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. enter image description here

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).
  • $\begingroup$ This is the condition for multiple N slit principle max nλ=dsin(θ). The min occur at sλ=Ndsin(θ) s = +/-1,2, when s/N is integer it is principle max. There are N-2 max between each principal max and N-1 min. As the N is increased the width of the principle max decreases so the spectral resolution increase. $\endgroup$
    – Anonymous
    Commented Apr 21, 2015 at 6:31

1 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.

  • $\begingroup$ Thanks for your answer, it is very helpful. Just one thing is my statement (which I will change slightly here) above that 'a beam that takes the shape of an arbitrary [continuous] aperture will propagate (and spread out) as if it had passed through that aperture? $\endgroup$ Commented Apr 21, 2015 at 17:57
  • $\begingroup$ Also you don't by any chance happen to have a source? $\endgroup$ Commented Apr 21, 2015 at 19:09
  • $\begingroup$ Well, you can consider a wide aperture itself as the limit of infinitely many closely spaced point sources, so that there is no gap at all. You certainly expect the behavior of a grating to approach that of a slit in this case. However, in practical cases, the pattern won't be exactly the same precisely because of the gaps. $\endgroup$
    – AV23
    Commented Apr 22, 2015 at 5:17
  • $\begingroup$ If you want some references, you can find a derivation of the resolving power of a grating, for example, here: hyperphysics.phy-astr.gsu.edu/hbase/phyopt/gratres.html#c4. The argument for the first minimum can be found, for example, in Wikipedia: en.wikipedia.org/wiki/Diffraction#Single-slit_diffraction. $\endgroup$
    – AV23
    Commented Apr 22, 2015 at 5:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.