nliu, they kinda sorta mildly touch this in Synchrotron detail study, but I agree. Explained muchhhh more in depth in A level (eg singapore) or highschool equivalent in china and other asian countries.
Anyways, for those interested from page 496:
A collimated beam of X-rays incident upon a layer of atoms will be scattered. Bragg’s law states that the beams will interfere constructively, producing maxima when the following relationship is satisfied:
2dsinθ = nλ where d is the distance between layers of atoms (m), θ is the angle the X-ray beam makes with the surface, λ is the wavelength of incident X-ray photons (m) and n is the number of the maxima occurring, 1, 2, 3, etc.
What nliu is trying to say is that as d (which can be considered as the slit width) and λ (wavelength) are approaching the same width, ie d/λ -> 1, the equation becomes:
2λsinθ = nλ -> sinθ= n/2.
n=1 is the centre band, so we look for n=2, the band next to it:
sinθ= n/2 -> sinθ= 2/2 -> θ= 90 degrees.
This implies that the wave has defracted so completely that the next band occurs at right angles to the path, ie parallel to the screen.
So, using a "similar" version of Bragg's Law (sadly barely touched upon in VCE) the max defraction occurs when the ratio of wavelength to slit with approaches one.
Ah Bragg's law. It's similar, although in that case you've got to remember where the two comes from; the path difference is 2 d sin theta, and that's because there are two layers in the crystal lattice and the 2 comes from a return journey from the first layer to the second layer. Look it up on wiki for a diagram.
Alwin, you'll confuse people xP
In double slit interference, the condition is d sin theta = m lambda for CONSTRUCTIVE interference because d sin theta is the path difference.
For single slit diffraction, the argument works differently. Again, look at some diagrams; they'll be better than what I can draw.
But essentially, if we split up the slit into half, to form two minislits of size d/2, then we have a double slit scenario with path difference d/2 * sin theta. This is equal to half a wavelength for minimum, so d/2*sin theta = w/2. d sin theta = w
If we split the slit into quarters, then quarter 1 can interfere destructively with quarter 2 and then 3 with 4. The slit width is now d/4, so the condition is d/4 * sin theta = w/2. d sin theta = 2w. Etc if we split the slit into 6, 8, 10, 12.
That's sort of where it comes from.
Define "max diffraction" though.