1. Descartes' Rainbow
The white ray light travels from air of index of refraction n1, through a water
droplet of index of refraction n2, and emerges to the air toward an observer.
What are the colors could we see? What is the minimum deviation of the
ray?. We will use the model of Descartes to answer these two questions. We will
see the two rainbows: the primary (one internal reflection and two refractions) and
the secondary (two internal reflections and two refractions).
Â Â
Î´ is the deviation between the incidence and emergence
of the sun light ray.
Î´ = (i  r) + (2Ï€  2r) + (i  r)
= 2Ï€ + 2i  4r
We have Î” = 2Ï€  Î´
Î” = 2Ï€  (2Ï€ + 2i  4r)
Î” = 4r  2i Â Â Â (1)
d(Î”)/di = d(4r  2i)/di = 0
Then: dr/di = 1/2 Â Â Â (2)
From Snell's law, we have:
n1 sin i = n2 sin r â‡‰ sin r = (n1/n2) sin i. The
ratio n1/n2 for airwater is generally taken equal to 3/4.
Therefore:
r = arcsin((n1/n2) sin i)
dr/di = [1  (n1/n2)^{2} sin^{2} i]^{1/2} x (n1/n2) cos i Â Â Â (3)
The relationship (2) becomes:
1/2 = [1  (n1/n2)^{2} sin^{2} i]^{1/2} x (n1/n2) cos i
Or
(1/4) [1  (n1/n2)^{2} sin^{2} i] = (n1/n2)^{2} cos^{2} i
= (n1/n2)^{2} (1  sin^{2} i)
Then:
(1/4) = (1/4) (n1/n2)^{2} sin^{2} i + (n1/n2)^{2} (1  sin^{2} i)
= (1/4) (n1/n2)^{2} sin^{2} i + (n1/n2)^{2}  (n1/n2)^{2} sin^{2} i
Thus:
(1/4)  (n1/n2)^{2} =  (3/4) (n1/n2)^{2} sin^{2} i
or
(n1/n2)^{2}  (1/4) = (3/4) (n1/n2)^{2} sin^{2} i
(4/3) [(n1/n2)^{2}  1/4] (n2/n1)^{2} = sin^{2} i
(4/3) [1  (1/4) (n2/n1)^{2}] = sin^{2} i
sin i = ((4/3) [1  (1/4) (n2/n1)^{2}])^{1/2}
sin i = ((1/3) [4  (n2/n1)^{2}])^{1/2}
i = arcsin [([4  (n2/n1)^{2}]/3)^{1/2}] Â Â Â (4)
i = arcsin [([4  (n2/n1)^{2}]/3)^{1/2}]
for n2/n1 = 4/3, we have :
i = arcsin [((1/3) [4  16/9])^{1/2}] = arcsin((20/27)^{1/2})= 59.40
i = 59.40 ^{o}
From Snell's law: n1 sin i = n2 sin r, we have:
r = arcsin[(n1/n2) sin i] = arcsin[(3/4) (20/27)^{1/2}] = arcsin[0.64] = 40.20
= 40.20 ^{o}
r = arcsin[(n1/n2) sin i]
The relationship (1) becomes:
Î” = 4r  2i
Î” = 4r  2i = 4(40.20)^{o}  2(59.40)^{o} =
160.80^{o}  118.80 ^{o} = 42^{o}.
Î” = 42^{o}
Then, the deviation is :
Î´ = 2Ï€  Î´ = 222 ^{o}
Note &delta is min, then Î” is max.
The relative index of refraction of a ray from
the menium n1 to the medium n2 is:
n = n2/n1
Thus, Snell's law is rewritten, if i and r are the angles od incidence
and refraction respectively, as:
sin i = n sinr â†’ sin r = (1/n) sin i
Using the relationship (2): dr/di = 1/2, we get:
cos r (dr/di) = (1/2) cos r = (1/n) cos i
Thus cos i = (n/2) cos r = (n/2) [1  sin^{2}r]^{1/2} =
(n/2) [1  (1/n)^{2} sin^{2}i]^{1/2}
= (n/2) [1  (1/n)^{2} (1  cos^{2}i)]^{1/2}
= (n/2) [1  (1/n)^{2} + (1/n)^{2} cos ^{2}i]^{1/2}
Then:
cos^{2} i = (n/2)^{2} [1  (1/n)^{2} + (1/n)^{2} cos^{2}i] =
(n/2)^{2}  (1/4)+ (1/4) cos^{2}i)
Or:
(3/4) cos^{2}i = (n/2)^{2}  (1/4).
Hence
3 cos^{2}i = n^{2}  1. Then
cos i = [(n^{2}  1)/3]^{1/2}
cos i = [(n^{2}  1)/3]^{1/2}
2. Properties of Descartes' Rainbow
Here is a table that gives the color, the wavelength in vacuum, the
related index of refraction in water, and the related Descartes rainbow max
deviation.
The index of refraction of air is taken equal to
1.0003 for all these wavelengths of the white light ray spectrum.
color  wavelength  index of refraction  max rainbow deviation 
violet  380–450  1.3455  40.35 
blue  450–475  1.3425  40.77 
green  495–570  1.3350  41.85 
yellow  570–590  1.3334  42.08 
orange  590–620  1.3330  42.14 
red  620–750  1.3250  43.32 
The values of the refractive indexes in water, in this table, are the average
values taken from the results of philip Laven .
3. Secondary rainbow
In this case have, we have two internal reflections and two
refractions:
Î´ = (i  r) + (2Ï€  2r) + (i  r) + 2Ï€  2r
Î´ = 2i  6r + 4Ï€
The condition of minimum is written as:
dÎ´/di = 0 = 2  6 (dr/di).
Hence:
dr/di = 1/3
Using Snell's law:
sin r = (1/n) sin i, we get:
cos r (dr/di) = (1/3) cos r = (1/n) cos i
Thus cos i = (n/3) cos r = (n/3) [1  sin^{2}r]^{1/2} =
(n/3) [1  (1/n)^{2} sin^{2}i]^{1/2}
= (n/3) [1  (1/n)^{2} (1  cos^{2}i)]^{1/2}
= (n/3) [1  (1/n)^{2} + (1/n)^{2} cos ^{2}i]^{1/2}
Then:
cos^{2} i = (n/3)^{2} [1  (1/n)^{2} + (1/n)^{2} cos^{2}i] =
(n/3)^{2}  (1/9)+ (1/9) cos^{2}i = 0.
Or:
(8/9) cos^{2}i = (n/3)^{2}  (1/9).
Hence
8 cos^{2}i = n^{2}  1. Then
cos i = [(n^{2}  1)/8]^{1/2}.
The condition of minimum is:
cos i = [(n^{2}  1)/8]^{1/2}
For the example of n = 4/3, we have:
i = arcos [(n^{2}  1)/8]^{1/2} = arcos [(16/9  1)/8]^{1/2} = 72.0 ^{o}
i = 72.0 ^{o}
r = arcsin [(1/n) sin i] = arcsin [(3/4) sin 72.0] = 45.45^{o}
r = arcsin [(1/n) sin i] = arcsin [(3/4) sin 72.0] = 45.45^{o}
r = 45.45^{o}
Î´ = 2i  6r + 4Ï€ = 2 x 72.0  6 x 45.45 + 00 = 144  272.7 + 720 = 591.3 ==
591.3  360 = 231.3
Î” = Î´  Ï€ = 231.3  180 = 51.3^{o}
Î” = 51.3^{o}
