Exercise
1: Obtain and use information from peer-reviewed literature
The
plot for the refractive index of water should look like:
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The
plot for the deviation of the refractive index of air from unity should
look like:

Exercise
2: Deflection angle for a light ray entering a spherical raindrop
To
calculate the deflection angle, one needs to know the refractive index of
the air and the water at the particular wavelength and also the angle of
incidence of the incoming ray. The laws of refraction and reflection are
then applied to obtain the angles of refraction and the angle of
reflection for the internal reflection. Students should be able to show
that the equalities shown in the figure illustrating the ray/raindrop
geometry.
Exercise
3: Compute the deflection angle versus incident angle
The
solution for Exercise 3 is:

The
minimum of the deflection function for λ=400 nm
(λ=650 nm)
is δ=139.3∘ (δ=137.6∘).
This means that you must look (with the sun at your back and the distant
raindrops in front of you) at an angle of 180∘−139.3∘=40.7∘ (180∘−137.6∘=42.4∘) relative
to the horizontal to see the bright violet (red) band of the rainbow. So
the red band appears above the violet band. (It really does!) It is worth
noting that the rays producing the primary rainbow enter the top half of
the raindrops.
Exercise
4: Where is the double (secondary) rainbow?
The
solution for Exercise 4 is:
The
minimum of the deflection function for λ=400 nm
(λ=650 nm)
is δ=233.3∘ (δ=230.2∘).
These rays are directed back up into the sky and the ground-based observer
won’t see them. However, if you realize that you have just computed the
deflection angles for rays entering the bottom half
of the raindrop, you realize that you must look (with the sun at your back
and the distant raindrops in front of you) at an angle of 233.3∘−180∘=53.3∘,,(230.2∘−180∘=50.2∘) relative
to the horizontal to see the bright violet (red) band of the rainbow. So
the red band of the secondary rainbow appears below the
violet band. (It really does!). Thus the rays producing the secondary
rainbow enter the bottom half of the raindrops and the order of the color
bands is reversed relative to
the primary rainbow. Comparing these angles to those from Exercise 3, we
see that the secondary rainbow appears above the primary rainbow. The
reduced brightness of the secondary rainbow is due to the additional
internal reflection and the corresponding loss of irradiance during that
reflection. It is posible to extend Exercises 3 and 4 to account for the
change in intensity occurring for every refraction or internal reflection.
For fun, it must be noted that fascination with the double rainbow became
an Internet meme in 2010 when a person living on the border of Yosemite
National Park recorded his reaction to a spectacular double rainbow. At
the time of writing, the video can be seen at: https://www.youtube.com/watch?v=OQSNhk5ICTI).
Exercise
5: Crude estimate of the irradiance versus deflection angle for a single
wavelength: Primary Rainbow
The
solution for Exercise 5 is:
Note
that irradiance is distributed over a large range of detection angles, and
that the irradiance has a peak near (but not at) the minimum of the
deflection function for these chosen parameters. If we use an angular
width smaller than 1∘ for
the assumed irradiance distribution of a ray, the irradiance peak
approaches the minimum in the deflection angle and the “background
irradiance” decreases. The main result is that irradiance accumulates in
the direction specified by the minimum in the deflection function because
several rays are deflected into essentially the same direction. This is
known as rainbow scattering, and it is also
observed in scattering events such as atomic/molecular collisions or ion
collisions with ordered surfaces.