# Lectures 20 and 21 Propagation of electromagnetic waves in

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```					PHY205 Electromagnetism Lectures 20 and 21

Lectures 20 and 21 Propagation of electromagnetic waves in non-
conducting materials
Introduction
In previous lectures we have studied electromagnetic waves travelling in vacuum. We now
extend this treatment to propagation in non-conducting materials. In particular we will be
interested in the propagation of waves from one media into a second different media
(reflection and refraction effects).

Summary of electromagnetic waves in vacuum
• Maxwell's equations in a vacuum lead to wave equations for E and B. The resultant
waves propagate with a velocity c=(ε0µ0)-1/2.
• E, B and the propagation direction are mutually perpendicular (TEM).
• E and B are in phase and have amplitudes related by E=cB.

Propagation in LIH non-conducting media

Propagation velocity and refractive index
The treatment in this case parallels that in a vacuum except that we must replace ε0 by ε0εr
and µ0 by µ0µr. Hence the speed of light in the medium, cm, is given by

cm=1/(ε0εrµ0µr)1/2 = c/(εrµr)1/2

Where c is the speed of light in a vacuum.

However the only materials that have a µr which is significantly different from 1 are non-
LIH ones (e.g. iron). Hence for most LIH non-conducting materials cm≈ c/(εr)1/2.

The refractive index n of a given material is defined as the speed of light in vacuum divided
by that in the material. Hence

n = c/cm = (εrµr)1/2 ≈ (εr)1/2

or              n2 ≈ ε r

Because εr always has a value greater than one, the speed of light in a material is always
less than in a vacuum.

As both εr and n may vary strongly with frequency, discrepancies may arise when
comparing values of εr (often the static, DC value is used) with n2 (generally the value
appropriate to optical frequencies).

Relationship between E and B
In vacuum E=cB.

In a material this is modified to

E=cmB=cB/n

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PHY205 Electromagnetism Lectures 20 and 21

and because

H=B/(µ0µr)≈B/µ0 as µr≈1

⇒ H=nE/(cµ0)

Reflection and refraction at the interface between two different materials
The aim is to establish the properties of electromagnetic waves when they encounter a
plane interface separating two different non-conducting materials (generally two different
dielectrics). In particular equations will be derived which give the fractions of the incident
wave which are reflected and transmitted at the interface.

Mathematical description of a plane wave not propagating along one of the principal axes
So far we have only considered waves which propagate along one of the principal axes.
For example a plane wave propagating along the x-axis is described by an equation of the
form E=E0sin(kx-ωt), one propagating along the y-axis by E=E0sin(ky-ωt) etc.

In the following we will need to consider plane waves which do not propagate along one of
these principal axes. However we can always resolve the direction of the wave onto two or
more of the principal axes.
y

E

θ

x
For example in the figure below the wave lies in the x/y-plane and propagates in a
direction which makes an angle θ to the y-axis. There are hence components cosθ along
the y-axis and sinθ along the x-axis. The equation for this plane wave is hence of the form

E=E0sin(k(xsinθ+ycosθ)-ωt)

The wave vector, k, in the above equations is given by 2π/λ where λ is the wavelength of
the wave. In addition if the wave propagates in a material with a velocity cm then

ω=cmk=(c/n)k

where ω is the angular frequency of the wave. Hence

k=nω/c

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PHY205 Electromagnetism Lectures 20 and 21

Boundary conditions for E, D, B and H at an interface
In the lectures on dielectrics and magnetic materials it was shown that, in the absence of
free surface charge and conduction currents, the following boundary conditions existed for
E, D, B and H

D and B: Normal components continuous
E and H: Tangential components continuous

These boundary conditions will be used below for electromagnetic waves incident at the
boundary between two materials.

Frequency and direction of reflected and refracted waves
In the figure below a plane electromagnetic wave travelling in a material of refractive index
n1 is incident on the plane boundary with a second material of refractive index n2. The
incident wave propagates at an angle θi to the normal of the boundary and has E- and H-
fields of magnitude Ei and Hi. The E-field of the incident wave lies in the plane of incidence
(the x/y-plane), this is known as the E parallel configuration.

y

Ei                      Er         For this configuration the B- and H-fields must be
normal to the plane of incidence.
Hr
Hi
In general there will be, in addition to the incident
θi θr                      wave, a reflected wave at an angle θr and a
n1                        Boundary
x   refracted or transmitted wave at an angle θt. Each
n2                                       of these waves will have E- and H-fields as
θt                      shown with amplitudes related by the expression
Ht              H=nE/µ0c where n is the refractive index of the
Et       appropriate material.

The E-fields of the three waves are given by the
following expressions

Ei=Ei0sin(k1(xsinθi-ycosθi)-ω1t)                  (A)

Er=Er0sin(k1(xsinθr+ycosθr)-ω1t)                  (B)

Et=Et0sin(k2(xsinθt-ycosθt)-ω2t)                  (C)

Where Ei0, Er0 and Et0 are the amplitudes of the E-fields and k1 and k2 and ω1 and ω2 are
the wavevectors and angular frequencies of the waves in the two materials.

The boundary condition for E requires that the tangential component (the component
parallel to the boundary) be continuous. Hence if we evaluate this component on both
sides of the boundary we must obtain the same result.

The tangential component of each E-field is given by the magnitude of the E-field
multiplied by cosθ, where θ is the appropriate angle.
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PHY205 Electromagnetism Lectures 20 and 21

On the side of the boundary in material 1 the total tangential component of the E-field is
the difference of the components due to the incident and reflected waves (see above
figure). In material 2 there is only the transmitted wave.

Hence the requirement that the tangential component of the E-field be continuous can be
written

Eicosθi⎜y=0 - Ercosθr⎜y=0=Etcosθt⎜y=0                    (D)

Where ⎜y=0 indicates that the preceding expression is evaluated at the boundary y=0.

Substituting in (D) the expressions for Ei, Er and Et given by (A), (B) and (C) and evaluated
for y=0

Ei0cosθisin(k1xsinθi-ω1t)-Er0cosθrsin(k1xsinθr-ω1t)=Et0cosθtsin(k2xsinθt-ω2t)       (E)

This equation must hold at all times and for all values of x. This is only possible if all the
coefficients of x and all the coefficients of t are equal. This requires

ω1=ω2 and k1sinθi=k1sinθr=k2sinθt

Hence
• The frequencies of the waves in the two materials are equal ω1=ω2
• The angles of incidence and reflection are equal k1sinθi=k1sinθr ⇒ θi=θr
• The incident and transmitted angles are related by k1sinθi=k2sinθt. However as
k1=n1ω1/c and k2=n2ω2/c = k1=n2ω1/c the expression k1sinθi=k2sinθt. can be written
as n1sinθi=n2sinθt. This is Snell’s law of refraction.

The above results would have been obtained if E were polarized normal to the plane of
incidence (H polarised parallel to the plane of incidence). These results hence apply to any
incident wave.

Amplitudes of the reflected and refracted waves
The above procedure provided information on the frequencies and directions of the
incident, reflected and transmitted waves. The next step is to derive expressions which
give their relative amplitudes.

Returning to equation (E) above, which is valid when E is parallel to the plane of
incidence, the spatial and time dependent components have been shown to be equal.
Hence (E) reduces to

Ei0cosθi - Er0cosθi=Et0cosθt        (F)

Where θr has been replaced by θi.

In addition the tangential component of the H-field must be continuous at the boundary
between the two materials. For the present case H is normal to the incident plane so that
the tangential component of H is simply H.
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PHY205 Electromagnetism Lectures 20 and 21

For the tangential component of H to be continuous we must therefore have

Hi0+Hr0=Ht0

But H=nE/µ0c so the above can be written as

n1Ei0+n1Er0=n2Et0                (G)

Equations (F) and (G) can now be used to eliminate either Er0 or Et0.

Eliminating Et0:

From (G) Et0=(Ei0+Er0)n1/n2

Substituting into (F)

Ei0cosθi-Er0cosθi=(Ei0+Er0)(n1/n2)cosθt

Ei0(cosθi-(n1/n2)cosθt)=Er0(cosθi+(n1/n2)cosθt)

Multiplying through by n2

Ei0(n2cosθi-n1cosθt)=Er0(n2cosθi+n1cosθt)

E r 0 n2 cosθ i − n1 cosθ t
r/ / =        =                                          (H)
Ei 0 n2 cosθ i + n1 cosθ t

Now eliminating Er0. From (G)

Er0=(n2Et0-n1Ei0)/n1

Substituting into (F)

Ei0cosθi-(n2Et0-n1Ei0)/n1cosθi=Et0cosθt

Ei0n1cosθi-(n2Et0-n1Ei0)cosθi=Et0n1cosθt

2Ei0n1cosθi=Et0(n1cosθt+n2cosθi)

Et 0      2n1 cosθ i
t // =        =                                          (I)
Ei 0 n2 cosθ i + n1 cosθ t

r// and t//, which relate the amplitudes of the reflected and transmitted E-fields to that of the
incident E-field, are known as the reflection and transmission coefficients for E parallel to
the plane of incidence.

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PHY205 Electromagnetism Lectures 20 and 21

The polarisation of E can also be aligned normal to the plane of incidence (H is now
parallel to the plane of incidence). The boundary conditions that the tangential components
of E and H are continuous now require:

Ei0+Er0=Et0

Hi0cosθi-Hr0cosθi=Ht0cosθt

Again eliminating either Er0 or Et0 from these two equations leads to expressions for the E
perpendicular reflection r⊥ and transmission t⊥ coefficients:

E r 0 n1 cosθ i − n2 cosθ t
r⊥ =        =                                    (J)
Ei 0 n1 cosθ i + n2 cosθ t

Et 0      2n1 cosθ i
t⊥ =        =                                    (K)
Ei 0 n1 cosθ i + n2 cosθ t

The equation (H)-(K) are known as the Fresnel relationships or equations.

The signs of these equations give the relative phases of the waves. If positive there is no
phase change, if negative there is a π phase change.

Properties of the Fresnel Equations
The properties can be more easily seen if we consider a special case where one of the
materials is air (n=1) and the other has a refractive index n.

Consider the case where the wave is incident from air. Hence n1=1 and n2=n. The Fresnel
equations become

n cosθ i − cosθ t        cosθ i − n cosθ t
r// =                      r⊥ =
n cosθ i + cosθ t        cosθ i + n cosθ t
2 cosθ i                 2 cosθ i
t // =                     t⊥ =
n cosθ i + cosθ t        cosθ i + n cosθ t
r//

r⊥
with sinθi/sinθt=n

The above figure (plotted for n=2) shows
a number of points:

For normal incidence (θi=θt=0) both r//
and r⊥ and t// and t⊥ have the same magnitudes

⎢r//⎢=⎢r⊥⎢=(n-1)/(n+1), ⎢t//⎢=⎢t⊥⎢=2/(1+n)

As θi→90° the magnitudes of the reflection coefficients tend to 1 (total reflection) and the
magnitude of the transmission coefficients tend to zero (zero transmission).

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PHY205 Electromagnetism Lectures 20 and 21

For a certain angle θB, known as the Brewster angle, r// becomes zero whereas r⊥ remains
non-zero. For this angle light can only be reflected with E perpendicular to the plane of
incidence.

If unpolarised light is incident at the Brewster angle then the reflected light will be
polarised.

The Brewster angle is found by setting r//=0.

n cosθ i − cosθ t
r// =                     =0
n cosθ i + cosθ t

⇒         ncosθi=cosθt               (L)

In addition Snell’s law gives us

nsinθt=sinθi                         (M)

dividing (L) by (M)

(ncosθi)/(nsinθt)=cosθi/sinθt=cosθt/sinθi

⇒ cosθisinθi= cosθtsinθt

and hence

sin2θi=sin2θt as 2sinθcosθ=sin2θ

The equality sin2θi=sin2θt implies that either θi=θt or θi=90-θt. The former can not be
correct as we also must have nsinθt=sinθi and n≠1.

Hence at the Brewster angle we must have θi=90-θt and Snell’s law becomes

nsinθt=sinθB nsin(90-θB)=sinθB
θi+ θt=90°
but sin(90-θ)=cosθ hence

θi θi                       ncosθB=sinθB                 ⇒      n=sinθB/cosθB=tanθB
Air
n
Waves propagating from a material into air
θt
We now have n1=n and n2=1. Fresnel’s equations
have the form

cosθ i − n cosθ t          n cosθ i − cosθ t
r// =                        r⊥ =
cosθ i + n cosθ t          n cosθ i + cosθ t
2 cosθ i                   2 cosθ i
t // =                       t⊥ =
cosθ i + n cosθ t          n cosθ i + cosθ t

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PHY205 Electromagnetism Lectures 20 and 21

and nsinθi=sinθt
In this case the equations for r// and r⊥ and for t// and t⊥ are interchanged compared to
those for when the wave is incident from air.
1.2                                             However because nsinθi=sinθt ⇒θt>θi.
1                                             For θi equal to a certain value, the
0.8                                             critical angle θc, sinθt becomes equal to
0.6
unity and hence θt=90°.
0.4
r⊥                                        At this point both reflection coefficients
θΒ                                    become equal to one and the waves
0.2
are totally reflected. As the value of
0
r//
sinθt can not exceed unity, for θi>θc the
-0.2          θC                                 reflection coefficients remain equal to
-0.4
0   10   20     30    40    50    60     70    unity.
80   90

Angle of Incidence (Deg)

Hence for θi>θc all the incident light is
reflected. This is known as ‘total internal reflection’.

θi=θc occurs when sinθt=1. Hence nsinθc=1 ⇒ sinθc=1/n

Power reflection and transmission coefficients
The reflection and transmission coefficients derived above give the amplitudes of the
electric fields associated with the waves.

However in general it is the power of the wave which is measured experimentally. From
lecture 19 the power density of an electromagnetic wave is given by the Poynting vector
N=ExH.

This has a magnitude EH = nE2/(cµ0), using H=nE/(cµ0).

Hence the energy density ∝nE2

The coefficients which give the fraction of energy reflected or transmitted at a boundary
between two materials equal the appropriate values of r2 or t2 with the inclusion of the
appropriate value(s) of n.

If R is the power reflection coefficient at an interface between a material of refractive index
n1 and one of n2 then

n E2   E2
R = 1 r0 = r0 = r 2
n1 Ei2
0 Ei2
0

where r is the appropriate reflection coefficient given by the Fresnel relationships.

Similarly if T is the power transmission coefficient
n E2        n
T = 2 t0 = 2 t 2
n1 Ei2 0
n1

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PHY205 Electromagnetism Lectures 20 and 21

because energy must always be conserved

R+T=1

Example:
What are the reflection and transmission power coefficients for light incident normally from
air into a material of refractive index n?

θi=θt=0° and ⏐r//⏐=⏐r⊥⏐=(n-1)/(n+1)

FG n − 1IJ 2
Hence Rn =
H n + 1K
⏐t//⏐=⏐t⊥⏐=2/(n+1)

Tn = n
FG 2 IJ 2 = 4n
H (n + 1) K (n + 1) 2
Conclusions
• Propagation velocity and refractive index
• Relationship between E and B in a material
• Mathematical description of a plane wave not propagating along one of the principal
axes
• Frequency and direction of reflected and refracted waves
• Amplitudes of the reflected and refracted waves: the Fresnel equations
• Properties of the Fresnel equations
• Brewster angle
• Critical angle and total internal reflection
• Power reflection and transmission coefficients

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