Covering a slit in a diffraction grating - what happens?

1. May 26, 2014

unscientific

1. The problem statement, all variables and given/known data

For an N = 1000 slit diffraction grating, the distance from a maxima to a minima at order p is given by:

$$\delta \theta = \frac{\lambda}{Np}$$

The centremost $\frac{N}{2}$ slit is covered, what happens to this width?

2. Relevant equations

3. The attempt at a solution

I'm tempted to say to think that one slit won't make a difference if N is large like 1000 slits. Physically, that would make the $\frac{N}{2}-1$ and $\frac{N}{2} +1$ slits have double the separation.

I have never come across any question quite like this, so I am at a loss.

2. May 27, 2014

Staff: Mentor

Just one slit is covered, or the central half?
A single slit won't make a large difference, half of them could.

3. May 27, 2014

unscientific

Just the centermost slit. I'm asked to find what happens to $\delta \theta$, the width from the maxima to the next minima. I'm given 10 marks to work out the ratio between $\frac{\delta \theta\space '}{\delta \theta}$.

4. May 27, 2014

unscientific

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:

$$u = u_0 e^{i(kz-\omega t)} \space \frac{1- e^{iN\delta}}{1- e^{i\delta}} = u_0 e^{i(kz-\omega t)} e^{i(\frac{N}{2}\delta - \frac{\delta}{2})} \frac{sin \left(\frac{N\delta}{2}\right)}{sin \left( \frac{\delta}{2}\right)}$$

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

Last edited: May 27, 2014
5. May 29, 2014

Staff: Mentor

You don't double the width just by covering one slit. To subtract the contributions from the 1000 slits and the single slit, you have to take into account the relative intensity, as it differs by a factor of 1000.

6. May 29, 2014

unscientific

I subtracted the angular displacement:

$$u = u_0 \left[e^{i(kz-\omega t)} e^{i(\frac{N}{2}\delta - \frac{\delta}{2})} \frac{sin \left(\frac{N\delta}{2}\right)}{sin \left( \frac{\delta}{2}\right)}\space - \space e^{i(\frac{N\delta}{2})}\right]$$

There should be a prefactor of $\frac{1}{r}$, but we asume that the distances are large so $r$ doesn't vary with angle.

7. May 29, 2014

Staff: Mentor

There should be a prefactor of 1000 as the first comes from 1000 slits and the second just from one.

8. May 29, 2014

unscientific

So It should be:

$$u = u_0 \left[e^{i(kz-\omega t)} e^{i(\frac{N}{2}\delta - \frac{\delta}{2})} \frac{sin \left(\frac{N\delta}{2}\right)}{sin \left( \frac{\delta}{2}\right)}\space - \frac{1}{1000}\space e^{i(\frac{N\delta}{2})}\right]$$

$$= I = I_0 \left[ \frac{sin^2 \left( \frac{N\delta}{2} \right)}{sin^2\left(\frac{\delta}{2}\right)} \space - 2\frac{1}{10^6} cos\left(\frac{\delta (2N+1)}{2})\right) \frac{sin \left( \frac{N\delta}{2} \right)}{sin \left(\frac{\delta}{2}\right)} + \frac{1}{10^6} \right]$$

Last edited: May 29, 2014
9. May 29, 2014

Staff: Mentor

How did you expand the small term, where does the sine come from and why does it have an N? The single slit does not "know" about the total number of slits.
I would expect a slight shift in the position of the minima - when the sine is zero, the cosine is not zero and the other sine should look different.

10. May 29, 2014

unscientific

To find the intensity:

$$I \propto uu* = u_0^2 \left[e^{-i(kz-\omega t)} e^{-i(\frac{N}{2}\delta - \frac{\delta}{2})} \frac{sin \left(\frac{N\delta}{2}\right)}{sin \left( \frac{\delta}{2}\right)}\space - \frac{1}{1000}\space e^{-i(\frac{N\delta}{2})}\right] \left[e^{i(kz-\omega t)} e^{i(\frac{N}{2}\delta - \frac{\delta}{2})} \frac{sin \left(\frac{N\delta}{2}\right)}{sin \left( \frac{\delta}{2}\right)}\space - \frac{1}{1000}\space e^{i(\frac{N\delta}{2})}\right]$$

$$= I = I_0 \left[ \frac{sin^2 \left( \frac{N\delta}{2} \right)}{sin^2\left(\frac{\delta}{2}\right)} \space - 2\frac{1}{10^6} cos\left(\frac{\delta (2N+1)}{2})\right) \frac{sin \left( \frac{N\delta}{2} \right)}{sin \left(\frac{\delta}{2}\right)} + \frac{1}{10^6} \right]$$

Now we see that if $\frac{N\delta}{2} = n\pi$, the intensity is no longer zero, but $I_0\frac{1}{10^6}$. Thus the minimum has shifted by a bit. How do I find the extent of the shift?

Last edited: May 29, 2014
11. May 29, 2014

Staff: Mentor

You can try to find the derivative of the intensity with respect to the position.

12. May 29, 2014

unscientific

I used small angles above to determine the shift (see above), do you think that's right?

Last edited: May 29, 2014
13. May 29, 2014

abitslow

Man, you're way over my head. Why isn't it just:
δΘ'/δΘ = (λ/(2N'p)) / (λ/(2Np)) = N/N' = 1000/999 ~ 1.001 ?
hard to believe the given equation is accurate to 1 ppk...

14. May 29, 2014

unscientific

because it is a 10 mark question.

15. May 29, 2014

unscientific

I think there's a simpler way..

16. May 29, 2014

Staff: Mentor

Sure, all angles are small.

I'm not sure if it works without finding the derivative.

It would be that easy if a slit on the side would get covered. I'm not sure a slit in the middle gives the same result.

17. May 29, 2014

unscientific

Differentiating it would be a huge mess and I'm not even sure if it will produce any result.