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.

  • $\begingroup$ 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. $\endgroup$ Commented May 27, 2014 at 12:10
  • $\begingroup$ I'm thinking of subtracting the overall diffraction pattern with the diffraction pattern just by that slit alone, would that work? $\endgroup$
    – user44840
    Commented May 28, 2014 at 2:12

1 Answer 1


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

  • $\begingroup$ 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. $\endgroup$
    – Rahul
    Commented Jun 5, 2014 at 13:45
  • $\begingroup$ You can do that, or you can subtract the contribution from the centermost slit. I'm not sure how to do this question. $\endgroup$
    – user44840
    Commented Jun 5, 2014 at 16:10

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