"Seeing" and "observing" a rainbow

You have certainly seen a rainbow many times. You were surely fascinated by the spectacle, but do you think you looked at it carefully? For example, how many of the following questions could you answer?

1. How big is a rainbow? That is, how high is it above the horizon?
2. What is the radius of the arc described by the rainbow?
3. What is the sequence of the colours? Is red on the inside or on the outside?
4. Sometimes a second arc can be seen. The previous questions apply to it too.
5. Is there also a third arc, and a fourth, etc.?
6. Is the sun behind us or in front of us?
7. Can anything be seen below the main arc?

It has been known since the 13th century that the rainbow is produced by water droplets suspended in the air, for example after thunderstorms. Isaac Newton, in his treatise on optics, showed that white light can be split by a prism into monochromatic components, and so gave an explanation of the phenomenon. His theory, however, was based on the idea that light has a corpuscular nature and, as we will see later, this makes it impossible to describe some special features of the phenomenon.

We will first show Newton's reasoning, and then try to find out what all this has to do with waves.

Answers from geometrical optics

If the water drops have a diameter of a few millimetres, the formation and the features of the rainbow can be explained entirely by applying geometrical optics.

Refraction by a drop

Let us first consider the refraction of white light by a drop. We choose the x axis along the direction from which the light comes.

Referring to the drawing, where the light ray indicates the path of a wavefront in the drop, the law of refraction (or Snell's law) holds

${\displaystyle \sin \alpha =n\sin \beta }$,

where

${\displaystyle n=\frac{n_{\mathrm{w}\mathrm{a}\mathrm{t}\mathrm{e}\mathrm{r}}}{n_{\mathrm{a}\mathrm{i}\mathrm{r}}}}$,

and the refractive index of air is about 1, while that of water varies with colour. On average it is 1.336, ranging from 1.331 for red to 1.343 for violet.

The exit angle φ from the drop with respect to the direction of entry can be expressed in terms of the angles of incidence α and of refraction β as

${\displaystyle \varphi =4\beta -2\alpha }$

Substituting Snell's law into the previous expression, and plotting φ as a function of the angle of incidence α (see figure), we observe that:

• for α=0 the light is reflected backwards.
• as α increases, φ initially grows. But for α≈60° φ reaches a maximum value that depends on the colour.
• For larger values of α the angle φ decreases again.

The value of α for which φ is maximum is found by solving

${\displaystyle \frac{d\varphi }{d\alpha }=0}$,

which gives

${\displaystyle {\cos }^2\alpha =\frac{n^2-1}{3}}$,

Now, for each colour we can calculate the angle of incidence for which the exit angle from the water drop is maximum (see table)

This means that the largest angle occurs for red, and the smallest for violet.

The drop has spherical symmetry, so these angles define cones, which bound the region within which each colour can be refracted.

For angles smaller than 40°39' all colours are refracted, and so the original white light stays white. Light, on the other hand, cannot leave the drop at angles larger than 42°22'.

Therefore:

• the maximum angular size of the arc is about 42°22'. This angle is measured with respect to the elevation of the sun. So at sunrise or sunset the rainbow is higher, while when the sun is higher than about 42° (for example around noon in summer) the rainbow stays below the horizon.
• red occupies the outermost arc, while violet lies inside, at an angle of 40°39'.

The light ray can also undergo two reflections inside the drop before leaving it. This produces the second, fainter rainbow. This time there is a minimum angle φ instead of a maximum, and the order of the colours is reversed.

• The second rainbow lies outside the first. Red is on the inside, at an angle of 50°22', while violet is on the outside, at an angle of 53°28'.

Beyond the rainbow: answers from wave optics

If the diameter of the drops is smaller (about 1 mm), the interference of the light waves inside the drops becomes important, and dark bands can appear inside the rainbow, because of the zones where the interference is destructive.

For drops with a diameter between 0.1 mm and 50 μm, diffraction effects become significant.

For drops with a diameter of less than 50 μm, diffraction prevents the separation of the colours, but not the reflection of the light, and the so-called "white rainbow" occurs.