Wave Fronts

Consider a swimming pool, with children equally spaced along a straight edge of the pool, each of the children tapping the water with a rod in tune (at the same time) to the beat of a song that is played over loud speakers.

Each child generates a circular wave, and the circular waves combine away from the side of the pool to form parallel straight waves. The straight waves are called the “envelope” of the circular waves, or simply “wave fronts”.

Now, consider if the children are staggered in their timing, so that a given amount of time passes between each successive child tapping the water:

That is the idea behind phased array antennas:

This explains some of the properties of light, according to Huygens:

A beam of light can be considered to be a set of consecutive parallel wave fronts. Each wave front is a plane that is perpendicular to the light beam direction:

Wave front surface normals could be modeled, uniformly spaced across a wave front. The line of travel of each wave front surface normal is called a “ray”:

The density of light rays (wave front surface normals) could represent the number of photons in the light beam, or more importantly the amount of energy of the light beam. We consider the latter.

The amount of energy in a light beam is referred to as “irradiance” (abbreviated I) and is measured as Watts per square meter (W/m²) on the wave front. Irradiance may also be referred to as “intensity”, and may be abbreviated E (for Energy).

Closer spacing of light rays (more surface normals per square meter) represents higher energy (more Watts per square meter).

Say that a flat surface is perpendicular to the direction of a light beam. In that case, a light beam wave front is parallel to the surface, and the beam irradiance (W/m² on the wavefront) equals the irradiance striking the surface (W/m² on the surface):

On the other hand, if the surface and the light beam direction are not perpendicular, the unit vectors strike the flat surface further apart than on the wave front, resulting in less surface irradiance (W/m²):

We refer to this geometric dissipation of energy as dihedral spreading, because the wave fronts are not parallel to the receiving surface, forming a dihedral angle between a wave front and the surface.

A dihedral angle is the angle between two planes.

Using the vectors of Figure 8 above, the dihedral angle is given by the following formula:

\[ \cos{\alpha} = \frac{ \mathbf{R} \cdot \mathbf{R'} } { \left|\mathbf{R}\right|\left|\mathbf{R'}\right| } \]

which is, in fact, the angle between any two vectors (in this case, the vectors R and R′). The vectors could be the surface normals of the two planes; that yields the same dihedral angle:

cos α = n · n′

This cosine (of the dihedral angle) is the “cosine factor” that is multiplied into the wavefront irradiance to give the (geometrically corrected) surface irradiance of the wavefront striking the surface:

Abbreviating I for Irradiance, S for surface, and B for beam wavefront, the formula of eq. 3 becomes:

Is = (Ib) (cos α)

This is the geometrically corrected surface irradiance of the wavefront striking the surface.

Dihedral spreading spreads the wavefront over a larger area of the receiving surface, the new area equal to the approaching beam (wavefront) width times the reciprocal of the cosine of the dihedral angle:

In this example, the distance  a  is always shorter than  a/cosθ  because the reciprocal of $\cos{\theta}$ will always be greater than one.

Problem: Show that Figure 9 above is correct, using Figure 9 of the Angles lesson o.

Solution: Here is Figure 9 of the Angles lesson again:

First, transpose the triangle (by reflection and rotation), maintaining all distances and angles (hence maintaining the trigonometric formulas):

Next, magnify the triangle, with constant magnification, to create a larger similar triangle o (with the same angles and with corresponding sides of the two triangles parallel), and collocate the similar triangles at the vertex of theta:

Applying the formulas of Figure 11 to the formulas of Figure 12 gives:

\begin{align} \mathbf{x'} & = \frac{ \mathbf{x} } { \cos{\theta} } \\ & \\ \mathbf{c'} & = \frac{ \mathbf{c} } { \cos{\theta} } \end{align}

Subtracting x′ from c′ gives:

\begin{align} \mathbf{a'} & = \mathbf{c'} - \mathbf{x'} \\ & \\ & = \frac{ \mathbf{c} } { \cos{\theta} } - \frac{ \mathbf{x} } { \cos{\theta} } \\ & \\ & = \frac{ (\mathbf{c} - \mathbf{x}) } { \cos{\theta} } \\ & \\ & = \frac{ \mathbf{a} } { \cos{\theta} } \end{align}

which is the projection of  a  to  a/cosθ  of Figure 9 above (showing that projection is correct).