8—Multivariable Calculus 239At(b 0 ,θ 0 ), Eq. ( 33 ) gives zero and Eq. ( 32 ) tells youθ 0. The coefficientγcomes from the second derivative of
Eq. ( 32 ) 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
sinθ1
2
√
γ(θ−θ 0 )+
b 0 −√
(θ−θ 0 )/γ
sinθ1
2
√
γ(θ−θ 0 )≈b 0
sinθ 0√
γ(θ−θ 0 )(34)
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 isπ−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 0db 0
dndn
dλ(35)
When you graph Eq. ( 34 ) 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