# In single slit diffraction, why does the minima occur at integer multiples of lambda, which is contrary to double slit?

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?

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 $$0$$.