Diffraction: The sinc function, live and in pure color

Some of the most interesting applications of the Fourier transform are in the field of optics, understood broadly to include most of the electromagnetic spectrum in its purview. An excellent book on the subject is Fourier Optics, by Stanford’s own J. W. Goodman — highly recommended.

The fundamental phenomenon associated with the wave theory of light is diffraction or interference. Som-merfeld says that diffraction is “any deviation of light rays from rectilinear paths which cannot be inter-preted as reflection or refraction.” Very helpful. Is there a difference between diffraction and interference?

In his Lectures on Physics, Feynman says “No one has ever been able to define the difference between interference and diffraction satisfactorily. It is just a question of usage, and there is no specific, important physical difference between them.” He does go on to say that “interference” is usually associated with patterns caused by a few radiating sources, like two, while “diffraction” is due to many sources. Whatever the definition, or nondefinition, you probably know what the picture is:

Such pictures, most notably the “Two Slits” experiments of Thomas Young (1773–1829), which we’ll analyze, below, were crucial in tipping the balance away from Newton’s corpuscular theory to the wave theory propounded by Christiaan Huygens (1629–1695). The shock of the diffraction patterns when first seen was that light + light could be dark. Yet the experiments were easy to perform. Spoke Young in 1803 to the Royal Society: ”The experiments I am about to relate . . . may be repeated with great ease, whenever the sun shines, and without any other apparatus than is at hand to every one.”²⁴

24Young also did important work in studying Egyptian hieroglyphics, completely translating a section of the Rosetta Stone.

We are thus taking sides in the grand battle between the armies of “light is a wave” and those of “light is a particle”. It may be that light is truly like nothing you’ve ever seen before, but for this discussion it’s a wave. Moreover, jumping ahead to Maxwell, we assume that light is an electromagnetic wave, and for our discussion we assume further that the light in our problems is:

• Monochromatic

◦ Meaning that the periodicity in time is a single frequency, so described by a simple sinusoid.

• Linearly polarized

◦ Meaning that the electric field vector stays in a plane as the wave moves. (Hence so too does the magnetic field vector.)

With this, the diffraction problem can be stated as follows:

Light — an electromagnetic wave — is incident on an (opaque) screen with one or more aper-tures (transparent openings) of various shapes. What is the intensity of the light on a screen some distance from the diffracting screen?

We’re going to consider only a case where the analysis is fairly straightforward, the Fraunhofer approxima-tion, or Fraunhofer diffraction. This involves a number of simplifying assumptions, but the results are used widely. Before we embark on the analysis let me point out that reasoning very similar to what we’ll do here is used to understand the radiation patterns of antennas. For this take on the subject see Bracewell, Chapter 15.

Light waves We can describe the properties of light that satisfy the above assumptions by a scalar -valued function of time and position. We’re going to discuss “scalar” diffraction theory, while more sophisticated treatments handle the “vector” theory. The function is the magnitude of the electric field vector, say a function of the form

u(x, y, z, t) = a(x, y, z) cos(2πνt − φ(x, y, z))

Here, a(x, y, z) is the amplitude as a function only of position in space, ν is the (single) frequency, and φ(x, y, z) is the phase at t = 0, also as a function only of position.²⁵

The equation

φ(x, y, z) = constant

describes a surface in space. At a fixed time, all the points on such a surface have the same phase, by definition, or we might say equivalently that the traveling wave reaches all points of such a surface φ(x, y, z) = constant at the same time. Thus any one of the surfaces φ(x, y, z) = constant is called a wavefront. In general, the wave propagates through space in a direction normal to the wavefronts.

The function u(x, y, z, t) satisfies the 3-dimensional wave equation

∆u = 1 c²

∂²u

∂t² where

∆ = ∂²

∂x² + ∂²

∂y² + ∂²

∂z²

25It’s also common to refer to the whole argument of the cosine, 2πνt − φ, simply as “the phase”.

is the Laplacian and c is the speed of light in vacuum. For many problems it’s helpful to separate the spatial behavior of the wave from its temporal behavior and to introduce the complex amplitude, defined to be

u(x, y, z) = a(x, y, z)e^iφ(x,y,z). Then we get the time-dependent function u(x, y, z, t) as

u(x, y, z, t) = Re u(x, y, z)e^2πiνt
.

If we know u(x, y, z) we can get u(x, y, z, t). It turns out that u(x, y, z) satisfies the differential equation

∆u(x, y, z) + k²u(x, y, z) = 0

where k = 2πν/c. This is called the Helmholtz equation, and the fact that it is time independent makes it simpler than the wave equation.

Fraunhofer diffraction We take a sideways view of the situation. Light is coming from a source at a point O and hits a plane S. We assume that the source is so far away from S that the magnitude of the electric field associated with the light is constant on S and has constant phase, i.e., S is a wavefront and we have what is called a plane wave field. Let’s say the frequency is ν and the wavelength is λ. Recall that c = λν, where c is the speed of light. (We’re also supposing that the medium the light is passing through is isotropic, meaning that the light is traveling at velocity c in any direction, so there are no special effects from going through different flavors of jello or something like that.)

Set up coordinates so that the z-axis is perpendicular to S and the x-axis lies in S, perpendicular to the z-axis. (In most diagrams it is traditional to have the z-axis be horizontal and the x-axis be vertical.) In S we have one or more rectangular apertures. We allow the length of the side of the aperture along the x-axis to vary, but we assume that the other side (perpendicular to the plane of the diagram) has length 1.

A large distance from S is another parallel plane. Call this the image plane.

The diffraction problem is:

• What is the electric field at a point P in the image plane?

The derivation I’m going to give to answer this question is not as detailed as is possible (for details see Goodman’s book), but we’ll get the correct form of the answer and the point is to see how the Fourier transform enters.

The basis for analyzing diffraction is Huygens’ principle which states, roughly, that the apertures on S (which is a wavefront of the original source) may be regarded as (secondary) sources, and the field at P is the sum (integral) of the fields coming from these sources on S. Putting in a little more symbolism, if E0

is the strength of the electric field on S then an aperture of area dS is a source of strength dE = E0dS.

At a distance r from this aperture the field strength is dE⁰⁰ = E₀dS/r, and we get the electric field at this distance by integrating over the apertures the elements dE⁰⁰, “each with its proper phase”. Let’s look more carefully at the phase.

The wave leaves a point on an aperture in S, a new source, and arrives at P sometime later. Waves from different points on S will arrive at P at different times, and hence there will be a phase difference between the arriving waves. They also drop off in amplitude like one over the distance to P , and so by different amounts, but if, as we’ll later assume, the size of the apertures on S are small compared to the distance between S and the image plane then this is not as significant as the phase differences. Light is moving so fast that even a small differences between locations of secondary point sources on S may lead to significant differences in the phases when the waves reach P .

The phase on S is constant and we might as well assume that it’s zero. Then we write the electric field on

S in complex form as

E = E0e^2πiνt

where E₀ is constant and ν is the frequency of the light. Suppose P is at a distance r from a point x on S. Then the phase change from x to P depends on how big r is compared to the wavelength λ — how many wavelengths (or fractions of a wavelength) the wave goes through in going a distance r from x to P . This is 2π(r/λ). To see this, the wave travels a distance r in a time r/c seconds, and in that time it goes through ν(r/c) cycles. Using c = λν that’s νr/c = r/λ. This is 2πr/λ radians, and that’s the phase shift.

Take a thin slice of width dx at a height x above the origin of an aperture on S. Then the field at P due to this source is, on account of the phase change,

dE = E₀e^2πiνte^2πir/λdx . The total field at P is

E = Z

apertures

E₀e^2πiνte^2πir/λdx = E₀e^2πiνt Z

apertures

e^2πir/λdx

There’s a Fourier transform coming, but we’re not there yet.

The key assumption that is now made in this argument is to suppose that r  x ,

that is, the distance between the plane S and the image plane is much greater than any x in any aperture, in particular r is large compared to any aperture size. This assumption is what makes this Fraunhofer diffraction; it’s also referred to as far field diffraction. With this assumption we have, approximately,

r = r₀− x sin θ ,

where r₀ is the distance between the origin of S to P and θ is the angle between the z-axis and P .

Plug this into the formula for E:

E = E₀e^2πiνte^2πir0/λ Z

apertures

e−2πix sin θ/λ

dx

Drop that constant out front — as you’ll see, it won’t be important for the rest of our considerations.

We describe the apertures on S by a function A(x), which is zero most of the time (the opaque parts of S) and 1 some of the time (apertures). Thus we can write

E ∝ Z ∞

−∞

A(x)e−2πix sin θ/λ

dx It’s common to introduce the variable

p = sin θ λ and hence to write

E ∝ Z ∞

−∞

A(x)e^−2πipxdx .

There you have it. With these approximations (the Fraunhofer approximations ) the electric field (up to a multiplicative constant) is the Fourier transform of the aperture! Note that the variables in the formula are x, a spatial variable, and p = sin θ/λ, in terms of an angle θ. It’s the θ that’s important, and one always speaks of diffraction “through an angle.”

Diffraction by a single slit Take the case of a single rectangular slit of width a, thus described by A(x) = Πa(x). Then the field at P is

E ∝ a sinc ap = a sinc

a sin θ λ

 .

Now, the intensity of the light, which is what we see and what photodetectors register, is proportional to the energy of E, i.e., to |E|². (This is why we dropped the factors E0e^2πiνte^2πir0/λmultiplying the integral.

They have magnitude 1.) So the diffraction pattern you see from a single slit, those alternating bright and dark bands, is

intensity = a²sinc²

a sin θ λ

 .

Pretty good. The sinc function, or at least its square, live and in color. Just as promised.

We’ve seen a plot of sinc² before, and you may very well have seen it, without knowing it, as a plot of the intensity from a single slit diffraction experiment. Here’s a plot for a = 2, λ = 1 and −π/2 ≤ θ ≤ π/2:

Young’s experiment As mentioned earlier, Thomas Young observed diffraction caused by light passing through two slits. To analyze his experiment using what we’ve derived we need an expression for the apertures that’s convenient for taking the Fourier transform.

Suppose we have two slits, each of width a, centers separated by a distance b. We can model the aperture function by the sum of two shifted rect functions,

A(x) = Π_a(x − b/2) + Π_a(x + b/2) .

(Like the transfer function of a bandpass filter.) That’s fine, but we can also shift the Π_a’s by convolving with shifted δ’s, as in

A(x) = δ(x − b/2) ∗ Πa(x) + δ(x + b/2) ∗ Πa(x)

= (δ(x − b/2) + δ(x + b/2)) ∗ Π_a(x) ,

and the advantage of writing A(x) in this way is that the convolution theorem applies to help in computing the Fourier transform. Namely,

E(p) ∝ (2 cos πbp)(a sinc ap)

= 2a cos

πb sin θ λ

 sinc

a sin θ λ



Young saw the intensity, and so would we, which is then intensity = 4a²cos²

πb sin θ λ

 sinc²

a sin θ λ



Here’s a plot for a = 2, b = 6, λ = 1 for −π/2 ≤ θ ≤ π/2:

This is quite different from the diffraction pattern for one slit.

Diffraction by two point-sources Say we have two point-sources — the apertures — and that they are at a distance b apart. In this case we can model the apertures by a pair of δ-functions:

A(x) = δ(x − b/2) + δ(x + b/2) .

Taking the Fourier transform then gives

E(p) ∝ 2 cos πbp = 2 cos

πb sin θ λ

 .

and the intensity as the square magnitude:

intensity = 4 cos²

πb sin θ λ

 .

Here’s a plot of this for b = 6, λ = 1 for −π/2 ≤ θ ≤ π/2:

Incidentally, two radiating point sources covers the case of two antennas “transmitting in phase from a single oscillator”.

An optical interpretation of F δ = 1 What if we had light radiating from a single point source?

What would the pattern be on the image plane in this circumstance? For a single point source there is no diffraction (a point source, not a circular aperture of some definite radius) and the image plane is illuminated uniformly. Thus the strength of the field is constant on the image plane. On the other hand, if we regard the aperture as δ and plug into the formula we have the Fourier transform of δ,

E ∝ Z ∞

−∞

δ(x)e^−2πipxdx

This gives a physical reason why the Fourier transform of δ should be constant (if not 1).

Also note what happens to the intensity as b → 0 of the diffraction due to two point sources at a distance b.

Physically, we have a single point source (of strength 2) and the formula gives intensity = 4 cos²

πb sin θ λ



→ 4 .