# Analysing the primary rainbow(using the Debye series) Fig. 1   Three ray paths through a sphere which result in a scattering angle of 141° for wavelength λ = 0.65 µm and a refractive index of 1.33257

Fig. 1 shows that there are two paths of order p = 2 (ray A and ray B) and one path of order p = 0 (ray C) which result in scattering at angle of 141°.   Interference between ray A and ray B causes the supernumerary arcs of the primary rainbow because the two rays have almost identical amplitudes:

• if the optical path difference between rays A and B is n λ (where n is an integer), the two rays add constructively.
• if the optical path difference between rays A and B is a (n + 0.5) λ (where n is an integer), the two rays cancel each other. Fig. 2    Primary rainbow:  Debye series calculation of scattering by a water drop of radius r = 100 µm  for wavelength λ = 0.65 µm (perpendicular polarisation)

Fig. 2 shows that the p = 2 rays are responsible for the smooth maxima and minima of the primary rainbow and its supernumeraries.  However, Mie calculations also show a high frequency ripple structure.  The red curve in Fig. 10 represents the vector sum of the p = 0 and p = 2 contributions, indicating that the ripples are caused by interference between the p = 0 rays (external reflections from the surface of the sphere) and the p = 2 rays.

Note that the intensity at the rainbow angle predicted by geometrical optics (in this case 137.9°) is about 50% of the maximum intensity (at about 138.8°) predicted by Mie and Debye methods.

MiePlot offers the option of using the Debye series.

Page updated on 13 October 2002