mfb said:
Just one slit is covered, or the central half?
A single slit won't make a large difference, half of them could.
I think I might have solved it.
I'm subtracted the diffraction pattern with the pattern produced just by that slit alone to find the new pattern.
Without covering the slit, we have the usual diffraction:
[tex]u = u_0 e^{i(kz-\omega t)} \space \frac{1- e^{iN\delta}}{1- e^{i\delta}} = u_0 e^{i(kz-\omega t)} e^{i(\frac{N}{2}\delta - \frac{\delta}{2})} \frac{sin \left(\frac{N\delta}{2}\right)}{sin \left( \frac{\delta}{2}\right)}[/tex]
After covering the slit, we simply have:
[tex]u = u_0 \left[e^{i(kz-\omega t)} e^{i(\frac{N}{2}\delta - \frac{\delta}{2})} \frac{sin \left(\frac{N\delta}{2}\right)}{sin \left( \frac{\delta}{2}\right)}\space - \space e^{i(\frac{N\delta}{2})}\right][/tex]
The intensity is given by ##I \propto uu^*##:
[tex]I = I_0 \left[ \frac{sin^2 \left( \frac{N\delta}{2} \right)}{sin^2\left(\frac{\delta}{2}\right)} \space -2 cos\left(\frac{N\delta}{2}\right) \frac{sin \left( \frac{N\delta}{2} \right)}{sin \left(\frac{\delta}{2}\right)} + 1 \right][/tex]
To find ##I_0##, simply substitute in ##\delta = 0##. We know that the central intensity is proportional to ##N^2## where N is number of slits, so we would expect ##I_0 = (N-1)^2## here.
Substituting ##\delta = 0##, we have ##I \propto (N^2 -2N + 1) = (N-1)^2##, so it is satisfied!
Assuming wave is normalized,
[tex]I = (N-1)^2 \left[ \frac{sin^2 \left( \frac{N\delta}{2} \right)}{sin^2\left(\frac{\delta}{2}\right)} \space -2 cos\left(\frac{N\delta}{2}\right) \frac{sin \left( \frac{N\delta}{2} \right)}{sin \left(\frac{\delta}{2}\right)} + 1 \right][/tex]
Maximas still occur at ##d sin \theta = n\lambda##, with intensity ##(N-1)^2##.
Minimas (non-zero intensity) now occur at ##\frac{N\delta}{2} = \frac{n\pi}{2}##. So minimas are at ##d sin \theta = \frac{n}{N}(\frac{\lambda}{2})##.
Comparing it to the uncovered case, this implies that ##\delta \theta \space ' = \frac{1}{2} \delta \theta##.