When diffracting light through a double slit, I understand that the two sources must arrive on the screen m(wavelengths) out of phase [where m is any whole number], in order for a bright fringe to form. This is due to the fact that a path difference of mλ would result in constructive interference, i.e. reinforcement, and hence maximum intensity would be detected at this point.

However, when diffracting light through a single slit, the situation is different. The two sources must arrive on the screen (m+1/2)λs out of phase, e.g., λ/2, 3λ/2, 5λ/2 etc, for constructive interference. I cannot understand why this is the case, and my physics teacher cannot either. Why would this result in constructive interference and not destructive?

The sources I'm referring to are: https://www.youtube.com/watch?v=7CmbItRjM-Y&ab_channel=khanacademymedicine https://www.youtube.com/watch?v=T-kgoxhFSmU&ab_channel=khanacademymedicine

Any help greatly appreciated!

  • $\begingroup$ Did you watch the video to see how he explained it? $\endgroup$ – John Jan 18 '16 at 21:55
  • $\begingroup$ It shows you how it considers the single slit as two sources, and they're only 15 minutes long $\endgroup$ – John Jan 18 '16 at 22:01
  • $\begingroup$ I watched the video. He is talking about destructive interference for the single slit versus constructive interference for the double slit. $\endgroup$ – Peter R Jan 18 '16 at 23:15
  • $\begingroup$ @AccidentalFourierTransform, ..."if there is only one slit, how can there be two sources?" Because the slit, no matter how small, has a certain width. It's similar to how light through a point-like aperture results in an Airy disk. $\endgroup$ – David Reishi May 13 '16 at 5:36

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