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A2: Electromagnetism and Optics
Toby Adkins

June 20, 2016

Contents
1 Electromagnetism I
1
...
2

1
...
4
1
...
6

1
...
8

Electric Fields in Matter
1
...
1 Dipoles
1
...
2 Polarisation
1
...
3 The Electric Displacement Vector
Magnetic Fields in Matter
1
...
1 Magnetic Dipoles
1
...
2 Magnetisation
1
...
3 The Auxiliary Magnetic Field
1
...
4 Ferromagnetism
Boundary Conditions
1
...
1 Perpendicular Fields
1
...
2 Parallel Fields
Potentials
1
...
1 Electric Potentials
1
...
2 Magnetic Potentials
Electromagnetic Waves in a Vacuum
1
...
1 Waves in a Vacuum
1
...
2 The Poynting Vector
1
...
3 Energy Conservation
1
...
4 Energy and Momentum Transport
Electromagnetic Waves in Linear Media
1
...
1 Normal Incidence
1
...
2 Parallel Polarisation
1
...
3 Perpendicular Polarisation
Electromagnetic Waves in Conductors
1
...
1 Dispersion Relation
1
...
2 Energy Transport
1
...
3 Wave Impedance
Dispersion and Plasmas
1
...
1 Plasmas

2 Electromagnetism II
2
...
2
2
...
1
...
1
...
1
...
2
...
2
...
3
...
3
...
1

3
...
3

Fraunhofer Diraction
3
...
1 Fourier Optics
3
...
2 Abbe's Theory of Imaging
3
...
3 Diraction Limited Imaging
Spectroscopic Instruments
3
...
1 Fringe Formation
3
...
2 Some Important Denitions
3
...
3 The Michelson Interferometer
3
...
4 The Fabry-Perot Interferometer
Polarisation
3
...
1 Types of Polarisation
3
...
2 Birefringence

2

48

49
50
53
55
60
60
60
61
66
70
70
71

1
...
It will be assumed that students are competent with
the use of vector calculus, and in the manipulation of partial dierential equations, as
well as being able to recall Maxwell's equations as covered in the CP2 course
...


3

Toby Adkins

A2

1
...
We will
now look at what happens when materials are eected by external electric elds, in the
particular case of dielectrics ; materials where the electrons are bound to their atoms
...
The application of an external electric eld will cause a separation
of these two components, as shown below
...
1: The original atom (left) and one with an external eld applied (right)
The nucleus of the atom reaches equilibrium a distance d from it's original position when
E = E 0 , where E is the eld due to the electrons at the position of the nucleus
...
Evidently, this means that the
stronger the external eld applied, the stronger the induced dipole eld that is created as
a result
...
1
...
We know that for some dipole
moment p that the potential is given by

V =

1 p·r
4π 0 r3

With careful manipulation (most easily done using index notation, and remembering that
∂i rj = δij ), it can be shown that the coordinate-free form of the electric eld is given by

E=

1 1
(3(p · r)r − p)
4π 0 r3
4

(1
...
Let the external electric eld have value
E + dE at r1 and E at r2
...
2)

(p · E)

F =

This gives us the energy of the ensemble for 'free', as we know that F = − U
...
p1 is orientated with it's dipole moment along x, while p2 is orientated with it's
dipole moment along y
...
Why are these torques not equal and opposite?
The easiest way to answer this question is to use the form of the electric eld given by
(1
...
We nd that in each case, the electric elds are given by

1 2p1
x
4π 0 r3
1 p2
E2 = −
y
4π 0 r3
E1 =

Evaluating the torques using these elds:

1 p1 p2
z
4π 0 d3
1 p1 p2
Γ2 = p2 × E 1 = −
z
2π 0 d3
Γ1 = p1 × E 2 =

These are evidently not equal and opposite as these are the torques of the dipoles around
their own centre, and not their total torque
...


1
...
2 Polarisation
Returning to the scenario above, suppose that we now have n small dipole moments per
unit volume
...
This means that we can dene the
polarisation as the number of dipole moments per unit volume :

P = np =

ni qi di
i

5

(1
...


• Polarisation Volume Charge - Consider a uniform cylinder of oriented surface area
element dΣ
...
The total charge that leaves the volume is given by
the integral of this over the surface, which means that the divergence theorem can
be used to show that
ρp = − · P
(1
...
5)

σp = P · n

• Polarisation Current - Each of the charges moves a distance di as a result of the
external electric eld at some characteristic velocity v i
...
This means that the polarisation
current is clearly
∂P
(1
...
This means that the
polarisation current is a macroscopic eect created from the average of all the microscopic
changes in the positions of the polarisation charges
...
This is why many of these
quantities are often referred to as bound quantities
...
1
...
We now have another charge type to take
account of; the polarisation charges ρp , as well as the free charges ρf
...
4), we can write that

· ( 0 E + P ) = ρf −→
We call the quantity D

· D = ρf

the electric displacement vector dened as
D=

0E

+P

(1
...


The electric eld inside a dielectric is E 0 and the polarisation is P , such that D0 =
0 E 0 + P
...
Calculate the displacement vector D for the cases where the cavity is needle shaped, and where it is a thin circular wafer perpendicular to the polarisation
...
We can model
the rst case as two small spheres of uniform charge separated by a distance d
...
For the ends

σp = P
q = σp A
= PA
The eld at the centre of the needle is that due to the point charges at either end
...
This means that we can assume
that the electric eld inside the cavity is not modied, and so we nd that D = D0 − P
...
We know that σp = P · n = P
...


Linear Materials
Dielectric materials are called linear, homogeneous, and
larisation varies linearly with the electric eld; that is,

P =

isotropic assuming that their po-

0 χe E

The constant of proportionality χe is known as the electric susceptibility of the material,
and is a measure of how the polarisation responds to the applied eld
...
In a vacuum, r = 1, as there is no
polarisation
...
8)

This means that in many cases, computation with D becomes identical to those readers
will be used to with E , up to a constant
...
9)

Always remember when calculating the energy of systems to take account of the energy
required to build up the system
...


Two long concentric, conducting cylinders of radii a and b, where a < b, are separated by
a dielectric of permittivity 1
...
The
dielectric is now removed and the cylinders are placed vertically with one end in the surface
of a non-conducting oil of permittivity 1 and mass density ρ
...

Throughout this question, we will use a dash to prefer to per-unit-length quantities
...
Finding the potential:

V =−

E · dr

Q
b
log
2π 1
a
Q
2π 1
C
=
=
b
V
log a
=

This is the capacitance per-unit-length of the system
...
Then, the total capacitance is given by

C=

2π
b
log a

0(

− z) +

fraction out of oil

2π
=
log

0
b
a

( + z(

2π
b
log a

1z

fraction in oil
r

− 1))

Now we have to turn to the result that we saw last year when dealing with moving capacitor
plates around, namely that

dW = dWCapacitor − dWBattery
For a battery, we know that WB = V Q
...
10)

Toby Adkins

A2

as the battery imposes a constant voltage on the system by denition
...


1 2 dC
V
= mg
2 0 dz
1 2 d
V
2 0 dz

2π
log

0
b
a

( + z(

r

− 1))

= π(b2 − a2 )ρzg
z=

V02 ( r − 1)
(b2 − a2 )ρg log

This result behaves sensibly for some easy limits, such as

9

r

b
a

= 1 and b → a+
...
2 Magnetic Fields in Matter
We are now going to look at what happens when materials are aected by external magnetic
elds
...
Each electron will complete vi /2πri
orbits per second, and so the current is given by

I=

evi
2πri

Then, the magnetic moment for each electron is given by
2
mi = πri I =

evi ri
2

The orbital angular momentum of the electron is given by Li = me ri × v i , which simplies
to Li = me ri vi in the case of a circular orbit
...
However, when an external magnetic eld is applied, these
angular momenta align in a particular direction, that creates a net magnetic moment
...
This
means that the magnetic susceptibility χm is negative
...

• Paramagnetic - These acquire a magnetic moment parallel to the eld
...
There are some permanently magnetic materials, which we will
cover in Section (1
...
4)
...
2
...
Suppose that we have some closed loop bounding an orientated surface
Σ that carries a current I
...
4
...
11)
4π r3
which is quite similar in form to that of the electric monopole; as is usual with magnetic
elds, we replace the scalar product by a vector product
...
Then,
the force on the dipole is given by

F =

(m · B)

Furthermore, the torque is given by

F =m×B
The torque will act in a sense such that it reduces its own magnitude, meaning that it
attempts to align the normal of m with the magnetic eld
...
2
...

Then, the magnetisation of the system is given by

M =n

dm

(1
...


Calculate the magnetisation of a spinning spherical shell of radius R that has charge density σ
...
Thus,

dI = ωσR sin θ(Rdθ)
A = π(R sin θ)2
dm = πωσR4 sin3 θ dθ z
4
m = πσωR4 z
3
This means that the magnetisation of the spherical shell is given by
M = σωR z
In a similar way to polarisation, magnetisation can create currents within the material
...
The resultant magnetisation of this section is the same as if the entire
surface where paved with small current loops
...
These are the surface currents due to
magnetisation, and are given by
Km = M × n

(1
...
Considering this, it can be shown that the magnetisation gives
rise to volume currents given by
Jm =

×M

(1
...
Furthermore, these are
not dissipative, as they merely result from the orbital motion and/or spin of the electrons
...
2
...
We now have another type of current to take
account of; the magnetisation current J m as well as the free current J f
...
14), we can write that

×
We call the quantity H the

B
−M
µ0

= Jf →

× H = Jf

auxiliary magnetic eld dened as
H=

B
−M
µ0

(1
...


Linear Materials
Magnetic materials are called linear, homogeneous, and isotropic assuming that their magnetisation varies linearly with the applied magnetic; that is,

M = χm H
The constant of proportionality χm is known as the magnetic susceptibility of the material,
and is a measure of how the magnetisation responds to the applied eld
...
In a vacuum, µr = 1, as there is no
magnetisation
...
16)

This means that in many cases, computation with H becomes identical to those readers
will be used to with B , up to a constant
...
There is a small air gap of width w in the solenoid, such that w R
...
Compare the magnitude of the magnetic energy
stored in the core, and in the air gap
...
This means that Bc = Bg
...
15),

Bg
Bc
(2πR − w) +
w = IN
µ
µ0
Bg (2πR + (µr − 1)w) = IN µ
Bg =

IN µ
2πR + w(µr − 1)

Evidently, this is a sensible expression; we re-obtain the result for the non-magnetisable
solenoid for µr = 1
...
17)
2
Suppose that the cross-section of the core is A
...
This is one of the reasons why solenoids sometimes have an air-gap in
them
...
2
...
This is due to domains in the material (regions where the local magnetic

moments are all in the same direction) that become aligned under the application of external magnetic eld
...
That is, ferromagnetic materials have non-zero magnetisation in the absence of an external magnetic eld
...
This leads to the concept of a hysteresis
curve
...
This has a few key features:

• Magnetisation Curve (1) - This shows the response of the material under the applied
eld when it is initially un-magnetised
• Magnetisation Energy - The shaded area gives the energy required to fully magnetise
the material
• Saturation eld Bsat - The strength of the magnetic eld in the material at saturation;
i
...
2: A typical hysteresis curve
It can be shown that the work per unit volume moving along the hysteresis curve is given
by

W =

H · dB

(1
...
Evidently, this can often not be computed exactly, but can easily be approximated through rough estimation of the area in terms of a
rectangle, for example
...
This is why soft iron is advantageous when it is used as a core in transformers,
for example
...
However, when this is used with
alternating currents, there is a frequency limit to how quickly the domains can re-align
with one another, limiting the eectiveness of the core
...
3 Boundary Conditions
Now that we have outlined the way in which electric and magnetic elds are modied in
dierent types of materials, we now turn to looking at the boundary conditions on these
elds between two dierent materials
...
Note that these are summarised at the start of Section
(1
...


1
...
1 Perpendicular Fields
Consider the surface below bounding two media characterised by 1 , µ1 and 2 , µ2 , and the
Gaussian surface of side length ε
...


Figure 1
...
For ε → 0
D⊥ − D⊥
2
1

dΣ = ρf n12

→ D⊥ − D⊥ = σf n12
2
1
Thus, the perpendicular D eld is discontinuous at the boundary according to the surface
charge density on that boundary
...

Similarly, consider

· B = 0
...


always continuous

1
...
2 Parallel Fields
Consider the surface below bounding two media characterised by 1 , µ1 and 2 , µ2 , and
the Amperian Loop of side length ε
...


15

Toby Adkins

A2

Figure 1
...
We then have
∂t
E2 − E1 d = −

∂
∂t

B · dΣ
S

where dΣ is the surface bounded by the Amperian loop
...

We are thus left with

→ E2 − E1 = 0
This means that the parallel component of the electric eld is always continuous at the
boundary between media
...


Then

H2 − H1 d = Kf d +

∂
∂t

D · dΣ
S

Through the same logic as before, the time-varying part of this expression is negligible,
and so we arrive at

→ H 2 − H 1 = K f × n12
This means that the parallel component of the auxiliary eld is discontinuous according
to the free surface currents
...


16

Toby Adkins

A2

1
...
In this light, we can write Maxwell's
equations in a new form:

· D = ρf

(1
...
20)

×E =−

∂B
∂t

× H = Jf +

(1
...
22)

By imposing certain conditions on these equations, we can write our E , D, B and H elds
as scalar or vector potentials, that will allow us to easily solve boundary value problems,
as we shall see in the following sections
...
They are as follows:

D⊥ − D⊥ = σf n12
2
1

(1
...
24)

E2 − E1 = 0

(1
...
26)

These are relatively easy to commit to memory, but can can always be recalled from their
derivations, as well as from Maxwell's Equations
...


1
...
1 Electric Potentials
The Electric eld (E ), as we saw in the First Year CP2 course, can be described by a scalar
potential
...

These conditions will be outlined in the following sections
...
This is assuming that there are no time dependant
˙
magnetic elds (B = 0) by (1
...
Assuming further that ρf = 0, meaning that by
denition the polarisation P is uniform, E will satisfy Laplace's equation
2

(1
...
In general, the result that proves the most useful is the solution to this
equation in spherical coordinates, assuming that there is azimuthal symmetry (i
...

∞

V (r, θ) =

Ar +
=0

17

B
r +1

P (cos θ)

(1
...
Inside the
hole there are two line charges of innite length with line charge densities +λ and −λ,
respectively
...
Find the potential everywhere
...


Figure 1
...
However, what does the eld look like for r
d? To answer this, we need to
consider the potential of a line charge

V =−

λ
log r + V0
2π 0

for some reference potential V0
...
Evidently, in order to
match the boundary conditions at r → 0 and r → ∞, we require only the = 1 polynomial
terms in (1
...
This means that the solution must be of the form
18

Toby Adkins

A2

V (r, φ) =

for r ≤ a
for r > a

A1 r + B1 cos φ
r
B2
cos φ
r

We have automatically done away with the linear term in r for r > a as we require that
the solution is properly bounded for large values, as well as the constant and logarithmic
terms for similar reasons
...
A note to the wise; always do some
'sense-checks' on results that you obtain when doing these sorts of questions; they are very
easy to do, and can often result in you nding an error, if it exists
...
From (1
...

19

Toby Adkins

A2

1
...
2 Magnetic Potentials
The Magnetic eld (H or B ), is usually described by a vector potential
...
These conditions will be
outlined in the following sections
...
22), it is clear that in order to be able to dene a scalar potential Ψ for the
˙
magnetic eld H , we require that J f = D = 0
...
29)

Ψ=0

assuming that · M = 0, which is always the case when there is a uniform polarisation
...


Consider a sphere of radius R made of a material with a permanent magnetisation M =
M0 z
...
There are no free currents
...

What is the magnetic eld inside the sphere?
One very useful result worth remembering is expressing z in polar coordinates as
(1
...

Turning our attention to the problem, we need to apply the boundary conditions (1
...
26)
...
The rst condition implies that

B⊥ − B⊥ = 0
2
1
H⊥ − H⊥ = M
2
1
⊥
⊥
H2 − H1 = M0 cos θ

using the fact that B = µ0 H + M
...
Evidently, the solutions to Laplace's equation of Ψ are going to be of the form
of (1
...

B0
B0
lim Ψ = − dz Hz = − z = − r cos θ
(1
...
Imposing the remaining boundary conditions, we obtain the equations

A1 = A2 +
A1 + A2 +

2B2
= M0
R3
20

B2
R3

Toby Adkins

A2

Solving these two equations, and nding A2 by (1
...


H in = − Ψin =
B in =

B0
external eld

B0 M 0
−
µ0
3
2
+
µ0 M
3
magnetisation eld

This would have been a horrible problem to solve if we had not written H as a scalar
potential
...
32)

Calculate the magnetic vector potential A at a point P a perpendicular distance r away
from the centre of a wire of length 2L that carries a steady current I
...

In this case, d is simply the line integral along the wire
...


21

Toby Adkins

A2

1
...
In such
a vein, let us now examine electromagnetic waves in a vacuum (*gasp* has Christmas come
early? )
...
5
...
Note that sometimes a tilde is used to denote complex quantities; this author
nds that this is generally a waste of time; instead, assume that everything has the potential
to be a complex quantity
...

Substitution of the solutions into the wave equations yields the dispersion relation of:

vp =

ω
1
=√
k
µ0

0

Electromagnetic waves thus travel at the speed of light in a vacuum; nothing particularly
surprising thus far
...
5
...
The Poynting Vector is dened
as
S =E×H
(1
...
It is thus
very common to have to evaluate a ux-integral over some bounding surface
...
By performing appropriate integration, show that the energy stored in the electromagnetic eld and the total work done are equal
...
Assuming that
r = µr = 1, we can write that

∂
B · dΣ
∂t
∂S
∂
µ0 Inπa2
E (2πa) = −
∂t
1
˙
E = − µ0 naIφ
2
E·d =−

We will quote the result for the magnetic eld around an innite solenoid as

B = µ0 Inz
meaning that the Poynting vector is given by

1
E×B
µ0
1
dI
= − µ0 n2 aI
r
2
dt

S=

Then the work done is given by

S · dΣ dt

U =

1
dI
= − µ0 n2 a dS dt I
2
dt
1
2 2
=−
(µ0 n a π)
2

I2

inductance per unit length, L

1
= − LI 2
2
We note that this is opposite and equal to the energy required to raise the current in a
magnetically inductive system from 0 to some value I , as required
...
5
...


dW = F · d = qE · d = qE · v dt
Assuming that this single charge is part of a larger ensemble of n charges per unit volume
that are only weakly interacting, then we can write that

dW
=
dt

dV nqv · E
V

dV J f · E

=
V

Thus, we nd that the work done on charges in some volume is equal to the Ohmic Heating
within that volume, as expected
...
22):

−J f · E = −
=−

∂D
∂t
∂D
×H −
∂t
×H −

·E
·E+

×E+
=0

23

∂B
∂t

·H

from Maxwell's equations

Toby Adkins

A2

∂D
∂B
+H ·
+H · ×E−E· ×H
∂t
∂t
∂ 1
1
=
(E × H)
E·D+ B·H + ·
∂t 2
2
=E·

Poynting Vector, S

internal energy density, u

Making the labelled substitutions, we arrive at the energy conservation result of

∂u
+
∂t

· S = −J f · E

(1
...


1
...
4 Energy and Momentum Transport
For plane electromagnetic waves, we know that B 2 = E 2 /c2
...
The intensity of a wave is the average power transported per
unit area, and is given by
I= S
(1
...

Let us consider the radiation pressure of the waves
...
The force (momentum per unit time) is then SdΣ/c, meaning
that the pressure (force per unit area) communicated to the surface is

p rad =

I
c

This corresponds to the component of the Lorentz force along the direction normal to the
surface at incidence
...
Note that the lower case 'p' has
been used so as not to confuse it with power; the power of the wave is simply the intensity
multiplied by the area
...
This is because, in general, the frequency of the electromagnetic
radiation will be on the order of
106 , and so the instantaneous value would vary so
quickly as to be almost meaningless to talk about!

24

Toby Adkins

A2

1
...
This means that Maxwell's
equations reduce to the same as those in a vacuum, but with r = 1 and µr = 1, and so we
obtain the same wave equations, except with the replacements 0 → and µ0 → µ
...


• Refractive Index - Most readers will already be familiar with this concept, but not
necessarily aware of how it is dened
...
As we know that it has to
be less than c, it would make sense to write vp = c/n for n > 1
...
36)

This is evidently a well-dened quantity as both µr and r are bounded > 1
...


• Impedance - Again, most readers may have seen this term in the context of circuit
theory, and not necessarily when applied to waves
...
37)
Hph
The subscript 'ph ' stands for 'phasor', and simply refers to the complex amplitudes
of the components after the time-dependant components have been factored out
...
We nd
that
µ
c
Z=
µ0
(1
...
This is quite a useful expression, and well worth remembering for the
coming sections
...

We rst derive the wave equation, then assume plane-wave solutions, and substitute them
in
...


1
...
1 Normal Incidence
Let us now turn to examining the reection and transmission of electromagnetic waves at
the boundaries between media
...

The rst case we will consider is that where the light is normally incident on the innite
planes bounding the various regions
...


Figure 1
...
We will treat the single
boundary case, and then work up to this slightly more involved example
...
We have thus reduced the problem to the single boundary at
z = − , which we can then shift to z = 0 without loss of generality
...
25) and (1
...
Then,
using the denition of Z

E1 + E1 = E2 and

E1 E1
E2
−
=
Z1
Z1
Z2

These are two equations in to unknowns that we can solve for the reection and transmission coecients of
E2
Z2 − Z1
E1
2Z2
=
and
=
(1
...
38) as required
...
6)
...
40)

E3
E2 E2
−
=
Z2
Z2
Z3

(1
...
If this is the case, then the second
set of equations become

E1 eik1 + E1 e−ik1 = −i(E2 − E2 )
E1 ik1
E
i
e
− 1 e−ik1 = − (E2 + E2 )
Z1
Z1
Z2

(1
...
43)

We can use (1
...
41) to eliminate E2 and E2 from (1
...
43) to obtain

Z2
Z3
Z1
= −iE3
Z2

E1 eik1 + E1 e−ik1 = −iE3
E1 eik1 − E1 e−ik1

By inspection, we can only have E1 = 0 in the above two equations if

Z2 =

Z1 Z3

(1
...
This also happens to be the 'zero reection condition' for similar problems
in dierent systems, such as transmission lines
...
6
...
Let us now delve into that whimsical
world
...
The concept of polarisation is covered fully in
Section (3
...

For the boundary to be properly dened, we require that the boundary conditions hold at
every point within the z = 0 plane, and for all times
...
As the boundary conditions are themselves essentially a set vector
equations, they must also hold component-wise
...
45)

From this, we can extract two familiar results
...
46)
• Snell's Law - Using the fact that k ∝ n, we nd that
ni sin θi = nt sin θt

(1
...
In general, it is only worth remembering how to show (1
...


27

Toby Adkins

A2

Figure 1
...
25) and (1
...
We then obtain

Ei cos θi + Er cos θr = Et cos θt
Ei Er
Et
−
=
Zi
Zi
Zt
As usual, we can simultaneously solve these equations to obtain the reection coecient
as
Er
Zt cos θt − Zi cos θi
(1
...
Instead, let us ask ourselves the question
of under what conditions we obtain no reection?

0 = Zt cos θt − Zi cos θi
= sin 2θt − sin 2θi
= cos(θi + θt ) sin(θi − θt )
This means that the condition is clearly

θi + θt =

π
2

Substituting this result into Snell's Law, we obtain an expression for

θB = tan−1

n2
n1

Do not get this confused with θc , the critical angle for total
dened, where as θc is only dened for n1 < n2
...
49)

internal reection ; θB is always

Toby Adkins

A2

1
...
3 Perpendicular Polarisation
This is very similar to the case above, except we reverse the orientations of the E and H
elds for each of the waves
...
50)

Again, let us ask ourselves the same question about the conditions under which we obtain
no reection
...


Figure 1
...
Rp corresponds to polarisation parallel to the plane of
incidence, while Rs corresponds to perpendicular
...
7 Electromagnetic Waves in Conductors
In a conductor, we cannot assume that ρf = 0 and Jf = 0
...
We have two limiting

• Perfect Conductor - In such a material, we have that σ → ∞, and so τ → 0
...

• Perfect Insulator - In such a material, we have that σ → 0, and so τ → ∞
...

For the remainder of this section, we will be working within the domain of the former case,
though interested readers are welcome to work through the following derivations in the
opposite limit to see if they can re-obtain the results of Section (1
...


1
...
1 Dispersion Relation
In conductors, we thus have Maxwell's Equations in their full form as given in Section (1
...
Taking the curl of (1
...
22), it follows quickly
that
∂2E
∂E
2
E=
µ
+
µσ
2
∂t
∂t
Displacement current

Conduction current

We obtain a similar wave equation for B
...
51)

It can be useful at this stage to write k = k + ik , such that we can make some simplications of this dispersion relation under certain conditions
...
This is because, for a good conductor, the magnitude of
the displacement current can be considered negligible in comparison to the Conduction
current, such that σ
ω
...
Under the good conductor condition, we have that

k2−k

2

0→k =k
k 2 = iµσω
k=
30

µσω
(1 + i)
2

Toby Adkins

A2

where we have chosen the solution that is appropriately bounded for r → ∞
...
52)

is known as the skin depth of the conductor, and is a measure of the distance over it
which it takes the amplitude of the signal to fall to 1/e of it's original value
...


1
...
2 Energy Transport
Given that the amplitude of the wave is decaying, it would be interesting to look at how
the energy is transported with the wave
...
Imposing σ
ω0 , we nd the interesting result that the magnetic contribution
always dominates; the energy is mostly carried by the magnetic eld in good conductors
...
This is because if λ = 2π/k = 2πδ δ , then the envelope
through which we dene group velocity becomes undened; it becomes meaningless to talk
about group velocity!

1
...
3 Wave Impedance
If we substitute the solutions for E and B into (1
...
53)

In the limit of a perfect conductor, δ → 0, meaning that Z → ∞
...


31

Toby Adkins

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1
...
One such way of doing this is by the Lorentz oscillator model, where we model
the electrons as being classically coupled to their atoms
...

Initially, let us assume that the oscillations of the electrons are small enough around their
2
equilibrium point such that they can be approximated by a linear spring force mω0 x, with
some damping term mγ x
...
Recalling that
p = qx:

P = np
=

nq 2
E
2 − ω 2 ) − iγω
m (ω0

Assuming that the material is linear, P =
r

=1+

0 χe E
...
more than one electron) per
atom
...
This allows us to arrive at the slightly more
comprehensive result of
r (ω)

=1+

nq 2
0m

j

2
(ωj

fj
− ω 2 ) − iγj ω

(1
...
This is
what we observe in most materials
• Anomalous dispersion - Higher frequencies travel slower than lower frequencies
...
We saw this in the case of the conductor in the previous section
...
8
...
There
are many interesting aspects of plasmas, such as Debye Shielding and Magnetic Drifts,
that the author would love to be able to go into here, but for the sake of both brevity
and sensibility, we shall concentrate on applying what we have learnt about dispersion to
plasmas
...


Collisionless Plasma
Diuse, high-temperature plasmas tend to be collisionless
...
Of course, this
is ignoring all electromagnetic interactions between the particles
...
This means that the equation of motion is simply
q
x=−
E
mω 2
At this stage, we actually have the decision to treat the plasma as either a dielectric,
or a conductor with complex conductivity (this is ne; there is nothing that says that
conductivity must be real)
...
This turns out to be
r (ω)

where ωp = ne2 / 0 m is known as the
(1
...
55)

plasma frequency
...

J f = nq x
˙
ne2
E
mω
Assuming that the conductor obeys Ohms law
=i

J f = σE
ne2
mω
From the results of Section (1
...
This
means that we can write quite generally that

k=

1
c

2
ω 2 − ωp

(1
...
This means that electromagnetic waves can
only travel in a plasma for ω > ωp , or else they are evanescent waves with a very short
skin depth
...
For frequencies below the plasma frequency, the electrons within the plasma
are able to move quickly enough through it to compensate for the external elds, but this
begins to break down as the frequency of the incident wave reaches and surpasses the
plasma frequency
...
The density
of free electrons is negligible below a height hp and above this height it increases with height
h as
n = a(h − hp )

A transmitter on the ground sends short pulses of radiation of frequency ν vertically upwards
...
If for ν1 = 5M Hz , ∆τ1 = 1
...
16ms, nd values for a and hp
...
First of all, let us look at the group velocity of waves in a plasma
...
57)

vp vg = c2

Now to the question
...
Through the plasma, we have that

tp = 2

dh
=2
vg (k)

1 dh
dk
vg dk

From above, we have that

dh
2c2 k 0 m
=−
dk
ae2
34

Toby Adkins

A2

Then

tp = −

4ω 0 m
ae2

kf

dk =
ki

4ω 2 0 m
ae2 c

By denition, kf is the wave-number at the point of reection, which is zero (cannot
propagate for n too large), and so kf = 0
...
From this, we are able to
complete our calculation
...
2 × 107 m−3

∆τ2 − ∆τ1 =

hp =

c
2

∆τ1 −

2
16 0 mπ 2 ν1
ae2 c

≈ 1 × 105 m

Collision-dominated Plasma
Dense, low temperature plasmas tend to be dominated by collisions
...
Once again, we can treat the system in two ways:

• As a dielectric - In this case, we obtain a complex dielectric constant of
r

=1+

ine2
0 mγω

• As a conductor - Through an identical calculation as before, we obtain
σ=

ne2
mγ

Once again, we obtain the same dispersion relation via substitution into the appropriate
expressions in both cases
...
Other than that, it is an interesting introduction
into the world of Plasma Physics
...


Electromagnetism II

This second chapter on Electromagnetism covers the following:

• Radiation
• Electromagnetism and Special Relativity
• Transmission Lines
The division of the material into two chapters is completely arbitrary
...
In fact, this author does not
understand the decision to include them as part of the course in the rst place, and hopes
to change the syllabus to reect this
...


36

Toby Adkins

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2
...
If both the magnetic and electric elds
fall of as the inverse square of the radius, the ux through such a sphere is zero
...
Note that
this does not mean that all such systems produce radiation; certain symmetries (such
as spherical) prevent radiation from actually being emitted, as the radiation elds may
interfere destructively
...
1
...
1)
6π 0 c3
That is, it is the statement that any accelerated charged particle radiates energy proportional
to the square of it's acceleration
...
From this equation, it is clear that Prad
does not depend on the radius r, nor the time t
...
For some dipole moment p
and some solid angle dΩ, the formula can be re-written as

dPrad
1
=
2 c3
dΩ
16π 0

d2 p
dt2

2

sin2 θ

An electron is released at rest and falls under the inuence of gravity
...
Give an order of magnitude
estimate
...

Assume that the only force eecting the acceleration is gravity, and that the force remains
roughly constant
...
Let the height fallen be h
...
The time taken to fall a
height h is simply given by th = 2h/g
...


Proving Larmor's Formula
Now that we know the result, the question is, how did we arrive here? Let us go about
proving Larmor's Formula, and in doing so consider some interesting eects of acceleration
on charges
...
1)
...
This nite travel time actually means that components of the eld
in the θ direction are created, as shown on the right of Figure (2
...


Figure 2
...
It is possible
to shown this all mathematically, but the knowledge that the elds scale with 1/r is the
most important information for deriving Larmor's Formula (up to numerical factors)
...

Consider an accelerating charge that has an eective dipole moment p
...
Suppose now that the dipole oscillates in the z direction, and dene
θ as normal in spherical coordinates
...
Substituting this in, we arrive at
¨
Larmor's Formula
...
1
...
This is known as
38

Toby Adkins
electromagnetic

A2

scattering
...
2)

This is essentially a measure of the amount of the incident power that is radiated
...
This means that we let ω0 = 0 in the equation of motion of the electron,
meaning that the scattering cross section is a constant; or, put in another way, it
does not depend on frequency
• Compton Scattering - This is the inelastic scattering of EM radiation, where the
frequency of the emitted wave is not the same as the incident wave
...
We can model the electron as
2
being classically bound to it's atom by a spring force mω0
...
Rayleigh scattering occurs
under the condition that ω
ω0
Note that we usually only consider the eect of the electric eld of the incident wave on
the electron, and not that of the magnetic eld
...


FB
qvB
=
FE
qE

Z0 ωB0
E0

ωZ0
c

This quantity is evidently much less than one in most materials, and for most frequencies
...
1
...
Assuming we
are in the Rayleigh Scattering regime, we can write the equation of motion of the electron
as
2
m¨ = qE − mω0 x
x
qE 0
x=
2
m(ω0 − ω 2 )

where we have assumed that the electric eld is of the form E = E 0 cos ωt
...
Writing it in this form is useful, as it allows us to see that the
radiated power is proportional to the incident power, and the scattering cross section
...

This limit is interesting to us as it allows us to explain a commonly mis-understood question; why is the sky blue? The light from the sun that is incident on atoms in the upper
atmosphere causes said atoms to radiate energy, as we have demonstrated
...
5 times the power of red light, making the sky appear blue
...

Conversely, for a sunset, the light from the sun has to travel through a much greater amount
of atmosphere to reach the observer, as the path of the light rays are almost tangential
to the surface of the Earth
...


40

Toby Adkins

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2
...
This author shall assume
that the reader is familiar with Einstein's Postulates of Special Relativity, as well as the
Lorentz Transformations, Time Dilation, Length Contraction and Relativistic Velocity
addition
...


2
...
1 Field Transformations
We will now examine how both E and B elds transform under Special Relativity
...
The derivations will not be covered here, as this author believes knowledge
of them is not required by the syllabus
...
3)

B =B

(2
...
5)

B⊥ = γ B⊥ −

1
v × E⊥
c2

(2
...
We can observe two interesting cases of
these equations:

• If B = 0 in frame F , then there is actually a magnetic eld in F given by
B =−

1
v×E
c2

• If E = 0 in frame F , then there is actually an electric eld in F given by
E =v×B
Both of these cases highlight the importance of the eects of Special Relativity on electrodynamics, as 'new' elds can appear if we previously had not been taking into account its
eects
...

Astute readers may be asking the question as to whether there are any frame invariant
products in relativistic electrodynamics, in the same way as (for example) E 2 − p2 c2 was
invariant for massive objects
...


1
1
E · (v × E ⊥ ) + (v × B ⊥ ) · B ⊥ − 2 (v × B ⊥ ) · (v × E ⊥ )
c2 ⊥
c
1
= γ 2 E ⊥ · B ⊥ − 2 ((v · v) ((E ⊥ · B ⊥ ) − (B ⊥ · v) (E ⊥ · v))
c
v2
= γ2 E⊥ · B⊥ − 2 E⊥ · B⊥
c
= E⊥ · B⊥

E⊥ · B⊥ = γ2 E⊥ · B⊥ −

Similarly, it can also be shown that E 2 − c2 B 2 is also frame invariant
...
2
...
7)

E · dΣ

0
∂V

Now consider a charge q that is at rest at the origin of frame F
...
Our aim is then to nd the eld
at some point P at (r , θ ) in a frame F moving at a velocity v = (v, 0, 0) relative to F
...
7), we can write the electric eld in frame F as

E =
Recalling that r = x 2 + y 2 + z

E =

2

1/2

q
4π

γq r
4π 0 r 3

, we can use the relations above to show that

1 − β2
0

1−

β2

3/2

2

sin θ

r
r2

where β = v/c as normal
...
Let us compute the ux of E through a sphere of radius r in frame F :

E · dΣ =
=
=

q
4π

0

q
2 0γ2
q
2 0γ2

1
γ2

dθ dφ
π

dθ
0
1

du
−1

sin θ
1 − β 2 sin2 θ
sin θ

1 − β 2 sin2 θ

3/2

3/2

1
(1 − β 2 u2 )3/2

where the substitution u = cos θ has been made
...
This means that charge is invariant across frames,
regardless of whether they are relativistic or not
...


42

Toby Adkins

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2
...
is all 'lumped' together in the components
...

A transmission line consists an active wire that carries the (time-varying) voltage/current
from the input circuit to the load, and another conductor that acts as a return path and
is usually earthed
...
Now consider the voltage and current at z
and z + dz as below
...


Figure 2
...
We can re-arrange the above two expressions to obtain

∂I
∂V
= −L
∂z
∂t
∂V
∂I
= −C
∂z
∂t

(2
...
9)

These equations are known as the telegraph equations that dene the behaviour of the
voltage and current
...
This means that both current and voltage behave as waves in the
circuit, propagating at
1
vp = √
(2
...

Substitute this into (2
...
11)

as the characteristic impedance of the transmission line
...


A transmission line is formed by a wire of radius a placed a distance d in a vacuum above
an innite conducting plane, where d a
...
Assume that the wire has charge per unit length λ, and carries a current I
We can treat this system using the method image charges
...
Then, the electric eld at some point satisfying z > 0 along the line joining
the two wires is given by

E(z) =

λ
λ
−
2π 0 (d − z) 2π 0 (d + z)

Integrating to nd the potential:
d−a

∆V = −

dz E(z)
0

d−a
λ
1
1
dz
−
2π 0 0
d−z d+z
λ
2d − a
=
log
2π 0
a
λ
2d
log
2π 0
a

=−

We have made use of the fact that d
per unit length is given by

C =

a in the last step
...
By Ampere's law, it is easy to show that

B(z) =

µ0 I
2π

1
1
−
d−z d+z

44

Toby Adkins

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Calculating the magnetic ux:

φ=

B · dΣ
d−a

=

µ0 I
2π
0
µ0 I
2d
log
2π
a

1
1
−
d−z d+z

dz

where we have again made use of d
is given by

L =

a
...
This means that, to a rough estimate, we require that d
3a,
meaning our assumption that d >> a is only valid for systems of much higher impedance
...
3
...


Figure 2
...
11)
...
From
this, it is easy to show that
Vr
ZT − Z0
r=
=
(2
...


V ref lected → −V incident
 ZT = Z0 - No energy is reected
...


V ref lected → 0
 ZT → ∞ - All the energy is reected but with no phase shift
...
13)

as being the input impedance of the system at z = −
...
Note that we can re-obtain the condition for impedance
matching by letting Zi = Z0 in this expression
...
The resistance of the leak is
equal to the characteristic impedance of the line
...

We can model this problem by considering the leak as being some impedance Z , as shown
in the following Figure
...
4: A leaky transmission line
We need to consider the waves in each of the regions (1) and (2)
...
Recalling that Pf =

Vf 2
Vi ,

we have shown the result
...
3
...
These are
typically used to replace capacitors and inductors in high frequency circuits, as under these
conditions these components perform poorly due to parasitic reactance
...
However,
at the higher frequencies, the associated wavelength is suciently small such that the stub
does not have to be very long
...
In this case, we can either use the condition
that I = 0 at the end (z = − ), or look at limiting cases of (2
...

ZT → ∞
r→1
Zopen → −iZ0 cot k
• Short Circuit Stub - Let us assume the end of the stub that is not connected to the
system is terminated by an short circuit
...
13)
...
14)

This actually has an interesting practical application; one can nd the characteristic
impedance of a transmission line by measuring Zopen and Zshort by attaching a stub to
it
...

Note that with stubs, it is generally useful to match admittances, dened as Y = 1/Z
...

47

3
...
Generally, it is not hard to gain an understanding of the
fundamentals of Optics, but students often nd it dicult to apply this learned knowledge
to problems
...


48

Toby Adkins

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3
...
5), light behaves as a superposition of orthogonal electric
and magnetic elds, and as such, must be a solution to the wave equation
...
As we have seen previously, we
take take the amplitudes of the waves from point sources, and add them together to give
the total resultant amplitude U (θ)
...
1)

any non-negligible phase dierence between the two point sources will lead to an interference pattern when properly imaged
...
The new wave-front is the tangential surface to all of these
secondary wavelets
...
This idea forms the basis of most wave optics calculations
...
The resultant intensity pattern is observed at innity,

or more conveniently in the focal plane of a lens that collects the diracted rays
...
Consider the gure below
...
1: Diraction by some general aperture
Let the coordinates (x, y, 0) represent a point P1 within the aperture, and the coordinates
(x , y , z ) represent some point P2 at a distance r from the centre of the aperture
...
Then

r=

(x − x)2 + z 2

r − x sin θ +

x2
2r

where we have expanded under the assumption that x
r , and observed that under the
small angle approximation, sin θ x/r
...
2)

R

Note that this is not how we dene Fraunhofer Diraction; it is a consequence of the
Fraunhofer condition, not the other way around
...
1
...
It can be shown that the Fresnel-Kircho diraction integral is of the form

i
U (θ) =
λ

incoming amplitude

U0

dS

eikr
r

obliquity factor

η(n, r)

Huygens' Wave

integral over aperture

The normalisation constant in-front of the integral aside, it is pretty clear that this expression is simply summing up individual Huygens waves across the entire surface of the
aperture
...
Its explicit form is not an important consideration here; just know
that it has value zero for waves travelling in the negative direction
...
Ignore the obliquity factor - For small angles around the optical axis, η(n, r)

1
...
Far Field Approximation - If we assume that we are viewing the diraction under
the condition given by Equation (3
...
This means that we can ignore the factor
of 1/r
...
Fraunhofer Condition - If we impose the condition that the phase dierence has to
be a linear function of position, we can separate out this from the exponential factor

eikr = eikr0 eikx sin θ ∝ eikx sin θ
We can absorb the remaining factor into our normalisation assuming that r0 is
roughly constant, which is the case in the Far Field Approximation
...
Region of Integration - As we are no longer integrating over all r, we can then restrict
the integral to one dimension: dS −→ dx
...
3)

Toby Adkins

A2

The function T (x) is the aperture's transmission function
...
Some common transmission functions include:

• Single Slit of width a:

T (x) =

• Double slits separated by d:

1 for |x| < a/2
0 otherwise

T (x) = δ(x − d/2) + δ(x + d/2)

• Diraction grating of spacing d:

T (x) =

N −1
m=0 δ(x

− md)

Once we start dealing with diraction as a Fourier transform, results from the mathematics
of Fourier transforms become useful
...

This means that we can nd the diraction pattern resulting from a convolution of two
known transmission functions by the product of the amplitude patterns resulting from each
transmission function
...
If we then
want two triangular functions as our nite amplitude proles for double slits, we can then
simply multiply this by the double split pattern
...

The factor β = k sin θ is known as the spatial frequency ; the number of cycles per visual
degree of angle
...
We
can intentionally use this by blocking o part of the Fourier plane if we want to limit our
imaging to high or low spatial frequencies
...
For example, consider T (x) = 1 + cos(ωs x) for ωs = 2π/ds i
...
a
periodic structure of size ds
...
Evidently, Blocking some of these will change the nature of the image
...


An Illustrative Example
Find the intensity pattern for a grating of N slits of width a, each separated by a distance
d
...
Find the
51

Toby Adkins

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resolving power of the grating
Using the apparatus that we have set-up above, we can write that

I(θ) = |U1 (θ) U2 (θ)|2
where U1 (θ) corresponds to the amplitude resulting from the single slit, and U2 (θ) corresponds to the amplitude resulting from the delta-function diraction grating
...

N −1

∞

dx eikx sin θ

U2 (θ) ∝
−∞

∝

δ(x − md) ∝ 1 + eikd sin θ + e2ikd sin θ + e3ikd sin θ +
...
4)

diraction pattern

The normalisation factor comes from the fact that we want the intensity pattern to have
value I(0) at θ = 0; expanding the sine functions for small values of their argument allows
us to see that this is the correct normalisation factor
...
On the other hand, the
minima occur both at the zeros of the sinc and sine functions:

N kd sin θ = 2πn and ka sin θ = 2πm
for integers n and m
...
What happens if we now let d = a?
It is quickly obvious that this means that every th maxima of the diraction pattern
coincides with a minima of the envelope, and so is missing from our diraction pattern
...
This is because the spatial
frequency of the grating is on the order of that of a single slit, and so one cannot distinguish
between the dierent slits
...
This system can add a phase dierence
depending on the incident angle
...
Evidently, for each individual
mirror surface, φ = θ
...
This
introduces a phase dierence of
(3
...
2: Phase dierence for a reection grating
This is known as
the source slit
...
Note that if θ = −φ, the light will return through

We obtain the same intensity pattern as the diraction grating, except with this new
phase term
...
Furthermore, the
intensity is signicantly reduced as a result of the fact that only half of the grating surface
is reective
...


3
...
2 Abbe's Theory of Imaging
Abbe's Theory of Imaging characterises the action of an optical system as a Fourier transform of the initial object, part of which is sampled and inverse Fourier transformed by
the imaging system, such as the lens
...

Let us now go about demonstrating this
...
The diracted
rays ass through a lens of focal length f placed at a distance u from the object, as shown
in Figure (3
...

Let r1 be the optical path from the object plane (x) to the image plane (x )
...
We will now make use of the Fresnel-Kircho
diraction integral as given by Equation (3
...
3: Demonstrating Abbe's Theory of Imaging
Then, considering the resultant amplitude in the image plane:

fI (x ) =

i
λ

dx eikr2 f (x ) =

−1
λ2

dx eik(r2 +ox ) F (k sin θ)

This means that the image plane at v is the conjugate plane of the source, and is the
inverse Fourier transform of the amplitude sampled by the lens, as stated above
...
Find an expression
for the intensity pattern as a function of the angle θ between the direction of observation
and the normal to the slits
...
In the case where d = 3a, make a sketch of the intensity
distribution: (i) in the plane of the slits, (ii) in the focal plane of the lens, and (iii) in the
image plane of the lens at a distance v behind the lens
...

Clearly, the intensity pattern will be as result of the convolution of the pattern due to a
single slit of width a, and two point sources separated by a distance d
...
1
...

a/2

U1 (θ) = U0

dx e−ikx sin θ ∝ sinc

−a/2

1
ka sin θ
2

and
∞

dx δ(x + d/2) + δ(x − d/2) ∝ cos

U2 (θ) = U0
−∞

1
kd sin θ
2

The intensity pattern is thus given by

I(θ) = I(0) sinc2

1
ka sin θ cos2
2
54

1
kd sin θ
2

Toby Adkins

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The intensity pattern in the slit is just the mod-square of the result of the convolution of
the two transmission functions; that is, two 'top-hat' functions, as shown above
...


Figure 3
...
There will
also be a reduction in clarity or detail due to the nite diameter of the lens, leading to
diraction limited imaging (see the following section)
...
5: The intensity distribution at a distance v from the slits
Now we can ask ourselves the question as to what occurs when the lens is removed
...
This is because
the Fraunhoer diraction pattern is formed by angles of equal inclination
...
This only occurs at v because there is phase information encoded in the waves
at the plane of the lens that takes into account where the object is (i
...
of the distance u)
...
1
...
6)

Note that the right-hand side of this equation should be multiplied by a factor of 1
...
This is also the minimum angular separation for which two point
sources resolved under the Rayleigh Criterion:

Two objects are considered distinguishable if and only if the limit of the
maxima of one falls on the minima of the other
...

In the subsequent sections, we will investigate some optical devices that are limited in this
way
...
The smallest spatial scales of
the object are only sampled if, under the small angle approximation:

θ=

λ
D
2λu
<
−→ dmin =
d
2u
D

Figure 3
...
7)

NA stands for numerical aperture, which is in fact dened by this equation; it is a dimensionless number that characterises the range of angles over which a system can accept or
emit light
...


56

Toby Adkins

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Figure 3
...
The angular size subtended at the near point
D 25 cm of the eye is α = xo /D
...
Then, the
total magnication is

M=

β
D xi
D v0
=
=
α
fe xo
fe u0

Thus, the total magnication of the system is the product of the magnications that result
from the individual lenses
...
This means that we can write the magnication in the form

M=

D L
fe f0

(3
...
This can either be done by reection or
refraction
...
7), except that the
rays are incident parallel to one another at some angle α, and that v0 = f0
...
9)

What about it's resolution? The easiest way to argue this is in reverse by asking the
question 'what would a point source look like if projected from a telescope
...
If the diameter of the telescope is D,
then the minimum resolvable angle is

θresolution

1
...
10)

Optical Fibre
These utilise total internal reection to keep the light travelling inside the bre
...


57

Toby Adkins

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Figure 3
...


nf sin θf = ni sin θi
When sin θf > ni /nf , that is, greater than the critical angle, then we have total internal
reection
...
Apply Snell's
law at the end of the bre:

n0 sin θ0 = nf sin

π
− θf
2

= cos θf =

1 − sin2 θf

But the condition for total internal reection is that sin θf > ni /nf , such that

nf cos θf = n0 sin θ0 <
We thus dene

NA =

n2 − n2
i
f

n2 − n2 = n0 sin θ0max
i
f

(3
...
What the two denitions have in common, however, is that they both set
a maximum angle at which the component can accept information
...
The F # (said 'Fnumber'), measures the amount of light that a lens can collect
...
The ux density is then related
both to the image area (∝ f 2 ) and the energy collected (∝ D2 )
...
12)

This means that the ux density is proportional to one over the square of the F #
...
This means
that the ux density is proportional to D4 , where D is the diameter of the lends
...

58

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Finite Sources
We are now going to look at the case where the source pertaining to an optical system
is nite, and thus limited by diraction
...
Then by
Equation (3
...
13)

for a lens of diameter D
...

It is also worth considering the angular shift in the diraction pattern that results from
the nite nature of the source slit
...
The diraction pattern is observed in the
focal plane of a focal length f2
...
For this eect to be neglected, we require that x0
xp , where xp is
the linear position of the pth principle maximum
...
2 Spectroscopic Instruments
We are now going to delve into the realm of spectroscopy (loosely, the measurement of the
intensity as a function of wavelength), and take a look at some spectroscopic instruments
...
2
...
As such, we are able to add wave amplitudes, as the single wavelength
can interfere with itself to produce the fringe pattern
...
In all cases, fringes will occur where the sum (over all sources and
wavelengths) of intensity is constructive
...
They are then non-localised if they can be seen everywhere the beams
cross
...
Extended sources can be
modelled as a series of incoherent point sources
...
This means that extended sources, in general, produce localised fringes
...


As a result, these will be localised at innity, and will be circular rings
...


3
...
2 Some Important Denitions
There are some important denitions that we need to cover before investigating two common interferometers in the coming sections
...
To calculate the free spectral range, nd
the change in ν that increases the phase dierence δ by 2π
...
To calculate the instrumental width, equate the
full-width-half-maximum (FWHM) for the peaks to the change in phase dierence
∆δ
...
For order p, it is dened by
RP =

λ
∆λ INST

=

ν
¯
∆¯ INST
ν

(3
...

60

Toby Adkins

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3
...
3 The Michelson Interferometer
The Michelson interferometer is common conguration for optical interferometry and was
invented by Albert Abraham Michelson
...


Figure 3
...
M2 represents the image of M2
in-front of M1
...
It thus ensures that the phase dierence between the two beams is
due solely to the mirror separation d, and so will be zero at zeroth order
...
This creates the symmetry required for
circular fringes
...
Straight, equally spaced
fringes are observed when there is a small angle between M1 and M2 , and when there
is only a small separation between the mirrors
...


Interference Pattern
To nd the interference pattern, we rst need to derive the phase dierence that results
from some mirror separation d
...
Beam (1) travels an extra AB +BC between the mirror surfaces
...
10: Equivalent Set-up for Michelson Interferometer
Assuming now that the medium separating the mirrors has a refractive index n, then the
total phase dierence between the two beams is given by

δ = 4πn¯d cos θ
ν

(3
...
16)
2
The normalisation has come from the fact that we have to re-obtain the incident intensity
at zero phase dierence
...
Let this have order p0
...
This is because the linear
displacement from the centre is very large for large angles, and so the phase dierence
between the beams becomes vanishingly small in comparison to their propagation
...
We shall dene our order relative to p0
...

2
2
θp d

62

= p0 λ − pλ
pλ

Toby Adkins

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where we have ignored terms of O(θ3 ) and higher
...
17)

Let us now investigate how the order changes when we vary some of the parameters of the
system
...


• Variation with d - From the denition of δ given by Equation (3
...

If this is less than the spread of wavelengths, then we will not see fringes
...
As stated in Section (3
...
1), we add the intensities
¯
of dierent wave-numbers
...
18)

interference

This means that we obtain the characteristic 'beat' pattern, as shown in the gure overleaf
...
11), but the
relationship above still approximately holds
...
This condition amounts to

4πd

∆¯ INST
ν
=π
2

meaning that the instrumental width of the Michelson is given by

∆¯ INST =
ν

63

1
2d

(3
...
11: Two wave-number beat pattern
This means that the resolving power is straightforwardly given by

ν
¯
2d

RP =

(3
...
The intensity of the lat¯ ¯
ν
¯
ν
ter two is half that of the rst
...

As usual, the resultant intensity is just the sum of the intensities from each of the components
...
Dierent velocities of the emitting
particles result in dierent Doppler shifts, the cumulative eect of which is the line broadening
...
This
creates a nite coherence length after which no interference pattern can be observed
...
Let ν0 be the original
¯
wave-number of the source, and that vz is the component of the velocity of an atom in the
source along z
...
21)

Toby Adkins

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Fourier Transform Spectroscopy
The Michelson Interferometer can be used for Fourier transform spectroscopy to nd the
power spectrum p(¯) of the source
...


I(x) = const
...
22)

d¯ p(¯) cos(2kx)
ν ν

(3
...
24)

where Imax − Imin is the dierence in intensity between the intensities at the light and dark
fringes; that is, the dierence between the cosine term of the interference pattern taking
values ±1
...

For a monochromatic source, this is simply unity
...
What is the
visibility of the pattern observed? Using Equation (3
...
+
= const
...
We make use of the result that
√
∞
π −b2 /4a2
−a2 x2
dx e
cos(bx + c) =
e
cos(c)
a
−∞
to nd that the intensity pattern is given by
√
π −(2πx)2 ( vth ν0 )2
¯
c
I(x) = const
...
This means that the visibility is given by
2

2 vth ¯
V = e−(2πx) ( c ν0 ) = e−

8π 2 kB T ν0 2
¯2
x
mc2

= e−x

2 /α2

Evidently, V → 0 for x → ∞ as the spacing will exceed the characteristic coherence length
α, and we expect that there is simply the mean average value left in the interference
pattern
...
If given data for the intensity as a function of the separation x, this
means that we can estimate α, and thus the temperature of the star
...
12: Path of light beam through the FPI

3
...
4 The Fabry-Perot Interferometer
Also known as the Fabry-Perot Etalon, the Fabry-Perot interferometer (FPI) consists of
two surfaces of reectivity R = |r|2 and transmittance T = |t|2
...
15)
...
17)
...


= U0 t2

1
1 − r2 eiδ

innite series

1
I(δ) ∝
∝
4 − 2r 2 cos δ
1+r
1+

1
4R
(1−R)2

sin2

δ
2

We can thus write the transmitted intensity of the FPI as

I(δ) =
where the quantity F is known as the

I0
1+

4F 2
π2

sin2

δ
2

(3
...
26)

Clearly, as 0 < R < 1, higher nesse corresponds to higher reectivity
...
To see why this, we need to consider the intensity of the outgoing rays
...
Suppose that the decrease in
intensity corresponding to the N -th ray is given by

R2N = e−6 −→ 2N log R = −6
Assuming that R 1 (which is valid in most cases), we can expand the logarithm in terms
of a small parameter β = 1 − R:

2N log(1 − β)

−2N β
66

−2N (1 − R)

Toby Adkins

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It follows that

N=

3
1−R

F

Evidently, this is a rough approximation; we have been quite arbitrary with our denition
of the decay in intensity required for us to neglect a ray
...


Figure 3
...


• Free Spectral Range - As stated in Section (3
...
2), we need to nd the change in ν
¯
that causes a change in δ by 2π
...
27)

• Half Width at Half Maximum - We need to let I(δ1/2 ) = I0 /2
...
28)

Toby Adkins

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• Instrumental Width - We can use the previous result to equate the FWHM to the
change in δ :
2π
= 4πn∆¯ INST d cos θ
ν
F
resulting in

∆¯ INST =
ν

1
∆¯ FSR
ν
F

(3
...
14), it is clear that
(3
...
We have used the fact that fringes are located at 2d cos θ = pλ = ν
...
31)

If this is satised, then the q th order peak from one wave-number will be between the pth
and the (p + 1)th orders of the other
...

¯
ν
¯

Figure 3
...
Of course, this
does not solve out ambiguity, but merely gives us two possible values
...

There are also other considerations that have to be taken into account when analysing a
spectrum, including:

• Etalon Design - The parameters that we have to tune are the optical thickness (nd)
and the reectivity (R) that will aect the Finesse, in order to satisfy the condition
given by Equation (3
...
Ideally, one would choose the spacing such that the fringes
corresponding to the second wave-number lie in the middle of an order
...

• Illumination Type - We should also deal with the case where the light passing through
the etalon is not a continuous beam, as we have thus far assumed, but a single laser
pulse that satises τ ∆ω 1, where ∆ω is the frequency width of the pulse
...
This places an upper
limit on the useful nesse, and thus reectivity, given by

F≤

πcτ
d

• Parallelism - If the reecting plates are not quite parallel, deviating at maximum by
h, then this introduces on average an error of 2h every time this mirror is visited
...
For coherence
to hold, we require that λ > 2hF
...
32)
2h¯
ν
The optimum set-up for the FPI will thus evidently depend on the spectrum being analysed,
and so the above points always need to be considered when conguring the apparatus
...
3 Polarisation
The polarisation of a light wave is dened as the direction of the electric eld as the wave
propagates
...
For light propagating along z :

Ex = E0x cos(kz − ωt)
Ey = E0y cos(kz − ωt + φ(t))
Note that we have made the arbitrary choice to absorb the phase factor into the y component; this could also be put in the x component, or even both
...
3
...
These will depend on both the
relative magnitude of E0x and E0y , as well as the behaviour of φ(t)
...
This means that the direction of the electric eld is not well behaved, and
so we cannot give it a determinate value
...


Linear Polarisation
Linear polarisation occurs where both the direction and amplitude of the electric eld
remains constant
...
The direction of polarisation is at
some angle α to the y -axis, given by

tan α =

E0x
E0y

Figure 3
...
For this, we require that the phase dierence is given by

φ=±

π
2

(3
...
Similarly, the negative
sign is known as right-hand polarised
...
This means that in general we can write

Ex = E0x cos(kz − ωt)
Ey = E0y cos(kz − ωt + φ) = E0y (cos(kz − ωt) cos φ − sin(kz − ωt) sin φ)
Then:
2
Ey
Ex
Ex
−
cos φ = sin(kz − ωt) sin φ = − 1 − 2
E0y
E0x
E0x

1/2

sin φ

Squaring both sides,

Ey
Ex
−
cos φ
E0y
E0x

2

=

1−

2
Ex
2
E0x

sin2 φ

Re-arranging, we arrive at

Ex
E0x

2

+

Ey
E0y

2

−2

Ex
E0x

Ey
E0y

cos φ = sin2 φ

(3
...
We can diagonalise this as a quadratic for to
nd the eigenvectors, which will give the directions of the axes of the ellipse
...

2

3
...
2 Birefringence
In some crystals, the relative permeability of the substance can depend on the orientation of
the electric within it, rather than being isotropic
...
As the wave propagates through such a material, this
71

Toby Adkins

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will introduce a phase dierence between the perpendicular and parallel components given
by
∆φ = |ne − n0 |k
(3
...
This fact is exploited by

wave-plates in order to change the polarisation of light
...
This
introduces a phase shift between the components of ∆φ = ±π/2
...
To
produce circularly polarised light from linearly polarised light, rst place a linear
polariser at 45◦ to the x-y components to ensure that equal intensities of each are
transmitted
...
Introducing a quarter-wave plate will then create the circularly polarised light
...
Then, a
quarter-wave plate aligned along one of these directions (corresponding to the minor
and major axes of the ellipse respectively) will create linearly polarised light
...


• Half-wave plate - As one could have guessed, ∝ λ/2
...
Suppose that the electric eld is initially at an angle α to the
optical axis, the component perpendicular will experience a phase shift of π
...
This means
that if we require a rotation by 2α, we need to place the wave-plate with the optical
axis at an angle α to the polarisation, which can be determined using a polariser
...
Do not forget to use linear polarisers to determine polarisation directions
...


•

Estimate the thickness of a zero-order quarter wave calcite plate for light of wavelength 589 nm (n0 = 1
...
486)
...

Evidently, we require ∆φ = π/2
...
856 µm

For a order-p plate, we make the transformation that ∆φ → ∆φ + 2πp
...
343 mm
...


•

A laser-light lter consists of a thin, plane-parallel quartz plate cut with its faces
parallel to the optic axis
...

The light is plane polarized in the plane of incidence and at 45◦ to the optic axis of the
72

Toby Adkins

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quartz
...
No maxima are found at any longer wavelengths
...
Given that the dierence in principal refractive indices
for quartz is 9 × 10−3 estimate the thickness of the plate
...
16: A diagram of the quartz plate
As θi = 55◦ , this is close to Brewster's angle, meaning that very little light is reected
...


k∆nd = 2π −→ λmax = d∆n −→ d

1
...
Using Snell's law:

nair sin θi = nglass sin θt
Using nair
this gives

•

1 and nglass
100 µm
...
5, we nd that θt

33◦
...

When using a polarising lter, maximum intensity is observed when the axis of the
lter is vertical, which is twice the minimum intensity (observed with axis horizontal)
...
It is now found that the maximum is at α 33
...

Find the ratio of the intensities of the elliptically polarised to un-polarised light
...

Let E1 be the component of the elliptically polarised light oriented along the vertical
(major axis), and E2 be the component orientated along the horizontal
...


Prisms
We can cut the uniaxial crystal into the form of a prism, which allows us to further
manipulate light based on its polarisation
...
17: A birefringent prism
The extraordinary ray can pass if θi < θc (critical angle), while the ordinary ray is totally
internally reected assuming that n0 > ne
...
Suppose that the light incident on it is
un-polarised
...


Figure 3
...
Let us
consider the two cases:
1
...
As n0 >
ne , it will refract towards the normal
...

2
...

This means that it will refract away from the normal
...

The angle θ = γ + is known as the angle of divergence of the two rays
...

If we change the optical axis in region (1) from to ↔, the polarisation direction that is
initially will remain undeviated, while the other will be refracted
...
There
is no dispersion for the polarised light, meaning that it would be good for separating a
particular polarisation from another for spectral analysis
...
If we are splitting an initially un-polarised beam, subsequent polarisations are uncorrelated, which causes fringes
to becomes washed out
...
This can be xed with the use
of wave-plates
...
One of the slits is covered with a polarising material
that transmits only horizontal polarisation, and the other vertical polarisation
...


Polarised Light and Brewster's Angle
Suppose that a beam of polarised light is incident on a boundary between two media at θB ,
as dened in Equation (1
...
We know that the reected and transmitted (refracted) rays
are created as a result of the action of electrons as dipoles within the material
...
This means
that the dipoles created by the light polarised in the plane of incidence are only able to
75

Toby Adkins

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radiate along the transmitted ray
...
This means
that the transmitted ray is partially polarised, while the reected ray only has polarisation
perpendicular to the plane of incidence
...


Figure 3

---