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I know this question has been asked quite a few times on the network but I still can't find a concrete answer. It's quite clear that intensity depends on the square of slit width in a single slit diffraction experiment and its explanation is easy to understand (Explanation). But when I check for the same relation for a double slit experiment, some sources claim intensity is proportional to the slit width (A similar question)(Source 2)(Source 3)

Why is this so? Shouldn't the intensity of each of the slits be proportional to the squares of their respective slit widths, just as it is in the single slit experiment? Or is it because we assume no diffraction occurs in such cases, and if so, how does the $I\propto W$ relation come about? Or is it just an experimental observation? Or is my assumption incorrect?

Please do explain the reasons behind whatever the right answer is. Thanks in advance.

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  • $\begingroup$ It's hard to give an accurate explanation without intensive mathematical analysis. However, I'll try to give some argument in favour of why it shouldn't be proportional to $\omega^2$ . Imagine a point for which the condition of constructive interference is satisfied in the limit of infinitesimal slit width (for a double slit situation). When you increase the slit width you obviously let more light through but you also slightly break the condition of constructive interference as now the phase difference is not exactly the multiple of $2\pi$ for every "pair of points" of two slits. $\endgroup$
    – Dario Barišić
    Commented Mar 3, 2023 at 16:17
  • $\begingroup$ @DarioBarišić Okay, so basically the interference that's involved in the double slit experiment is what's causing the intensity to rise less rapidly, right? $\endgroup$
    – RedMiner2005
    Commented Mar 4, 2023 at 5:08
  • $\begingroup$ Basically as you increase the slit width your interference pattern will become more "blurry", i.e. it will not stay as sharp as it would for a very small slit and therefore the intensity will not just simply scale quadratically with slit width. Here, specifically I'm refering to intensity of the interference fringes. $\endgroup$
    – Dario Barišić
    Commented Mar 4, 2023 at 16:46

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