Waves in composites and metamaterials/Rainbows

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
Rainbows and other atmospheric effects related to rainbows are some of the simplest effects of waves interacting with inhomogeneous media. In this case, the waves are light waves and the medium is an inhomogeneous mixture of air and water droplets.

An example of a rainbow forming from the spray of a waterfall can be seen in the image below.

One can also often see a primary and a secondary rainbow with a dark Alexander's band in between the two. See the image below for an example. Notice that the red colors are on the outside of the primary rainbow but on the inside of the secondary rainbow.

Another interesting phenomenon is that of the supernumerary rainbow. An example can be seen in the image below. In these rainbows we can see alternating bands of green and purple on the inside of the arc. These bands occur because of the wave nature of light.

If a rainbow forms near a water body, multiple rainbows may be observed due to reflection from the water surface. An example can be seen in the image below.

A related phenomenon is the  fogbow which forms in fogs where the size of the water droplets is very small. Fogbows are usually white with alternating bands of dark and light regions in some cases. An example of a fogbow can be seen in the image below.

Interactions between light waves and water droplets also lead to the  glory phenomenon, two examples of which can be seen below. The effect is that of a halo around the shadow of an object such as an aeroplane or a human head.

We can also observe similar phenomena in wet grass. See for example the image of a  Heiligenschein below.

A  dewbow on grass can be seen in the following photo. There's also a glory around the head of the shadow that's not very clearly visible in the image.

Another example is the formation of  coronas around objects. Coronas may be formed due to water droplets, ice, or even dust and pollen particles. A good sample can be seen in the image below.

 Icehalos are formed by the diffraction of light by ice particles. A beautiful example can be seen in the image below.

One can also often see  sundogs at the sides of icehalos where the intensity of the light is high. An image of sundogs adjacent to a halo can be seen below.

More complicated patterns such as  tangent arcs may also be observed around ice halos. A good example are the arcs and rainbows around the ice halo in the image below.

Among other interesting phenomena, there is the  sun pillar phenomenon where a pillar of light appears to extend upward from the ground. See the image below.

Basic Features of a Rainbow
A rainbow is a classic example of wave propagation in inhomogeneous media. In this case the medium is water droplets and air. The basic features of a rainbow are (see Figure 1):


 * 1) The occurrence of a primary and a secondary rainbow. The zenith of the primary rainbow is inclined at an angle of 42° to the line joining the sun and the antisolar point.  The secondary rainbow is inclined at 51° to that line.
 * 2) The sky is darker inside the Alexander's band that separates the primary and the second rainbows.
 * 3) The colors in the primary and the secondary rainbows are in reverse order.

In order to arrive at a theory of rainbows, we start with the following assumptions:


 * 1) The speed of water droplets in the atmosphere is negligible compared to the speed of light. We assume that the droplets are stationary.
 * 2) The droplets are sufficiently far from each other that the effect of multiple scatter can be neglected.
 * 3) It suffices to use geometric optics to explain rainbows.

A quick and easy explanation of the rainbow is that each droplet acts like a prism. This is true to the extent that the refractive index of water depends on the frequency of light and hence its color. However, the prism explanation is unsatisfactory because the surface of a droplet is curved and light gets refracted by a variety of angles. The key to understanding a rainbow is the dark band.

Recall Snell's law for refraction at an interface: $$ {  \cfrac{\sin i}{\sin r} = \cfrac{n_2}{n_1} } $$

where $$i$$ is the angle of incidence, $$r$$ is the angle of refraction, $$n_1$$ is the refractive index of the incident material and $$n_2$$ is the refractive index of the refractive material. For air, $$n_1 \approx 1$$ and for water $$n_2 \approx 1.33$$ (but depends on the frequency of the light). If the incident surface is curved, the light rays will be refracted at different angles and therefore different frequencies of light might overlap (unlike in a prism which has a planar interface).

The curvature of the water droplets causes some light rays to reach the observers whereas other rays are reflected away from the observer. This creates the darker Alexander's band (see Figure 2).

Consider rays of light getting refracted by and reflected inside a spherical water droplet (see Figure 3). The impact distance is defined as the distance between a ray and the equatorial plane of the raindrop (assuming that all rays are horizontal). From the figure we can also see that the rays form a denser cluster at a certain impact distance. The cluster of rays at which this phenomenon occurs is called the  rainbow ray because the intensity of the scattered light appears to be the brightest along this ray. The primary rainbow is formed from rays that undergo one reflection inside the drop. The secondary rainbow is from rays that undergo two reflections.

The scattering angle is defined as the angle between the incident and the transmitted rays. For the primary rainbow, the scattering angle is the angle between the incident rays and the rays transmitted after one internal reflection. For the secondary rainbow, the scattering angle is the angle after two internal reflections. A plot of the scattering angle as a function of the impact parameter is shown in Figure 4. The scattering angle reaches a peak at a value of 180°-42° for the primary rainbow and at an angle of 180°-51° for the secondary rainbow.

In fact, many more than the first two rainbows have been observed in experiments. A distribution of the first twenty rainbows (Walker, 1976) is shown in Figure 5.

Using ray optics (also called Descartes theory), we expect to get a peak in the intensity of scattered light at the rainbow angle. The intensity then drops off exponentially as the scattering angle increases. However, this is not what is observed in nature (because of the wave nature of light).

Descartes Theory of a Rainbow
Figure 6 shows the scattering of a light ray by a raindrop.

Let $$p$$ be the number of times the ray traverses inside the droplet. That means that if $$p$$ = 2 then the ray is reflected once.

From the figure, we can see that the scattering angle $$D$$ is given by

D = 2~(i-r) + (p-1)~(\pi - 2~r) = 2~(i-p~r) + (p-1)~\pi ~. $$

At the critical angle, $$D$$ is stationary, i.e.,

\cfrac{d D}{d b} = 0 \qquad \implies \qquad \cfrac{d D}{d i}~\cfrac{d i}{d b} = 0 ~. $$ Now, $$i$$ is a monotonic function of the impact parameter $$b$$. Therefore,

\cfrac{d i}{d b} \ne 0 \qquad \implies \qquad \cfrac{d D}{d i} = 0 ~. $$ Plugging in the expression for $$D$$, we get

2~\left(1 - p~\cfrac{d r}{d i}\right) = 0 \qquad \implies \qquad \cfrac{d r}{d i} = \cfrac{1}{p} ~. $$ From Snell's law, assuming $$n_1 \approx$$ 1 for air and $$n_2 = n$$ for water, we have

\sin i = n~\sin r \qquad \implies \qquad \cos i = n~\cos r~\cfrac{d r}{d i} = n~\cos r~\cfrac{1}{p} = \cfrac{n}{p}~\cos r~. $$ Squaring the two equations and adding, we have

\begin{align} & \sin^2 i + \cos^2 i = n^2~\sin^2 r + \cfrac{n^2}{p^2}~\cos^2 r \\ \text{or} \qquad & 1  = n^2~\sin^2 r - \cfrac{n^2}{p^2}~\sin^2 r + \cfrac{n^2}{p^2}~\sin^2 r + \cfrac{n^2}{p^2}~\cos^2 r \\ \text{or} \qquad & p^2  = n^2~(p^2 -1)~\sin^2 r + n^2 \\ \text{or} \qquad & \cfrac{p^2 - n^2}{n^2~(p^2 -1)} = \sin^2 r ~. \end{align} $$ Therefore,

\sin r = \cfrac{1}{n}~\sqrt{\cfrac{p^2-n^2}{p^2 - 1}} ~. $$ Similarly, we get

\sin i = \sqrt{\cfrac{p^2-n^2}{p^2 - 1}} ~. $$ Substituting into the expression for the scattering angle, we get

D_{\text{crit}} = 2~\left[\sin^{-1}\left(\sqrt{\cfrac{p^2-n^2}{p^2 - 1}}\right) - p~\sin^{-1}\left(\cfrac{1}{n}~\sqrt{\cfrac{p^2-n^2}{p^2 - 1}}\right) \right] + (p-1)~\pi ~. $$ The critical angle for the primary rainbow is given when $$p$$ = 2. The value is 42.5$$^o$$ below the equatorial plane of the droplet. For the secondary rainbow, the critical angle is 50.1$$^o$$ above the equatorial plane of the droplet.

The Descartes theory explains some of the observed phenomena. However, the size of the droplet is not included in the computation. Neither are supernumerary arcs explained. An improvement over the ray theory of rainbows is the  Airy theory of the rainbow which consider the wave properties of light.

Are raindrops spherical?
We have assumed that raindrops are spherical in the above calculations. But are they really spherical? The answer depends on the size (and therefore the velocity) of the raindrops. The motion of air around the raindrop leads to low pressure regions on the sides and for small raindrops conditions similar to Stokes flow exist. There is also a region of low pressure at the top of the raindrop due to flow separation at higher Reynolds numbers. As a result a raindrop evolves from a spherical shape to a bean-shaped form and finally breaks apart at sizes above 5 mm (see Figure 7).

What about the reflected light from the lower half of the raindrop?
The reflected light does play a small role in the colors seen in a rainbow. However, the effect is small compared to the intensity of light observed at the rainbow angle due to reflection inside a raindrop.

Do fish see bubblebows? (i.e., a rainbow caused by bubbles)
Figure 8 shows the scattering of light in an air bubble inside water. Clearly the light is scattered away from the observer and a region of high intensity is not observed. Therefore fish will not be able to see bubblebows.

Sundogs
Sundogs are bright light sources found (under the right conditions) on both sides of the sun when it is low on the horizon (near sunset or sunrise) at an angle of 22$$^o$$ to the observer. They are caused by light refracted from ice crystals (ice has $$n \approx$$ 1.31) which are flat with hexagonal shape, falling with the flat size nearly horizontal (see Figure 9).

Consider light refracted from a single hexagonal ice crystal as shown in Figure 10.

From the figure, we observe that

A = B = r + r' \qquad \text{and} \qquad D = (i- r) + (i' - r') ~. $$ Therefore, the scattering angle is

D = (i + i') - A ~. $$ To determine where the scattering angle has a point of inflection with respect to the angle of incidence, we take the derivative of $$D$$ with respect to $$i$$ to get

\cfrac{d D}{d i} = 1 + \cfrac{d i'}{d i}~. $$ Now, from Snell's law, with $$n$$ being the refractive index of ice, we have

\sin i = n~\sin r \qquad \text{and} \qquad \sin i' = n~\sin r' ~. $$ Therefore,

\cos i = n~\cos r~\cfrac{d r}{d i} \qquad \text{and} \qquad \cos i'~\cfrac{d i'}{d i} = n~\cos r'~\cfrac{d r'}{d i}     = n~\cos r'~\cfrac{d (A-r)}{d i}      = -n~\cos r'~\cfrac{d r}{d i} $$ or,

\cfrac{d r}{d i} = \cfrac{\sqrt{1 - n^2~\sin^2 r}}{n~\cos r} \qquad \text{and} \qquad \cfrac{d i'}{d i} = -\cfrac{n~\cos r'}{\sqrt{1 - n^2~\sin^2 r'}}~\cfrac{d r}{d i}   =  -\cfrac{\cos r'~\sqrt{1 - n^2~\sin^2 r}} {\cos r~\sqrt{1 - n^2~\sin^2 r'}} ~. $$ Therefore,

\cfrac{d D}{d i} = 0 \qquad \implies \qquad \cfrac{\cos r'~\sqrt{1 - n^2~\sin^2 r}} {\cos r~\sqrt{1 - n^2~\sin^2 r'}} = 1 $$ or,

(1 - \sin^2 r')(1 - n^2~\sin^2 r) = (1 - \sin^2 r)(1 - n^2~\sin^2 r')~. $$ After expanding out and simplifying, we get

\sin^2 r = \sin^2 r' \qquad \implies \qquad r = r' = A/2 = \pi/6 ~. $$ To determine whether this point is a minimum or a maximum, we take the second derivative of $$D$$ with respect to i to get

\cfrac{d^2 D}{d i^2} = \cfrac{d^2 i'}{d i^2} = - \cfrac{d }{d r}\left(     \cfrac{\cos r'~\sqrt{1 - n^2~\sin^2 r}}            {\cos r~\sqrt{1 - n^2~\sin^2 r'}} \right)~\cfrac{d r}{d i} ~. $$ At $$r = r'$$, we can use Maple to show that for all $$n > 1$$, we have

\cfrac{d^2 D}{d i^2} = \cfrac{8~(n^4 - 5~n^2 + 4)}{3~n~(4 - n^2)\sqrt{4 -n^2}} ~. $$ We can show that the above is negative for values of $$n$$ between 0 and less than 2. Hence the deflection $$D$$ is a minimum at $$r = r' = \pi/6$$.

Also,

r = r' \qquad \implies \qquad \sin i' = n~\sin r = \sin i \qquad \implies \qquad i = i' ~. $$ Therefore,

D + A = (i + i') = 2~i \qquad \implies \qquad \sin \cfrac{D+A}{2} = \sin i = n \sin r = n~ \sin \cfrac{A}{2} ~. $$ For ice with a refractive index of 1.31, this gives us an angle of deflection ($$D$$) of around 22$$^o$$.

Remark:
In general the edges of the ice crystals are at all angles, especially at the windy cirrus level, and when the crystals are of the short columnar type, then we see a halo rather than sundogs around the sun (or moon) at 22$$^o$$.