8—Multivariable Calculus 210the relation between the impact parameterband the scattering angleθ. Then compute the differential
scattering cross sectiondσ/dΩ.
Finally compute the total scattering cross section, the integral of this overdΩ.
8.35 Modify the preceding problem so that the incoming object is a ball of radiusR 1 and the fixed
billiard ball has radiusR 2.
8.36 Find the differential scattering cross section from a spherical drop of water, but instead of Snell’s
law, use a pre-Snell law:β=nα, without the sines. Is there a rainbow in this case? Sketchdσ/dΩ
versusθ.
Ans:R^2 sin 2β
/[
4 sinθ| 1 − 2 /n|
]
, whereθ=π+ 2(1− 2 /n)β
8.37 From the equation (8.43), assuming just a singlebfor a givenθ, what is the integral over alldΩ
ofdσ/dΩ? Ans:πb^2 max
8.38 Solve Eq. (8.47) forbwhendθ/db= 0. Forn= 1. 33 what value ofθdoes this give?
8.39 If the scattering angleθ=π 2 sin(πb/R)for 0 < b < R, what is the resulting differential scattering
cross section (with graph). What is the total scattering cross section? Start by sketching a graph ofθ
versusb. Ans: 2 R^2
/[
π^2 sinθ
√
1 −(2θ/π)^2
]
8.40 Find the signs of all the factors in Eq. (8.49), and determine from that whether red or blue is on
the outside of the rainbow. Ans: Look
8.41 If it suddenly starts to rain small, spherical diamonds instead of water, what happens to the
rainbow?n= 2. 4 for diamond.
8.42 What would the rainbow look like forn= 2? You’ll have to look closely at the expansions in this
case. For smallb, where does the ray hit the inside surface of the drop?
8.43 (a) The secondary rainbow occurs because there can be two internal reflections before the light
leave the drop. What is the analog of Eqs. (8.44) for this case? (b) Repeat problems8.38and8.40for
this case.
8.44 What is the shortest distance from the origin to the plane defined byA~.(~r−~r 0 ) = 0? Do this
using Lagrange multipliers, and then explain why of course the answer is correct.
8.45 The U.S. Post Office has decided to use a norm like Eq. (6.11)(2)to measure boxes. The size is
defined to be the sum of the height and the circumference of the box, and the circumference is around
the thickest part of the package: “length plus girth.” What is the maximum volume you can ship if
this size is constrained to be less than 130 inches? For this purpose, assume the box is rectangular, not
cylindrical, though you may expect the cylinder to improve the result. Assume that the box’s dimensions
area,a,b, with volumea^2 b.
(a) Show that if you assume that the girth is 4 a, then you will conclude thatb > aand that youdidn’t
measure the girth at the thickest part of the package.
(b) Do it again with the opposite assumption, that you assumebis big so that the girth is 2 b+ 2a.
Again show that it is a contradiction.
(c) You have two inequalities that you must satisfy: girth plus length measured one way is less than