13.2: Beams

Making a Beam

Consider a system with an opaque barrier in the \(z = 0\) plane. If it is illuminated by a plane wave traveling in the +\(z\) direction, the barrier absorbs the wave completely. Now cut a hole in the barrier. You might think that this would produce a beam of light traveling in the direction of the initial plane wave. But it is not that simple. This is actually the same problem that we considered in the previous section, (13.7)-(13.14), with the function, \(f(x, y)\), given by \[f(x, y) e^{-i \omega t}\]

where \[f(x, y)=\left\{\begin{array}{l}
1 \text { inside the opening } \\
0 \text { outside the opening }
\end{array}\right.\]

In fact, it will be useful to think about the more general problem, because the the function, (13.16), is discontinuous. As we will see later, this leads to more complicated diffraction phenomena than we see with a smooth function. In particular, we will assume that \(f(x, y)\) is signifigantly different from zero only for small x and y and goes to zero for large \(x\) and \(y\). Then we can talk about the position of the “opening” that produces the beam, near \(x = y = 0\).

We can think of this problem as a forced oscillation problem. It is much easier to analyze the physics if we ignore polarization, so we will discuss scalar waves. For example, we could consider the transverse waves on a flexible membrane or pressure waves in a gas. Equivalently, we could consider light waves that depend only on two dimensions, \(x\) and \(z\), and polarized in the \(y\) direction. We will not worry about these niceties too much, because as usual, the basic properties of the wave phenomena will be determined by translation invariance properties that are independent of what it is that is waving!

Caveats

It is worth noting that there are other approaches to the diffraction problem besides the ones we discuss here. The physical setup we are considering is slightly different from the standard setup of Huygens-Fresnel-Kirchhoff diffraction, because we are studying a different problem. In Huygens-Fresnel-Kirchhoff diffraction,² you consider the diffraction of a plane wave from a finite object, whereas, our opaque screen is infinite in the \(x\)-\(y\) plane. In the Huygens-Fresnel case, the appropriate boundary condition is that there are no incoming spherical waves coming back in from infinity toward the object that is doing the diffracting. The diffraction produces outgoing spherical waves only. We will not discuss this alternative physical setup in detail because it leads deeper into Bessel functions³ than we (and probably the reader as well) are eager to go. The advantage of our formulation is that we can set it up entirely with the plane wave solutions that we have already discussed. We will simply indicate the differences between our treatment and Huygens-Fresnel diffraction. For diffraction in the forward region, at large z and not very far from the z axis, the diffraction is the same in the two cases.

The reader should also notice that we have not explained exactly how the oscillation, (13.15), \[f(x, y) e^{-i \omega t}\]

in the \(z = 0\) plane is produced. This is by no means a trivial problem, but we will not discuss it in detail. We are concentrating on the physics for \(z > 0\). This will be quite interesting enough.

Boundary at \(\infty\)

To determine the form of the waves in the region \(z > 0\) (beyond the barrier), we need boundary conditions both at \(z = 0\) and at \(z=\infty\). At \(z = 0\), there is an oscillating amplitude given by (13.15).⁴ At \(z=\infty\), we must impose the condition that there are no waves traveling in the −\(z\) direction (back toward the barrier) and that the sodwlutions are well behaved at \(\infty\).

The normal modes have the form \[e^{i \vec{k} \cdot \vec{r}-i \omega t}\]

where \(\vec{k}\) satisfies the dispersion relation \[\omega^{2}=v^{2} \vec{k}^{2}.\]

Thus given two components of \(\vec{k}\), we can find the third using (13.18). So we can write the solution as \[\psi(\vec{r}, t)=\int d k_{x} d k_{y} C\left(k_{x}, k_{y}\right) e^{i \vec{k} \cdot \vec{r}-i \omega t} \text { for } z>0\]

where \[k_{z}=\sqrt{\omega^{2} / v^{2}-k_{x}^{2}-k_{y}^{2}}.\]

Note that (13.20) does not determine the sign of \(k_{2}\). But the boundary condition at \(\infty\) does. If \(k_{z}\) is real, it must be positive in order to describe a wave traveling to the right, away from the barrier. If \(k_{z}\) is complex, its imaginary part must be positive, otherwise \(e^{i \vec{k} \cdot \vec{r}}\) would blow up as \(z\) goes to \(\infty\). Thus, \[\text { if } \operatorname{Im} k_{z}=0, \text { then } \operatorname{Re} k_{z}>0 \text { ; otherwise } \operatorname{Im} k_{z}>0 \text { . }\]

We discussed the physical signifigance of the boundary condition, (13.21), in our discussion of tunneling starting on page 274. There is real physics in the boundary condition at infinity. For example, consider the relation between this analysis and the discussion of path lengths in the previous section. In the language of the last chapter, we cannot describe the effects of the waves with imaginary \(k_{z}\). However, the boundary condition, (13.21), ensures that these components of the wave will go to zero rapidly for large \(z\).

Boundary at \(z=0\)

All we need to do to determine the form of the wave for \(z > 0\) is to find \(C(k_{x}, k_{y})\). To do that, we implement the boundary condition at \(z = 0\) by using (13.19) \[\psi(\vec{r}, t)=\int d k_{x} d k_{y} C\left(k_{x}, k_{y}\right) e^{i \vec{k} \cdot \vec{r}-i \omega t} \text { for } z>0\]

and setting \[\left.\psi(\vec{r}, t)\right|_{z=0}=f(x, y) e^{-i \omega t}\]

to get (13.15). Taking out the common factor of \(e^{-i \omega t}\), this condition is \[f(x, y)=\int d k_{x} d k_{y} C\left(k_{x}, k_{y}\right) e^{i\left(k_{x} x+k_{y} y\right)}.\]

If \(f(x, y)\) is well behaved at infinity (as it certainly is if, as we have assumed, it goes to zero for large \(x\) and \(y\)), then only real \(k_{x}\) and \(k_{y}\) can contribute in (13.23). A complex \(k_{x}\) would produce a contribution that blows up either for \(x \rightarrow+\infty\) or \(x \rightarrow-\infty\). Thus the integrals in (13.23) run over real k from −\(\infty\) to \(\infty\).

(13.23) is just a two-dimensional Fourier transform. Using arguments analogous to those we used in our discussion of signals, we can invert it to find \(C\). \[C\left(k_{x}, k_{y}\right)=\frac{1}{4 \pi^{2}} \int d x d y f(x, y) e^{-i\left(k_{x} x+k_{y} y\right)}\]

Inserting (13.24) into (13.19) with (13.20) and (13.21) \[k_{z}=\sqrt{\omega^{2} / v^{2}-k_{x}^{2}-k_{y}^{2}}\]

\[\text { if } \operatorname{Im} k_{z}=0 \text { , then } \operatorname{Re} k_{z}>0 \text { ; otherwise } \operatorname{Im} k_{z}>0\]

gives the result for the wave, \(\psi(\vec{r}, t)\), for \(z > 0\). This result is really very general. It holds for any reasonable \(f(x, y)\).

____________________________

²For example, see Hecht, chapter 10.
³See the discussion starting on page 314.
⁴Note that in a real physical situation, the boundary conditions are often much more complicated than (13.16), because the physics of the boundary matters. However, this often means that diffraction in a real situation is even larger.