This happens in double slit, single slit and grating experiments with light.

Looking for an intuitive answer.

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    $\begingroup$ Light gets more spread out. So the same patch of intensity is now falling to a larger area as the angle increases. $\endgroup$ Feb 17 '20 at 19:39
  • $\begingroup$ @FellowTraveller So why would the peak be at the middle? $\endgroup$
    – XXb8
    Feb 17 '20 at 19:50
  • $\begingroup$ Because light has to travel the least distance there. $\endgroup$ Feb 17 '20 at 19:54
  • $\begingroup$ Have a look at this answer. $\endgroup$
    – Farcher
    Feb 17 '20 at 22:57

According to the Huygens-Fresnel Principle every point on the wavefront across a single slit becomes a source of secondary waves. At each point on the screen the waves from these secondary sources travel different distances, arriving at slightly different times with slightly different phases.

At the centre of the screen the secondary waves arrive symmetrically from the slit so there will be the narrowest range of phases (the highest coherence) and therefore the greatest amount of constructive interference, leading to a bright fringe.

As we move further away from the centre of the screen the difference in distances travelled from either end of the slit increases, so the waves arrive with a larger and larger range of phases (reduced coherence). There is less constructive interference than at the centre of the screen so the fringe isn't as bright as at the centre.

At some point on the screen depending on the width of the slit the range of phases covers a whole cycle. It isn't a uniform distribution so there isn't complete destructive interference. But as we move even further out from the centre the range of phases covers several cycles (at most, the width of the slit divided by the wavelength of light). Only phases in the range $\pm 90^{\circ}$ matter for interference, so mapping them into this range produces a more uniform distribution which tends towards complete destructive interference.

  • $\begingroup$ so nothing to do with the inverse square law of intensity? $\endgroup$
    – XXb8
    Feb 18 '20 at 8:57
  • $\begingroup$ That has a much smaller effect. There is also another geometrical factor - that the screen is increasingly oblique to the rays from the slit, so the intensity is reduced for the same reason as higher latitudes on Earth get less sun than lower latitudes. $\endgroup$ Feb 18 '20 at 9:42
  • $\begingroup$ reduced coherence? I don't get it. They have the same phase at that point (just n x \lambda shifted. Why is there reduced coherence? @sammygerbil $\endgroup$
    – Pandian Le
    Apr 5 '21 at 17:51

I will be using the word rays, by which I will mean the rays produced by wavelets.

As the angle of the rays emitted by the wavelets to the screen increases, there will be more and more of the rays interfering destructively so that the resultant amplitude will be smaller (superposition).

We also know that intensity is proportional to amplitude squared.

So, since when the angle of the rays to the screen increases, it means that the rays are approaching a farther place on the screen. And, as we know, more destructive interference occurs as the angle to the screen increases, making the resultant amplitude smaller, and as a result, this decreases the intensity.

I hope that this helped!


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