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2.3. REFLECTION AND TRANSMISSION BY A STACK OF LAYERS 15

R = RL + BTL : (2.28)

This is all we need to solve our problem. The system of equations (2.25)-(2.28) consists

of four linear equations with four unknowns A, B, R and T. We could solve this system

of equations by brute force, but some thinking will make life easier for us. Note that the

last two equations immediately give T and R once A and B are known. The rst two

equations give A and B.

Problem b: Show that

A = (1 TL R ) (2.29)

RL R

;

B = (1 TLRR ) : (2.30)

RL RR

;

This is a puzzling result, the right-going wave A between the layers does not only contain

the transmission coe cient of the left layer TL but also and additional term 1=(1;RL RR ).

Problem c: Make a series expansion of 1=(1 RLRR ) in the quantity RLRR and show

;

that this term accounts for the waves that bounce back and forth between the two

stacks. Hint: use that RL gives the re ection coe cient for a wave that re ects from

the left stack, RR gives the re ection coe cient for one that re ects from the right

stack so that RL RR is the total re ection coe cient for a wave that bounces once

between the left and the right stack.

This implies that the term 1=(1;RL RR ) accounts for the waves that bounce back and forth

between the two stacks of layers. It is for this reason that we call this term a reverberation

term. It plays an important role in computing the response of layered media.

Problem d: Show that the re ection and transmission coe cient of the combined stack

of layers is given by:

2

TL RR

R = RL + (1 R R ) (2.31)

LR

;

T = (1 TL TR ) : (2.32)

RL RR

;

In the beginning of this section we conjectured that the transmission coe cient of the

combined stacks is the product of the transmission coe cient of the separate stacks.

Problem e: Is this correct? Under which conditions is it approximately correct?

Equations (2.31) and (2.32) are very useful for computing the re ection and trans-

mission coe cient of a large stack of layers. The reason for this is that it is extremely

simple to determine the re ection and transmission coe cient of a very thin layer using

the Born approximation. Let the re ection and transmission coe cient of a single thin

layer n be denoted by rn respectively tn and let the re ection and transmission coe cient

of a stack of n layers be denoted by Rn and Tn respectively. Suppose the left stack consists

on n layers and that we want to add an (n + 1)-th layer to the stack. In that case the

CHAPTER 2. SUMMATION OF SERIES

16

right stack consists of a single (n + 1)-th layer so that RR = rn+1 and TR = tn+1 and the

re ection and transmission coe cient of the left stack are given by RL = Rn , TL = Tn .

Using this in expressions (2.31) and (2.32) yields

2

Tn rn+1

Rn+1 = Rn + (1 R r ) (2.33)

n n+1

;

Tn+1 = (1 Tn tn+1 ) : (2.34)

Rn rn+1

;

This means that given the known response of a stack of n layers, one can easily compute

the e ect of adding the (n + 1) th layer to this stack. In this way one can recursively

;

build up the response of the complex re ector out of the known response of very thin

re ectors. Computers are pretty stupid, but they are ideally suited for applying the rules

(2.33) and (2.34) a large number of times. Of course this process has to be started when

we start with a medium in which no layers are present.

Problem f: What are the re ection coe cient R0 and the transmission coe cient T0

when there are no re ective layers present yet? Describe how one can compute the

response of a thick stack of layers once we know the response of a very thin layer.

In developing this theory, Lord Rayleigh prepared the foundations for a theory that later

became known as invariant embedding which turns out to be extremely useful for a number

of scattering and di usion problems 6] 61].

The main conclusion of the treatment of this section is that the transmission of a com-

bination of two stacks of layers is not the product of the transmission coe cients of the two

separate stacks. Paradoxically, Berry and Klein 8] showed in their analysis of \transparent

mirrors" that for a large stacks of layers with random transmission coe cients the total

transmission coe cients is the product of the transmission coe cients of the individual

layers, despite the fact that multiple re ections play a crucial role in this process.

Chapter 3

Spherical and cylindrical

coordinates

Many problems in mathematical physics exhibit a spherical or cylindrical symmetry. For

example, the gravity eld of the Earth is to rst order spherically symmetric. Waves

excited by a stone thrown in water usually are cylindrically symmetric. Although there

is no reason why problems with such a symmetry cannot be analyzed using Cartesian

coordinates (i.e. (x y z)-coordinates), it is usually not very convenient to use such a

coordinate system. The reason for this is that the theory is usually much simpler when one

selects a coordinate system with symmetry properties that are the same as the symmetry

properties of the physical system that one wants to study. It is for this reason that

spherical coordinates and cylinder coordinates are introduced in this section. It takes a

certain e ort to become acquainted with these coordinate system, but this e ort is well

spend because it makes solving a large class of problems much easier.

3.1 Introducing spherical coordinates

In gure (3.1) a Cartesian coordinate system with its x, y and z-axes is shown as well as

the location of a point r. This point can either be described by its x, y and z-components

or by the radius r and the angles and ' shown in gure (3.1). In the latter case one uses

spherical coordinates. Comparing the angles and ' with the geographical coordinates

that de ne a point on the globe one sees that ' can be compared with longitude and

can be compared with co-latitude, which is de ned as (latitude - 90 degrees). The angle '

runs from 0 to 2 , while has values between 0 and . In terms of Cartesian coordinates

the position vector can be written as:

r =x^+y^+z^

xyz (3.1)

where the caret (^) is used to denote a vector that is of unit length. An arbitrary vector

can of course also be expressed in these vectors:

u =uxx+uy y+uz^ :

^^z (3.2)

We want to express the same vector also in basis vectors that are related to the spherical

coordinate system. Before we can do so we must rst establish the connection between

the Cartesian coordinates (x y z) and the spherical coordinates (r ').

17

CHAPTER 3. SPHERICAL AND CYLINDRICAL COORDINATES

18

z-axis

^

Ï• ^

r

(x,y,z)

.

^

Î¸

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