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I know the formula which proves fringe separation is inversely proportional to slit separation in the double slits, but is there an intuitive explanation to show that as slit separation increases, fringe separation decreases?

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The bright fringes in a double slit interference pattern occur when the beams from the two slits are in phase. The phase difference depends on the extra distance which one beam travels relative to the other. If the slits get farther apart, an extra distance of one wavelength occurs at a smaller angle from the forward direction.

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To see this, we need to understand first that the pattern on the screen is the norn-squared of the Fourier transform of the slits (when the illumination is via plane-wave).

It follows the property of the scaling-property of Fourier transform that any dimension of real space is scaled inversely in the Fourier domain. A well-known example is a Gaussian function: a narrower Gaussian function in real-space is wider in Fourier-space.

The following animation shows the fact you mentioned, namely the separation scales inversely for the interference pattern on the screen. The Blue curve is the norm-squared of the Fourier transform of a single slit (a single rectangular function), and the orange curve is the norm-squared of the Fourier transform of two rectangles separated by a distance $d$. For the purpose of aesthetic, I multiplied the blue curve by a factor of two ( thanks to Pieter for pointing out that )

enter image description here

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    $\begingroup$ Nice animation! But if you talk about "norm squared", should not the total area of two slits be twice as large (conservation of energy) and the central maximum be four times as high (square of the sum of two in-phase fields) as for the single slit? $\endgroup$ – Pieter Feb 17 at 15:14
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    $\begingroup$ Pieter, yes. The area is twice as big. I should have mentioned that the orange is divided by a factor of two for visual purpose. $\endgroup$ – Mehedi Hasan Feb 17 at 15:31

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