applet-magic.com
Thayer Watkins
Silicon Valley USA

 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:

• i = 59.585°
• r =46.6164°
• Δ = 42.516°

For k=2 the values are:

• i = 71.94°
• r =45.63°
• Δ = 50.10°
And for k=3
• i = 76.91°
• r =47.08°
• Δ = 171.663°

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

From Snell's Law

But

This means that

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. 