In a previous post I mentioned that an optical plane wave—or any plane wave for that matter—when viewed in the frequency domain has a spatial frequency that depends on its arrival angle. Oftentimes, the data you see in the frequency domain is referred to as a *spectrum*, and plane waves arriving from different angles form an *Angular Plane Wave Spectrum*.

Looking at unit conversion, going from the time domain to its corresponding frequency domain converts from seconds to 1/seconds (or cycles per second or Hertz). Space, millimeters, goes to 1/mm (or sometimes line pairs per mm). But how do we go from an angle in radians to 1/mm?

First, consider a shoreline with waves coming in. The waves are comprised of a series of peaks and valleys with some spacing—the wave period—and they are traveling along a line that intersects the shoreline at some angle. If you were to freeze time and walk down the shoreline, you’ll also see peaks and valleys, and the spacing here is what defines the spatial frequency of the wave in the plane of the shoreline. (In optical terms, the “shoreline” could be any plane of interest in the system: an aperture, a lens, a sensor, etc.)

The simplest case, of course, contradicts what I just said. If the waves are coming in directly at the shore, as you walk along, the amplitude does not have peaks and valleys. It is constant, or has zero frequency (or is DC if you are an electrical engineer). As the angle of incidence increases, peaks and valleys appear, but they are spaced very far apart from each other. That is they have a low frequency, and the frequency increases as the angle increases. (In the limit, the angle is 90 degrees, the wave is traveling parallel to the shore, and the apparent frequency is equal to the wave frequency.)

Finally, a little trigonometry will give us the frequency of the wave in the angular plane wave spectrum as a function of its angle and period. Look at the same figure with a superimposed triangle. The wave (with period lambda) arrives at an angle, theta, relative to the dashed line which is perpendicular to the plane of interest (i.e. theta is relative to the *plane normal*). The spacing between peaks in the plane of interest is the hypotenuse of the triangle, and the length of this is the inverse of the spatial frequency, f, in the plane.

So f is given by:

$$f=\frac{\sin(\theta)}{\lambda}$$

This has units of 1/length which is expected for spatial frequency, and the sine function takes care of the angle units.