Waves in composites and metamaterials/Airy theory

The content of these notes is based on the lectures by Prof. Graeme W. Milton (University of Utah) given in a course on metamaterials in Spring 2007.

Introduction
The Descartes theory of rainbows cannot explain a number of observed phenomena, one of which is the existence of supernumerary bands of alternating green and purple.

The Airy theory of rainbows attempts to address that issue by considering the wave nature of light. Figure 1 shows a figure of a water droplet with the incident planar wavefront getting converted into a curved wavefront due to the spherical shape of the raindrop. Because the curved wavefront turns back on itself, there is interference between preceding and succeeding wavefronts. The regions of constructive and destructive interference give rise to supernumerary bands.

The outgoing wave front in the figure has been computed at the locus of points on the outgoing rays that are at a constant distance from the incident wave front. Airy determined that the outgoing wave appears to come from a cubic wavefront as shown in the figure.

An image of the interference patterns observed in a rainbow can be seen in Figure 2.

Airy's Theory
In Airy's theory, we start with the cubic wavefront and choose the point of inflection on the cubic wavefront at the origin of coordinates. The wavefront is oriented tangent to the $$x$$-axis as shown in Figure 3.

The equation of the cubic wavefront is (see Humphreys (1964) for details)
 * $$\text{(1)} \qquad

y = \cfrac{h}{3~a^2}~x^3 ~; h = \cfrac{(p^2 -1)^2}{p^2~(n^2-1)}~\sqrt{\cfrac{p^2-n^2}{n^2-1}} $$ where $$a$$ is the drop radius, $$n$$ is the refractive index of water, and $$p$$ is the number of times the ray traverses the raindrop.

Airy made the following approximations:


 * 1) The virtual wavefront $$A'B'$$ can be treated as a cubic.
 * 2) The amplitude along the wavefront is constant.
 * 3) The virtual wavefront is cylindrical.
 * 4) The wave can be treated as a scalar wave.

Note that coherence of natural light is assumed when we think of a planar wavefront with all the rays in phase.

To calculate the emergent radiation at large distances, Airy employed Huygen's principle, i.e., each point on the wavefront is treated as a source of radiation. The change in phase due to the different distances traveled by the waves is then used to compute the intensity of the rainbow. To determine the change in phase we start of with the diagram shown in Figure 4. We wish to know the difference in phase between the waves starting at point $$O$$ and those starting at point $$M$$ when they reach a distant observer at point $$P$$, i.e., we want ($$\varphi_M - \varphi_O$$) at $$P$$. The point $$P$$ is far enough away that the rays $$MP$$ and $$OP$$ may be assumed to be parallel.

From the geometry in Figure 4, we see that $$TP \approx RP$$. Therefore,

\varphi_M - \varphi_O = \cfrac{2~\pi}{\lambda} (OR - MT) $$ where $$\lambda$$ is the wavelength. Now, $$OR = x~\sin\theta$$ and $$MT = y~\cos\theta$$. Therefore,

\varphi_M - \varphi_O = \cfrac{2~\pi}{\lambda} (x~\sin\theta - y~\cos\theta)~. $$ Substituting equation (1) into the above equation and rearranging, we get

\varphi_M = \varphi_O - \cfrac{2~\pi}{\lambda} \left(\cfrac{h~x^3}{3~a^2}~\cos\theta - x~\sin\theta\right) ~. $$ If $$t$$ is the time it takes for the wave starting from $$O$$ to reach point $$P$$, and if $$T$$ is the period of the wave, then the phase of the wave at $$P$$ is

\varphi_O = \cfrac{2~\pi~t}{T} ~. $$ Therefore,

\varphi_M = 2~\pi~\left[\cfrac{t}{T} - \cfrac{1}{\lambda} \left(\cfrac{h~x^3}{3~a^2}~\cos\theta - x~\sin\theta\right)\right] ~. $$ Let us assume that the waveform is a sine wave (we could alternatively use a cosine waveform). Then the amplitude at point $$P$$ due to the wave starting at point $$M$$ is

A_M = K_M~\sin\varphi_M = K_M~\sin\left\{2~\pi~\left[\cfrac{t}{T} - \cfrac{1}{\lambda} \left(\cfrac{h~x^3}{3~a^2}~\cos\theta - x~\sin\theta\right)\right] \right\} ~. $$ Integrating contributions over the wavefront, we find that the amplitude at $$P$$ is

V = \int A_M~ds = \int K_M~\sin\left\{2~\pi~\left[\cfrac{t}{T} - \cfrac{1}{\lambda} \left(\cfrac{h~x^3}{3~a^2}~\cos\theta - x~\sin\theta\right)\right] \right\}~\text{d}s $$ where $$\text{d}s$$ is an differential element along the wavefront. Assuming that $$K = K_M$$ is constant and that $$y \ll x$$ along the effective portion of the front, we can write

V \approx K~\int \sin\left\{2~\pi~\left[\cfrac{t}{T} - \cfrac{1}{\lambda} \left(\cfrac{h~x^3}{3~a^2}~\cos\theta - x~\sin\theta\right)\right] \right\}~\text{d}x ~. $$ Define

\delta(x) := \cfrac{2~\pi}{\lambda} \left(\cfrac{h~x^3}{3~a^2}~\cos\theta - x~\sin\theta\right) ~. $$ Then,

V = K~\int \sin\left(\cfrac{2~\pi~t}{T} - \delta\right)~\text{d}x ~. $$ Using the identity $$\sin (a-b) = \sin a~\cos b - \cos a~\sin b$$, we have

V = K~\int \left(       \sin\cfrac{2~\pi~t}{T}~\cos\delta -         \cos\cfrac{2~\pi~t}{T}~\sin\delta\right)~\text{d}x $$ or,
 * $$\text{(2)} \qquad

V = A~\sin\cfrac{2~\pi~t}{T} - B~\cos\cfrac{2~\pi~t}{T} $$ where

A := K~\int \cos\delta~\text{d}x ~; B := K~\int \sin\delta~\text{d}x ~. $$ Because the wavelength of light is approximately 550 nm and the diameter of a raindrop is approximately 1 mm, the value of $$\cos\delta$$ changes rapidly with $$x$$ beyond a very small distance around $$O$$. So the net contribution from most of the wavefront (except from a small region around $$O$$) is essentially zero.

Hence, only a small error is incurred by extending the limits of integration to $$[-\infty,\infty]$$. Therefore,

A \approx K~\int_{-\infty}^{\infty} \cos\delta~\text{d}x ~; B \approx K~\int_{-\infty}^{\infty} \sin\delta~\text{d}x ~. $$ Recall that the series expansion

\sin\delta = \delta - \cfrac{\delta^3}{3!} + \cfrac{\delta^5}{5!} + \dots ~. $$ Since $$\delta$$ is an odd function of $$x$$, the series expansion of $$\sin\delta$$ involves odd powers of $$x$$. Hence $$\sin\delta$$ is an odd function of $$x$$ and its integral from $$-\infty$$ to $$\infty$$ is zero, i.e., $$B$$ = 0.

To evaluate the integral for $$A$$ we can write

A \approx 2~K~\int_0^{\infty} \cos\delta~\text{d}x $$ or

A \approx 2~K~\int_0^{\infty} \cos\left[ \cfrac{2~\pi}{\lambda} \left(\cfrac{h~x^3}{3~a^2}~\cos\theta - x~\sin\theta\right) \right] ~\text{d}x ~. $$ Apply a change of variables

z = \cfrac{4~x}{\lambda~u}~\sin\theta ~; u = \left(\cfrac{4~h}{3~a^2~\lambda}~x^3~\cos\theta\right)^{1/3} $$ to get

A \approx 2~K~\int_0^{\infty} \left(\cfrac{3~a^2~\lambda}{4~h~\cos\theta}\right)^{1/3}~ \cos\left[\cfrac{\pi}{2}~(u^3 - z~u)\right]~du ~. $$ We can write the above equation as

A \approx M~f(z) $$ where

M := 2~K\left(\cfrac{3~a^2~\lambda}{4~h~\cos\theta}\right)^{1/3} ~; f(z) := \int_0^{\infty}\cos\left[\cfrac{\pi}{2}~(u^3 - z~u)\right]~du ~. $$ The function $$f(z)$$ is called the  Airy function. Hints on how these integrals can be evaluated in terms of Bessel functions can be found in Adams (2002a).

Next, plugging back into equation (2), the amplitude at point $$P$$ is

V = M~f(z)~\sin\cfrac{2~\pi~t}{T}~. $$ Since a cylindrical waveform is assumed, we can compute the intensity at point P using the relation

I = A^2 = M^2~[f(z)]^2 ~. $$

A plot of the intensity versus the scattering angle is shown in Figure 5 Khare74.

The dark band occurs between the primary and the secondary rainbows because of the interference predicted by Airy's theory (see Figure 6). The figure shows the intensity for a single frequency of light. If multiple frequencies are plotted on the same graph, the peaks and troughs for each frequency occur at different locations and with different wavelengths (see for example Humphreys, 1964). However, when the water droplets are very small, the peaks for all the frequencies overlap (as do the troughs) and alternating bands of white and dark are observed. This accounts for the behavior of fogbows. Similar behavior is also observed for particles scattered by atoms.

For light that is polarized in certain directions, Airy's theory does not predict the correct behavior (see Khare, 1974 for examples). This is problematic because light from a rainbow is almost completely polarized. Luckily, for the dominant polarization which constitutes most of natural light, Airy's theory works remarkably well. For the opposite polarization Airy's theory predicts minima where there should be maxima. This has to do with  Brewster's angle.

When an unpolarized wave of light is incident on the interface between two materials with different refractive indices, some of the energy is reflected at the boundary and some is transmitted. However, if the incident wave is polarized such that the electric field is parallel to the plane containing a ray of light and the normal to the interface, an interesting thing occurs. For a certain angle of incidence, no light is reflected from the interface. This angle is called Brewster's angle ($$\theta_B$$) (see Figure 7).

If $$\theta_t$$ is the angle made by the transmitted wave to the normal to the interface, then at the Brewster angle

\theta_B + \theta_t = \cfrac{\pi}{2} ~. $$ By Snell's law

\cfrac{\sin\theta_t}{\sin\theta_B} = \cfrac{n_1}{n_2} ~. $$ Therefore, at Brewster's angle,

\cot\theta_B = \cfrac{n_1}{n_2}~. $$ For an air-water interface with $$n_1 = 1$$ and $$n_2 = 1.33$$, we get a value of $$\theta_B$$ = 53$$^o$$. This angle is very close to the angle that the rainbow ray makes with the normal to the surface of a droplet at the point where it is reflected inside the droplet. Hence only certain polarizations get reflected at this point and other polarizations are transmitted. At angles greater than the Brewster angle, the radiation is 90$$^o$$ out of phase with light reflected at less than the Brewster angle.

Airy's assumption that the amplitude is constant along the cubic wavefront breaks down due to this polarization.