8—Multivariable Calculus 203middle one has equal angles of incidence and reflection. The dashed lines are from the center of the
sphere. The three terms in Eq. (8.44) for the evaluation ofθcome from the three places at which the
light changes direction, and they are the amount of deflection at each place. The third equation simply
relatesbto the radius of the sphere.
From these three equations, eliminate the two variablesαandβto get the single relation between
bandθthat I’m looking for. When you do this, you find that the resulting equations are a bit awkward.
It’s sometimes easier to use one of the two intermediate angles as a parameter, and in this case you will
want to useβ. From the picture you know that it varies from zero toπ/ 2. The third equation givesb
in terms ofβ. The first equation givesαin terms ofβ. The second equation determinesθin terms of
βand theαthat you’ve just found.
The parametrized relation betweenbandθis then
b=Rsinβ, θ=π+ 2β−4 sin−^1
(
1
n
sinβ
)
, (0< β < π/2) (8.45)
or you can carry it through and eliminateβ.
θ=π+ 2 sin−^1
(
b
R
)
−4 sin−^1(
1
n
b
R
)
(8.46)
The derivativedb/dθ= 1
/
[dθ/db]. Compute this.
dθ
db
=
2
√
R^2 −b^2
−
4
√
n^2 R^2 −b^2
(8.47)
In the parametrized form this is
db
dθ
=
db/dβ
dθ/dβ
=
Rcosβ
2 −4 cosβ/
√
n^2 −sin^2 β
In analyzing this, it’s convenient to have both forms, as you never know which one will be easier to
interpret. (Have you checked to see if they agree with each other in any special cases?)
0
R
b
0 90 180
θ
n= 1to 1. 5 , left to right
0 90 180
θ
dσ/dΩ
These graphs are generated from Eq. (8.45) for eleven values of the index of refraction equallyspaced from 1 to 1.5, and the darker curve corresponds ton= 1. 3. The key factor that enters the
cross-section calculation, Eq. (8.43), isdb/dθ, because it goes to infinity when the curve has a vertical
tangent. For water, withn= 1. 33 , theb-θcurve has a vertical slope that occurs forθa little less than
140 ◦.Thatis the rainbow.