PHYS 263

Physical Optics

Lecture Notes

Jakob J. Stamnes

—————————————————————–Department of Physics, University of Bergen, 5007 Bergen.

Tel: 55 58 28 18. Fax: 55 58 94 40. E-mail: JakobJ.Stamnes@fi.uib.no

Spring 2003

FYS 263 1

Contents

I Elementary electromagnetic waves 3

1 Maxwell’s equations, the material equations, and boundary conditions 41.1 Maxwell’s equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.2 The continuity equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.3 The material equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.4 Boundary conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2 Poynting’s vector and the energy law 8

3 The wave equation and the speed of light 9

4 Scalar waves 114.1 Plane waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114.2 Spherical waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124.3 Harmonic (monochromatic) waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134.4 Complex representation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144.5 Linearity and the superposition principle . . . . . . . . . . . . . . . . . . . . . . . . . 154.6 Phase velocity and group velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154.7 Repetition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

5 Pulse propagation in a dispersive medium 18

6 General electromagnetic plane wave 21

7 Harmonic electromagnetic waves of arbitrary form - Time averages 24

8 Harmonic electromagnetic plane wave – Polarisation 26

9 Reflection and refraction of a plane wave 299.1 Reflection law and refraction law (Snell’s law) . . . . . . . . . . . . . . . . . . . . . . 299.2 Generalisation of the reflection law and Snell’s law . . . . . . . . . . . . . . . . . . . 319.3 Reflection and refraction of plane electromagnetic waves . . . . . . . . . . . . . . . . 32

9.3.1 Reflectance and transmittance . . . . . . . . . . . . . . . . . . . . . . . . . . 359.3.2 Brewster’s law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 389.3.3 Unpolarised light (natural light) . . . . . . . . . . . . . . . . . . . . . . . . . 389.3.4 Rotation of the plane of polarisation upon reflection and refraction . . . . . . 399.3.5 Total reflection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

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List of Figures

1.1 A plane interface with unit normal n separates two different dielectric media. . . . . 7

4.1 A plane wave that propagates in direction s, has no variation in any plane that isnormal to s. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

5.1 A plane wave propagates in the positive z direction in a dispersive medium that fillsthe half space z ≥ 0. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

6.1 The vectors E, H, and s for an electromagnetic plane wave represent a right-handedCartesian co-ordinate system. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

8.1 Instantaneous picture of the electric vector of a plane wave that propagates in the zdirection. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

8.2 The end point of the electric vector describes an ellipse that is inscribed in a rectanglewith sides 2a1 and 2a2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

9.1 Reflection and refraction of a plane wave at a plane interface between two differentmedia. Illustration of propagation directions and angles of incidence, reflection, andtransmission. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

9.2 Reflection and refraction of a plane wave. Illustration of the co-ordinate system (n,b, t). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

9.3 Generalisation of Snell’s law and the reflection law to include non-planar waves thatare incident upon a curved interface. . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

9.4 Reflection and refraction of a plane electromagnetic wave at a plane interface betweentwo different media. Illustration of TE and TM components of the electric field. . . 35

9.5 Illustration of the angle αq between the electric vector Eq and the plane of incidencespanned by kq and eTMq. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

9.6 Illustration of Brewster’s law. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 389.7 Illustration of the refraction of a plane wave into an optically thinner medium, so that

θi < θt. When θi → θic, then θt → π/2, and we get total reflection. . . . . . . . . . . 41

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Part I

Elementary electromagnetic waves

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Chapter 1

Maxwell’s equations, the materialequations, and boundaryconditions

In this course we consider light to be electromagnetic waves of frequencies ν in the visible range, sothat ν � (4 − 7.5) × 1014 Hz. Since λ = c

ν , where c is the speed of light in vacuum (c � 3 × 108

m/s), we find that the corresponding wavelength interval is λ � (0.4 − 0.75) µm. Thus, to studythe propagation of light we must consider the propagation of the electromagnetic field, which isrepresented by the two vectors E and B, where E is the electric field strength and B is the magneticinduction or the magnetic flux density. To enable us to describe the interaction of the electromagneticfield with material objects we need three additional vector quantities, namely the current density J,the displacement D, and the magnetic field strength H.

1.1 Maxwell’s equations

The five vectors mentioned above are linked together by Maxwell’s equations, which in Gaussianunits are

∇× H =1cD +

4π

cJ, (1.1.1)

∇× E = −1cB. (1.1.2)

In addition we have the two scalar equations

∇ · D = 4πρ, (1.1.3)

∇ · B = 0, (1.1.4)

where ρ is the charge density. Equation (1.1.3) can be said to define the charge density ρ. Similarly,we can say that (1.1.4) implies that free magnetic charges do not exist.

1.2 The continuity equation

The charge density ρ and the current density J are not independent quantities. By taking thedivergence of (1.1.1) and using that ∇ · (∇× A) = 0 for an arbitrary vector A, we find that

∇ · J +14π

∇ · D = 0,

FYS 263 5

which on using (1.1.3) gives

∇ · J + ρ = 0. (1.2.1)

This equation is called the continuity equation, and it expresses conservation of charge. By inte-grating (1.2.1) over a closed volume V with surface S, we find

∫∫∫

V

∇ · Jdv = −∫∫∫

V

∂ρ

∂tdv, (1.2.2)

which by use of the divergence theorem gives

©∫∫

S

J · nda = − d

dt

∫∫∫

V

ρdv = − d

dtQ. (1.2.3)

Here n is the unit surface normal in the direction out of the volume V , so that (1.2.3) shows thatthe integrated current flux out of the closed volume V is equal to the loss of charge in the samevolume.

Digression 1: Notation

• Bold face is used to denote vector quantities, e.g.

E = Exex + Eyey + Ezez,

where ex, ey, and ez are unit vectors along the axes in a Cartesian co-ordinate system.

• A dot above a symbol is used to denote the time derivative, e.g.

B =∂

∂tB.

• E, B, D, H, ρ, and J are functions of the position r and the time t, e.g.

D = D(r, t).

• The connection between Gaussian and other systems of units, e.g. MKS units, follows fromJ.D. Jackson, ”Classical Electrodynamics”, Wiley (1962), pp. 611-621. For conversion betweenGaussian units and MKS units, we refer to the table on p. 621 in this book.

1.3 The material equations

Maxwell’s equations (1.1.1)-(1.1.4), which connect the fundamental quantities E, H, B, D, and J,are not sufficient to uniquely determine the field vectors (E, B) from a given distribution of currentsand charges. In addition we need the so-called material equations, which describe how the field isinfluenced by matter.

In general the material equations can be relatively complicated. But if the field is time harmonicand the matter is isotropic and at rest , the material equations have the following simple form

Jc = σE, (1.3.1)

D = εE, (1.3.2)

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B = µH, (1.3.3)

where σ is the conductivity, ε is the permittivity or dielectric constant, and µ is the permeability.Equation (1.3.1) is Ohm’s law, and Jc is the conduction current density, which arises because

the material has a non-vaninishing conductivity (σ �= 0). The total current density J in (1.1.1) canin addition consist of an externally applied current density J0, so that

J = J0 + Jc = J0 + σE. (1.3.4)

Digression 2: General material considerations

• A material that has a non-negligible conductivity σ is called a conductor, while a material thathas a negligible conductivity is called an insulator or a dielectric.

• Metals are good conductors.

• Glass is a dielectric; ε � 2.25; σ = 0; µ = 1.

• In anisotropic media (e.g. crystals) the relation in (1.3.2) is to be replaced by D = εE, whereε is a tensor, dyadic or matrix.

• In a plasma (1.3.1) is to be replaced by J = σE, where the conductivity is a tensor.

• There are also magnetically anisotropic media, in which (1.3.3) is to be replaced by B = µH.Thus, in this case the permeability is a tensor. Such materials are not important in optics.

• In dispersive media ε is frequency dependent, i.e. ε = ε(ω). Maxwell’s equations and thematerial equations are still valid for each frequency component or time harmonic componentof the field. For a pulse consisting of many frequency components, one must apply Fourieranalysis to solve Maxwell’s equations and the material equations separately for each timeharmonic component, and then perform an inverse Fourier transformation.

• In non-linear media there is no linear relation between D and E (equation (1.3.2) is not valid).Most media become non-linear when the electric field strength becomes sufficiently high.

1.4 Boundary conditions

Hitherto we have assumed that ε and µ are continuous functions of the position. But in opticswe often have systems consisting of several different types of glass. At the transition between airand glass or between two different types of glass the material parameters are discontinuous. Let ustherefore consider what happens to the electromagnetic field at the boundary between two media.

Consider two media that are separated by an interface, as illustrated in Fig. 1.1. From Maxwell’sequations, combined with Stokes’ and Gauss’ theorems, one can derive the following boundaryconditions

n · (B(2) − B(1)) = 0, (1.4.1)

n · (D(2) − D(1)) = 4πρs, (1.4.2)

n × (E(2) − E(1)) = 0, (1.4.3)

n × (H(2) − H(1)) =4π

cJs, (1.4.4)

where n is a unit vector along the surface normal. According to (1.4.1) the normal component ofB is continuous across the boundary, while (1.4.2) says that if there exists a surface charge density

FYS 263 7

ε 1

ε 2

^

1

, µ 2

n

, µ

Figure 1.1: A plane interface with unit normal n separates two different dielectric media.

ρs at the boundary, then the normal component of D is changed by 4πρs across the boundarybetween the two media. According to (1.4.3) the tangential component of E is continuous acrossthe boundary, while (1.4.4) implies that if there exists a surface current density Js at the boundary,then the tangential component of H, i.e. of n × H, is changed by 4π

c Js.

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Chapter 2

Poynting’s vector and the energylaw

The electric energy density we and the magnetic energy density wm are defined by

we =18π

E · D, (2.1)

wm =18π

H · B, (2.2)

and the total energy density is the sum of these, i.e.

w = we + wm. (2.3)

The energy flux of the field is represented by Poynting’s vector S, given by

S =c

4πE × H. (2.4)

Here S represents the amount of energy that per unit time crosses a unit area that is parallel withboth E and H.

In a non-conducting medium (σ = 0) we have the conservation law

∂w

∂t+ ∇ · S = 0, (2.5)

which expresses that the change of the energy density in a small volume is equal to the energy fluxout of the same volume [cf. (1.2.2) and (1.2.3)]. In optics the Poynting vector is very important,because its absolute value is proportional to the light intensity, i.e.

|S| ∝ light intensity. (2.6)

The direction of the Poynting vector, defined by the unit vector

s =S|S| , (2.7)

points in the direction of light propagation.

FYS 263 9

Chapter 3

The wave equation and the speedof light

The electric and magnetic fields E and H are connected through Maxwell’s equations (1.1.1)-(1.1.4),which are simultaneous, first-order partial differential equations. But in those parts of space wherethere are no sources, so that J = 0 and ρ = 0, we can through differentiation obtain second-orderpartial differential equations that E and H satisfy individually. We assume that the medium isnon-dispersive, so that D = εE, where ε = 0, and B = µH, where µ = 0. Then we have from (1.1.1)and (1.1.2)

∇× H =1cD =

1cεE, (3.1)

∇× E = −1cB = −1

cµH. (3.2)

Next, we assume that the medium is homogeneous, so that ε and µ do not vary with position. Bytaking the curl of (3.2) and combining the result with the time derivative of (3.1), we find that

∇× (∇× E) = −1cµ∇× H = −1

cµ

1cεE = −εµ

c2E. (3.3)

Now we use the vector relation

∇× (∇× A) = ∇(∇ · A) −∇2A, (3.4)

which applies to an arbitrary vector A, to obtain

∇(∇ · E) −∇2E = −εµ

c2E, (3.5)

which since ∇ · E = 0, gives

∇2E − εµ

c2E = 0. (3.6)

In a similar manner we find

∇2H − εµ

c2H = 0. (3.7)

By comparing these results with the scalar wave equation

∇2V − 1v2

V = 0, (3.8)

we see that in a source-free region of space each Cartesian component of E and H satisfies the scalarwave equation with phase velocity

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v =c√εµ

. (3.9)

Note that this derivation is valid only in a non-dispersive medium in which both the permittivityand the permeability do not depend on the frequency.

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Chapter 4

Scalar waves

Scalar waves are solutions of the scalar wave equation (3.8), which is given by

∇2V (r, t) − 1v2

∂2

∂t2V (r, t) = 0. (4.0.1)

4.1 Plane waves

Any solution of (4.0.1) of the form

V (r, t) = V (r · s, t), (4.1.1)

is called a plane wave, since V at any time t is constant over any plane

r · s = constant, (4.1.2)

which is normal to the unit vector s (see Fig. 4.1).To show that (4.1.1) is a solution of (4.0.1), we introduce a new variable

ζ = r · s = xsx + ysy + zsz, (4.1.3)

so that

∂ζ

∂x= sx ;

∂ζ

∂y= sy ;

∂ζ

∂z= sz. (4.1.4)

Further we find that

∂V

∂x=

∂V

∂ζ· ∂ζ

∂x= sx

∂V

∂ζ. (4.1.5)

∂2V

∂x2=

∂

∂x

(sx

∂V

∂ζ

)= sx

∂

∂x

(∂V

∂ζ

)= sx

∂

∂ζ

(∂V

∂ζ

)∂ζ

∂x= s2

x

∂2V

∂ζ2. (4.1.6)

In a similar way we find

∂2V

∂y2= s2

y

∂2V

∂y2;

∂2V

∂z2= s2

z

∂2V

∂z2. (4.1.7)

When we substitute (4.1.6) and (4.1.7) in (4.0.1) and take into account that s2x + s2

y + s2z = 1, since

s is a unit vector, the wave equation becomes

∂2V

∂ζ2− 1

v2

∂2V

∂t2= 0. (4.1.8)

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nθ

φ

y

x

z

e

e

e

^

^

^θ

φ

r^

z

e

e

e

θ

φ

r

^

^

^

θ

θ

n^

n

Figure 4.1: A plane wave that propagates in direction s, has no variation in any plane that is normalto s.

By introducing two new variables p and q, defined by

p = ζ − vt ; q = ζ + vt, (4.1.9)

we find (Exercise 2) that the wave equation in (p, q) variables can be written

∂2V

∂p∂q= 0. (4.1.10)

This equation has the following general solution

V = V1(p) + V2(q), (4.1.11)

where V1 and V2 are arbitrary functions. By substitution from (4.1.3) and (4.1.9) in (4.1.11), wefind the following general plane-wave solution

V (r, t) = V1(r · s − vt) + V2(r · s + vt). (4.1.12)

Note that

ζ − vt = ζ + vτ − v(t + τ), (4.1.13)

so that

V1(ζ, t) = V1(ζ + vτ, t + τ). (4.1.14)

Equation (4.1.14) shows that during the time τ , V1 is displaced a length s = vτ in the positiveζ direction, i.e. V1 propagates with velocity v in the positive ζ direction. The conclusion is thatV (ζ ± vt) represents a plane wave that propagates at velocity v in the positive ζ direction (lowersign) or in the negative ζ direction (upper sign).

4.2 Spherical waves

Consider now solutions of the scalar wave equation (4.0.1) of the form

V = V (r, t), (4.2.1)

where

r = |r| =√

x2 + y2 + z2, (4.2.2)

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is the distance from the origin (0, 0, 0). Since we have no angular dependence in this case, theLaplacian operator has the following form in spherical coordinates (Exercise 3)

∇2V =1r

∂2

∂r2(rV ), (4.2.3)

which upon substitution in the wave equation (4.0.1) gives

∂2

∂r2(rV ) − 1

v2

∂2

∂t2(rV ) = 0. (4.2.4)

Since (4.2.4) is of the same form as (4.1.8), the solution becomes (cf. (4.1.12))

rV = V1(r − vt) + V2(r + vt). (4.2.5)

Thus, we have obtained the following result: V (r±vt)r represents a spherical wave that converges

towards the origin (upper sign) or diverges away from the origin (lower sign). Thus, V (r+vt)r prop-

agates towards the origin with velocity v, whereas V (r−vt)r propagates away from the origin with

velocity v.

4.3 Harmonic (monochromatic) waves

At a given point r in space the solution of the wave equation is a function only of time, i.e.

V (r, t) = F (t), (4.3.1)

where F (t) can be an arbitrary function. If F (t) has the simple form

F (t) = a cos(ωt − δ), (4.3.2)

then we have a harmonic wave in time. The quantities in (4.3.2) have the following meaning: a is theamplitude (positive), ω is the angular frequency, and ωt − δ is the phase. A harmonic wave is alsocalled a monochromatic wave because it consists of only one frequency or wavelength component.

The frequency ν and the period T follow from

ν =ω

2π=

1T

. (4.3.3)

The harmonic wave in (4.3.2) has period T because

F (t + T ) = a cos(ω(t + T ) − δ) = a cos(ωt − δ + 2π) = F (t). (4.3.4)

From (4.1.12) we see that the general expression for a wave that propagates in the s directioncan be written

V = V1(r · s − vt) = V1

[−v

(t − r · s

v

)]= V ′

1

(t − r · s

v

), (4.3.5)

where both V1 and V ′1 are arbitrary functions. By replacing t with t− r·s

v in (4.3.2) we get a harmonicplane wave

V (r, t) = a cos[ω

(t − r · s

v

)+ δ

]= a cos[kr · s − ωt + δ], (4.3.6)

wherek =

ω

v(4.3.7)

is the wave number. We see that that V (r, t) remains unchanged if we replace r · s with r · s + nλ,where n = 1, 2, . . ., and λ is given by

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λ =2π

k= v

2π

ω= vT =

v

ν. (4.3.8)

The quantity λ is called the wavelength. Note that for t = constant, V (r, t) in (4.3.6) is periodicwith wavelength λ, i.e.

V (r · s, t) = V (r · s + nλ, t) ; n = 1, 2, 3, . . . . (4.3.9)

Now we introduce the wave vector or propagation vector k, defined by

k = ks. (4.3.10)

so that the expression (4.3.6) for a plane, harmonic wave can be written

V (r, t) = a cos(k · r − ωt + δ). (4.3.11)

In a similar way the expression for a converging or a diverging harmonic spherical wave becomes

V (r, t) = acos(∓kr − ωt + δ)

r, (4.3.12)

where the upper sign corresponds to a converging spherical wave and the lower sign to a divergingspherical wave.

Consider now a plane, harmonic wave that propagates in the positive z direction, so that [cf.(4.3.11)]

V (z, t) = a cos(kz − ωt + δ). (4.3.13)

A wave front is defined by the requirement that the phase shall be constant over it, i.e.

φ = kz − ωt + δ = constant. (4.3.14)

Hence it follows that on a wave front we have

z = vt + constant ; v =ω

k. (4.3.15)

Thus, the wave front propagates at the velocity

v =ω

k, (4.3.16)

which is called the phase velocity.

4.4 Complex representation

Alternatively we can express (4.3.11) and (4.3.12) in the following way

V (r, t) = Re{U(r)e−iωt}, (4.4.1)

where Re{. . .} stands for the real part of {. . .}, and where the complex amplitude U(r) is given by

U(r) = aei(k·r+δ), (4.4.2)

for a plane wave, and by

U(r) =a

rei(±kr+δ), (4.4.3)

for a diverging (upper sign) or converging (lower sign) spherical wave.

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Note that when we perform linear operations, such as differentiation, integration or summation,we can drop the ’Re’ symbol during the operations, as long as we remember to take the real part ofthe result in the end.

By substituting

V (r, t) = U(r)e−iωt, (4.4.4)

in the wave equation (4.0.1), we get

(∇2 + k2)U(r) = 0, (4.4.5)

which shows that the complex amplitude U(r) is a solution of the Helmholtz equation.

4.5 Linearity and the superposition principle

For any linear equation the sum of two or several solutions is also a solution. This is called thesuperposition principle. Since Maxwell’s equations are linear, the superposition principle is validfor electromagnetic waves as long as the material equations are linear. The superposition principleimplies that we can construct general solutions of the wave equation or Maxwell’s equations byadding elementary solutions in the form of harmonic plane or spherical waves. We will discuss thisin detail in Part II.

4.6 Phase velocity and group velocity

Consider a harmonic wave of the form [cf. (4.3.11)]

V (r, t) = Re[U(r)e−iωt

], (4.6.1)

where the complex amplitude U(r) is a solution of the Helmholtz equation (4.4.5), i.e.

(∇2 + k2)U(r) = 0. (4.6.2)

The wave number k can be written

k =ω

v=

ω

c

( c

v

)= k0n, (4.6.3)

where k0 is the wave number in vacuum, i.e.

k0 =ω

c, (4.6.4)

and n is the refractive index given by

n =c

v=

√εµ. (4.6.5)

A general wave V (r, t) can always be expressed as a sum of harmonic components. We will returnto this later. If ε depends on ω, i.e. ε = ε(ω), then the phase velocity also will depend on ω, sincev = c

n = v(ω). This means that different harmonic components will propagate at different phasevelocities. A polychromatic wave or a pulse, which is comprised of many harmonic components,therefore will change its shape during propagation, and the energy will not propagate at the phasevelocity, but at the group velocity, which is defined as

vg =dω

dk. (4.6.6)

If n(ω) = constant, we have a non-dispersive medium. Since

ω = vk, (4.6.7)

FYS 263 16

where the phase velocity v = cn now is constant, we have in this case

vg =d

dk(vk) = v.

Thus, the phase velocity and the group velocity are equal in a non-dispersive medium where n =constant. In dispersive media we have

vg =d

dk(vk) = v + k

dv

dk= v − λ

dv

dλ= v + ν

dv

dν, (4.6.8)

where the last two results follow from the relation k = 2πλ = ω

v .

4.7 Repetition

From Maxwell’s equations in source-free space (J = 0 ; ρ = 0) we find

∇2E − εµ

c2E = 0 ; ∇2H − εµ

c2H = 0. (4.7.1)

Comparison of (4.7.1) with the scalar wave equation

∇2V − 1v2

V = 0, (4.7.2)

shows that any Cartesian component of E and H satisfies the scalar wave equation with phasevelocity v given by

v =c√εµ

=c

n. (4.7.3)

The scalar wave equation (4.7.2) has simple solutions in the form of plane waves or spherical waves.

Plane waves

For a plane wave V is given by

V (r, t) = V1(r · s − vt) + V2(r · s − vt), (4.7.4)

where V (ζ ∓ vt) represents a plane wave that propagates in the positive ζ direction (upper sign) orin the negative ζ direction (lower sign).

Spherical waves

For a spherical wave V is given by

V (r, t) =V1(r − vt)

r+

V2(r + vt)r

, (4.7.5)

where V (r∓vt)r represents a spherical wave that propagates away from the origin (upper sign) or

towards the origin (lower sign).

Harmonic (monochromatic) waves

A plane harmonic wave that propagates in the direction k = ks is given by

V (r, t) = a cos(k · r − ωt + δ), (4.7.6)

and the corresponding spherical wave is

V (r, t) =a

rcos(±kr − ωt + δ), (4.7.7)

where the upper sign represents a diverging spherical wave and the lower sign represents a convergingspherical wave.

FYS 263 17

Complex representation of harmonic waves

In complex notation we haveV (r, t) = Re[U(r)e−iωt]. (4.7.8)

For a plane wave the complex amplitude U(r) is given by

U(r) = aei(k·r+δ),

and for a diverging or converging spherical wave it is given by

U(r) =a

rei(±kr+δ).

By substituting (4.7.8) into the wave equation (4.7.2), we find that U(r) satisfies the Helmholtzequation, i.e.

(∇2 + k2)U(r) = 0. (4.7.9)

FYS 263 18

Chapter 5

Pulse propagation in a dispersivemedium

z=0

z

n( )ω

Figure 5.1: A plane wave propagates in the positive z direction in a dispersive medium that fills thehalf space z ≥ 0.

Consider a polychromatic, plane wave that propagates in the positive z direction in a linear,homogeneous, isotropic, and dispersive medium that fills the half space z > 0 (Fig. 5.1). Thepolychromatic, plane wave u(z, t) is comprised of different harmonic components, which implies thatwe can represent u(z, t) by the following inverse Fourier transform

u(z, t) =12π

∫ ∞

−∞u(z, ω)e−iωtdω, (5.1)

where the frequency spectrum u(z, ω) is given as the Fourier transform of u(z, t), i.e.

u(z, ω) =∫ ∞

−∞u(z, t)eiωtdt. (5.2)

Thus, u(z, t) and u(z, ω) constitute a Fourier transform pair. Since u(z, ω) can be any Cartesiancomponent of the frequency spectrum of the electric or magnetic field, it satisfies the Helmholtzequation, i.e.

[∇2 + k2(ω)]u(z, ω) = 0, (5.3)

where

k(ω) =ω

v(ω)=

ω

c

c

v(ω)=

ω

cn(ω). (5.4)

FYS 263 19

Suppose now that u(z, t) is known for all values of t in the plane z = 0, and that u(0, t) vanishes fort < 0.

Since there is no variation in the x and y directions, the Helmholtz equation (5.3) can be writtenas

[d2

dz2+ k2(ω)

]u(z, ω) = 0, (5.5)

which has the following general solution

u(z, ω) = u+(ω)eik(ω)z + u−(ω)e−ik(ω)z. (5.6)

If we consider propagation in the positive z direction only, then u−(ω) = 0, so that (5.1) gives

u(z, t) =12π

∫ ∞

−∞u+(ω)ei(k(ω)z−ωt)dω. (5.7)

Now we put z = 0 i (5.7), take an inverse Fourier transform, and use (??) to obtain

u+(ω) =∫ ∞

−∞u(0, t)eiωtdt = u(0, ω), (5.8)

so that (5.7) gives

u(z, t) =12π

∫ ∞

−∞u(0, ω)ei(k(ω)z−ωt)dω, (5.9)

or

u(z, t) =12π

∫ ∞

−∞u(0, ω)ei z

c f(ω)dω, (5.10)

where

f(ω) = ω[n(ω) − θ] ; θ =ct

z. (5.11)

Consider first the special case in which n(ω) = cv(ω) = constant, which implies that we have a

non-dispersive medium. Since k = ωv , where v now is constant, we have from (5.1) and (5.9)

u(z, t) =12π

∫ ∞

−∞u(0, ω)e−iω(− z

v +t)dω = u(0, t − z

v

). (5.12)

This result shows that the pulse propagates in the positive z direction at velocity v without changingits shape.

Suppose now that the medium is dispersive and that the frequency spectrum g(ω) of the pulsein (5.10) does not contain singularities and that it is sufficiently wide. Then the main contributionto the pulse in (5.10) comes from frequencies ωs for which the phase f(ω) in (5.11) is stationary, i.e.from ωs that satisfy the equation

f ′(ωs) = n(ωs) − θ + ωsn′(ωs) = 0. (5.13)

A model that is commonly used to study propagation in dispersive media, is the so called Lorentz-medium. For such a medium with one single resonance frequency the refractive index n(ω) is givenby the following expression

n(ω) =[1 − b2

ω2 − ω20 + 2δiω

]1/2

, (5.14)

where b is a constant, ω0 is the resonance frequency, and δ represents the damping (attenuation) inthe medium.

FYS 263 20

Equation (5.9) shows that when the medium is dispersive, then u(z, t) (for any z > 0) is a sumof harmonic plane waves of the form u(0, ω)exp[i(k(ω)z − ωt)] = u(0, ω)exp[−ki(ω)z]exp[i(kr(ω)z −ωt)], where kr(ω) and ki(ω) are the real and the imaginary part, repectively, of k(ω). Thus, theamplitude u(0, ω)exp[−ki(ω)z], is damped exponentially as z increases, and the phase velocity isgiven by v(ω) = ω

kr(ω) , where k(ω) = (ω/c)n(ω) = (ω/c)[nr(ω) + ini(ω)] = kr(ω) + iki(ω). Sincethe phase velocity v depends on the frequency ω, plane waves of different frequencies will arrive ata given position z at different times and thus cause a distortion of the pulse, i.e. the shape of thepulse will get changed. Also, the damping factor ki(ω) depends on ω, so that different frequencycomponents will have different amplitudes when they arrive at a given position z.

FYS 263 21

Chapter 6

General electromagnetic planewave

A general electromagnetic plane wave can be written in the form

E = E(k · r − ωt) ; H = H(k · r − ωt), (6.1)

where k = ks, with s pointing in the direction of propagation. We introduce a new variable u =k · r − ωt, so that

∂u

∂x= kx ;

∂u

∂y= ky ;

∂u

∂z= kz ;

∂u

∂t= −ω. (6.2)

In source-free space Maxwell’s equations (1.1.1)-(1.1.2) are given by

∇× H =1cD =

ε

cE, (6.3)

∇× E = −1cB =

µ

cH. (6.4)

By using the chain rule, we find that the x component of ∇× E can be expressed as follows

(∇× E)x = ∇yEz −∇zEy =∂Ez

∂y− ∂Ey

∂z=

dEz

du

∂u

∂y− dEy

du

∂u

∂z

= E′zky − E′

ykz = (k × E′)x =ω

v(s × E′)x, (6.5)

where E′ = dEdu . By proceeding in a similar manner, we find that

∇× E =ω

vs × E′, (6.6)

∇× H =ω

vs × H′, (6.7)

where

E′ =dEdu

; H′ =dHdu

. (6.8)

Further, we have

E =∂E∂t

=dEdu

∂u

∂t= −ωE′ ; H = −ωH′. (6.9)

FYS 263 22

By substitution of (6.6)-(6.9) into Maxwell’s equations (6.3)-(6.4) the result is

s × H′ =ε

c(−v)E′ = −ε

c

c√εµ

E′ = −√

ε

µE′, (6.10)

s × E′ = −µ

c(−v)H′ =

õ

εH′, (6.11)

where we have used (3.9). Thus, we have

E′ = −√

µ

εs × H′ ; H′ =

√ε

µs × E′. (6.12)

By integrating over u in (6.12) and setting the integration constant equal to zero, we get

E = −√

µ

εs × H ; H =

√ε

µs × E. (6.13)

Scalar multiplication of the equations in (6.13) with s gives

s · E = s · H = 0, (6.14)

which shows that both E and H are transverse waves, i.e. both E and H are normal to the propa-gation direction s, as illustrated in Fig. 6.1. Thus, the vectors s, E, and H represent a right-handedCartesian co-ordinate system.

s

E

H

Figure 6.1: The vectors E, H, and s for an electromagnetic plane wave represent a right-handedCartesian co-ordinate system.

For the electric and the magnetic energy density we find

we =18π

E · D =ε

8πE2 ; E = |E|, (6.15)

wm =18π

B · H =µ

8πH2 ; H = |H|. (6.16)

Since√

µH =√

εE (cf. (6.13)), we get we = wm, and the total energy density becomes

w = we + wm = 2we =14π

εE2 = 2wm =14π

µH2, (6.17)

and the Poynting vector (2.4) becomes

S =c

4πE × H =

c

4πEH s =

c

4πE

√ε

µEs =

(14π

εE2

) (c√εµ

)s = wvs. (6.18)

FYS 263 23

Thus, we have

S = wvs, (6.19)

which shows that the Poynting vector represents the energy flow, both with respect to absolute valueand direction. A dimensional analysis of (6.19) shows that

[|S|] = [w][v] =Energi

m3· m

s=

Energim2 · s =

Wm2

. (6.20)

Thus, S represents the amount of energy per unit time that passes through a unit area of the planethat is spanned by E and H, as asserted previously in chapter 2.

FYS 263 24

Chapter 7

Harmonic electromagnetic waves ofarbitrary form - Time averages

The E and H fields for a harmonic wave of arbitrary form can be written

E = Re{E0(r)e−iωt

}; H = Re

{H0(r)e−iωt

}, (7.1)

where E0(r) and H0(r) are complex vectors. Thus, we have

E0(r) = ER0 (r) + iEI

0((r), (7.2)

H(r) = HR0 (r) + iHI

0(r), (7.3)

where ER0 , EI

0, HR0 , and HI

0 are real vectors.Since optical frequencies are very high (ω � 1015 s−1), we can only observe averages of we, wm,

and S, taken over a time interval −T ′ ≤ t ≤ T ′, where T ′ is much larger than the period T = 2πω .

For the time average of the electric energy density we have [cf. (2.1)]

〈we〉 =1

2T ′

∫ T ′

−T ′

ε

8π|E|2dt. (7.4)

For any complex number z, we have Rez = 12 (z + z∗), where z∗ is the complex conjugate of z.

Therefore, we may write

E = Re[E0(r)e−iωt] =12[E0e

−iωt + E∗0e

+iωt],

so that we get

|E|2 = E ·E =14[E0e

−iωt + E∗0e

iωt] · [E0e−iωt + E∗

0eiωt] =

14[E2

0e−2iωt + 2E0 ·E∗

0 + E∗20 e2iωt]. (7.5)

Further, we have

12T ′

∫ T ′

−T ′e−2iωtdt =

12T ′

[e−2iωt

−2iω

]T ′

−T ′=

12T ′

(1ω

)sin(2ωT ′) =

14π

T

T ′ sin(2ωT ′). (7.6)

Since T ′ T , the integral that includes the factor e−2iωt can be neglected. Similarly, the integralthat includes the factor e2iωt can be neglected, and we get

〈we〉 =ε

16πE0 · E∗

0. (7.7)

FYS 263 25

By proceeding in a similar manner, we find that the time average of the magnetic energy densitybecomes

〈wm〉 =µ

16πH0 · H∗

0. (7.8)

The time average of the Poynting vector is given by [cf. (2.4)]

〈S〉 =1

2T ′

∫ T ′

−T ′

c

4π(E × H)dt, (7.9)

where E × H can be written

E × H =12[E0e

−iωt + E∗0e

iωt] × 12[H0e

−iωt + H∗0e

iωt]

=14{E0 × H0e

−2iωt + E0 × H∗0 + E∗

0 × H0 + E∗0 × H∗

0e2iωt}. (7.10)

By substituting (7.10) into (7.9) and performing time averaging, we find that the time average ofthe Poynting vector becomes

〈S〉 =c

16π{E0 × H∗

0 + E∗0 × H0} =

c

8πRe(E0 × H∗

0). (7.11)

FYS 263 26

Chapter 8

Harmonic electromagnetic planewave – Polarisation

For an electromagnetic plane wave that is time harmonic, each Cartesian component of E and H isof the form

a cos(τ + δ) = Re[aei(τ+δ)] ; a > 0, (8.1)

where

τ = k · r − ωt. (8.2)

Let the z axis point in the s direction. Then only the x and y components of E and H are non-zero,since the electromagnetic field of a plane wave is transverse.

Now we want to determine that curve which the end point of the electric vector describes duringpropagation. This curve consists of points that have co-ordinates (Ex, Ey) given by

Ex = a1 cos(τ + δ1) ; a1 > 0, (8.3)

Ey = a2 cos(τ + δ2) ; a2 > 0, (8.4)

Ez = 0. (8.5)

In order to determine that curve which E(τ) describes (Fig. 8.1), we eliminate τ from (8.3)-(8.4).We let β = τ + δ1 and get

Ex = a1 cos β, (8.6)

Ey = a2 cos(β + δ) = a2[cos β cos δ − sinβ sin δ], (8.7)

where δ = δ2 − δ1. We substitute from (8.6) into (8.7) and get

Ey

a2=

Ex

a1cos δ −

√1 −

(Ex

a1

)2

sin δ, (8.8)

which upon squaring gives

(Ex

a1

)2

+(

Ey

a2

)2

− 2Ex

a1

Ey

a2cos δ = sin2 δ. (8.9)

This is the equation of a conic section. The cross term implies that it is rotated relative to theco-ordinate axes (x, y). By letting δ = π

2 , we get

FYS 263 27

E ( )τx

E ( )τy

E( )τ

x

y

z

Figure 8.1: Instantaneous picture of the electric vector of a plane wave that propagates in the zdirection.

(Ex

a1

)2

+(

Ey

a2

)2

= 1, (8.10)

which shows that the equation describes an ellipse. In a co-ordinate system (ξ, η), which coincideswith the axes of the ellipse, the equations for the field components become

Eξ = a cos(τ + δ0), (8.11)

Eη = ±b sin(τ + δ0), (8.12)

which upon squaring gives

(Eξ

a

)2

+(

Eη

b

)2

= 1. (8.13)

When τ +δ0 = 0, we have Eξ = a; Eη = 0, and when τ +δ0 = π2 , we have Eξ = 0; Eη = ±b. This

shows that when the upper or lower sign in (8.12) applies, the electric vector rotates against or withthe clock, respectively, if we view the xy plane from the positive z axis. Rotation against the clockis called left-handed polarisation, and rotation with the clock is called right-handed polarisation.

The relation between the two co-ordinate systems (x, y) and (ξ, η) is shown in Fig. 8.2, where(cf. Exercise 7)

a2 + b2 = a21 + b2

2, (8.14)

tan 2ψ = tan(2α) cos δ ; tanα =a2

a1(0 ≤ α ≤ π

2), (8.15)

sin 2ψ = sin(2α) sin δ ; tanψ = ± b

a. (8.16)

Since sin δ < 0 when the upper sign in (8.12) applies, we have left-handed polarisation when sin δ < 0.We consider now some special cases of (8.6)-(8.7).

Linear polarisation. If the phase difference δ is equal to an integer times π, i.e. if

δ = mπ (m = 1,±1,±2, . . .), (8.17)

then we get from (8.6)-(8.7)

Ex = a1 cos β, (8.18)

FYS 263 28

2a1

2a2

ξη

x

y

b

a

ψ

Figure 8.2: The end point of the electric vector describes an ellipse that is inscribed in a rectanglewith sides 2a1 and 2a2.

Ey = a2 cos(β + mπ) = a2(−1)m Ex

a1, (8.19)

which shows that the ellipse degenerates into a straight line, i.e.

Ey

Ex= (−1)m a2

a1. (8.20)

Circular polarisation. If the amplitudes are equal and the phase difference is ±π2 plus a multiple

of 2π, i.e. if

a1 = a2, (8.21)

δ = ±π

2+ 2mπ (m = 0,±1,±2, . . .), (8.22)

then the ellipse in (8.6)-(8.7) degenerates into a circle, i.e.

Ex = a cos β, (8.23)

Ey = a cos(β + 2mπ ± π

2

)= ∓a sinβ. (8.24)

By squaring these two equations, we get

E2x + E2

y = a2. (8.25)

We have right-handed circular polarisation when Ey = −a sinβ and left-handed circular polarisationwhen Ey = +a sinβ.

FYS 263 29

Chapter 9

Reflection and refraction of a planewave

ki

kr

kt

n = e^ ^z

θ t

θ i

θ r

1ε , µ 1ε , µ

22

Figure 9.1: Reflection and refraction of a plane wave at a plane interface between two differentmedia. Illustration of propagation directions and angles of incidence, reflection, and transmission.

We let a plane wave be incident upon a plane interface between two different media, as shown inFig. 9.1. The incident wave gives rise to a reflected wave and a transmitted wave, which we assumeare plane waves as well. Thus, each component of E or H can be written

Aqj = Re{aq

jei(kq·r−ωt)} (j = x, y, z) , (9.0.1)

where A stands for E or H and q = i, r, t, so that ki, kr, and kt are the wave vectors of the incident,reflected, and transmitted waves, respectively.

9.1 Reflection law and refraction law (Snell’s law)

The existence of continuity conditions that E and H must satisfy at the interface between the twomedia in Fig. 9.1, implies that when r represents a point at the interface, the argument in theexponential function in (9.0.1) must be the same for the reflected and transmitted waves as for theincident wave. Thus, we have

FYS 263 30

θi

k

k

i

i___________

| |

^

^

^b =

n

n

x

x

^ ^^t n b= x

ki

n

Figure 9.2: Reflection and refraction of a plane wave. Illustration of the co-ordinate system (n, b,t).

ki · r − ωt = kr · r − ωt = kt · r − ωt, (9.1.1)

or

ki · r = kr · r = kt · r. (9.1.2)

Now we introduce a Cartesian co-ordinate system in which the unit vectors n, b, and t representa right-handed system (Fig. 9.2). Let n point along the interface normal into the medium of therefracted wave, and let b and t be defined by

b =ki × n|ki × n| ; t = n × b. (9.1.3)

In this co-ordinate system we have

ki = kitt + ki

nn ; kib = ki · b = 0, (9.1.4)

kr = krt t + kr

nn + krb b, (9.1.5)

kt = ktt t + kt

nn + ktbb, (9.1.6)

r = rtt + rbb. (9.1.7)

Note that the co-ordinate system is defined such that ki has no component along b, i.e. b is normalto the plane of incidence, which is spanned by the vectors ki and n.

Since

ki · r = (kitt + ki

nn) · (rtt + rbb) = kitrt, (9.1.8)

kr · r = (krt t + kr

nn + krb b) · (rtt + rbb) = kr

t rt + krbrb, (9.1.9)

kt · r = (ktt t + kt

nn + ktbb) · (rtt + rbb) = kt

trt + ktbrb, (9.1.10)

it follows from the continuity condition (9.1.2) that

kitrt = kr

t rt + krbrb = kt

trt + ktbrb. (9.1.11)

FYS 263 31

But since (9.1.11) shall apply to any point at the interface, i.e. to all values of rt and rb, we musthave

krb = kt

b = 0. (9.1.12)

Thus, both kr and kt must lie in the plane of incidence spanned by ki and n. Therefore, we have

kit = kr

t = ktt = kt, (9.1.13)

which implies that the components of ki, kr, and kt parallel to the interface are equal. By using

n × kq = n × (ktt + kqnn) = −bkt ; q = i, r, t, (9.1.14)

we find that

n × ki = n × kr, (9.1.15)

n × kt = n × ki. (9.1.16)

Further, we use the relation |a× b| = |a||b| sin θ, where θ is the angle between the vectors a and b.Thus, we find from (9.1.15) and Fig. 9.1 that

ki sin θi = kr sin θr. (9.1.17)

Also, we know that ki = kr = n1k0, where n1 is the refractive index in medium 1, and k0 is thewave number in vacuum. The reflection law therefore becomes

θi = θr, (9.1.18)

which in (9.1.15) is given in vectorial form.From (9.1.16) and Fig. 9.1 we get the refraction law or Snell’s law

ki sin θi = kt sin θt. (9.1.19)

which by using ki = n1k0 and kt = n2k0, becomes

n1 sin θi = n2 sin θt. (9.1.20)

Equation (9.1.16) represents Snell’s law in vector form. Note that (9.1.15) and (9.1.16) contain moreinformation than (9.1.18) and (9.1.20). From the vector equations it is clear that kr and kt lie inthe plane of incidence.

9.2 Generalisation of the reflection law and Snell’s law

The reflection law and Snell’s law (the refraction law) can be generalised to include non-planar wavesthat are incident upon a non-planar interface. This is illustrated in Fig. 9.3, where the field from apoint source propagates towards a curved interface. Suppose now that the distance from the pointsource to the interface is much larger than the wavelength. Then at each point on the interface wemay consider the incident wave to be a plane wave locally, and we may replace the interface locallyby the tangent plane through the point in question. Then we can use Snell’s law and the reflectionlaw as derived for a plane wave that is incident upon a plane interface, as illustrated in Fig. 9.3.

FYS 263 32

k

k

k

θ

θ

θ

n n

t

r

1 2

^

i

r

t

i

n

local tangent plane

Point source

Figure 9.3: Generalisation of Snell’s law and the reflection law to include non-planar waves that areincident upon a curved interface.

9.3 Reflection and refraction of plane electromagnetic waves

Note that the reflection law and the refraction law apply to all types of plane waves, i.e. to acoustic,electromagnetic, and elastic waves. In the derivation we have only used that kq · r − ωt (q = i, r, t)shall be the same for q = i, q = r, and q = t. Now we take a closer look at the reflection andrefraction of plane electromagnetic waves in order to determine how much of the energy in theincident wave that is reflected and transmitted.

We know that a plane electromagnetic wave is transverse, i.e. that both E and B = µH arenormal to the propagation direction k = ks. In Fig. 9.1 we have chosen the z axis in the direction ofthe interface normal. If E is normal to the plane of incidence, we have s polarisation (from German,“Senkrecht”) or TE polarisation (“transverse electric” relative to the plane of incidence or the zaxis). And if E is parallel with the plane of incidence, we have p polarisation or TM polarisation,since in this case B is normal to the plane of incidence or the z axis; hence the use of the term TMor “transverse magnetic”.

A general time-harmonic, plane electromagnetic wave consists of both a TE and a TM compo-nent. With the time dependence e−iωt suppressed, we have for the spatial part of the field

E = ETE + ETM ; B = BTE + BTM , (9.3.1)

ETE = ETE kt × ez

kteik·r, (9.3.2)

ETM = ETM k × (kt × e)z

kkteik·r, (9.3.3)

BTE =1k0

k × ETE = ETE k × (kt × ez)k0kt

eik·r, (9.3.4)

BTM =1k0

k × ETM = ETM 1k0kkt

k × [k × (kt × ez)]eik·r. (9.3.5)

But since k × [k × (kt × ez)] = k[k · (kt × ez)] − (kt × ez)k · k = −k2kt × ez, we get

BTM =−k

k0ETM kt × ez

kteik·r. (9.3.6)

FYS 263 33

Note that the vectors

eTE =kt × ez

kt; eTM =

k × (kt × ez)kkt

, (9.3.7)

are unit vectors in the directions of ETE and ETM , respectively.We represent each of the incident, reflected, and transmitted fields in the manner given above,

so that (q = i, r, t)

Eq = ETEq + ETMq ; Bq = BTEq + BTMq, (9.3.8)

ETEq = ETEq kt × ez

kteikq·r, (9.3.9)

ETMq = ETMq kq × (kt × ez)kqkt

eikq·r, (9.3.10)

BTEq =kq

k0ETEq kq × (kt × ez)

kqkteikq·r, (9.3.11)

BTMq =−kq

k0ETMq kt × ez

kteikq·r, (9.3.12)

where

ki = kt + kz1ez ; kt = kxex + kyey, (9.3.13)

kr = kt − kz1ez ; kt = kt + kz2ez, (9.3.14)

kq ={

k1 = n1k0 for q = i, rk2 = n2k0 for q = t.

(9.3.15)

The continuity conditions that must be satisfied at the interface z = 0 are that the tangentialcomponents of E and H = 1

µB be continuous, i.e.

ez ×{ETEi + ETEr − ETEt + ETMi + ETMr − ETMt

}= 0, (9.3.16)

ez ×{

1µ1

(BTEi + BTEr

)− 1

µ2BTEt +

1µ1

(BTMi + BTMr

)− 1

µ2BTMt

}= 0. (9.3.17)

Further, we have

ez × [kq × (kt × ez)] = (kq · ez) ez × kt, (9.3.18)

ez × (kt × ez) = kt. (9.3.19)

By substituting from (9.3.9)-(9.3.12) into the boundary conditions (9.3.16)-(9.3.17) and using (9.1.2)and (9.3.18)-(9.3.19), we get

kt

{ETEi + ETEr − ETEt

}+ ez × kt

{kz1

k1ETMi − kz1

k1ETMr − kz2

k2ETMt

}= 0, (9.3.20)

ez × kt

{1µ1

(kz1

k0ETEi − kz1

k0ETEr

)− 1

µ2

kz2

k0ETEt

}

+kt

{1µ1

(−k1

k0ETMi − k1

k0ETMr

)− −k2

k0ETMt

}= 0. (9.3.21)

FYS 263 34

Since kt and ez × kt are orthogonal vectors, the expression inside each of the {} parentheses in(9.3.20) and (9.3.21) must vanish, i.e.

ETEi + ETEr = ETEt, (9.3.22)

kz1µ2

(ETEi − ETEr

)= kz2µ1E

TEt, (9.3.23)

kz1k2

(ETMi − ETMr

)= kz2k1E

TMt, (9.3.24)

k1µ2

(ETMi − ETMr

)= k2µ1E

TMt. (9.3.25)

Now we define reflection and transmission coefficients as

RTE =ETEr

ETEi; TTE =

ETEt

ETEi, (9.3.26)

RTM =ETMr

ETMi; TTM =

ETMt

ETMi, (9.3.27)

so that (9.3.22)-(9.3.25) give

1 + RTE = TTE ; 1 − RTE =kz2µ1

kz1µ2TTE , (9.3.28)

1 − RTM =kz2k1

kz1k2TTM ; 1 + RTM =

k2µ1

k1µ2TTM . (9.3.29)

The two equations in (9.3.28) have the following solution

RTE =µ2kz1 − µ1kz2

µ2kz1 + µ1kz2; TTE =

2µ2kz1

µ2kz1 + µ1kz2, (9.3.30)

whereas the two equations in (9.3.29) give

RTM =k22µ1kz1 − k2

1µ2kz2

k22µ1kz1 + k2

1µ2kz2; TTM =

2k1k2µ2kz1

k22µ1kz1 + k2

1µ2kz2. (9.3.31)

The interpretation of the reflection and transmission coefficients follow from (9.3.26)-(9.3.27). Thus,the reflection coefficient represents the amplitude ratio between the reflected and the incident Efield, whereas the transmission coefficient represents the amplitude ratio between the transmittedand the incident E field.

Note that (9.3.22)-(9.3.23) and (9.3.28) contain only TE quantities, whereas equations (9.3.24)-(9.3.25) and (9.3.29) contain only TM quantities. This implies that these two wave types areindependent or de-coupled upon reflection and refraction. Thus, an incident TE plane wave producesa reflected TE plane wave and a transmitted TE plane wave, whereas an incident TM plane waveproduces a reflected TM plane wave and a transmitted TM plane wave. Upon reflection andrefraction there is no coupling between TE and TM waves.

From Fig. 9.1 it follows that

kz1 = ki · ez = k1 cos θi ; kz2 = kt · ez = k2 cos θt, (9.3.32)

so that if µ1 = µ2 = 1 the reflection and transmission coefficients become

TTM =2n1 cos θi

n2 cos θi + n1 cos θt; RTM =

n2 cos θi − n1 cos θt

n2 cos θi + n1 cos θt, (9.3.33)

TTE =2n1 cos θi

n1 cos θi + n2 cos θt; RTE =

n1 cos θi − n2 cos θt

n1 cos θi + n2 cos θt, (9.3.34)

FYS 263 35

n1 n2

ez

= = =

z = 0

θ i

θ = θir

θ t

^ TMie

e TMr

e TMt

k ie TEi e TEr e TEt

kr k t k k

ti

^ TEe

Figure 9.4: Reflection and refraction of a plane electromagnetic wave at a plane interface betweentwo different media. Illustration of TE and TM components of the electric field.

These expressions are called the Fresnel formulas. By using Snell’s law (9.1.20), we can rewrite themas (Exercise 9)

TTM =2 sin θt cos θi

sin(θi + θt) cos(θi − θt); RTM =

tan(θi − θt)tan(θi + θt)

, (9.3.35)

TTE =2 sin θt cos θi

sin(θi + θt); RTE = − sin(θi − θt)

sin(θi + θt). (9.3.36)

At normal incidence where θi = θt = 0, we get from (9.3.33) and (9.3.34)

TTE = TTM =2

n + 1; RTM = −RTE =

n − 1n + 1

; n =n2

n1. (9.3.37)

The fact that RTM = −RTE at normal incidence follows from the way in which ETE and ETE

are defined. From Fig. 9.4 we see that these two vectors point in opposite directions at normalincidence.

9.3.1 Reflectance and transmittance

Fig. 9.4 shows the polarisation vectors eTMq (q = i, r, t) and eTE for TM and TE polarisation.These unit vectors are parallel with the electric field and follow from (9.3.9)-(9.3.12)

eTEi = eTEr = eTEt = eTE =kt × ez

kt; |eTE | = 1, (9.3.38)

eTMq =kq × (kt × ez)

kqkt; |eTMq| = 1. (9.3.39)

Let the angle between Eq and the plane of incidence spanned by kq and eTMq, be αq [see Fig. 9.5],so that

Eq = eTEEq sinαq + eTMqEq cos αq. (9.3.40)

FYS 263 36

E

e

q

E

E

e TMq

^ TEq

TEq

k q

TMq

αq

Figure 9.5: Illustration of the angle αq between the electric vector Eq and the plane of incidencespanned by kq and eTMq.

Further, we let J i, Jr, and J t denote the energy flows of respectively the incident, reflected, andtransmitted fields per unit area of the interface. Then we have

Jpq = Spq cos θq ; p = TE, TM ; q = i, r, t, (9.3.41)

where Spq is the absolute value of the Poynting vector, given by

Spq =c

4π|Epq × Hpq| =

c

4πEpqHpq =

c

4π

√εq

µq(Epq)2 . (9.3.42)

Here we have used the relation√

εqEpq =√

µqHpq. The reflectance Rp (p = TE, TM is the ratiobetween the reflected and incident energy flows. From (9.3.41)-(9.3.42) we have

RTM =JTMr

JTMi=

|ETMr|2|ETMi|2 = (RTM )2. (9.3.43)

RTE =JTEr

JTEi=

|ETEr|2|ETEi|2 = (RTE)2, (9.3.44)

Thus, the reflectance Rp is equal to the square of reflection coefficient Rp.The transmittance T p (p = TE, TM) is the ratio between the transmitted and incident energy

flows, and (9.3.41)-(9.3.42) give

T TM =JTMt

JTMi=

n2

n1

µ1

µ2

cos θt

cos θi(TTM )2, (9.3.45)

T TE =JTEt

JTEi=

n2

n1

µ1

µ2

cos θt

cos θi(TTE)2. (9.3.46)

Thus, the transmittance T p is proportional to the square of the transmission coefficient T p (p =TE, TM). When µ2 = µ1 = 1, we find on substitution from (9.3.35)-(9.3.36) into (9.3.43)-(9.3.46)the following expressions for the reflectance and the transmittance

RTM =tan2(θi − θt)tan2(θi + θt)

, (9.3.47)

FYS 263 37

RTE =sin2(θi − θt)sin2(θi + θt)

, (9.3.48)

T TM =sin 2θi sin 2θt

sin2(θi + θt) cos2(θi − θt), (9.3.49)

T TE =sin 2θi sin 2θt

sin2(θi + θt). (9.3.50)

By use of these formulas one can show that

RTM + T TM = 1 ; RTE + T TE = 1, (9.3.51)

so that for each of the two polarisations the sum of the reflected energy and the transmitted energyis equal to the incident energy.

From (9.3.41) and (9.3.42) we have

Jpi =c

4π

√ε1

µ1|Epi|2 cos θi, (9.3.52)

which by the use of (9.3.40) gives

JTEi = cos θi c

4π

√ε1

µ1|ETEi|2 = cos θi c

4π

√ε1

µ1E2 sin2 αi, (9.3.53)

JTMi = cos θi c

4π

√ε1

µ1|ETMi|2 = cos θi c

4π

√ε1

µ1E2 cos2 αi. (9.3.54)

But since the total incident energy flow is given by

J i = cos θi c

4π

√ε1

µ1E2, (9.3.55)

we find

JTEi = J i sin2 αi ; JTMi = J i cos2 αi. (9.3.56)

Thus, we have

R =Jr

J i=

JTMr + JTEr

J i=

JTMr

JTMicos2 αi +

JTEr

JTEisin2 αi, (9.3.57)

which gives

R = RTM cos2 αi + RTE sin2 αi, (9.3.58)

and similarly we find

T = T TM cos2 αi + T TE sin2 αi. (9.3.59)

At normal incidence, θi = θt = 0, and the distinction between TE and TM polarisation disappears.From (9.3.43)-(9.3.46) combined with (9.3.33)-(9.3.34), we find (when µ1 = µ2 = 1)

R = RTM = RTE = (RTE)2 =(RTM

)2=

(n − 1n + 1

)2

; n =n2

n1, (9.3.60)

T = T TM = T TE =(TTE

)2=

(TTM

)2=

4n

(n + 1)2; n =

n2

n1. (9.3.61)

When n → 1, we see that R → 0 and T → 1, as expected. Similarly, we find from (9.3.47)-(9.3.50)that R‖ → 0, R⊥ → 0, T‖ → 1, T⊥ → 1 when n → 1.

FYS 263 38

tB

1

2

k

k

k

θi

r

t

θ

θ iB iB

Figure 9.6: Illustration of Brewster’s law.

9.3.2 Brewster’s law

From (9.3.47) it follows that RTM = 0 when θi + θt = π2 , since then tan(θi + θt) = ∞. We call this

particular angle of incidence θiB and the corresponding refraction or transmission angle θtB . Byusing Snell’s law (9.1.20), we find

n2 sin θtB = n2 sin(π

2− θiB

)= n2 cos θiB = n1 sin θiB , (9.3.62)

so that RTM = 0 when θi = θiB , where θiB is given by

tan θiB =n2

n1= n. (9.3.63)

The angle θiB is called the polarisation angle or the Brewster angle. When the angle of incidenceis equal to θiB , the E vector of the reflected light has no component in the plane of incidence (Fig.9.6). This fact is exploited in sunglasses with polarisation filter. The filter is oriented such thatonly light that is polarised vertically (Fig. 9.6) is transmitted. Thus, one avoids to a certain degreeannoying reflections from e.g. a water surface.

Note that kr · kt = 0, i.e. kr and kt are normal to one another when θi = θiB , as shown inFig. 9.6.

9.3.3 Unpolarised light (natural light)

For natural light, e.g. light from an incandescent lamp, the direction of the E vector varies veryrapidly in an arbitrary or irregular manner, so that no particular direction is given preference. Theaverage reflectance R is obtained by averaging over all directions α. Since the average value of bothsin2 α and cos2 α is 1

2 , we find from (9.3.56) that

JTMi

= J icos2 αi = JTEi

= J isin2 αi =12J i. (9.3.64)

For the reflected components we find

JTMr

=J

TMr

JTMi

· JTMi=

JTMr

JTMi

· 12J i =

12RTMJ i, (9.3.65)

FYS 263 39

JTEr

=J

TEr

JTEi

· 12J i =

12RTEJ i, (9.3.66)

which shows that the degree of polarisation for the reflected light can be defined as

P r =∣∣∣∣R

TM −RTE

RTM + RTE

∣∣∣∣ =|JTMr − JTEr|JTMr + JTEr

. (9.3.67)

The average reflectance is given by

R =J

r

Ji

=J

TMr+ J

TEr

Ji

=J

TMr

2JTMi

+J

TEr

2JTEi

=12

(RTM + RTE

), (9.3.68)

so that the degree of polarisation becomes

P r =1R

12|RTM −RTE |, (9.3.69)

where |RTM −RTE | is called the polarised part of the reflected light.Similarly, we find for the transmitted light

T =12(T TM + T TE) ; P t =

1T

12|T TM − T TE |. (9.3.70)

9.3.4 Rotation of the plane of polarisation upon reflection and refraction

Note that if the incident light is linearly polarised, then also the reflected and the transmitted lightwill be linearly polarised, since the phases only change by 0 or π. This follows from the fact that thereflection and transmission coefficients are real quantities [cf. (9.3.33)-(9.3.36)]. But the planes ofpolarisation for the reflected and the transmitted light are rotated in opposite directions relative tothe polarisation plane of the incident light. The angles αi, αr, and αt that the planes of polarisationof the incident, reflected, and transmitted light form with the plane of incidence, are given by [cf.Fig. 9.5]

tanαi =ETEi

ETMi, (9.3.71)

tanαr =ETEr

ETMr=

ET Er

ET Ei

ET Mr

ET Mi

ETEi

ETMi=

RTE

RTMtanαi, (9.3.72)

tanαt =ETEt

ETMt=

ET Et

ET Ei

ET Mt

ET Mi

ETEi

ETMi=

TTE

TTMtanαi. (9.3.73)

By use of the Fresnel formulas (9.3.35)-(9.3.36) we can write

tanαr = −cos(θi − θt)cos(θi + θt)

tanαi, (9.3.74)

tanαt = cos(θi − θt) tanαi. (9.3.75)

Since 0 ≤ θi ≤ π2 and 0 ≤ θt ≤ π

2 , we get

| tanαr| ≥ | tanαi|, (9.3.76)

| tanαt| ≤ | tanαi|. (9.3.77)

In (9.3.76) the equality sign applies at normal incidence (θi = θt = 0) and at grazing incidence (θi =π2 ), whereas in (9.3.77) the equality sign applies only at normal incidence. These two inequalities

FYS 263 40

imply that upon reflection the plane of polarisation is rotated away from the plane of incidence,whereas upon transmission it is rotated towards the plane of incidence. Note that when θi = θiB ,so that θiB + θtB = π

2 , then tanαr = ∞. Thus, we have αr = π2 in accordance with Brewster’s law.

9.3.5 Total reflection

Snell’s law (9.1.20) can be written in the form

sin θt =sin θi

n; n =

n2

n1=

√ε2µ2

ε1µ1. (9.3.78)

Hence, it follows that if n < 1, then we get sin θt = 1 when θi = θic, where

sin θic = n. (9.3.79)

This implies that when θi = θic, we get θt = π2 , so that the transmitted light propagates along the

interface. If θi ≥ θic, we have total reflection, i.e. no light will pass into the other medium. Alllight is then reflected. There exists a field in the other medium, but there is no energy transportthrough the interface. When θi > θic, then sin θt > 1, which means that θt is complex. We havefrom (9.3.78)

cos θt = ±√

1 − sin2 θt = ±i

√sin2 θi

n2− 1 =

±i√

sin2 θi − n2

n. (9.3.80)

The lower sign in (9.3.80) must be discarded. Otherwise the field in medium 2 would grow expo-nentially with increasing distance from the interface. The electric field in medium 2 is

Ept = T pEpieptei(kt·r−ωt) (p = TE, TM), (9.3.81)

where

kt · r = kxx + kyy + kz2z, (9.3.82)

with (cf. Fig. 9.1 and (9.3.80) with upper sign)

kz2 = k2 cos θt = ik21n

√sin2 θi − n2 ; n =

n2

n1, (9.3.83)

so that

eikt·r = ei(kxx+kyy)e−|kz2|z ; |kz2| =k2

n

√sin2 θi − n2 . (9.3.84)

We see that Ept represents a wave that propagates along the interface and is exponentially dampedwith the distance z into medium 2.

From ∇ · Dt = ε2∇ · Et = 0 it follows that

kt · Et = 0, (9.3.85)

which gives

Etz = −

(kxEtx + kyEt

y)kz2

. (9.3.86)

If we let the plane of incidence coincide with the xz plane, we have (cf. Fig. 9.7)

kx = −k1 sin θi ; ky = 0, (9.3.87)

Ety = ETEtei(kxx−ωt)e−|kz2|z = TTEETEiei(kxx−ωt)e−|kz2|z, (9.3.88)

FYS 263 41

ir

z

x

= =

θ t

E TMt

E TMr

θ i kk

kt

θ

θ t

i

E TMik i E TEi E ETEr TEt

Figure 9.7: Illustration of the refraction of a plane wave into an optically thinner medium, so thatθi < θt. When θi → θic, then θt → π/2, and we get total reflection.

Etx = −ETMt cos θtei(kxx−ωt)e−|kz2| = −TTMETMi cos θtei(kxx−ωt)e−|kz2|z, (9.3.89)

Etz = − kx

kz2Et

x =kx

k2TTMETMiei(kxx−ωt)e−|kz2|z. (9.3.90)

From these expressions for the components of Et and corresponding expressions for the componentsof Ht one can show (Exercise 11) that the time average of the z component of the Poynting vectoris zero, which implies that there is no energy transport through the interface, as asserted earlier.

The reflection coefficients in (9.3.35)-(9.3.36) can be written as follows

RTM =sin θi cos θi − sin θt cos θt

sin θi cos θi + sin θt cos θt, (9.3.91)

RTE = − sin θi cos θt − sin θt cos θi

sin θi cos θt + sin θt cos θi. (9.3.92)

By combining Snell’s law (9.3.78) and (9.3.80) with the upper sign with (9.3.91)-(9.3.92), we get

RTM =n2 cos θi − i

√sin2 θi − n2

n2 cos θi + i√

sin2 θi − n2, (9.3.93)

RTE =cos θi − i

√sin2 θi − n2

cos θi + i√

sin2 θi − n2. (9.3.94)

Since both reflection coefficients are of the form z/z∗, where z is a complex number, it follows that

|RTM | = |RTE | = 1, (9.3.95)

FYS 263 42

which shows that for each polarisation the intensity of the totally reflected light is equal to theintensity of the incident light.

But the phase is altered upon total reflection. Letting

Rp =Epr

Epi= eiδp

=zp

zp∗ = e2iαp

(p = TE, TM), (9.3.96)

where [cf. (9.3.93)-(9.3.94)]

zTM = n2 cos θi − i√

sin2 θi − n2 = |zTM |eiαT M

, (9.3.97)

zTE = cos θi − i√

sin2 θi − n2 = |zTE |eiαT E

, (9.3.98)

we find

tanαTM = tan(

12δTM

)= −

√sin2 θi − n2

n2 cos θi, (9.3.99)

tanαTE = tan(

12δTE

)= −

√sin2 θi − n2

cos θi. (9.3.100)

The relative phase difference

δ = δTE − δTM , (9.3.101)

is determined by

tan(

12δ

)=

tan(

12δTE

)− tan

(12δTM

)1 + tan

(12δTE

)tan

(12δTM

) , (9.3.102)

which upon substitution from (9.3.99)-(9.3.100) gives

tan(

12δ

)=

cos θi√

sin2 θi − n2

sin2 θi. (9.3.103)

We see that δ = 0 for θi = π2 (grazing incidence) and θi = θic (critical angle of incidence). Between

these two values there is an angle of incidence θi = θim which gives a maximum phase differenceδ = δm, where θim is determined by

dδ

dθi

∣∣∣∣θim

= 0. (9.3.104)

From (9.3.104) we find

sin2 θim =2n2

1 + n2, (9.3.105)

which upon substitution in (9.3.103) gives (Exercise 10)

tan(

12δm

)=

1 − n2

2n. (9.3.106)

If the phase difference δ is equal to ±π2 and in addition ETMi = ETEi, the totally reflected light

will be circularly polarised. By choosing the angle αi between the polarisation plane and the planeof incidence equal to 45◦, we make ETMi equal to ETEi. In order to obtain δ = π

2 we must haveδm

2 ≥ π4 . This means that tan

(δm

2

)≥ tan

(π4

)= 1, which according to (9.3.106) implies that

n2 + 2n − 1 ≤ 0. (9.3.107)

By completing the square on the left-hand side of (9.3.107), we find that

FYS 263 43

n ≤√

2 − 1 ;1n

=n1

n2≥

√2 + 1 = 2.41. (9.3.108)

Thus, n1n2

must exceed 2.41 in order that we shall obtain a phase difference of π2 in one single

reflection.