Thayer Watkins
Silicon Valley

Refractive Angles for Rainbows

What is meant by refractive angle for a rainbow is the angle between the light ray that enters a rain droplet and the light ray leaving that droplet. This angle determines the angle at which the top of the rainbow appears in the sky; that angle being the difference between the refractive angle and the elevation angle of the sun in the sky, as is shown in the diagram below. (For reflected rainbows a different relation applies.)

The deviation angle resulting from a ray of light entering a rain droplet and being reflected k times bebore leaving the droplet is given by the formula:

Δ = 180 -[2(i-r) + k(180 - 2r)]

where i is the incident angle of the ray relative to the perpendicular to the droplet surface (measured in degrees) and r is the refracted angle of the ray relative to the perpendicular to the droplet surface. Δ is the deviation of the ray leaving the droplet from the ray entering the droplet.

The rainbow is formed from the light rays of incidence angle i that produce the maximum value for Δ. The Δ maximizing value of i is found from the first order condition

dΔ/di = -2[1 - (k+1)(dr/di)] = 0
which means that
(k+1)(dr/di) = 1

Let n denote the index of refraction of water. The value of dr/di is found from Snell's Law; i.e.,

sin(i) = nsin(r)
which upon differentiation with respect to i gives
cos(i) = ncos(r)(dr/di)

The above relation for (dr/di) and the firtst order condition together imply that

(k+1)cos(i) = ncos(r)
and hence
(k+1)2cos2(i) = n2cos2(r)

Snell's Law gives

sin2(i) = n2sin2(r)
or, equivalently2(i) = n2(1 - cos2(r))

This relationship combined with the previous relationship derived from the first order condition gives:

1 - cos2(i) = n2 - (k+1)2cos2(i)
and thus
cos2(i) = (n2 - 1)/[(k+1)2 - 1]

This determines i and with i and Snell's Law r can be determined and hence the value of Δ. For an index of refraction of 1.33, a roughly average value for visible light in water and a value of k of 1:

For k=2 the values are:

And for k=3

Since the index of refraction is dependent upon the wavelength of the light it is of interest to determine dΔ/dn.

From the equation for Δ

dΔ/dn = -2(di/dn) + 2(k+1)(dr/dn)
= -2[(di/dn) - (k+1)(dr/dn)]

From Snell's Law

cos(i)(di/dn) = sin(r) + ncos(r)(dr/dn)


ncos(r) = (k+1)cos(i)
so the previous relationship reduces to
cos(i)(di/dn) = sin(r) + (k+1)cos(i)(dr/dn)
and hence
cos(i)[(di/dn) - (k+1)(dr/dn)] = sin(r)
and thus
[(di/dn) - (k+1)(dr/dn)] = sin(r)/cos(i)

This means that

dΔ/dn = -2sin(r)/cos(i).

From Snell's Law

sin(r) = sin(i)/n
so the above relationship becomes
Δ/dn = -2tan(i)/n

From the condition for determining i we have

tan2(i) = [(k+1)2 - n2]/(n2-1)
and hence
dΔ/dn = -2[(k+1)2-n2]1/2/(n2-1)1/2n

The angular width of a rainbow is proportional to dΔ/dn so the above formula says that the secondary rainbow (k=2) should be wider than the primary rainbow (k=1). According to the formula the ratio of the widths should be about 1.8.

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