8—Multivariable Calculus 204To complete this I should finish withdσ/dΩ. The interesting part of the problem is near the
vertical part of the curve. To see what happens near such a point use a power series expansion near
there. Notb(θ)butθ(b). This has zero derivative here, so near the vertical point
θ(b) =θ 0 +γ(b−b 0 )^2
At(b 0 ,θ 0 ), Eq. (8.47) gives zero and Eq. (8.46) tells youθ 0. The coefficientγcomes from the second
derivative of Eq. (8.46) atb 0. What is the differential scattering cross section in this neighborhood?
b=b 0 ±
√
(θ−θ 0 )/γ, so db/dθ=±
1
2
√
γ(θ−θ 0 )
dσ
dΩ
=
∑
ibi
sinθ
∣
∣∣
∣
dbi
dθ
∣
∣∣
∣
=
b 0 +
√
(θ−θ 0 )/γ
sinθ
1
2
√
γ(θ−θ 0 )
+
b 0 −
√
(θ−θ 0 )/γ
sinθ
1
2
√
γ(θ−θ 0 )
=
b 0
sinθ
√
γ(θ−θ 0 )
≈
b 0
sinθ 0
√
γ(θ−θ 0 )
(8.48)
In the final expression, because this is nearθ−θ 0 and because I’m doing a power series expansion of
the exact solution anyway, I dropped all theθ-dependence except the dominant factors. This is the
only consistent thing to do because I’ve previously dropped higher order terms in the expansion ofθ(b).
Why is this a rainbow? (1) With the sun at your back you see a bright arc of a circle in
the direction for which the scattering cross-section is very large. The angular radius of this circle is
π−θ 0 ≈ 42 ◦. (2) The value ofθ 0 depends on the index of refraction,n, and that varies slightly with
wavelength. The variation of this angle of peak intensity is
dθ 0
dλ
=
dθ 0
db 0
db 0
dn
dn
dλ
(8.49)
When you graph Eq. (8.48) note carefully that it is zero on the left ofθ 0 (smallerθ) and large on
the right. Large scattering angles correspond to the region of the sky underneath the rainbow, toward
the center of the circular arc. This implies that there is much more light scattered toward your eye
underneath the arc of the rainbow than there is above it. Look at your next rainbow and compare the
area of sky below and above the rainbow.
There’s a final point about this calculation. I didn’t take into account the fact that when light
hits a surface, some is transmitted and some is reflected. The largest effect is at the point of internal
reflection, because typically only about two percent of the light is reflected and the rest goes through.
The cross section should be multiplied by this factor to be complete. The detailed equations for this
are called the Fresnel formulas and they tell you the fraction of the light transmitted and reflected at
a surface as a function of angle and polarization.
This is far from the whole story about rainbows. Light is a wave, and the geometric optics
approximation that I’ve used doesn’t account for everything. In fact Eq. (8.43) doesn’t apply to waves,
so the whole development has to be redone. To get an idea of some of the other phenomena associated
with the rainbow, see for example