A Calculation of Electric Field Strengthsfor Light in a Multilayer Thin Film Structure

by Brent Royuk, Bachelor of Science

An Advanced Research Paper Submitted in PartialFulfillment of the Requirements

For the Master of Science Degree

Department of Physicsin the Graduate School

Southern Illinois University at EdwardsvilleEdwardsville, IL

July, 1996

ii

Abstract

Equations are derived that give electric field strengths for light incident on a multilayer thin film

structure. The equations assume the structures have semi-infinite incident layers and substrates,

and apply to structures with any number of intermediate layers. All layers are separated by

boundaries that are planar and parallel. The form of the equations is recursive, and no matrices

are used to find the reflectance and transmittance of a multilayer. The equations are used to

generate electric field strength vs. depth plots for a variety of thin film examples that illustrate

some familiar design characteristics.

iii

Table of Contents

ABSTRACT ii

ACKNOWLEDGMENTS iv

LIST OF FIGURES v

Chapter Page

I. INTRODUCTION 1

II. BACKGROUND THEORY 3

Plane Electromagnetic Waves 3Optical Constants 4Polarization States 5Boundary Conditions 6Fresnel’s Coefficients 7Rouard’s Method 8

III. CALCULATION OF THE ELECTRIC FIELD 10

Notational Conventions 10S-Polarized Light 11

Incident mediumSubstrateIntermediate Phases

P-Polarized Light 22Incident mediumSubstrateIntermediate Phases

Tangential ComponentNormal Component

Equation Summary 34

IV. NUMERICAL ANALYSIS 36

Two Layers on an Absorbing Substrate 36Single Layer AR Coating 40Dielectric HR Mirror 44Critical Angle 46Plasmon Resonance Angle 50Transmittance Properties of Copper 54

V. SUMMARY 56

REFERENCES 58

iv

Acknowledgments

First and foremost, I wish to thank my thesis advisor, Dr. Arthur Braundmeier. The derivation

that is the heart of this paper follows a roughed-out version that Dr. Braundmeier completed

more than ten years before this paper was written. I will be forever grateful to him for the many

hours of help, advice and encouragement that he has given me, most of it long distance. One of

my best memories from this project is when Dr. Braundmeier and I utilized a ten-hour problem

solving session to remove the glitches from my numerical calculation process.

I would like to thank the members of my thesis committee, Drs. George Henderson and Jerry

Pogatshnik, and I am also indebted to all the fine teachers in the SIUE Physics Department who

taught my classes.

Finally, I thank my dear wife, Sandra, without whom I never could have survived our “three

year plan.” It has been difficult but rewarding to raise two babies and a Master’s Degree

contemporaneously.

Soli Deo Gloria

v

List of Figures

Figure Description Page

2.1 The total electric field, composed of a transmitted and a reflected part 32.2 Labeling angles at the interface between medium j-1 and medium j 52.3 S- and p-polarized light 63.1 Conventions used to label optical media and interfaces 103.2 Notation used for labeling field strengths at interfaces 113.3 Geometric relationships for s-polarized light 123.4 Finding the direction of the wave vector 143.5 The transmitted portion of the electric field at an interface 213.6 Geometric relationships for p-polarized light 224.1 Diagram of a thin film 374.2 Electric field strengths for a thin film consisting of 400 nm Si, 400 nm Al2O3 38

on a GaAs Substrate, with light incident at 30o and λo = 550 nm4.3 Detail of Figure 4.2 394.4 Single layer quarter wavelength AR coating: 72 nm layer HfO2 on Si 42

substrate with light at normal incidence and λo = 550 nm, R = 0.17%.Uncoated silicon substrate: R = 36%.

4.5 Single layer quarter wavelength AR coating: 72 nm layer HfO2 on Si 43substrate with light incident at 30o and λo = 550 nm. RS = 1.4% andRP = 0.74%. An uncoated silicon substrate would have reflectances ofRS = 41.0% and RP = 30.6%

4.6 Imperfect single-layer AR coating: 150 nm layer HfO2 on Si substrate 44with light incident at 30o and λo = 550 nm. RS = 41.0% and RP = 30.6%.

4.7 Schematic of a design for a dielectric HR mirror 454.8 A dielectric high reflectance structure: Eight alternating layers of quarter 46

wavelength layers of TiO2 and MgF2 on glass substrate with light incident at 30o and λo = 550 nm. RS = 95.9% and RP = 90.4%.

4.9 Reflectance vs. angle for light traveling from glass into air 474.10 Electric field strengths for glass to air at 40 owith λo = 550 nm 484.11 Electric field strengths for glass to air at 45 owith λo = 550 nm 494.12 Reflectance vs. incident angle for light traveling from quartz through a 51

50 nm layer of silver into air with λo = 550 nm4.13 Electric field strengths for light traveling from quartz through a 50 nm 52

layer of silver into air at 40o incident angle with λo = 550 nm4.14 Electric field strengths for light traveling from quartz through a 50 nm 53

layer of silver into air at 45.3o incident angle with λo = 550 nm4.15 Electric field strengths for a 50 nm layer of copper on glass at 30owith 54

λo = 500 nm4.16 Electric field strengths for a 50 nm layer of copper on glass at 30owith 55

λo = 1000 nm

1

Chapter I

Introduction

The purpose of this paper is to derive equations that allow one to calculate electric field

strengths within a multilayer thin film structure. After the equations have been derived,

calculations will be made to demonstrate their utility in finding electric field strengths for light

incident on several thin film optical stacks.

With the equations in hand, one needs to know a) the strength of the electric field for the light

at the first interface of the incident medium, b) the angle of incidence at the first interface, c)

the thickness of each element of the stack, and d) the optical constants (n and k values) for each

optical element, which are a function of the frequency of the incident light.

It is not difficult to find the electric field strength for light propagating through a single

medium. The most arduous task in this derivation is to keep track of the field strengths as the

light travels across each interface of the thin film stack. The Fresnel coefficients provide a

mathematical description of the electric fields as they cross these interfaces, with separate

treatments being necessary for the p- and s-polarized varieties of light. Furthermore, when

dealing with a multilayer structure, one must take into account all the multiple reflections that

may occur between the layers. This is accomplished through the use of Rouard’s method.

In order to “divide and conquer” the problem of calculating electric field strengths in a thin

film, a separate mathematical treatment is necessary for each of three different regions of the

thin film structure: the incident medium, the intermediate layers, and the substrate. We will

2

assume the incident medium and the substrate to be semi-infinite in extent in order to eliminate

any reflections that would occur from a “front” surface in the incident medium or a “back”

surface in the substrate. Only one equation will be necessary for each of these regions when

considering s-polarized light, while there will need to be two equations in each region for p-

polarized light, one for the electric field component tangential to the interface and one for the

normal component. This strategy will yield a total of nine equations that are derived in Chapter

III.

Chapter IV will be occupied with the application of the equations to some common thin-film

structures, showing some electric field strength vs. depth plots. We will investigate several thin

film designs that are chosen to demonstrate some of the more interesting and educational

electric field strength scenarios that occur in thin film structures.

The electric field profile for a thin film structure can be useful to a thin film designer, especially

if the structure is to be used in an environment where the electric field strengths are high, such

as high energy laser applications. High index optical materials are especially susceptible to

electric field damage, since energy density scales as n2, the square of the refractive index. When

the electric field strength profile is solved for a given structure, the designer can see where in the

stack the field would be especially high. Designs can be altered in order to insure that regions

of high field do not occur at interfaces, since a thin film is most vulnerable at the interfaces,

where different materials adhere to each other. The forces on the charges in the film materials

that are due to the electric field of the incident light can pull the layers apart or produce a

shearing force, which can cause films to disintegrate, de-laminate or crack.

3

Chapter II

Background Theory

Plane Electromagnetic Waves

In order to describe the electric field for light propagating through any optical element in a thin

film stack, we will use the exponential wave equation, E = Eoe i k ⋅r −ωt( ), which is a solution of

Maxwell’s equations1. In all phases of a thin film stack except the semi-infinite substrate, this

field will consist of two parts: a transmitted part and a reflected part. The equation for the total

field in layer j will be written as:

E j = E j0( t ) e

i( k jt ⋅r −ω t) + E j

0(r ) ei(k j

r ⋅r −ω t). (2.1)

Figure 2.1: The total electric field, composed of a transmitted and a reflected part

Layer j

E j0(r) e

i( k jr ⋅r− ω t)

E j0(t ) e

i(k jt ⋅r− ω t)

E j

4

This wave equation is explicitly time-dependent, but in our treatment of the equations the time-

dependence will tend to become irrelevant since time-averages are calculated by taking the

absolute squares of the measurable quantities, leaving expressions that are constant in time. It

should be noted that in all equations in this paper vectors will be denoted with boldfaced letters

and scalars will be normal type. A caret (^) over a boldfaced variable denotes a unit vector and a

tilde (~) over a variable indicates a complex number.

Optical Constants

The wave vector, kj, can be written as k j =2π˜ n jλo

ˆ S j , where ˆ S j is a unit vector in the direction of

the wave’s energy propagation2 and λo is the wavelength of the incident light in vacuum. The

constant ˜ n j is the complex index of refraction for the jth layer. It has a real and a complex part:

˜ n j = n j − ik j .3 In the wave equation, nj is the index of refraction and determines the phase of

the wave as it propagates through the medium, while kj is an absorption coefficient that will

cause the wave to “die out” as it propagates through the medium. These constants are

characteristic of the material through which the wave is propagating and the frequency of the

incident wave. Also useful is Snell’s law:

˜ n j sin ˜ θ j = ˜ n j +1 sin ˜ θ j +1 . (2.2)

Snell’s law can be derived by applying conservation of momentum for light incident at a

boundary4. This equation gives the refraction angles for the light as it passes into media with

different optical constants. Note that ˜ θ is, in general, complex. Figure 2.2 shows light

refracting at a layer boundary with the angles labeled.

5

Figure 2.2: Labeling angles at the interface between medium j-1 and medium j

Polarization States

Electromagnetic waves can have two different polarization states, s- and p-polarization. S-

polarized waves have their electric field vectors normal to the plane of incidence and are thus

tangential to the interface . P-polarized light consists only of waves that have their electric field

vectors parallel to the plane of incidence. The p-polarized field vectors will therefore have

components tangential to the interface and normal to the interface5. These two components

require separate treatments in the derivation. Figure 2.3 illustrates the two polarization states.

The small double-headed arrows arranged along the propagation vectors show the direction of

the electric field vectors for the light.

Interface j-1

˜ n j − 1

Interfaces

˜ n j

˜ θ j −1

˜ θ j

6

Figure 2.3: S- and p-polarized light

Boundary Conditions

In order to obtain a full description of electric field strengths for light propagating through a

medium, one needs to make use of both the electric field, E, and the magnetic field, H. These

are related by6

H j =˜ ε jµ j

1/2

ˆ S j × E j . (2.3)

In this equation, ˜ ε j is the permittivity of the medium and µ j is the permeability of the medium,

which measure the tendency of the material to allow propagation of electric and magnetic fields,

respectively. These values are related to the complex index of refraction for a given material by

˜ n j = µ j˜ ε j( )1 / 2

. The cross product in equation 4.2 shows us that the electric and magnetic

fields are perpendicular to each other. As these fields propagate across an interface, we note

that Maxwell’s equations predict that the tangential portion of the fields must be continuous

across the interface7. These boundary conditions will be enforced during the derivation and will

have markedly different implications for the treatment of s- and p-polarized light.

7

Fresnel’s Coefficients

As light propagates through an interface into a new medium, we need a way to relate the

amplitudes of the transmitted and reflected portions of the light. We can do this by making use

of Fresnel’s coefficients. We will consider the (j-1) interface to be the interface between the

(j-1)th and the jth media, and will define the Fresnel reflection coefficient for that interface as ˜ r j − 1.

For the (j-1)th interface, the Fresnel reflection coefficient is defined as ˜ r j − 1 ≡E j −1

( r )

E j−1( t ) . So

considering only reflections that take place at a single interface, the field reflected at that

interface will be related to the incident field by E j −1(r ) = ˜ r j−1E j −1

( t ) .

We note that all three quantities in the preceding equation are complex numbers, so ˜ r j − 1 must

provide information about the phase relationship between the incoming and reflected wave,

along with relating their field strengths8.

The Fresnel reflection coefficients come in two varieties, one for s-polarized light and one for p-

polarized light. We define these as:

˜ r j-1S ≡ s-polarized Fresnel reflection coefficient, and

˜ r j-1P ≡ p-polarized Fresnel reflection coefficient.

The following equations are derived by conserving the tangential components of the E and H-

fields at an interface and are given by9

˜ r j-1S =

˜ n j cos ˜ θ j − ˜ n j − 1 cos ˜ θ j −1

˜ n j cos ˜ θ j + ˜ n j −1 cos ˜ θ j − 1

and (2.4)

8

˜ r j -1P =

˜ n j cos ˜ θ j −1 − ˜ n j−1 cos ˜ θ j˜ n j cos ˜ θ j −1 + ˜ n j −1 cos ˜ θ j

. (2.5)

These coefficients are defined so that there is no phase change difference between s- and p-

polarized light that is reflected from an interface at normal incidence. A phase change

difference of π2

is sometimes “built into” the equations when problems of this type are

considered. We will not adopt that convention in this paper.

The amplitude of light transmitted across an interface can be described by the Fresnel

transmission coefficients, which are defined by10 ˜ t j −1 ≡E j

( t )

E j −1( t )

, and are given as

˜ t j-1S =

2˜ n j − 1 cos ˜ θ j −1

˜ n j cos ˜ θ j + ˜ n j −1 cos ˜ θ j − 1

and (2.6)

˜ t j -1P =

2 ˜ n j cos ˜ θ j−1

˜ n j cos ˜ θ j −1 + ˜ n j−1 cos ˜ θ j. (2.7)

Rouard’s Method

Equations 4.4 - (2.7) apply only to a single interface. When analyzing light that is propagating

through a multilayer optical stack, it is necessary to take into account light that may reflect back

and forth between layer interfaces, thus increasing the effective reflective properties of the stack.

Rouard’s expression sums these multiple reflections and gives a single coefficient that

represents the reflection due to all materials beyond a given interface. This method sums the

multiple bounce contributions as part of an infinite mathematical series that produces the

exponential terms in the equations below. The series is convergent since the more bounces that

occur, the weaker they are. The result is11:

9

′ ˜ r j -1 =

˜ r j -1 + ′ ˜ r j exp4πil j

˜ n j cos ˜ θ j

λo

1+ ′ ˜ r j ˜ r j-1 exp4πil j ˜ n j cos ˜ θ j

λ o

. (2.8)

The prime (′) denotes the “total” coefficient that takes into account all multiple reflection

possibilities beyond the interface. We note that equation (2.8) is a recursive relation that

necessitates knowledge of the “total” coefficient at the interface beyond the one in question.

So in practice it is necessary to “work your way back” from the substrate to the incident

medium, finding the total reflection coefficient at each interface in turn. This process will

eventually yield a total combined reflection coefficient for the whole stack, ˜ ′ r 1 . This constant

will be featured prominently when we derive the equations for the incident medium, where we

don’t care what the light is “doing” inside the thin film, but only need to know its total

reflectance. The total absolute reflectanceof the stack, R, which gives the percentage of energy

reflected by the stack, is related to ˜ ′ r 1 by R = ˜ ′ r 12.

Instead of following this recursive process, it is also quite common to use matrix notation to

reduce multiple layer contributions to a single total reflection coefficient in matrix form.

The corresponding treatment of the transmission coefficients yields12

˜ ′ t j -1 =

˜ ′ t j ˜ t j−1 exp2πil j

˜ n j cos ˜ θ j

λo

1+ ′ ˜ r j ˜ r j-1 exp4πil j ˜ n j cos ˜ θ j

λo

. (2.9)

10

Chapter III

Calculation of the Electric Field

Notational Conventions

As we consider a multilayer thin film system, the following conventions will be used to label the

interfaces and the optical indices for the stacked media:

n1 ˜ n 2 ˜ n 3 ˜ n 4 ..... ˜ n j -1 ˜ n j ˜ n j +1 ..... ˜ n m −1 ˜ n m

Interface: 1 2 3 4 j-2 j-1 j j+1 m-2 m-1

Figure 3.1: Conventions used to label optical media and interfaces

In all diagrams in this paper, the incident medium will be to the left of the substrate, so the

incident light will always travel left to right.

It will often be important for us to refer to the electric field just at the edge of a given medium.

The notation that will be used to label electric field strengths at the edges of interfaces is

summarized in Figure 3.2:

IncidentMedium

Substrate

11

ˆ n j ˆ n j +1

Ej-(r ) E

j +(r )

Ej -

(t ) Ej +( t )

interface: j j+1

Figure 3.2: Notation used for labeling field strengths at interfaces

The superscript “t” or “r” indicates whether the variable represents the transmitted or

reflected electric field. The subscript refers to the interface number, with the raised “+” or “-”

denoting the “outgoing” or “incoming” sides of the interface.

S-Polarized Light

For s-polarized light, we have the following geometry for the incident and reflected fields:

ˆ k

ˆ i

12

Figure 3.3: Geometric relationships for s-polarized light

INCIDENT MEDIUM

For the incident medium, we assume, necessarily and without loss of generality, that n1 is real.

There is therefore no absorption in the incident medium. The electric field in the incident

medium can then be written as E1 = E10(t ) e i ( k1

t ⋅r −ω t ) + E10( r ) e i( k1

r ⋅ r− ω t ) , where E10( r ) = ˜ ′ r 1E1

0( t) . We

recall that ˜ ′ r 1 is the total reflection coefficient at the first interface for all layers beyond the first

interface and is calculated with Rouard’s method. We next define R to be the energy reflected

from the multilayer stack. We note that R = ˜ ′ r 1 2 and the inverse relationship is ˜ ′ r 1 = R1/2 e iδ r

,

where δ r is the phase change that occurs when the light is reflected off the first interface of the

thin film structure. Thus we can write E1 = E10( t ) ei ( k1

t ⋅r −ω t ) + R1/2eiδ r

ei( k1r ⋅r −ω t )[ ] . This

expression gives E1 as a vector quantity. Since we have assumed the incident medium to be

transparent, E1 is real, which will not generally be true in layers beyond the first interface.

While it is mathematically necessary for the electric field to be a complex vector quantity, we

will ultimately be interested in calculating real measurable field strengths, i.e. the intensity of the

H0

13

light, which can be found by calculating the time-average of the electric field vector. This is

defined as E 2 ≡1

2E

2. The absolute value for a complex quantity is found by taking E⋅E*,

so the time-average will always be real.

The time-average of the electric field in the first medium is

E12 =

1

2E1

2=

1

2E1 ⋅ E1

*( )

=1

2e i(k 1

t ⋅r− ω t ) + R1/2e iδ r

ei ( k1r ⋅r− ω t )[ ] e -i( k1

t ⋅ r− ω t ) + R1/2e−iδ r

e−i ( k1r ⋅r −ω t )[ ] E1

0( t )( )2

=1

21+ R + R1/2 e -iδ r

e i(k1t ⋅ r −k1

r ⋅ r ) + e iδ r

e i(k1r ⋅r −k1

t ⋅r )( )[ ] E10( t )( )2

We can simplify the exponential terms if we let Φ = k1t ⋅ r − k1

r ⋅ r −δ r , so

E12 =

1

21 + R + R1/2 e iΦ + e-iΦ( )[ ] E1

0(t )( )2.

But e iΦ + e- iΦ = 2cosΦ , so

E12 =

1

21+ R( ) + R1/2 cosΦ

E10( t)( )2

=1

21 + R( ) + R1/2 cos k1

t ⋅ r − k1r ⋅ r −δ r( )

E10( t)( )2

.

The cosine is an even function, i.e. cosΦ = cos −Φ( ), so we can rewrite the argument of the

cosine in a more pleasing form:

E12 =

1

21+ R( ) + R1/2 cos δ r + k1

r ⋅r − k1t ⋅ r( )

E10( t )( )2

.

The argument of the cosine in the above equation can be simplified if we note that

k1r =

2πn1

λ o

−sinθ1ˆ i − cosθ1

ˆ k ( ) and k1t =

2πn1

λo

−sinθ1ˆ i + cosθ1

ˆ k ( ) .

These relations can be found by analyzing the direction of the wave vector, k, in terms of the

direction vectors we have defined. Figure 3.4 illustrates this analysis for the transmitted portion

of the wave.

14

Figure 3.4: Finding the direction of the wave vector

The scalar product is distributive over addition and subtraction, so we can write

k1r ⋅r − k1

t ⋅r = k1r − k1

t( )⋅r

=2πn1

λo

−2cosθ1ˆ k ( )⋅r .

Now r = xˆ i + yˆ j + zˆ k , so

k1r − k1

t( ) ⋅r =−4πn1

λo

cosθ1ˆ k ⋅ xˆ i + yˆ j + z ˆ k ( )

=−4πn1

λo

cosθ1z .

Therefore:

E12 =

1

21+ R( ) + R1/2 cos δ r −

4πn1

λo

cosθ1z

E10( t )( )2

. (3.1)

−k1t sin θ1

ˆ i

θ1

k1t

k1t cosθ1

ˆ k

Interface 1(top view)

θ1

ˆ k

ˆ i

15

This equation gives the time-averaged field intensities for the electric field in the incident

medium for s-polarized light. It is the first of nine equations that will be derived in this chapter

that will, together, completely describe electric field strengths in a thin film structure.

SUBSTRATE

We assume the substrate is semi-infinite in extent, so there is no reflected field. Therefore the

electric field in the substrate is just Em = Em0(t )e

i k mt ⋅r−ω t( )

,where Em0( t) is the total electric field

transmitted all the way through the stack, just inside the substrate.

We now need to make use of the total transmittance, ˜ ′ t 1 . This is the “total” transmittance

calculated using equation (2.9) that takes multiple reflections into account. The subscript of this

variable is 1, which may seem odd since it represents the amount transmitted to the substrate

(layer m), but we adopt this convention since the recursive nature of equation (2.9) requires us

to work our way from “back to front” when calculating ˜ ′ t 1 .

Now Em0( t) = ˜ ′ t 1E1

0( t ) , and the time-average for the electric field in the substrate is

Em2 =

1

2Em ⋅ Em

*( ) =1

2˜ ′ t 1E1

0( t ) e i( k mt ⋅r−ω t ) ⋅ ˜ ′ t 1E1

0( t ) ei( k mt ⋅r−ω t )( )*

=1

2˜ ′ t 1

2E1

0( t )( )2ei (k m

t ⋅r −ω t) ⋅ ei(k m

t ⋅r −ω t )[ ]∗.

Let us examine e i (k mt ⋅r−ω t) ⋅ e

i (kmt ⋅ r−ω t )[ ]∗ to simplify.

e i(k mt ⋅r −ω t ) ⋅ e

i (kmt ⋅ r −ω t)[ ]∗ = e

i( k mt ⋅r −ω t )+ i( k m

t ⋅r −ω t )[ ]*

= ei k m

t ⋅r − iω t + i(k mt ⋅ r )[ ] *

− iωt( ) *

.

But ωt is real, so iωt( )* = −iωt , and therefore

e i (k mt ⋅r−ω t ) ⋅ e

i (kmt ⋅ r−ω t )[ ]∗ = e

i ( kmt ⋅r)+ i(km

t ⋅ r)[ ]*

.

Now let us simplify i km ⋅ r( ). Analyzing the direction of vector km, we can write

16

km =2πλo

nm − ikm( ) sin ˜ θ mˆ i + cos ˜ θ m ˆ k ( ) , and

r = xˆ i + y j + z ˆ k , so

i km ⋅ r( ) = i2πλo

nm − ikm( ) sin ˜ θ mˆ i + cos ˜ θ m ˆ k ( )⋅ xˆ i + yˆ j + zˆ k ( ) .

To simplify the derivation, we will now assume x = y = 0 . The x and y coordinates can be

chosen arbitrarily since z measures depth through the optical stack, and is thus the only

“meaningful” variable. So with this assumption we get

i km ⋅ r( ) =2πi

λo

nm − ikm( )cos ˜ θ mz .

We now make a useful substitution that will only be used for a moment in this section but will

be quite helpful later in this chapter.

Let ˜ ξ m ≡ ˜ n m cos ˜ θ m , so

i km ⋅r( ) =2πi

λ o

˜ ξ mz .

We also need the complex conjugate of the argument, so we write

i km ⋅r( ) =2πi

λ o

ℜe ˜ ξ m + iℑm ˜ ξ m[ ]z =2πλo

iℜe ˜ ξ m − ℑm ˜ ξ m[ ]z . Then

i km ⋅r( )[ ]*=

2πλo

−iℜe˜ ξ m − ℑm ˜ ξ m[ ]z , so

i km ⋅ r( ) + i km ⋅ r( )[ ]*= −

4πλo

ℑm˜ ξ m( )z ; therefore

Em2 =

1

2˜ ′ t 1

2E1

0( t)( )2e

− 4πλo

ℑm ˜ ξ m( ) z

Em2 =

1

2˜ ′ t 1

2E1

0(t )( )2e

− 4πλo

ℑm ˜ n m cos ˜ θ m( ) z

. (3.2)

This is an expression for the electric field at depth z inside the substrate. In this equation we

can see that if ˜ n m has an imaginary component, the electric field strength will decay

17

exponentially with distance into the substrate. This will occur when the substrate is an

absorbing medium. If the substrate is transparent, the exponential term is equal to 1.0, so the

field in the substrate is constant. Both cases will be investigated in Chapter IV.

INTERMEDIATE PHASES

Now let us consider the intermediate layers between the incident medium and the substrate. At

any interface, the E-field is transverse to the interface, so the total electric field is continuous

across each interface, i.e.

Ej −( t) + E

j−( r) = E

j +

(t ) + Ej +

(r ) . (3.3)

The transverse component of the H-field is also continuous across each interface, which

corresponds to the x direction in our notation, so

Hj− x

(t ) + Hj− x

(r) = Hj+ x

(t ) + Hj+ x

( r) .

The fields are related by H j =˜ ε jµ j

1 / 2

ˆ S j × E j ,where ˆ S j is a unit propagation vector pointing

in the direction of the wave flow. So in the jth medium,

H j =˜ ε jµ j

1 / 2

ˆ S j(t ) × E j

(t )( ) +˜ ε jµ j

1 / 2

ˆ S j(r ) × E j

( r )( ) , since Ej is the sum of the transmitted and

reflected wave. From the geometry of the incident and reflected waves we see that

ˆ S j( t ) = + cos ˜ θ j ˆ k −sin ˜ θ jˆ i ˆ S j

(r ) = −cos ˜ θ j ˆ k − sin ˜ θ j ˆ i ;

E j( t) = E j

(t )ˆ j E j( r ) = E j

(r )ˆ j .

ˆ S j( t ) × E j

( t) = ˆ i Sy(t) Ez − Sz

(t ) Ey( ) + ˆ j Sz( t )Ey − Sx

(t ) Ez( ) + ˆ k S x(t )E y − Sy

(t )E x( )

= −ˆ i cos ˜ θ j E j( t) − ˆ k sin ˜ θ j E j

(t ) .

Likewise,

ˆ S j( r) × E j

(r ) = ˆ i cos ˜ θ jE j(r ) − ˆ k sin ˜ θ jE j

(r ) .

18

So H jx =˜ ε jµ j

1 / 2

− cos ˜ θ jE j( t) + cos ˜ θ j E j

(r )[ ]

=˜ ε jµ j

1/2

− E j(t) + E j

( r)[ ]cos ˜ θ j

H jx =˜ n j cos ˜ θ j

µ j

−E j( t ) + E j

(r )[ ] , the transverse component of the H-field in the jth medium.

At the interface between the jth and the (j+1)th medium (the jth interface), Hj− x

= Hj + x

. So

˜ n j cos ˜ θ jµ j

−E j(t ) + E j

(r )[ ] =˜ n j +1cos ˜ θ j +1

µ j +1

−E j +1(t) + E j +1

(r )[ ].

Again using ˜ ξ j ≡ ˜ n j cos ˜ θ j , we get

˜ ξ jµ j

E j(r ) − Ej

( t)[ ] =˜ ξ j+1

µ j+1

E j+1( r) − E j +1

(t )[ ] .

But the electric fields are equal on either side of the interface so we can define

E j +1( r) ≡ E

j +

( r) , Ej +1(t ) ≡ E

j+(t ) and E j

(r ) ≡ Ej −(r ), E j

( t ) ≡ Ej −(t ) .

In this new notation, the subscript “j” refers to the jth interface, with j − referring to the

incoming side of the interface, and j + referring to the “far” side of the interface.

In this notation we have

˜ ξ jµ j

Ej −( r ) − E

j −( t)[ ] =

˜ ξ j+1

µ j+1

Ej+( r) − E

j+( t )[ ] .

We combine this result with equation (3.3) to eliminate Ej +

( r ) and get

Ej +

(t ) =1

21 +

˜ ξ j µ j +1

˜ ξ j+1µ j

E j −

( t ) + 1 −˜ ξ jµ j +1

˜ ξ j+1µ j

E j −

( r)

. (3.4)

In like manner we can eliminate Ej +

( t ) to get

19

Ej +

( r) =1

21 −

˜ ξ j µ j +1

˜ ξ j +1µ j

E j −

( t) + 1 +˜ ξ jµ j +1

˜ ξ j +1µ j

E j −

( r)

. (3.5)

These expressions allow one to propagate the transverse components of the E-field across an

interface if we know the reflected and transmitted E-fields on the “incoming” side of the jth

interface.

Using the fact that Ej +

( r) = ′ ˜ r j

Ej −( t ) , where ′ ˜ r j is the reflection coefficient at the jth interface, and

equation 4.4, we can write

Ej +

( t ) =1

21 +

˜ ξ j µ j +1

˜ ξ j +1µ j

+ 1−

˜ ξ jµ j + 1

˜ ξ j+1µ j

′ r

j

E j −

( t ) . (3.6)

Similarly, with equation (2.7),

Ej +

( r) =1

21−

˜ ξ j µ j+1

˜ ξ j+1µ j

+ 1+

˜ ξ j µ j +1

˜ ξ j+1µ j

′ r

j

E j −

( t ) .

These equations are very useful since they take the transmitted portion of the field across the

interface.

We now need to describe the behavior of the field as it propagates through the jth medium from

the (j-1)th interface to the jth interface. The field inside the jth medium is composed of the initial

transmitted and reflected fields attenuated by exponential terms:

E j+1 = Ej +

( t )ek j+1

t ⋅r−ω t( ) + Ej+( r)e

k j+1r ⋅r− ω t( )

.

Analyzing the geometry of the s-polarized light, we get

k j +1t ⋅ r =

2π˜ n j +1

λo

−x sin ˜ θ j +1 + zcos ˜ θ j+1( ) and

k j +1r ⋅ r =

2π ˜ n j +1

λ o

−x sin ˜ θ j +1 − z cos ˜ θ j+1( ).

20

Therefore

E j +1 =1

21 +

˜ ξ j µ j +1

˜ ξ j+1µ j

+ 1 −

˜ ξ j µ j+ 1

˜ ξ j +1µ j

′ r

j

e

2πi ˜ n j +1

λo

z cos ˜ θ j +1

+1

21 +

˜ ξ j µ j +1

˜ ξ j +1µ j

+ 1−

˜ ξ jµ j+ 1

˜ ξ j+1µ j

′ r

j

e

−2π i ˜ n j +1

λo

z cos ˜ θ j +1

∗e −iω te

−2 πi ˜ n j+1

λo

x sin ˜ θ j +1

Ej −(t ) .

We will now ignore e

−2πi ˜ n j +1

λo

x sin ˜ θ j +1 −iωt

since any numerical use of these equations necessitates

taking the absolute square of these fields which will cause us to calculate

e

−2πi ˜ n j +1

λo

x sin ˜ θ j +1 −iωt

* e

−2πi ˜ n j +1

λo

x sin ˜ θ j +1 −iωt

*

, which equals 1.

We will now substitute ˜ ϕ j +1 =2π˜ n j +1 cos ˜ θ j +1z

λo

=2π ˜ ξ j+1 z

λo

and then use

ei ˜ ϕ j+1 = cos ˜ ϕ j +1 +i sin ˜ ϕ j+1 to get

E j +1 = cos ˜ ϕ j +1 1 + ˜ ′ r j( ) + i

˜ ξ j µ j+ 1

˜ ξ j +1µ j

sin ˜ ϕ j +1 1 − ˜ ′ r j( )

Ej −

(t ) .

Recalling ˜ ϕ j +1 =2π˜ n j +1 cos ˜ θ j +1z

λo

, ˜ ξ j = ˜ n j cosθ j , and ˜ ξ j +1 = ˜ n j +1 cosθ j +1 , we get

E j +1 = cos2π˜ n j +1 cos ˜ θ j +1z

λo

1 + ˜ ′ r

j( ) + i˜ n j cos ˜ θ j µ j +1

˜ n j +1 cos ˜ θ j+ 1µ j

sin

2π ˜ n j +1 cos ˜ θ j+1z

λo

1 − ˜ ′ r

j( )

Ej −

(t ) .(3.7)

21

This equation gives the electric field within any intermediate layer. It requires knowledge of

Ej −( t ) , which is the transmitted portion of the electric field, just before the interface that precedes

the desired intermediate layer. We will now derive an expression for Ej −( t ) . The notation being

used is illustrated in the figure below.

Ej −t( ) . .E

j +t( ) E

j +1( ) −t( ) .

Layer: j j+1

Interface: j j+1

Figure 3.5: The transmitted portion of the electric field at an interface

Starting with equation (3.6), we have

Ej +

( t ) =1

21 +

˜ ξ j µ j +1

˜ ξ j +1µ j

+ 1−

˜ ξ jµ j + 1

˜ ξ j+1µ j

′ r

j

E j −

( t )

. (3.6)

This takes the transmitted portion of the field across an interface. Then we let Ej +t( ) propagate to

the (-) side of the next interface using Ej +1( ) −t( ) = E

j +t( ) e

2πi ˜ n j +1 cos ˜ θ j+1

λo

lj +1

Ej −t( ) , where lj+1 is the thickness

of the (j+1) layer, to get

22

Ej +1( ) −

( t) =1

21 +

˜ n j cos ˜ θ jµ j + 1

˜ n j +1 cos ˜ θ j +1µ j

+ 1 −

˜ n j cos ˜ θ jµ j +1

˜ n j +1 cos ˜ θ j + 1µ j

′ r

j

e

2πi ˜ n j +1 cos ˜ θ j+ 1

λ o

l j + 1

Ej −( t ) . (3.8)

P-Polarized Electric Field

For p-polarized light, the geometry of the incident and reflected fields looks like this:

Figure 3.6: Geometric relationships for p-polarized light

INCIDENT MEDIUM

First, we consider the incident medium, where we again assume n1 is real, i.e. k1 = 0. The total

field is the vector sum of the incident and reflected plane waves. We define E1 as the total

electric field in medium 1 and E10(t) as the incident p-polarized electric field intensity to get

E1 = E10( t )e

i k1t ⋅r −ω t( ) + E1

0( r )ei k1

r ⋅r −ω t( ).

Sr

Er

Hr

E0

S0

H0

ˆ k

ˆ j ˆ i

Et

23

The components of E10(t) are E1x

( t ) = cosθ1E1( t) and E1z

(t ) = sinθ1E1( t ) . The x-component of E1

0( t ) is

continuous across the interface since it is tangential to the interface. Since the magnetic field is

always tangential to the interface for p-polarized light we can write H1( r) = ˜ ′ r 1 H1

(t) , where ˜ ′ r 1 is

the “total” reflection coefficient for p-polarized light at the first interface and includes all layers

and multiple bounces beyond the first interface.

We also know

E1( t) =−

µ1

˜ ε 1

1/2

ˆ S 1( t ) × H1

( t )( ) and

H1( t ) =

˜ ε 1µ1

1/2

ˆ S 1( t ) ×E1

(t )( ) =˜ ε 1µ1

1/2

E1(t ) +ˆ j ( ) .

We now want to calculate the total field.

E1( t) = cosθ1

ˆ i +sinθ1ˆ k ( )E1

( t )

E1( r) = −cosθ1

ˆ i +sinθ1ˆ k ( )E1

( r ) . Now,

E1( r) = −

µ1

˜ ε 1

1/2

ˆ S 1( r) × H1

( r)( ) with

ˆ S 1(r ) = −sinθ1

ˆ i − cosθ1ˆ k

= − sinθ1ˆ i + cosθ1

ˆ k ( ) and

H1( r ) = H1

( r)ˆ j . So

ˆ S 1(r ) × H1

( r)( ) = ˆ i SyH z − SzH y( ) + ˆ j SzH x − S xH z( ) + ˆ k Sx Hy − SyH x( )

= ˆ i cosθ1H1( r )( ) + ˆ k −sinθ1H1

(r )( ). Thus

24

E1( r) = −

µ1

˜ ε 1

1/2

cosθ1( )ˆ i − sinθ1( )ˆ k [ ]H1( r) , where we have used

H1( r) = ˜ ′ r 1 H1

(t) = ˜ ′ r 1˜ ε 1µ1

E1

(t) . So

E1( r ) = ˜ ′ r 1 − cosθ1

ˆ i +sinθ1ˆ k [ ]E1

(t ) .

Therefore, the total electric field can now be written as

E1 = cosθ1ˆ i +sinθ1

ˆ k ( )E1(t )e

i k1t ⋅r −ω t( ) +˜ ′ r 1 − cosθ1

ˆ i +sinθ1ˆ k ( )E1

(t )ei k1

r ⋅ r−ω t( ).

Now we write the x and z components of E1, E1xand E1z .

E1x = cosθ1 ei k 1

t ⋅r −ω t( ) − ˜ ′ r 1ei k1

r ⋅ r −ω t( )

E1

t

E1z = sinθ1 ei k1

t ⋅r −ω t( ) + ˜ ′ r 1ei k1

r ⋅r−ω t( )

E1

t .

As we did with s-polarized light, we substitute ˜ ′ r 1 = R1/2 e iδ r

and find the time-average:

E1x2 =

1

2cos2 θ1 1 + R − R1/2 e

i k1t ⋅r − k1

r ⋅r−δ r( ) + e− i k 1

t ⋅r− k1r ⋅r −δr( )

E10( t )( )2

.

Substituting cosΦ =1

2e iΦ + e− iΦ( ) we obtain

E1x2 =

1

2cos2 θ1 1 + R − 2R1/2 cos k1

t ⋅ r − k1r ⋅ r − δ r( )[ ] E1

0( t )( )2.

We simplify the argument of the cosine by using the geometry of the p-polarized electric field

components to get

k1t ⋅ r − k1

r ⋅ r =4πn1

λo

cosθ1z .

Therefore, we can find the final form for the tangential (x) component of the electric field,

25

E1x2 =

1

2cos2 θ1 1+ R − 2R1/2 cos

4πn1

λo

cosθ1z − δ r

E10(t )( )2

. (3.8)

The normal (z) component of the electric field is found by taking

E1z2 =

1

2E1z ⋅ E1 z

∗( )

which similarly yields

E1z2 =

1

2sin 2 θ1 1+ R + 2R1/2 cos

4πn1

λo

cosθ1z − δ r

E10( t )( )2

. (3.9)

These equations describe the tangential and normal components of the p-polarized fields in the

incident medium. If one wishes to know the combined p-polarized field strengths, this can be

found by simply taking E12 = E1 x

2 + E1 z2 . This equation also applies in the substrate and

intermediate layers, though this is the only time it will be mentioned.

SUBSTRATE

The electric field in the substrate consists only of a traveling wave moving away from the last

interface, since we are assuming that the substrate is a semi-infinite medium. Anywhere in the

substrate the electric field can be written as Em = Emt( )e

i k mt ⋅r−ω t( )

, where Emt( ) is the component of

the electric field transmitted across the last interface and r is a position vector, measured from

the last interface pointing into the substrate.

We now need to calculate Emt( ) . The H-field vector is continuous across all interfaces since it is

always tangential for p-polarized light. We can then define a transmittance coefficient, ˜ ′ t 1H as

H mt( ) = ˜ ′ t 1 H H1

0 t( ) , where H10 t( ) is the amplitude of the H-field in the incident beam at the first

interface.

26

We recall that

H10(t ) =

˜ ε 1µ1

1 / 2

ˆ S 10(t ) × E1

0(t )( ) =˜ ε 1µ1

1 / 2

E10(t ) +ˆ j ( ) .

Thus

H m(t ) = ˜ ′ t 1H

˜ ε 1µ1

1 / 2

E10(t )ˆ j .

Since

Em(t ) = −

µm

˜ ε m

1/2

ˆ S mt( ) × Hm

( t)( ) and

ˆ S mt( ) =− sinθm

ˆ i +cosθmˆ k , then

ˆ S mt( ) × Hm

(t )( ) = −˜ ′ t 1H˜ ε 1µ1

1/2

E10(t ) cos ˜ θ mˆ i +sin ˜ θ m ˆ k [ ] . Therefore

Em( t ) = −

µm˜ ε 1

˜ ε m µ1

1 / 2

˜ ′ t 1H E10(t ) cos ˜ θ mˆ i +sin ˜ θ m ˆ k [ ] ,

just across the boundary inside the substrate.

The field in the substrate a distance z from the last interface is

Em( t ) = −

µm˜ ε 1

˜ ε m µ1

1 / 2

˜ ′ t 1 H E10(t ) cos ˜ θ mˆ i +sin ˜ θ m ˆ k [ ]e i km

t ⋅r -ω t( ).

Breaking this field into its tangential component, Emx, and its normal component, Emz, we get

Emx =µm

˜ ε 1˜ ε m µ1

1/2

˜ ′ t 1 H E10(t ) cos ˜ θ me

i kmt ⋅r-ω t( )

Emz =µm

˜ ε 1˜ ε mµ1

1 / 2

˜ ′ t 1 H E10(t ) sin ˜ θ me

i kmt ⋅r -ω t( )

.

Now

27

µm˜ ε 1

˜ ε m µ1

1 / 2

˜ ′ t 1 H = ˜ ′ t 1E ,

due to the relations between the E- and H-fields for p-polarized light. So we get

Emx = ˜ ′ t 1E E10(t ) cos ˜ θ me

i k mt ⋅r -ω t( )

and

Emz = ˜ ′ t 1E E10( t) sin ˜ θ m e

i k mt ⋅r- ω t( )

.

We wish to calculate real measurable quantities, so we want to find time averages, i.e.

Emx2 =

1

2Emx ⋅ Emx

∗( ) and Emz2 =

1

2Emz ⋅ Emz

∗( ) .

Emz ⋅ Emz∗ = ˜ ′ t 1E E1

0( t ) cos ˜ θ mei k m

t ⋅r -ω t( ){ }* ˜ ′ t 1 E( )*E1

0( t ) cos ˜ θ m( )∗e

i kmt ⋅r-ω t( )[ ]*

= ˜ ′ t 1 E

2E1

0( t )( )2cos ˜ θ m

2

ei k m

t ⋅ r-ω t( ) + i kmt ⋅ r-ω t( )[ ]* .

The argument of the exponential can be analyzed with a method identical to that which we used

to simplify the argument of the exponential in the s-polarized substrate equation, where we

derived the result:

i km ⋅ r( ) + i km ⋅ r( )[ ]*= −

4πλo

ℑm˜ ξ m( )z .

Thus:

Emx2 =

1

2˜ ′ t 1E

2cos ˜ θ m

2

E10( t)( )2

e− 4π

λo

ℑm ˜ n m cos ˜ θ m( ) z

(3.10)

Emz2 =

1

2˜ ′ t 1E

2sin ˜ θ m

2

E10(t )( )2

e− 4π

λ o

ℑm ˜ n m cos ˜ θ m( ) z

. (3.11)

INTERMEDIATE PHASES

28

We now want to calculate the electric field at any depth in an intermediate layer.

We write the total electric field anywhere in the (j+1) medium as

E j +1 = Ej+(t )e

i k j +1t ⋅ r−ω t( ) + E

j +

( r)ei k j+1

r ⋅r −ω t( ).

We wish to make use of the fact that the H-field is continuous across the interface, so we need

to calculate Ej+( t) and E

j +

( r ) in terms of H.

Ej +

(t ) = −µ j+1

˜ ε j+1

1 / 2

ˆ S j +t( ) × H

j+(t )( ) and

Ej +

(r ) = −µ j +1

˜ ε j +1

1 / 2

ˆ S j +r( ) × H

j +

( r)( ).

ˆ S j +t( ) × H

j +

( t ) = ˆ i − cos ˜ θ j+1Hj +

( t )( ) − ˆ k sin ˜ θ j +1Hj +

( t)( ) and

ˆ S j +r( ) × H

j +

(r ) = ˆ i cos ˜ θ j+1Hj +

( r)( ) − ˆ k sin ˜ θ j +1Hj+(r )( ) , so

Ej+( t) =

µ j+1

˜ ε j +1

1 / 2

cos ˜ θ j+1ˆ i +sin ˜ θ j+1

ˆ k [ ]H j+( t) and

Ej +

(r ) =µ j +1

˜ ε j +1

1/2

− cos ˜ θ j +1ˆ i +sin ˜ θ j +1

ˆ k [ ]H j +

( r) .

Thus

E j+1 =µ j +1

˜ ε j +1

1 / 2

Hj +

(t )ei k j +1

t ⋅r −ω t( ) − Hj +

( r )ei k j+1

r ⋅r −ω t( )

cos ˜ θ j +1

ˆ i

+µ j +1

˜ ε j +1

1 / 2

Hj +

( t )ei k j +1

t ⋅r −ω t( ) + Hj+(r )e

i k j+1r ⋅r −ω t( )

sin ˜ θ j+1

ˆ k .

Note that Ej+1 has a tangential (x) component and a normal (z) component:

29

E j+1( ) x =µ j+1

˜ ε j+1

1 / 2

Hj +

( t )ei k j +1

t ⋅r −ω t( ) − Hj+(r )e

i k j+1r ⋅r −ω t( )

cos ˜ θ j+1 (3.12)

E j+1( ) z =µ j +1

˜ ε j +1

1/2

Hj+(t )e

i k j +1t ⋅r −ω t( ) + H

j +

( r)ei k j+1

r ⋅r−ω t( )

sin ˜ θ j +1. (3.13)

Just inside the jth interface,

E j+1( ) x = Ej +x

(t ) + Ej+ x

( r) and E j+1( )z = Ej +z

( t ) + Ej +z

( r ) .

Let us define the sum of E-fields from incoming and outgoing light on one side of an interface

as

Ej +x

≡ Ej+ x

( t) + Ej +x

(r ) and Ej − x

≡ Ej − x

( t ) + Ej −x

(r ) .

The tangential component of the electric field is continuous across an interface, so Ej +x

= Ej −x

,

which can be related to equation 4.5 Single layer quarter wavelength AR coating: 72 nm layer HfO2

substrate with light incident at 30o and λo = 550 nm. RS = 1.4% andRP = 0.74%. An uncoated silicon substrate would have reflectances ofRS = 41.0% and RP = 30.6%

4.6 with

Ej +x

(new notation) = E j+1( ) x (old notation) and

Ej −x

(new notation) = E jx (old notation) to yield

µ j

˜ ε j

1 / 2

Hj −(t ) − H

j −(r )[ ]cos ˜ θ j =

µ j +1

˜ ε j +1

1/2

Hj +( t) − H

j +(r )[ ]cos ˜ θ j +1 .

We rewrite with ˜ ξ j = ˜ n j cos ˜ θ j , ˜ ξ j+1 = ˜ n j+1 cos ˜ θ j +1 and ˜ n j2 = µ j

˜ ε j :

˜ ξ j˜ ε j

Hj −(t ) − H

j−( r)[ ] =

˜ ξ j +1

˜ ε j +1

Hj +(t ) − H

j+( r)[ ].

The magnetic field is all tangential, so

Hj +(t ) + H

j +( r) = H

j −(t ) + H

j −( r ) , and we can eliminate one of these four variables Let us use

Hj+(r ) = H

j −( t ) + H

j −(r ) − H

j +( t ) to get

30

˜ ξ j˜ ε j

Hj −(t ) − H

j−( r)[ ] =

˜ ξ j +1

˜ ε j +1

Hj +(t ) − H

j−(t ) + H

j −(r ) − H

j+(t )( )[ ], which we re-write as

˜ ξ j˜ ε j

+˜ ξ j +1

˜ ε j+1

H j−

( t ) +˜ ξ j +1

˜ ε j +1

−˜ ξ j˜ ε j

H j−

( r) =2 ˜ ξ j +1

˜ ε j +1

Hj +(t ) .

Now Hj−(r ) = ˜ ′ r j H j −

(t ) , where ˜ ′ r j is the total reflection coefficient due to all interfaces beyond the jth

interface, so

˜ ξ j˜ ε j

+˜ ξ j +1

˜ ε j+1

H j−

( t ) + ˜ ′ r j

˜ ξ j+1

˜ ε j+1

−˜ ξ j˜ ε j

H j −

( t ) =2 ˜ ξ j+1

˜ ε j+1

Hj +(t ) .

We multiply by ˜ ε j+1

2 ˜ ξ j+1

and collect terms to get

Hj +(t ) =

1

21 +

˜ ε j +1˜ ξ j

˜ ε j˜ ξ j+1

+ 1 −

˜ ε j +1˜ ξ j

˜ ε j˜ ξ j +1

′ r j

H j −

(t ) . (3.14)

We can similarly eliminate Hj+(t ) from the same equation to get

Hj +(r ) =

1

21−

˜ ε j +1˜ ξ j

˜ ε j˜ ξ j+1

+ 1 +

˜ ε j+1˜ ξ j

˜ ε j˜ ξ j +1

′ r j

H j −

(t ) . (3.15)

Tangential Component

These equations for Hj+(t ) and H

j+(r ) can now be substituted into equations 4.5 Single layer quarter wavelength AR coating: 72 nm layer HfO

substrate with light incident at 30o and λo = 550 nm. RS = 1.4% andRP = 0.74%. An uncoated silicon substrate would have reflectances ofRS = 41.0% and RP = 30.6%

4.6 and (3.13). First, the tangential component:

E j +1( ) x =µ j +1

˜ ε j +1

1/2

1

21 +

˜ ε j +1˜ ξ j

˜ ε j˜ ξ j +1

+ 1−

˜ ε j+1˜ ξ j

˜ ε j˜ ξ j+1

′ r j

e

i k j +1t ⋅r −ω t( )

31

−1

21 −

˜ ε j +1˜ ξ j

˜ ε j˜ ξ j +1

+ 1+

˜ ε j+1˜ ξ j

˜ ε j˜ ξ j+1

′ r j

e

i k j +1r ⋅ r− ω t( )

cos ˜ θ j +1H j −(t ) .

Let A =1 +˜ ε j+1

˜ ξ j˜ ε j

˜ ξ j +1

and B =1 −˜ ε j +1

˜ ξ j˜ ε j ˜ ξ j+1

:

E j+1( ) x =1

2

µ j +1

˜ ε j +1

1/2

cos ˜ θ j+1Hj −( t) A+ B˜ ′ r j( )e i kj+1

t ⋅r−ω t( ) − B + A˜ ′ r j( )e i k j +1r ⋅r−ω t( )

.

And as we had before,

k j +1t ⋅ r =

2π˜ n j+1

λo

−xsin ˜ θ j +1 + zcos ˜ θ j +1( ) and

k j+1r ⋅r =

2π ˜ n j +1

λo

−x sin ˜ θ j+1 − zcos ˜ θ j+1( ), so

E j +1( ) x =1

2

µ j +1

˜ ε j+1

1 / 2

cos ˜ θ j+1 Hj −( t )e

− 2πi ˜ n j +1

λo

x sin ˜ θ j +1 −iωt

A + B ˜ ′ r j( )e2πi ˜ n j +1

λo

z cos ˜ θ j +1

− B + A˜ ′ r j( )e−2 πi˜ n j+1

λo

z cos ˜ θ j +1

.

We can again ignore e

−2πi ˜ n j +1

λo

x sin ˜ θ j+1 − iωt

, for the same reasons given before.

Let ˜ ϕ j+1 =2π˜ n j+1

λo

zcos ˜ θ j +1, and use ei ˜ ϕ j+ 1 = cos ˜ ϕ j +1 + i sin ˜ ϕ j +1 to get

E j+1( ) x =1

2

µ j +1

˜ ε j +1

1/2

cos ˜ θ j+1Hj −( t) cos ˜ ϕ j+1 A + B˜ ′ r j( ) − B + A ˜ ′ r j( )[ ]{

+i sin ˜ ϕ j+1 A + B˜ ′ r j( ) + B + A˜ ′ r j( )[ ]} .

A + B = 2 and A - B=2 ˜ ε j +1

˜ ξ j˜ ε j

˜ ξ j +1

, so

32

E j +1( ) x =µ j +1

˜ ε j +1

1 / 2

cos ˜ θ j+1 Hj −( t)

˜ ξ j ˜ ε j +1

˜ ξ j +1˜ ε j

cos ˜ ϕ j +1 1 − ˜ ′ r j( ) + i sin ˜ ϕ j +1 1 + ˜ ′ r j( )

.

We now need to rewrite Hj−(t ) in terms of E

j −( t ).

Hj −( t) =

˜ ε jµ j

1/2

ˆ S (t) × Ej −(t )( ) =

˜ ε jµ j

1 / 2

Ej −( t )ˆ j . Thus

E j +1( ) x = cos ˜ θ j +1

µ j+1

µ j

1 / 2˜ ε j+1

˜ ε j

1 / 2 ˜ ξ j

˜ ξ j +1

cos ˜ ϕ j +1 1− ˜ ′ r j( )

+iµ j+1

µ j

1 / 2˜ ε j˜ ε j+1

1 / 2

sin ˜ ϕ j+1 1 + ˜ ′ r j( )

Ej −( t ) .

Now ˜ n j = µ j˜ ε j( )1/2

and ˜ ξ j = ˜ n j cos ˜ θ j , so

E( j +1) x = cos ˜ θ j cos2π˜ n j +1 cos ˜ θ j +1z

λo

1− ˜ ′ r j( ) + i

˜ n j cos ˜ θ j+1µ j +1

˜ n ( j+1)µ j

sin

2π ˜ n j +1 cos ˜ θ j+1z

λo

1 + ˜ ′ r j( )

Ej −

(t )

(3.16)

This result gives the p-polarized electric field all through any intermediate layer. The field

intensities for the tangential component are continuous across each interface.

Normal Component

To calculate the normal component of the p-polarized E-field, we start with equation (3.13):

E j +1( ) z =µ j +1

˜ ε j +1

1 / 2

Hj +(t )e

i k j +1t ⋅r −ω t( ) + H

j +( r )e

i k j+1r ⋅r −ω t( )

sin ˜ θ j +1 . (3.13)

We now use equations (3.14) and 4.1 with the result

Hj −( t ) =

˜ ε jµ j

1/2

Ej −( t )ˆ j .

33

E j +1( ) z =1

2

µ j +1

˜ ε j +1

1 / 2˜ ε jµ j

1 / 2

Ej −( t ) sin ˜ θ j +1 cos ˜ ϕ j +1 A + B˜ ′ r j( ) + B + A ˜ ′ r j( )[ ]{

+i sin ˜ ϕ j+1 A + B˜ ′ r j( ) − B + A˜ ′ r j( )[ ]} ,

with A =1 +˜ ε j +1

˜ ξ j˜ ε j

˜ ξ j+1

, B =1 −˜ ε j+1

˜ ξ j˜ ε j

˜ ξ j +1

, and ˜ ϕ j+1 =2π˜ n j+1

λo

zcos ˜ θ j +1.

Simplifying as before, the result is

E j+1( ) z =1

2

˜ ε jµ j+1

˜ ε j+1µ j

1/2

Ej−( t ) sin ˜ θ j +1 cos ˜ ϕ j+1 A+ B( ) 1+ ˜ ′ r j( )[ ]{ +i sin ˜ ϕ j+1 A − B( ) 1− ˜ ′ r j( )[ ]} .

Substituting A + B =2; A - B =2 ˜ ε j +1

˜ ξ j˜ ε j

˜ ξ j+1

and using ˜ n j = µ j˜ ε j( )1/2

and ˜ ξ j = ˜ n j cos ˜ θ j , we obtain

E j+1( )z = sin ˜ θ j +1

˜ n j µ j +1

˜ n j +1µ j

cos ˜ ϕ j +1 1 + ˜ ′ r j( ) +i

cos ˜ θ jcos ˜ θ j+1

sin ˜ ϕ j +1 1 − ˜ ′ r j( )

E j−

( t) .

Now ˜ n j+1 sin ˜ θ j+1 = n1 sinθ1 , which is real, so we substitute this into the expression and get

E j +1( ) z = n1 sinθ 1( )˜ n j µ j+1

˜ n j+1

2 µ j

cos

2π ˜ n j +1 cos ˜ θ j+1 z

λ o

1 + ˜ ′ r j( )+ i

cos ˜ θ j˜ n j +1 cos ˜ θ j+1

sin

2π ˜ n j +1 cos ˜ θ j +1z

λo

1− ˜ ′ r j( )

E j −

(t)

(3.17)

This result gives the p-polarized electric field through any intermediate layer. The field

intensities for the normal component are not continuous across each interface.

We need an expression for Ej −t( ) , as we did for s-polarized light. Starting with equation (3.14),

Hj +(t ) =

1

21 +

˜ ε j +1˜ ξ j

˜ ε j˜ ξ j+1

+ 1 −

˜ ε j +1˜ ξ j

˜ ε j˜ ξ j +1

′ r j

H j −

(t )

. (3.14)

We let Hj +

t( ) propagate to the (-) side of the (j+1) interface using Hj+ 1( )−t( ) = e

2 πi˜ n j+1 cos ˜ θ j +1

λ o

l j +1

Hj−t( ) ,

where lj+1 is the thickness of the (j+1) layer, to get

34

Hj+ 1( )−

(t ) =1

21+

˜ ε j +1˜ ξ j

˜ ε j ˜ ξ j+1

+ 1 −

˜ ε j+1˜ ξ j

˜ ε j ˜ ξ j +1

′ r j

e

2πi ˜ n j +1 cos ˜ θ j+ 1

λo

l j +1

Hj −( t) .

Substituting as above and with Hj −(t ) = ˜ n j E

j −( t ) and H

j+ 1( )−(t ) = ˜ n j +1E

j +1( ) −( t ) , we get

Ej +1( ) −

( t ) =1

2

˜ n j˜ n j +1

1 +

˜ n j +1 cos ˜ θ j˜ n j cos ˜ θ j+1

+ 1−

˜ n j +1 cos ˜ θ j˜ n j cos ˜ θ j + 1

′ r j

e

2πi ˜ n j +1 cos ˜ θ j+ 1

λo

l j +1

Ej −( t ) . (3.18)

EQUATION SUMMARY

We have now derived nine equations that allow one to calculate electric field strengths for s- or

p-polarized light propagating through a multilayer thin film structure. The table of equations on

the following page lists these equations in the order they were derived.

35

Equation Summary

S-POLARIZED

incident

E12 =

1

21+ R( ) + R1/2 cos δ r −

4πn1

λo

cosθ1z

E10( t )( )2

(3.1)

substrate

Em2 =

1

2˜ ′ t 1

2E1

0(t )( )2e

− 4πλo

ℑm ˜ n m cos ˜ θ m( ) z

(3.2)

intermediate4.1

(3.7)

P-POLARIZED

incident4. 4.1

E1z2 =

1

2sin 2 θ1 1+ R + 2R1/2 cos

4πn1

λo

cosθ1z − δ r

E10( t )( )2

(3.8)

substrate4.1 4.

Emz2 =

1

2˜ ′ t 1E

2sin ˜ θ m

2

E10(t )( )2

e− 4π

λ o

ℑm ˜ n m cos ˜ θ m( ) z

(3.11)

intermediate

E( j +1) x = cos ˜ θ j cos2π˜ n j +1 cos ˜ θ j +1z

λo

1− ˜ ′ r j( ) + i

˜ n j cos ˜ θ j+1µ j +1

˜ n ( j+1)µ j

sin

2π ˜ n j +1 cos ˜ θ j+1z

λo

1 + ˜ ′ r j( )

Ej −

(t )

(3.16)

E j+1( )z = n1 sin θ1( )˜ n j µ j+1

˜ n j+1

2 µ j

cos

2π ˜ n j+1 cos ˜ θ j +1 z

λ o

1 + ˜ ′ r j( ) + i

cos ˜ θ j˜ n j +1 cos ˜ θ j+1

sin

2π ˜ n j+1 cos ˜ θ j +1 z

λo

1 − ˜ ′ r j( )

E j −

(t)

(3.17)

36

Chapter IV

Numerical Analysis

In this chapter we will use the equations just derived to generate graphs illustrating electric field

strengths for several thin film structures. We will choose examples that illustrate several

common thin film design goals and strategies.

Two Layers on an Absorbing Substrate

For an introduction to the form and function of electric field plots, we first consider a thin film

structure consisting of two intermediate layers. For incident light of wavelength 550 nm, we

will let air be the incident medium, the first layer will be crystalline silicon ( ˜ n = 3.9822 - 0.0334

i) with a thickness of 400 nm, the second layer will be a 400 nm layer of Al2O3 (n = 1.6203),

and the substrate will be GaAs ( ˜ n = 4.047 - .324 i). The light is incident on the first interface at

an angle of 45o.

37

Figure 4.1: Diagram of a thin film

The graph of the electric field strengths for this example is given in Figure 4.2.

Crystalline Silicon

Al2O3

GaAs Substrate

Air

45o

400 nm

400 nm

38

0

1

2

3

Pz4Pz3Pz2

Pz

Px

S

Figure 4.2: Electric field strengths for a thin film consisting of 400 nm Si, 400 nm Al2O3 on aGaAs Substrate, with light incident at 30o and λo = 550 nm

To see the detail of the field strength plot in the intermediate layers and substrate, we enlarge the

vertical scale:

Depth (nm)

E 2

E10 t( )( )2

Air Silicon Al2O3 GaAs

39

0

0.1

0.2

0.3

0.4

Pz4Pz3Pz2

Pz

Px

S

Figure 4.3: Detail of Figure 4.2

To generate the data graphed above, the following sequence was followed:

1. Propagation angles for light at each layer were calculated using Snell’s law.

2. Fresnel’s reflection and transmission coefficients were calculated at each interface, for both

s- and p-polarized light.

3. Rouard’s expression was used to find a total reflectance and transmittance for the stack.

4. δ r was calculated for the s- and p-polarized cases.

5. The electric field equations were programmed and numerical values for the electric field

strengths were generated.

All of the above numerical calculations were performed using Mathematica. Data were then

exported to CA-Cricket Graph III to generate the graphs in this paper.

E 2

E10 t()( )2

Depth (nm)

Air Silicon Al2O3 GaAs

40

The field strengths are given as relative values, which is equivalent to graphing E2 with the

assumption that the incident intensity of the light E10 t( )( )2

is equal to one.

In the graph above, one can note that the field strengths of the s-polarized light and the x

component of the p-polarized light are both continuous at the interfaces, as Maxwell’s equations

require for tangential field components. The z component of the p-polarized light is obviously

discontinuous.

In all layers that precede the substrate, the electric field strength plot is oscillatory. This is not a

“stop action” picture of the light waves, but rather the time average of the electric field. The

plot is oscillatory because of interference between the reflected and transmitted portions of the

incident light. There is no oscillation in the substrate because there is no reflected field to

interfere with the transmitted field, so the time-averaged plot of the transmitted light merely

decays because of the absorptive properties of GaAs.

The field strength plot in the silicon layer shows an oscillation that is attenuated because silicon

is an absorbing material. Since ˜ n silicon ≠ ℜe ˜ n silicon( ) , the exponential terms in our equations “die

out” as z increases. The field also “dies out” in the substrate, but does not do so in the air and

Al2O3, which are transparent.

Single Layer AR Coating

41

One of the simplest common thin film structures is a quarter wavelength single layer anti-

reflective (AR) coating. AR coatings are commonly used on optical instruments where it is

important to increase the transmission of light. They work through the well-known mechanism

of destructive interference that occurs between light waves that reflect off the front surface of the

coating and those that reflect off its back surface. If the layer thickness is equal to a quarter

wavelength, the destructive interference is maximized. It is to be noted that the quarter

wavelength thickness is determined by the wavelength of the light in the material, which in this

example we label as λ1.

In Figure 4.4, we see the electric field strength plot for a single layer AR coating with the

following parameters: λo = 550 nm, incident medium is air (n = 1.00), the single intermediate

layer is HfO2 (n = 1.916), substrate is silicon ( ˜ n = 3.98 - .0334 i). The thickness of the

intermediate layer is 72 nm (λ1 / 4). The incident angle of the light is 0o. At normal incidence,

there is no z component of the p-polarized light. Furthermore, the x component of the p-

polarized light becomes identical to s-polarized light so the graph contains only a single line.

The reflectance for this design is quite low, R = 0.17%. In comparison, an uncoated silicon

substrate would have a reflectance of R = 36% at normal incidence.

42

0

0.25

0.5

0.75

1

1.25

Pz

Px

S

Figure 4.4: Single layer quarter wavelength AR coating: 72 nm layer HfO2 on Si substratewith light at normal incidence and λo = 550 nm, R = 0.17%. Uncoated siliconsubstrate: R = 36%.

We note that the electric field strength in the incident medium oscillates about a value of

approximately 1.0, which is the incident intensity. This makes sense, since the reflectance is

low enough that there should not be much of a reflected field to increase the field strength

values in the air. If the reflectance had been very high, we would expect the field to oscillate

about a value of approximately two, since that would be the value of the sum of the incident and

reflected fields.

E 2

E10 t( )( )2

Depth (nm)

Air HfO2 Silicon

43

Next, we use the same design, but make the incident angle 30o. These parameters yield total

reflectances of RS = 1.4% and RP = 0.74%. An uncoated silicon substrate would have

reflectances of RS = 41.0% and RP = 30.6% at this angle of incidence.

0

0.25

0.5

0.75

1

1.25

Pz3Pz2

Pz

Px

S

Figure 4.5: Single layer quarter wavelength AR coating: 72 nm layer HfO2 on Si substratewith light incident at 30o and λo = 550 nm. RS = 1.4% and RP = 0.74%. Anuncoated silicon substrate would have reflectances of RS = 41.0% and RP = 30.6%

The anti-reflective properties of the film become much less effective if the layer thickness is not

equal to a quarter wavelength. Figure 4.6 was created with parameters identical to those used to

produce Figure 4.5, except the layer thickness was set at 150 nm. This thickness is close to a

half wavelength, which is a very bad AR design. The reflectances for this design are RS =

41.0% and RP = 30.6%, which are nearly identical to an uncoated silicon substrate.

Air HfO2 Silicon

E 2

E10 t( )( )2

Depth (nm)

44

0

0.5

1

1.5

2

2.5

3

Pz3Pz2

Pz

Px

S

Figure 4.6: Imperfect single-layer AR coating: 150 nm layer HfO2 on Si substrate with lightincident at 30o and λo = 550 nm. . RS = 41.0% and RP = 30.6%.

In comparing this plot to Figure 4.5, we note the much higher reflected field in the incident

medium due to the imperfect layer thickness. Since the HfO2 layer is not a quarter-wave, the

multiple reflections are not out of phase and they add, generating a large reflected wave.

Dielectric HR Mirror

Another interesting example to consider is a high reflectance structure composed only of

dielectric materials. One common design that is used for this purpose is a thin film stack

Air HfO2 Silicon

E 2

E10 t( )( )2

Depth (nm)

45

composed of quarter-wavelength layers that alternate between high and low-index materials, as

illustrated in the following figure.

Figure 4.7: Schematic of a design for a dielectric HR mirror

Thin film structures with a large number of repeated high-low stacks can be made to maximize

reflectance for a given wavelength.

Figure 4.8 shows the electric field within an eight layer stack containing alternating layers of

TiO2 (n = 2.2303) and MgF2 (n = 1.3862) on a glass substrate for an incident angle of 30o and

light with a wavelength of 550 nm. Each layer is a quarter wavelength thick at 30o incidence (60

nm for TiO2 and 100 nm for MgF2), which yields reflectances of RS = 95.9% and RP = 90.4%,

for the given wavelength and angle of incidence.

IncidentMedium

HighInde

x

LowIndex

HighInde

x

LowInde

x

Repeated Pattern ... Substrate

λH/4 λL/4 λH/4 λL/4

46

0

1

2

3

4

Pz10Pz9Pz8Pz7Pz6Pz5Pz4Pz3Pz2

Pz

Px

S

Figure 4.8: A dielectric high reflectance structure: Eight alternating layers of quarterwavelength layers of TiO2 and MgF2 on glass substrate with light incident at 30o

and λo = 550 nm. RS = 95.9% and RP = 90.4%.

Here we see the s-polarized field with a maximum value near 4.0, so its average value is almost

2.0, which is the value we would obtain for a perfect mirror.

Critical Angle

We now consider the critical angle. As light passes from a higher index material into one that

has a lower index, the beam is refracted away from the normal, as predicted by Snell’s Law. As

the incident angle is increased, an angle is eventually reached where the beam is refracted so

much that the refracted angle becomes greater than ninety degrees. At this point, the beam is

Depth (nm)

E 2

E10 t( )( )2

Air TiO2 MgF2 TiO2 MgF2 TiO2 MgF2 TiO2 MgF2 Glass

47

“trapped” inside the higher index incident medium, and does not pass through the interface at

all, but instead exhibits total internal reflection. We consider a system where light travels from

glass (n = 1.5) into air with an incident wavelength of 550 nm (in air). First, Figure 4.9 shows a

graph of the reflectance of the light plotted vs. the incident angle. The reflectance goes to 1.0 at

the critical angle of 41.8o.

0

10

20

30

40

50

60

70

80

90

100

P

S

Figure 4.9: Reflectance vs. angle for light traveling from glass into air

Next, we graph the electric field strengths for light at an incident angle of 40o, which is just

lower than the critical angle.

Incident Angle (degrees)

Reflectance(%)

48

0

1

2

3

4

Pz2

Pz

Px

S

Figure 4.10: Electric field strengths for glass to air at 40o with λo = 550 nm

In the graph above, we note the constant electric field values in the substrate. The field strength

is constant because there is only a transmitted field, with no reflected field to interfere with it

and cause oscillations. There is also no absorption in the substrate.

In Figure 4.11, the parameters are the same, except the incident angle is 45o. This is greater

than the critical angle, so the light is totally reflected at the interface.

E 2

E10 t( )( )2

Depth (nm)

Glass Air

49

0

1

2

3

4

5

6

Pz2

Pz

Px

S

Figure 4.11: Electric field strengths for glass to air at 45 owith λo = 550 nm

The field plots in Figure 4.11 are quite interesting. The s-polarized field intensity in the

incident medium has maxima at exactly 4.0, which tells us that the average value of the field is

2.0. This corresponds to 100% reflection. Furthermore, the sum of the components of the p-

polarized light is also two, so the p-polarized light is 100% reflected as well.

In the substrate, we see that the field intensities “die off” exponentially past the interface, even

though air is a transparent medium. Mathematically, this occurs because the refracted angle is

imaginary, so the argument of the exponential term in the substrate equations is not zero. But in

this case the imaginary portion of the argument is not a consequence of the physical properties

of the substrate, but results from the geometric relationships that occur due to the refraction of

the light. To further elaborate, we have Snell’s law, Equation (2.2): ˜ n 1 sin ˜ θ 1 = ˜ n 2 sin ˜ θ 2 . In this

Glass Air

E 2

E10 t( )( )2

Depth (nm)

50

example, n1 = 1.5, θ1 = 45o, and n2 = 1. Substituting these values into Snell’s Law, we get

sin ˜ θ 2 = 2 ≅1.414 . The only way the value of the sine can be greater than one is for the

refracted angle to be imaginary, particularly ˜ θ 2 ≅ 1.57 − .881i( ). Thus we acquire the non-zero

argument for the exponential in the substrate equations.

It may seem odd that there is any field in the substrate medium at all, since the reflection at the

interface is 100%. But energy is conserved since no energy propagates away from the

interface. The energy in the substrate medium propagates parallel to the interface, and is thus

present as a sort of “phantom field.”

Finally, we note that the z component of the p-polarized electric field is very high on the

outgoing side of the interface, exceeding a value of 6.5 in our example. Thin film designers

must be aware of this when working with designs intended for high-intensity applications, such

as high energy lasers. Problems can occur at the interface in a number of ways. For instance, if

there is a particle impurity in the film near the interface, such as a dust mote, the high fields at

the critical angle can vaporize the particle in a “flash” of energy, possibly damaging the film.

Damage can also be magnified if the film contains cracks or scratches, since electric fields are

further increased by “edge effects.”

Plasmon Resonance Angle

In this section we will consider a system where the incident medium is quartz, the intermediate

medium is a 50 nm layer of silver ( ˜ n = 3.7681 - .04115 i at λo = 550 nm), with an air substrate.

Figure 4.12 is a plot of the reflectance of this structure vs. the incident angle of the light.

51

0

25

50

75

100

P

S

Figure 4.12: Reflectance vs. incident angle for light traveling from quartz through a 50 nmlayer of silver into air with λo = 550 nm

The sharp “dip” in the p-polarized reflectance occurs at the plasmon resonance angle. This

occurs because at this particular angle the free electrons in the silver at the interface experience

oscillations that are greatly augmented by the geometrical relationship that exists between them

and the p-polarized electric field vectors. The electrons oscillate and absorb the light, thus

decreasing the reflectance of the thin film. These free electrons are referred to as a surface

“plasmon” since they are part of a crystal lattice but still have some freedom of motion and are

thus similar to a plasma. The s-polarized light is unable to excite these exaggerated oscillations

because its field vector is tangential to the interface and it cannot “couple” to the plasmon field.

First, we plot the field intensities for the above system at an incident angle of 40o:

Reflectance(%)

Incident Angle(degrees)

52

0

1

2

3

4

Pz3Pz2

Pz

Px

S

Figure 4.13: Electric field strengths for light traveling from quartz through a 50 nm layer ofsilver into air at 40o incident angle with λo = 550 nm

The above plot is consistent with a fairly ordinary high reflectance structure, with absorptive

attenuation in the silver layer and relatively low transmitted intensities in the air substrate.

We now examine the electric field distribution at the surface plasmon resonance. At the

plasmon resonance angle (θ1 ≅ 45.3o), we see much greater penetration of the field for the p-

polarized light.

Quartz Silver Air

E 2

E10 t( )( )2

Depth (nm)

53

0

5

10

15

20

25

Pz2

Pz

Px

S

Figure 4.14: Electric field strengths for light traveling from quartz through a 50 nm layer ofsilver into air at 45.3o incident angle with λo = 550 nm

In Figure 4.14, we note that the s-polarized light is highly reflected, but the p-polarized light is

transmitted into the silver quite efficiently. The plot of the z component of the p-polarized light

is not visible in the substrate because the relative value of the field strength exceeds 300 just

across the interface. At the plasmon resonance angle, the geometrical relationships between the

p-polarized electric fields and the free electrons in the silver are such that there is a very efficient

coupling between the two, because of energy and momentum conservation for the p-polarized

light. The resonant oscillations of the free electrons cause the very strong fields that occur at

the interface, which a thin film designer would need to be aware of when working with designs

of this sort.

Quartz Silver Air

E 2

E10 t( )( )2

Depth (nm)

54

Transmittance Properties of Copper

For our final example, we briefly observe how the transmittance properties of copper vary with

wavelength. We will compare two electric field plots for light incident on a 50 nm layer of

copper on a glass substrate, one where the incident wavelength is 500 nm, and one with an

incident wavelength of 1000 nm.

When λo = 500 nm, the optical constant for copper is ˜ n = 1.1411 - 2.5718 i. In this

configuration, TS = 2.4% and TP = 3.1%.

0

1

2

3

4

Pz3Pz2

Pz

Px

S

Figure 4.15: Electric field strengths for a 50 nm layer of copper on glass at 30o with λo =500 nm

The reflectivity of the above structure is RS = 62% and RP = 53%. We note that R + T ≠ 100%

since copper is an absorbing medium.

Air Copper Glass

Depth (nm)

E 2

E10 t( )( )2

55

For Figure 4.16, when the incident wavelength of the light is 1000 nm and ˜ n = 0.3616 - 6.3181

i, we have TS = .51%, TP = .74%, RS = 96% and RP = 94%.

0

1

2

3

4

Pz3Pz2

Pz

Px

S

Figure 4.16: Electric field strengths for a 50 nm layer of copper on glass at 30o with λo =1000 nm

The plot shows that the field in the incident medium is much higher than in the previous graph,

indicating a reflectivity that is nearing 100%. We have found that the reflection and

transmission characteristics for a structure can be very dissimilar for light of different incident

wavelengths.

Air Copper Glass

E 2

E10 t( )( )2

Depth (nm)

56

Chapter V

Summary

In this paper we have derived equations that calculate electric field strengths in a multilayer thin

film structure. Starting with several important equations that describe the electric field, we made

use of several physical principles to motivate the derivation, especially reflection and

transmission coefficients at interfaces, and the continuity of tangential field components. The

equations were simplified using basic rules of algebra and vector calculus, and an effort was

made to make the notation logical and consistent throughout the paper. The equations are

recursive, and can be used for any arbitrary number of intermediate layers. They are in a form

that is well-suited to numerical analysis with a computer.

After the equations were derived, they were used to calculate the fields for a variety of thin film

designs, showing the behavior of the electric field in several interesting situations. The

oscillatory nature of the time-averaged electric field strengths in thin films that have a reflected

field was evident in our graphs. We investigated an AR and a HR design. We observed the

interesting field penetration effects that occur near the plasmon resonance angle for a thin metal

layer. We saw the field that exists beyond an interface that is exhibiting total internal reflection

for light incident at angles greater than the critical angle. The transmittance properties of two

wavelengths of light passing through a thin copper layer were compared.

The numerical analysis in this paper was a mere “dip” into the enormous field of thin film

optical design applications. The examples included in this paper were chosen to exemplify a

few simple examples that make use of the derived equations, but do not represent a thorough

57

study of thin film optics. While the main work of this paper was to actually derive the

equations “from scratch,” the equations represent a useful tool to further investigate deeper and

more interesting questions of thin film design.

58

References

1. Jackson, John David, Classical Electrodynamics, Second Edition; John Wiley and Sons:

New York, 1975, p. 271.

2. Klein, Miles V., Optics; John Wiley and Sons: New York, 1970, p. 534

3. Anders, Hugo, Thin Films in Optics; The Focal Press: London, 1965, p. 15

4. Ref. 1, p. 280

5. Macleod, H.A., Thin-Film Optical Filters; American Elsevier: New York, 1969, p. 14

6. Ref. 5, p. 10

7. Ref. 2, p. 568

8. Ref. 3, p. 19

9. Born, Max and Wolf, Emil, Principles of Optics; Third Edition; Pergamon Press: Oxford,

1965, p. 40.

10. Ref. 9, p. 40

11. Ref. 3,p. 59

12. Hansen, Wilford N., Opt. Soc. Am. 58; 381 (1968).