\begingroup$ look at this answer, that shows single photon at a time physics.stackexchange.com/questions/285142/… . no symmetry. it is all quantum mechanics probabilities. $\endgroup$
    – anna v
    Nov 10, 2022 at 5:36
  • $\begingroup$ @annav, I did not see it talking about asymmetry. $\endgroup$ Nov 10, 2022 at 7:39
  • $\begingroup$ did you look at the data? it is obvious in the radndom positions of the dots on the left $\endgroup$
    – anna v
    Nov 10, 2022 at 8:00

1 Answer 1


Let's say situation is otherwise symmetrical, but areas of slits differ by 100%.

Now for each photon that gets through the screen with the slits, 66.66% of the photon comes out of the larger slit, and 33.33% comes out of the other slit.

This produces a low-contrast interference pattern. The brightness varies only by 300%.

When the same percentage of photon comes out of both slits, then the brightness varies by infinite %, as there exists a perfect black as a result of perfect cancellation.

Now let's say slits are identical, one slit is movable, and we start at symmetrical situation. We move the movable slit until it receives 50% of the amount of light that the other slit receives. Now we have the same low-contrast interference pattern as in the previous case.

  • $\begingroup$ This really isn't a good way to approach an interference problem. Forget photons, use waves. Then, the wave intensity tells you where to expect photons to show up. $\endgroup$
    – John Doty
    Nov 10, 2022 at 16:13

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