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Why is no interference pattern in the vertical axis in a double-slit ? Isn't a vertical slit just thousands of small slits one next to the other ? So shouldn't this create an interference pattern also along the vertical axis ?

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    $\begingroup$ The vertical dimension is usually too big. Had it been comparable to the slit width you would have obtained an interference along that dimension too. $\endgroup$ – Raziman T V Jan 8 '17 at 3:44
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The diffraction pattern of a pair of slits can be obtained from Fourier theory. This is because the far field is given by Fraunhofer diffraction which is essensially a Fourier transform of the near field. So to understand why one has a particular diffraction pattern, one simply needs to see how the Fourier transform of the near field (two slit pattern) looks.

The question is only concerned with one direction, so we'll focus on that direction. The argument is that the vertical direction can be seen as an infinite number of slits lying against each other. Well, the Fourier theorems tell us that in that case, we won't see fringes.

We can construct such a one-dimensional function as a square function that represents the slit convolved by a comb function $$ h(y/w) \otimes C_d(y) $$ where $h(y/w)$ is the is the square function of width $w$ and $C_d(y)$ is a comb function consisting of impulse functions separated by $d$.

The Fourier transform of this is given by the product of the Fourier transform of the square function, which is a sinc function $$ {\rm sinc}(wu) = \frac{\sin(\pi wu)}{\pi wu} $$ and the Fourier transform of the comb function, which is again a comb function, but where the impulses are separated by $1/d$. Note that the sinc function has zeros where $u=n/w$ for any integer $n$ not equal to zero.

Now it is easy to see what happens when the slits touch each other. It means that the width of the square function is equal to the separation between the impulses. For the Fourier transform, it implies that the locations of the impulses in the diffraction pattern would coincide with the zeros of the since. So the only impulse that survives is the one at the origin. As a result there are no fringes.

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