Triple Rainbow

A Triple Rainbow?
Archibald W. Hendry,

Indiana University, Bloomington IN 47405

I

t’s a good question — is there such a thing as a
triple rainbow? No doubt we’ve all admired
spectacular rainbows, including some double
rainbows. I must also admit that most times when I’m
out watering the yard on a sunny day, I cannot help
myself from trying to trace out the primary and secondarycircular bows in the spray. But a triple rainbow?
This question appeared recently in a national magazine.1 Unfortunately the columnist gave a wrong answer. It’s certainly natural to expect that the third bow
will lie above the secondary just as the secondary lies
above the primary, but this is not the case. At least
since the time of Edmond Halley2 (of comet fame), it
has been known that, incontrast to the first and second bows, the third bow lies in a direction toward the
Sun, at an angle of about 41 around the Sun.3

Fig. 1. Path of a ray through a spherical raindrop, with
one internal reflection. The total angle through which
the direction of the ray is changed is (i – r) + ( – 2r) +
( i – r), that is, ( + 2i – 4r).

Rainbows
Let us recall how this comes about.4 Rainbows areformed by light being refracted into raindrops, reflected one or more times inside the droplet, then refracted
back out again. By this process, a ray of light is bent
through an angle (i – r) at the initial refraction (i, r being the incident, refracted angle, respectively),
through an angle ( – 2r) at each internal reflection,
and through an angle (i – r) at the final refraction.Altogether, therefore, such a ray is deflected through a
total deviation angle D given by
D = (i – r) + N( – 2r) + (i – r)
= N + 2i – 2(N + l)r,
460

Fig. 2. Paths of several rays through a raindrop, with
one internal reflection. Ray 6 is the ray that experiences
the least change of direction (that is, minimum deviation); all the other rays experience a greater change of
direction. There is anaccumulation of rays close to the
outgoing direction of ray 6, resulting in a significantly
greater intensity of light close to this direction.

(1)
DOI: 10.1119/1.1625204

THE PHYSICS TEACHER ◆ Vol. 41, November 2003

where N is the number of internal reflections
(N=1,2,3...) and radians = 180 . (See Fig. 1 for
the case where there is just one internal reflection.)
The angles i, r arerelated through Snell’s law
sin i = n sin r,

(2)

where n is the refractive index of water relative to air;
n of course depends on the color (wavelength) of the
individual ray.
The successive bows of a rainbow occur in directions where the total deviation angle D has a turning
point (see for example Fig. 2); there is an accumulation of rays in these directions. To find these directions, weneed to obtain the derivative dD/di and set it
equal to zero. From Eq. (1), we have
dD
dr
= 2 – 2(N + 1) .
di
di
The derivative dr/di in turn can be found by differentiating Snell’s law:
cos i = n cos r

dr
,
di

where cos r itself can be written in terms of cos i:
cos r = (1 – sin2r)1/2 = (1 –
= (1 –

1
1
+
cos2i )1/2.
n2 n2

n2 – 1
,
N(N + 2)

(3)

where N =1,2,3,... . Note that, as N increases, cos i
decreases; that is, i approaches 90 .
Condition (3) can also be understood (see Fig. 3)
in terms of another quantity, the impact parameter b,
defined as the perpendicular distance between the ray
under consideration and the parallel direction
through the center of the droplet. If a is the radius of
the droplet, then b/a = sin i. Condition (3) impliesthat the incident rays that define the directions of the
successive bows correspond to impact parameters that
take continually increasing values (the limit occurring
T HE PHYSICS TEACHER ◆ Vol. 41, November 2003

for grazing rays when b/a = 1).
To determine whether the directions of the successive bows correspond to maximum or minimum turning points of the deviation D, we need to take one...