438 Week 13: Interference and Diffraction
The lines projected on the screen, however, are not infinitely sharp (even if the sodium
wavelengths themselves are)! Thewidthsof the first principle maxima atλ 1 orλ 2 are:∆θ≈2 λ 1
N d≈
2 λ 2
N d(1061)
If the two maxima are too close together, their lines willoverlapand we won’t be able
to tell that there are two lines there at all! On the other hand, if they are far enough
apart, the lines won’t overlap at all (except out in the irrelevant morass of secondary
maxima and higher order minima) and we’ll be able to easily see two lines. We need a
criterionfor the minimal resolution of two spectral lines (or anything else) cast as an
“image” onto a screen, or a piece of film, or the retina. EnterRayleigh’s Criterion for
Resolution.13.5.1: Rayleigh’s Criterion for Resolution
Lord Rayleigh was yet another eponymous physicist who studied thewave properties of
“rays” and things such as the resolving power of spectral gratings or optical instruments.
We have encountered him before in the context of “Rayleigh scattering”, the original
blue-sky theory. He established a very simple criterion for when twospectral lines from
a diffraction grating or diffraction maxima from e.g. circular apertures are marginally
resolved. It is this:Two lines are said to bemarginally resolvedif the principlemaximumfor one
line is outside of thefirst minimumof the other.That’s it! Nothing to it. It is really slightly more general than this, however. We will
also use it below to determine whether two point-likeimages, when focussed on a screen
through a circular aperture, are marginally resolved, where instead of “lines” we simply
talk about the diffraction maxima of the dots, but the idea is exactly the same. For
us to be able to determine that there are two instead of one, they cannot overlap, and
“overlap” is defined to be the maximum of each further away than the first minimum of
the other.13.5.2: Resolving Power
With that criterion in hand, we can talk about and derive theresolving powerof a grating
and see how we can determine whether or not any given grating will beable to resolve
any given pair of closely spaced lines.
In order for our grating to resolve two lines the angular separationof their maxima has
to be larger than the angle of the first minimum of each maximum. Thatis:θm(λ 2 ) =mλ^2
d>mλ^1
d+λ^1
N d=θminm (λ 1 ) (1062)or
∆λ 21 =m(λ 2 −λ 1 )
d>
λ 1
N d(1063)
We can rearrange this, noting its symmetry under exchange of 1 and 2 and defining
λ≈λ 1 ≈λ 2 (the whole point is that they are very close together, right?) to define the
resolving powerof the grating:
R=mN=λ
∆λ(1064)
Note well thatR=λ/∆λis a measure of the relative resolution of the grating at any
wavelengthλ. R=mNtells you what this resolving power is, given the order of the