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Rainbow maths!
A. French. March 2004
Rain
cloud
Parallel rays of incident sunlight
Sun


Rainbow
Horizon
Observer
Close up of rainbow showing deflection of incident rays of sunlight
by internal reflection within a spherical raindrop of radius a
Incident ray


a
a
a




Spherical
raindrop


Deflected
ray
Figure 1.
Schematic of rainbow seen by ground based observer. The rainbow
is formed by the deflection and focus of incident sunlight by spherical
raindrops into a narrow range of elevation  .
Assumptions:
(1)
All raindrops are perfectly spherical. Since water molecules are polar, electric fields can
distort the spherical shape of rain droplets so this assumption may become invalid
during thunderstorms.
(2)
Assume raindrops have identical liquid content (water) whose refractive index variation
with optical frequency follows a defined semi-empirical model. This implies the
raindrops are fairly uniform in temperature, pressure and impurity content. We will be
using a model of ‘pure water’ to compute the refractive index, so we neglect the latter.
(3)
The refractive index of the ambient air is unity.
(4)
Assume a ray model of sunlight propagation through the raincloud . i.e. don’t include
any interference effects from the wave nature of light. For this assumption to be valid
the wavelength of light within the raindrop   a . Note wavelength of light varies
between 0.4 and 0.8 microns.
Colour
Red
Orange
Yellow
Green
Blue
Violet
Table 1.
Wavelength in vacuo /nm
780-622
622-597
597-577
577-492
492-455
455-390
Approximate refractive index
1.330
1.341
Primary colours related to ranges of optical wavelength (measured in
vacuo).
Hypothesis: Descartes Theory of the Rainbow
Assert a rainbow is formed by deflection and focus of incident
sunlight by spherical raindrops into a narrow range of elevation  .
The deflection occurs via internal reflection of incident sunlight and
the focusing effect results from the observation that the total angle of
deflection passes through an extremum as the angle between
incidence and raindrop surface normal  is varied.
The large multitude of raindrops within a raincloud allows us to
assume that all possible angles of  are explored within the
ensemble. If a mathematical model can determine  ( ) then the
postulate of ‘extremum deviation’ implies the first derivative of 
must pass through zero.
When
d
is small, this can mean  varies little for large deviations
d
in  . Hence a wide range of sunlight rays (each with a different 
value) will result in similar  angles. This is another way of saying
sunlight is focused in a narrow range of elevation angle.
A mathematical model of the deflection of light by spherical raindrops.
We can compute an expression for  ( ) using the geometrical construction at the bottom of
figure 1. The construction was created noting all surface normals on a sphere point radially
outwards and thus emanate from the center of the sphere. Thus the reflected ray paths bound
two joined isosceles triangles. Also assumed is the Law of Reflection, that is the angle of
incidence of light upon a surface equals the angle of reflection from that surface. (Angles
measured from the ray directions to the surface normal).
Decomposing the total deflection into (A) surface refraction at incidence, (B) internal reflection
& (C) surface refraction at ray emergence:
          2     
A
Incident ray

B
(1)
C
A

a
a
B



a


C
Spherical
raindrop

Deflected
ray
Hence:
  4  2
(2)
Let the refractive index of water within the raindrop be n and let n vary with optical frequency
f by some known relation n( f ) . This is described in equation (?) below.
Snell’s law relates angle  to in terms of n .
sin  
sin 
n
(3)
Therefore:
 sin  
  2
 n 
 ( )  4 sin 1 
(4)
 sin 
  2
 n 
 ( )  4 sin1 
Angle of rainbow
Gradient of 
From (2) we can differentiate to yield:
d
d
4
2
d
d
(5)
Implicit differentiation of Snell’s law (3) gives:
cos   n cos 
d
d
cos 


d
d n cos 
(6)
Using the trigonometric identity sin 2   cos 2   1 and (3) we can write:
sin 2 
cos   1  2
n
(7)
d
cos 

d
n 2  sin 2 
(8)
d
4 cos 

2
d
n 2  sin 2 
(9)
Hence:
This gives:
The minimum of  is found using the angle  min which solves the equation
4 cos  min  2 n 2  sin 2  min
i.e.
d
 0.
d
(10)

4 1  sin 2  min  2 n 2  sin 2  min
 41  sin 2  min   n 2  sin 2  min
 3 sin 2  min  4  n 2



 min  sin 1  
Therefore:
4  n2
3




(11)
The minimum value of  is thus



   4 sin 1  
since from (3)
4  n2
3n 2
2


  2 sin 1   4  n


3






(12)
 sin  
.
 n 
  sin 1 
Hence one expects to see rings of focused light (a separate ring for each frequency since n
varies with frequency) and elevations   .
Note the land based observer shown in Figure 1 will only see the upper arc. A flying observer,
experiencing illumination above and below will see both upper and lower arcs and therefore
see a circular rainbow.
Rain cloud
Incident sunlight
Sun
Rainbow
Incident sunlight
Refractive index of water (for optical frequencies)
Hecht [1] gives a formula for the frequency variation of refractive index of a transparent
medium. Using data from Kay & Layby [2] we can deduce the following semi-empirical
relationship between the refractive index of water and (optical) frequencies of light.
n
2

1
2
f


     15

 10 Hz 
2
(13)
where
(n*n-1)^-2
  1.7587 and   0.3612 .
2.0
1.8
1.6
1.4
1.2
1.0
0.8
0.6
0.4
0.2
0.0
y = -0.3612x + 1.7587
R2 = 0.9876
0.0
0.5
1.0
Refractive index of water
Linear (Refractive index of water)
1.5
(frequency/10^15 Hz)^2
2.0
2.5
The positive roots of (13) were used to compute the graph below for water at 20 degrees
celsius.
In summary, (12) combined with (13) can be used to give the elevation of each colour
component of the rainbow. Note that the + and - solutions of (12) yield the same numerical
result.
So a rainbow is observed at a mean angle of about 41.7o with an angular width of about 1.6o.
Moving outwards in angle, colours go from violet (higher frequency, smaller wavelength)
to red (lower frequency, larger wavelength)
Rain cloud
1.6o
Light from
rain cloud
Rainbow
Horizon
41.7o
Final notes
A full wave-theoretic description is necessary to explain all the properties of the rainbow, but
Descartes' simple model is sufficient for the majority of observations. Rare phenomenon such
as 'multiple rainbows' and 'fog bows' result from (respectively) multiple internal reflections (i.e.
more than just one) and interference effects when the wavelength of light is similar to the
raindrop size. Computation of the light intensity of the rainbow requires a wave-theoretic
approach.
References:
[1]
[2]
HECHT, E. Optics. 3rd Edition, 1998. Addison, Wesley, Longman Inc.
LAY & LAYBY. pp95.
A. French. March 2004.