Resolving power of a diffraction grating?

Question

I was working thru a derivation of the resolving power of a diffraction grating at the $n$th order. Here is that derivation:

Let us say we have a diffraction grating width $W$ and are looking at the $n$th order. The light from the diffraction grating is of width $Wcos(\theta)$ where $\theta$ is the angle to the normal that the $n$th order makes. The $n$th order occurs when $n\lambda=dsin(\theta)$ differentiating this we get $n\delta \lambda=d\delta \theta cos(\theta)$. Subbing in $\delta \theta=\frac{\lambda}{W cos(\theta)}$ which is the resolution from a single slit and with rearangment we get resolving power $=\lambda/ \delta \lambda=nN$. Where $N$ is the total number of slits.

My question is why have we used the minimum resolvable angle from a single slit? Here are two of my thoughts:

1. The beam takes the shape of a single slit and therefore will propagate as if it had passed thru a single slit even though it has not. (this does not account for the 'gaps' in the beam though i.e. when there are no 'holes' in the grating). So what I am saying is that the a beam of any shape will spread out as if it has been diffracted by a aperture the shape of that beam (is this right?).
2. We are viewing the pattern with some sort of aperture and it is this that we are recording (I think this is unlikely as we could equally just view it on a screen and we would have to use the diameter of the aperture for this and not that of the beam).

Answer 1

We first consider the relation: $$n\delta{\lambda} = d\delta{\theta}\cos{\theta}$$ It's content is that the $n^{th}$ order maximum of a wavelength $\lambda + \delta{\lambda}$ is displaced from the corresponding maximum for a wavelength $\lambda$ by the angle $\delta{\theta}$, related to $\delta{\lambda}$ by the above equation.

Now, we can ask the question, "for what (minimum) value of $\delta{\lambda}$ can we clearly distinguish between the $n^{th}$ order maxima of $\lambda$ and $\lambda + \delta{\lambda}$?" The answer is that we can certainly do this (using the Rayleigh criterion) when the angular width $\delta{\phi}$ of the $n^{th}$ order maximum of light of wavelength $\lambda$, on either side of the maximum, is less than the separation of the maxima, $\delta{\theta}$ i.e. when $$\delta{\phi} \le \delta{\theta}$$ or, the minimum value of $\delta{\lambda}$ that can be just resolved is one for which $$\delta{\phi} = \delta{\theta}$$

Now, what about the spread $\delta{\phi}$ of the $n^{th}$ order maxmimum? When considering the grating as a series of a large number of slits $N \gg 1$ with separation $d\cos{\theta}$, you can see that a minimum occurs at an angle for which the contribution from a slit of position $m \le \frac{N}{2}$ is out of phase with that of position $m + \frac{N}{2}$, so that each of these pairs have a net zero contribution (note that we can always consider $N$ to be even, when it is large, by neglecting the contribution from one slit if necessary). Therefore, with the diffraction grating width of $W = Nd$, we see that the required criterion is that slits at a separation of $\frac{W\cos{\theta}}{2}$ are out of phase i.e. that (for the first minima from the centre) $$\frac{2\pi}{\lambda} \frac{W\cos{\theta}}{2} \delta{\phi} = \pi$$ $$\delta{\phi} = \frac{\lambda}{W\cos{\theta}}$$ Note that a similar argument can be used for diffraction from a single, continuous wide slit (which is comparable to this case as both deal with a large number of point sources).

Thus, we now have, on equating $\delta{\phi}$ and $\delta{\theta}$, $$\delta{\lambda} = \frac{\lambda d}{nW} = \frac{\lambda}{nN} \implies \frac{\lambda}{\delta{\lambda}} = nN$$

This result is independent of your methods of observation (aperture or otherwise) so long as you take care to observe all parallel rays inclined at an angle $\theta$, focused at a point.