For an N-slit diffraction grating, the distance from a maxima to a minima at order p is given by
$$\delta \theta = \frac{\lambda}{Np}$$
What happens to this width when the centremost $\frac{N}{2}$ slit is covered?
I'm tempted to think that one slit won't make a difference if N is large like 1000 slits.
I think I have solved it.
Without covering the slit, we have the usual diffraction:
$$u = u_0 e^{i(kz-\omega t)} \space \frac{1- e^{iN\delta}}{1- e^{i\delta}} = u_0 e^{i(kz-\omega t)} e^{i(\frac{N}{2}\delta - \frac{\delta}{2})} \frac{\sin \left(\frac{N\delta}{2}\right)}{\sin \left( \frac{\delta}{2}\right)}$$
After covering the slit, we simply have:
$$ u = u_0 \left[e^{i(kz-\omega t)} e^{i(\frac{N}{2}\delta - \frac{\delta}{2})} \frac{\sin \left(\frac{N\delta}{2}\right)}{\sin \left( \frac{\delta}{2}\right)} \space - \space e^{i(\frac{N\delta}{2})}\right] $$
The intensity is given by $I \propto uu^*$:
$$I = I_0 \left[ \frac{\sin^2 \left( \frac{N\delta}{2} \right)}{\sin^2\left(\frac{\delta}{2}\right)} \space -2 \cos\left(\frac{N\delta}{2}\right) \frac{\sin \left( \frac{N\delta}{2} \right)}{\sin \left(\frac{\delta}{2}\right)} + 1 \right]$$
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,
$$I = (N-1)^2 \left[ \frac{\sin^2 \left( \frac{N\delta}{2} \right)}{\sin^2\left(\frac{\delta}{2}\right)} \space -2 \cos\left(\frac{N\delta}{2}\right) \frac{\sin \left( \frac{N\delta}{2} \right)}{\sin \left(\frac{\delta}{2}\right)} + 1 \right]$$
Maximas still occur at $d \sin\theta=n\lambda$, with intensity $(N−1)^2$.
Minimas now occur at $\frac{N\delta}{2}=\frac{n\pi}{2}$. So minimas are at $d\sin\theta= \frac{n}{N}\frac{λ}{2}$.
Comparing this to the uncovered situation, we have $\delta \theta \space ' = \frac{1}{2} \delta \theta$.
