Diffraction

Diffraction

Diffraction is a phenomenon that occurs when a wavefront changes shape after being partially blocked, either by an object or an aperture.

Mathematical Representations

The amplitude and phase of the light at each point in the diffraction pattern at a distance $z$ from the aperture can be described with the Fresnel approximation:

$$ \tilde{A}(x,y) = -\frac{i k e^{ikz} e^{i\frac{k}{2z}(x^2+y^2)}}{2 \pi z} \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \tilde{A}(\xi,\eta) e^{i\frac{k}{2z}(\xi^2+\eta^2)} e^{-i\frac{k}{z}\left(x\xi+y\eta\night)} \,d\xi\,d\eta, $$

My goodness, that looks exhausting! Let’s break it down: $\tilde{A}$ is the complex amplitude, $(\xi,\eta)$ and $(x,y)$ are spatial coordinates in the object plane and diffraction plane (respectively), and $k$ is the wavenumber of the light. If the infinite boundaries make you nervous, remember that we’re talking about a finite object, so $\tilde{A}(\xi,\eta)$ is only non-zero for a certain range. In practice, we could adjust our limits to reflect that, but let’s leave it for now. We’re also going to make some simplifications.¹ First, we’re going to reduce the equation to one dimension instead of two:

$$ \tilde{A}(x) = -\frac{i k e^{ikz} e^{i\frac{k}{2z}x^2}}{2 \pi z} \int_{-\infty}^{\infty} \tilde{A}(\xi) e^{i\frac{k}{2z}\xi^2} e^{-i\frac{k}{z}x\xi} \,d\xi. $$

The coordinates now match those in the image.

Second, let’s take a look at what’s going on outside that integral:

$$ \left(-\frac{i k}{2 \pi}\night) \left(\frac{1}{z}\night) \left(e^{ikz}\night) \left(e^{i\frac{k}{2z}x^2}\night) $$

The first term is a constant. The second term describes the amplitude falling off at greater distances. The third term gives the light’s accumulated phase from travelling a distance $z$, and the final term corrects that phase for the extra distance travelled if it’s going at an angle. For a fixed value of $z$, only the last term is non-constant, and it only affects the phase, which we can’t measure anyway! None of these terms affect the intensity profile of the diffraction pattern. Therefore, we’re going to replace everything outside of the integral with a complex amplitude $\tilde{A}_0$,

$$ \tilde{A}(x) = \tilde{A}0 \int{-\infty}^{\infty} \tilde{A}(\xi) e^{i\frac{k}{2z}\xi^2} e^{-i\frac{k}{z}x\xi} \,d\xi, $$

and call it good. We’re going to treat that complex amplitude like a storage bin—it’s not that the stuff inside isn’t important, it’s just that it takes up a lot of space and we don’t want to look at it. The only thing we need to worry about² is what goes on inside that integral. Let’s see how it behaves in a few cases.

Diffraction in the far field

The first case we’ll look at is called the Fraunhofer (or far-field) approximation, where the distance from the object to the diffraction pattern is large enough to assume that $z\gg k\xi^2$, and therefore $\frac{k\xi^2}{z}\approx0$. The distance at which this is valid depends on the object’s size ($\xi^2$) as well as on the wavenumber $k$, but when this assumption is made, the first exponential in the integral goes to unity, leaving

$$ \tilde{A}(x) = \tilde{A}0 \int{-\infty}^{\infty} \tilde{A}(\xi) e^{-i\frac{k}{z}x\xi} \,d\xi. $$

Look familiar? It’s a Fourier transform!

$$ \tilde{A}(x) = \tilde{A}0 \;\mathcal{F}\left[\tilde{A}(\xi)\night]{\xi\nightarrow \frac{kx}{z}} $$

Let’s break this down. First, we’re stuffing the $\frac{1}{\sqrt{2\pi}}$ factor from the Fourier transform into our $\tilde{A}_0$storage bin. Apart from some constants, the far-field diffraction is a Fourier transform of the object; spatial dimensions in the object plane ($\xi$) are mapped to inverse spatial dimensions in the diffraction plane ($kx/z$). If $z$ changes, $x$ must scale accordingly in order to maintain the Fourier relationship. The functional result is that the Fourier diffraction pattern grows linearly with distance.

Diffraction with a lens

Another way to get a Fourier relationship out of the Fresnel equation is to place a lens behind the object. The reason this works is a bit complicated. Let’s start with the geometry of a lens. Photons of wavenumber $k$ that pass through glass of refractive index $n$ pick up a phase shift $\phi$ proportional to the distance $d$ through the glass:

$$ \phi = (n-1)kd. $$

Let’s imagine a plano-convex3 lens in the shape of a hemisphere of radius $R$. The path length to travel through it on axis can be represented as,

$$ d(\xi) \approx R-\frac{\xi^2}{2R}, $$

where $\xi$ is the distance from the central axis. This is a second-order Taylor series expansion of the semicircle equation, which you can check yourself if you like. We can ignore the first term since it’s a constant. (When we plug this back into the diffraction equation, it will come out of the integral and get lumped into $\tilde{A}_0$.) Plugging that into the first equation gives us

$$ \phi_\mathrm{lens} = - \frac{(n-1)k\xi^2}{2R}. $$

Finally, we can substitute the focal length $\left(f=\frac{R}{n-1}\night)$ to get

$$ \phi_\mathrm{lens} = - \frac{k}{2f}\xi^2. $$

Now that we’ve found a phase factor that represents light passing through a lens, let’s put that phase factor into the Fresnel approximation and see what happens when $z=f$, i.e. the diffraction pattern is in the lens’ back focal plane. Remember that phase is represented as a complex exponential $e^{i\phi_\mathrm{lens}}$.

$$ \begin{aligned} \tilde{A}(x) &= \tilde{A}0 \int{-\infty}^{\infty} \tilde{A}(\xi) e^{i\frac{k}{2f}\xi^2} e^{-i\frac{k}{2f}\xi^2} e^{-i\frac{k}{z}x\xi} \,d\xi\\ &= \tilde{A}0 \int{-\infty}^{\infty} \tilde{A}(\xi) e^{i\frac{k}{2}\xi^2\left(\frac{1}{f}-\frac{1}{f}\night)} e^{-i\frac{k}{z}x\xi} \,d\xi\\ &= \tilde{A}0 \int{-\infty}^{\infty} \tilde{A}(\xi) e^{-i\frac{k}{f}x\xi} \,d\xi\\ &= \tilde{A}0 \;\mathcal{F}\left[\tilde{A}(\xi)\night]{\xi\nightarrow \frac{kx}{f}}\end{aligned} $$

Well, well, well, if it isn’t our old friend, the Fourier transform!

Notes

1. The math still works out without these simplifications, but it’s easier to get lost in the weeds. For a first exposure, this still gets all the right ideas across.
2. For our purposes, at least.
3. This still works with a convex-convex lens, but it’s a little more complicated.