# Covering centeremost slit of a N slit diffraction grating - what happens?

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.

• It won't make much of a difference, but the whole point here (I'm guessing this is homework?) is to evaluate the pattern generated by that one slit and see what happens to the overall pattern. May 27, 2014 at 12:10
• I'm thinking of subtracting the overall diffraction pattern with the diffraction pattern just by that slit alone, would that work? May 28, 2014 at 2:12

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 $\delta \theta \space ' = \frac{1}{2} \delta \theta$.

• If a single centermost slit is close then $e^{-iN\delta/2}$ term should not be minus. please check it. you can think that we have a aperture of double width at center. Jun 5, 2014 at 13:45
• You can do that, or you can subtract the contribution from the centermost slit. I'm not sure how to do this question. Jun 5, 2014 at 16:10