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Single slit diffraction is also a kind of interference, so why does the minima do not occur at a path difference with half the wavelength, but occur at integer multiples? In my lecture notes, it says divide the gap into halves, and the ray from these different halves would somehow cancel out because they are half a wavelength out of phase. Then, a/2sinΘ = lambda/2, wichi gives back asinΘ = m lambda. But it doesn't make sense to me. Is it merely because of some difference in geometry compared to the double slit? Or is there a more physical meaning?

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The first $0$ of the diffraction pattern of a wide slit corresponds to an angle $\theta$ such that $sin(\theta) = \lambda/a$. The path difference between the extreme rays is therefore $a sin(\theta) = \lambda$

To understand why the total amplitude is zero in this case, you can effectively decompose the wide slit into pairs of thin slits $a/2$ apart.

For each pair, the path difference between the rays is $(a/2) sin(\theta) = \lambda/2$: the waves emitted are in phase opposition and give a zero sum.

In total, by summing the amplitudes of all the pairs, we find $0$.

Hope it can help and sorry for my poor english.

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  • $\begingroup$ Wait, I thought the first zero (i.e. m= 0) is a maxima? So the amplitude at the centre is not zero? Sorry if I misunderstood. $\endgroup$ Commented May 11, 2023 at 1:14
  • $\begingroup$ Maybe I expressed myself badly: I wanted to talk about the first dark fringe, where the amplitude is zero. (English is not my native language). $\endgroup$ Commented May 11, 2023 at 6:24

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