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In this derivation, I can understand the formulas stated for 1D diffraction for both x andy (with the phase difference $\delta= ksin\theta y$) in the far-field limit, but how can we just sum the phases from both the x and y components in such a simple manner?

I would expect a formula like pythagoras would be needed instead of just $\delta = k(sin(\phi) x + sin(\theta)y)$.

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You would need a more complicated formula if you were doing Fresnel diffraction, but here you are making the Fraunhofer approximation that all rays are parallel.

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  • $\begingroup$ Even with this approximation, I'm not sure how the total path length difference can just be equal to the path length difference in the x direction + the path length difference in the y direction. Its obvious geometrically for the 1D case, but I've not seen a geometrical diagram to show it for 2D. $\endgroup$
    – Alex Gower
    Commented Aug 27, 2020 at 8:50
  • $\begingroup$ @AlexGower Did you ever figure this out? I've run into this exact problem as well, all the derivations I have seen are 1D and I understand perfectly well that it works in 1D, but when considering off-axis points in 2D my attempts show that the path length differences across the aperture start having terms of the form $\sqrt{x^2 +y^2}$ (coefficients omitted) which makes the subsequent integrals intractable. Am I missing something obvious or is this just being handwaved away? $\endgroup$
    – Thomas
    Commented Dec 26, 2022 at 3:50

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