Light in Planar Stratified Media - part 1

ECE 5368/6358                                            han q le - copyrighted
Use solely for students registered for UH ECE 6358/5368 during courses - DO NOT distributed (copyrighted materials)

Part 1

1. Introduction

1.1 Reflection and refraction

You are a scientist studying light behavior at the planar interface of two homogenous dielectric media. 

Below represent experiments you can do:
- you can select polarization: TE or TM
- you can vary the medium properties: refractive index (Subscript[n, 1]) for the top medium and (Subscript[n, 2])+i (Subscript[κ, 2]) for the bottom medium. ((Subscript[κ, 2]) is related to the medium absorption or gain coefficient, depending on being positive or negative, respectively)
- you can vary the angle of incident

In return you can observe the following: (which can be said a whole lot more than could 17th-19th century physicists, such as Snell, Huygens, Newton, Descartes, Brewster,.. who investigated this phenomenon without being aware of Maxwell EM-wave theory of light):
- seeing animated EM waves of various E and H components of both polarizations.
- see vectorial representation of E and H of both polarizations.
- Observation of the above as a function of parameters that you can vary.

Your assigment:
You will write a scientific report of what you discover. It should include at the minimum the following:
1- Your experiments (you can do many experiments)
2- For each experiment, describe the experimental result.
3- Your theory to explain the result. If you don’t have a theory and can’t explain, it’s OK - that’s what we will learn in class.
4- Most important: What you think we should discuss in class in order to explain everything this App shows.

1.2 General planar stratified media

Why should we bother to study reflection and refraction? we know what it is, why does it matter if we understand or don’t understand?
We are studying more than just simply reflection and refraction.

As it turns out, optical technology involves problems like the above, except that the number of layer media is more than just 2. This is what we call more generally “planar stratified media”. Sometimes, it can be 100’s of layer. As an example:

Or

Another example:

->  ->

All are important optical technologies around us and enable modern economy. As diverse as the various technologies are shown above, they share a common element that involves planar stratified media. We will learn how to solve the problem in this chapter with the transfer matrix technique.

2. Reflection and transmission (refraction)

2.1 Formulating the problem in wave optics

Formulate general wave solutions in each medium: there will be unknown quantities (or coefficients)

Match the solutions at the boundary(ies): Obtain boundary condition equations that involve the unknown quantities

Solve for the unknown quantities.

Substitute the unknown quantities to obtain specific and unique solutions.

2.2 Formulate general wave solutions in each medium

Let’s denote the incident medium (source of original light) semi-infinite medium 1 with refractive index (Subscript[n, 1]); its dielectric constant is          (Subscript[ϵ, 1])=((Subscript[n, 1])^2)                     (2.2.1a)
Let semi-infinite medium 2 have refractive index:
                                                    (Subscript[n, 2])=(Subscript[n, 2;R])+i (Subscript[κ, 2]);               (2.2.1b)
its dielectric constant is:                (Subscript[ϵ, 2])=((Subscript[n, 2])^2)=((((Subscript[n, 2;R])+i (Subscript[κ, 2])))^2)     (2.2.1c)
(If confused, you can ignore (Subscript[κ, 2]) term for now.)
For both, we assume the magnetic permeability μ=1. (We will carry both as (Subscript[μ, 1]), (Subscript[μ, 2]) just to show the correct formulation).

2.2.1 Polarization and geometry

But first, we have to decide: what polarization?
By geometry, we see that there are two principal (basis) polarizations: Transverse electric (TE) and transverse magnetic (TM). A general polarization is simply a linear combination of the two basis polarizations. Hence, it is advantageous to solve for the principal polarizations, and any general polarization can be solved with a linear combination of the two.

Let’s start with TE.

2.2.2 General wave solutions in incident medium

What we know (or assume as given): an single-frequency incident plane wave, with TE polarization. We can write the incident wave as:
                                     (Subscript[E, inc])=   (OverHat[y]) (Subscript[E, inc]) (e^(i  ((Subscript[k, 1]) x + β z -ω t )))                            (2.2.2a)
where:                            ((Subscript[k, 1])^2)+(β^2) =(Subscript[ϵ, 1])(Subscript[μ, 1])((Subscript[k, 0])^2)=((Subscript[n, 1])^2)((Subscript[k, 0])^2)     ;     (Subscript[k, 0])≡((2 π)/λ)              (2.2.2b) 
Note:                              (Subscript[k, 1])=(Subscript[k, 0])(Subscript[n, 1]) Cos[θ]  ;   β=(Subscript[k, 0]) (Subscript[n, 1])Sin[θ]                              (2.2.2c)
For simplicity, we will choose (Subscript[E, inc]) to be equal to 1 (unit amplitude). We control the angle θ.

Source code Reflection graphics

Manipulate[
If[RUN== 1,
Manipulate[ θr=θ*Pi/180. ; k0=0.707;
β=k0 n1 Sin[θr] ; k1=k0 n1 Cos[θr] ;kp=k0 n1{-Sin[θr],Cos[θr]};
kr=k0 n1{Sin[θr],Cos[θr]};βV={β,0};k1V={0,-k1} ;
Graphics[{Hue[0.4,1,1,0.3],Rectangle[{-1,-0.5},{1,0}]
,Black, Opacity[1],Arrow[{{-1,0},{1,0}}],Arrow[{{0,1},{0,-0.5}}]
,Inset["x",{0.1,-0.5},BaseStyle->{20,Black,Bold,FontFamily-> "TimesNewRoman"}] 
,Inset["z",{1,0.1},BaseStyle->{20,Black,Bold,FontFamily-> "TimesNewRoman"}]
,{Hue[0.16,1,1,1],EdgeForm[Black],Disk[{0,0},0.25,{Pi/2,θr+Pi/2}]}
, Opacity[0.8],Purple,Thickness[0.01],Arrowheads[0.06],Arrow[{kp,{0,0}}]
,Blue,Arrow[{kp,kp+βV}],Red,Arrow[{{-0.02,k1},{-0.02,0}}]
,Inset["(Subscript[k, inc])=(Subscript[n, 1]) (Subscript[k, 0])",0.5kp+{-0.35,0.1},BaseStyle->{20,Black,Bold,FontFamily-> "TimesNewRoman"}] 
,Inset["β",0.5 kp+ {0,0.5 k1+0.2},BaseStyle->{20,Black,Bold,FontFamily-> "TimesNewRoman"}]
,Inset["(Subscript[k, 1])", {-0.1,0.85 k1},BaseStyle->{20,Black,Bold,FontFamily-> "TimesNewRoman"}]
,Inset["(Subscript[θ, inc])", 0.35{-Sin[θr/2],Cos[θr/2]},BaseStyle->{20,Black,Bold,FontFamily-> "TimesNewRoman"}]
,If[REFL==1, {Blue,Arrow[{{0,k1},{0,k1}+βV}]
,Hue[0.35,1,0.5,0.7], Arrow[{{0.02,0},{0.02,k1}}]
, Hue[0.1,1,0.4,0.8],Arrow[{{0,0},kr}]
,{Hue[0.8,0.6,1,0.6],EdgeForm[Black],Disk[{0,0},0.25,{Pi/2-θr,Pi/2}]}
,Inset["(Subscript[k, refl])",0.5kr+{0.2,0},BaseStyle->{20,Black,Bold,FontFamily-> "TimesNewRoman"}] 
,Inset["β",0.5 kr+ {0,0.5 k1+0.2},BaseStyle->{20,Black,Bold,FontFamily-> "TimesNewRoman"}]
,Inset["-(Subscript[k, 1])", {0.1,0.85 k1},BaseStyle->{20,Black,Bold,FontFamily-> "TimesNewRoman"}]
,Inset["(Subscript[θ, refl])", 0.35{Sin[θr/2],Cos[θr/2]},BaseStyle->{20,Black,Bold,FontFamily-> "TimesNewRoman"}]},Point[{0,0}]]} ]
,{{REFL,0},{0,1}},{{θ,45},0,90},{n1,1,1.5}]
,TextCell["Not running"
,Black,20,Italic,FontFamily->"Calibri",Background->GrayLevel[0.5]]]
,Row[{Button["Toggle box to start, and turn off when not running",BaseStyle->{16,White, Bold,FontFamily-> "Calibri"}
,Background->Black],Control[{{RUN,0,"->"},{0,1}}]
}]
]

Demo: Reflection graphics

What we cannot assume: knowledge of reflection wave and refraction wave. How do we know they exist and what forms they have?
However, we will assume that:
Condition a: Since we want to solve for ONLY one incident monochromatic wave, we assume that there is NO other optical waves of any other frequency besides ω that propagates in the direction from top down at any other angle besides θ.
Condition b: At the interface, which we choose to set as x-origin, x=0, the incident E field is:
                          (Subscript[E, inc])[x=0]=   (OverHat[y]) (Subscript[E, inc]) (e^(i  (β z -ω t )))                              (2.2.3)
Can there be other E-field of this form:      (e^(i (β' z-ω t))) where β'≠ β ? Or even:
                              ∫a[β'] (e^(i (β' z-ω t)))dβ'  ?
The answer is no, because if there is other β’, it must come from some other fields, while we assume to solve for ONLY one field.  In a more elaborate proof, we can prove that   ∫a[β'] (e^(i (β' z-ω t)))dβ'  cannot exist except for component (e^(i  (β z -ω t ))). Thus, on the incident side at the interface x=0, there can only be E field of the form (e^(i  (β z -ω t ))), but the total amplitude (not just (Subscript[E, inc])) remains to be unknown for now.

Condition c:  one thing we can assume from basic physics: A passive linear medium  (no internal energy or power to create photons) cannot generate a new frequency, hence, there is no other frequency besides ω

Thus, what are all other possible waves in medium 1 within the given conditions a, b and c above, besides the incident?
- the wave can have:             (e^(-i ω t))
- then, it can be:
                           (Subscript[(OverHat[ϵ]), r]) r (e^(i ((Subscript[k, r;1]) x +(Subscript[k, r;2]) y +β z-ω t)))                              (2.2.4a)
where:  (Subscript[k, r])=(({{(Subscript[k, r;1]), (Subscript[k, r;2]), β}}))   with |(Subscript[k, r])|=(Sqrt[(Subscript[ϵ, 1])])(Subscript[k, 0])=(Subscript[n, 1])(Subscript[k, 0])           (2.2.4b)
- but (Subscript[k, r;1]) cannot be >0, because it would mean another beam coming down from top, hence, (Subscript[k, r;1])<0.

But wait, can (e^(i (Subscript[k, r;2]) y )) exist if (Subscript[k, r;2])≠ 0? 
The problem is y-independent! If we translate y continuously, the problem has to be the same, in other words:
                                            (e^(i (Subscript[k, r;2]) y ))=(e^(i (Subscript[k, r;2]) (y+T) ))  
 or any arbitrary value of T. The only way this is true is (Subscript[k, r;2])=0. Thus, the general wave in (2.2.4a) is:
                                (Subscript[(OverHat[ϵ]), r]) r (e^(i ((Subscript[k, r;1]) x +β z-ω t)))                              (2.2.5a) 
                                ((Subscript[k, r;1])^2)+(β^2) =(Subscript[ϵ, 1])(Subscript[μ, 1])((Subscript[k, 0])^2)                          (2.2.5b)

But we also see that    ((Subscript[k, 1])^2)+(β^2) =(Subscript[ϵ, 1])(Subscript[μ, 1])((Subscript[k, 0])^2)=((Subscript[n, 1])^2)((Subscript[k, 0])^2)  from (2.2.2c)
Hence,                      ((Subscript[k, r;1])^2)=((Subscript[k, 1])^2)
But they cannot have the same sign:  (Subscript[k, r;1])=-(Subscript[k, 1])                    (2.2.6)

Thus, the only other wave allowed in medium 1 is:
             (Subscript[E, refl])=(  (OverHat[x]) (Subscript[r, 1]) + (OverHat[y]) (Subscript[r, 2]) +  (OverHat[z]) (Subscript[r, 3]) ) (e^(i  (-(Subscript[k, 1]) x + β z -ω t )))                                (2.2.7)

The angle of reflection must be equal and opposite to the angle of incident. (Subscript[θ, refl])=(Subscript[θ, inc])

We still know nothing about the  r coefficients.

However, applying:
          ∇. (Subscript[E, refl]) =0  ⇒ (Subscript[k, r]) . (Subscript[E, refl])=(({{-(Subscript[k, 1]), 0, β}}))  . (({{(Subscript[r, 1])}, {(Subscript[r, 2])}, {(Subscript[r, 3])}}))= -(Subscript[k, 1]) (Subscript[r, 1])+β (Subscript[r, 3]) =0   (2.2.8)
it shows that there are only two independent terms of the r coefficients that we must solve.

2.2.3 General wave solutions in transmission medium

In medium 2, we will assume that the waves can only move from top down, not bottom up, because if it does then, it becomes an incident wave from medium 2 to medium 1, which is a different problem, not the one we are solving.

- the wave can have:             (e^(-i ω t))
- then, it can be:
           (Subscript[(OverHat[ϵ]), t]) t (e^(i ((Subscript[k, t]) . x-ω t))) where:  (Subscript[k, t])=(({{(Subscript[k, t;1]), (Subscript[k, t;2]), (Subscript[β, t])}}))   with |(Subscript[k, t])|=(Subscript[n, 2])(Subscript[k, 0])           (2.2.8)
and (Subscript[k, t;1])>0.

But the same argument about translation in y applies here, hence:
                 (Subscript[(OverHat[ϵ]), t]) t (e^(i ((Subscript[k, t]) . x-ω t))) where:  (Subscript[k, t])=(({{(Subscript[k, t;1]), 0, (Subscript[β, t])}}))   with |(Subscript[k, t])|=(Subscript[n, 2])(Subscript[k, 0])           (2.2.8a)
and:                  ((Subscript[k, t;1])^2)+((Subscript[β, t])^2)=(Subscript[ϵ, 2])(Subscript[μ, 2]) ((Subscript[k, 0])^2)=((Subscript[n, 2])^2)((Subscript[k, 0])^2)                                           (2.2.8b)
But how do we know there is only one (Subscript[β, t])? Thus, the general wave of the second medium should be:
               (Subscript[E, 2])=∫((OverHat[x])(Subscript[t, 1])[(Subscript[β, t])]+(OverHat[y])(Subscript[t, 2])[(Subscript[β, t])]+(OverHat[z])(Subscript[t, 3])[(Subscript[β, t])]) (e^(i ((Subscript[k, t;1]) x+(Subscript[β, t]) z-ω t)))d(Subscript[β, t])                 (2.2.9)

2.3 Boundary conditions: E field

Formulate general wave solutions in each medium: there will be unknown quantities (or coefficients)

Match the solutions at the boundary(ies): Obtain boundary condition equations that involve the unknown quantities

Solve for the unknown quantities.

Per Maxwell’s equations, at the boundary:
                                                    (Subscript[(OverVector[E]), //;1])=(Subscript[(OverVector[E]), //;2])  ;                           (2.3.1a)
                                                   (Subscript[(OverVector[H]), //;1])=(Subscript[(OverVector[H]), //;2])  ;                            (2.3.1b)
                                                (Subscript[ϵ, 1]) (Subscript[E, ⊥;1])=(Subscript[ϵ, 2])(Subscript[E, ⊥;2])  ;                          (2.3.1c)        
                                                (Subscript[μ, 1])(Subscript[H, ⊥;1])=(Subscript[μ, 2])(Subscript[H, ⊥;2])  ;                       (2.3.1d)

We will see that not all conditions are independent, depending on the symmetry of a problem. Here, we will first consider E field condition, then H field condition for the next subsection.

Review: Prove these boundary conditions

2.3.1 Parallel E field condition

First, let’s consider E field. At the boundary x=0:
          (Subscript[(Subscript[E, 1])|, x=0])= (OverHat[y])   (e^(i  (β z -ω t )))  +  ((OverHat[x])   (Subscript[r, 1])  + (OverHat[y]) (Subscript[r, 2])  +(OverHat[z])  (Subscript[r, 3]) )(e^(i  (β z -ω t )))              (2.3.2) 
          (Subscript[(Subscript[E, 2])|, x=0])= ∫((OverHat[x])(Subscript[t, 1])[(Subscript[β, t])]+(OverHat[y])(Subscript[t, 2])[(Subscript[β, t])]+(OverHat[z])(Subscript[t, 3])[(Subscript[β, t])]) (e^(i ((Subscript[β, t]) z-ω t)))d(Subscript[β, t])                          (2.3.3)           
Hence:                             (Subscript[(Subscript[(Subscript[E, //;1])|, x=0])=(Subscript[E, //;2])|, x=0])                                       (2.3.1a)
gives:          (OverHat[y])  (1+(Subscript[r, 2])) (e^(i  (β z -ω t )))   = (OverHat[y])∫(Subscript[t, 2])[(Subscript[β, t])] (e^(i ((Subscript[β, t]) z-ω t)))d(Subscript[β, t])        (2.3.4a)
                    (OverHat[z])  (Subscript[r, 3]) (e^(i  (β z -ω t )))            = (OverHat[z])∫(Subscript[t, 3])[(Subscript[β, t])] (e^(i ((Subscript[β, t]) z-ω t)))d(Subscript[β, t])        (2.3.4b)
Or:                  (1+(Subscript[r, 2])) (e^(i  β z ))   =∫(Subscript[t, 2])[(Subscript[β, t])] (e^(i (Subscript[β, t]) z))d(Subscript[β, t])        (2.3.5a)
                               (Subscript[r, 3]) (e^(i  β z ))   =∫(Subscript[t, 3])[(Subscript[β, t])] (e^(i (Subscript[β, t]) z))d(Subscript[β, t])        (2.3.5b)
Or:                (1/(2 π))(1+(Subscript[r, 2]))∫dz (e^(i  (β-(Subscript[β, t]))z ))=(Subscript[t, 2])[(Subscript[β, t])]           (2.3.6a)
                            (1/(2 π))(Subscript[r, 3])∫dz (e^(i  (β-(Subscript[β, t]))z ))  =(Subscript[t, 3])[(Subscript[β, t])]          (2.3.6b)
Or:                          (1+(Subscript[r, 2]))δ[β-(Subscript[β, t])]   =(Subscript[t, 2])[(Subscript[β, t])]              (2.3.7a)
                             (Subscript[r, 3])δ[β-(Subscript[β, t])]                 =(Subscript[t, 3])[(Subscript[β, t])]             (2.3.7b)

This means that there the only valid (Subscript[β, t]) is:
                                 (Subscript[β, t])=β                                      (2.3.8)

Snell law

Prove Snell’s law from (2.3.8) above.

Thus, we drop (Subscript[β, t]) as a continuous variable and we obtain 2 equations:
                         (1+(Subscript[r, 2]))  =(Subscript[t, 2])             (2.3.9a)
                                   (Subscript[r, 3])= (Subscript[t, 3])            (2.3.9b)

2.3.2 Wave in medium 2 and normal E field condition

Now, we can rewrite wave in medium 2 in simpler form:
               (Subscript[E, 2])=((OverHat[x])(Subscript[t, 1])+(OverHat[y])(Subscript[t, 2])+(OverHat[z])(Subscript[t, 3])) (e^(i ((Subscript[k, 2]) x+β z-ω t)))                  (2.3.10a)
where:               (Subscript[k, 2])= (Sqrt[(Subscript[ϵ, 2]) (Subscript[μ, 2]) ((Subscript[k, 0])^2) -(β^2) ])                           (2.3.10b)

We now apply the normal E field condition:
                              (Subscript[ϵ, 1]) (Subscript[E, ⊥;1])=(Subscript[ϵ, 2])(Subscript[E, ⊥;2])                              (2.3.1c)
                               (Subscript[ϵ, 1])(Subscript[r, 1]) =(Subscript[ϵ, 2]) (Subscript[t, 1])                                    (2.3.11)
Substitute (2.3.11), (2.3.9a, b) in (2.3.10a):
               (Subscript[E, 2])=((OverHat[x])((Subscript[ϵ, 1])/(Subscript[ϵ, 2])) (Subscript[r, 1])+(OverHat[y])(1+(Subscript[r, 2]))+(OverHat[z])(Subscript[r, 3])) (e^(i ((Subscript[k, 2]) x+β z-ω t)))       (2.3.12)
Then:
    ∇.(Subscript[E, 2]) =0  ⇒ (Subscript[k, 2]) . (Subscript[E, 2])=(({{(Subscript[k, 2]), 0, β}}))  . (({{((Subscript[ϵ, 1])/(Subscript[ϵ, 2])) (Subscript[r, 1])}, {1+ (Subscript[r, 2])}, {(Subscript[r, 3])}}))=((Subscript[ϵ, 1])/(Subscript[ϵ, 2])) (Subscript[k, 2]) (Subscript[r, 1])+ β (Subscript[r, 3]) =0   (2.3.13)

2.3.3 Vanishing r and t coefficients

From (2.2.8) and (2.3.13):
                                                 (({{-(Subscript[k, 1]), β}, {((Subscript[ϵ, 1])/(Subscript[ϵ, 2]))(Subscript[k, 2]), β}})) . (({{ (Subscript[r, 1])}, {(Subscript[r, 3])}})) =(({{0}, {0}}))                (2.3.14a)
Since                                  |Det[(({{-(Subscript[k, 1]), β}, {((Subscript[ϵ, 1])/(Subscript[ϵ, 2]))(Subscript[k, 2]), β}}))]|=β ((Subscript[k, 1])+((Subscript[ϵ, 1])/(Subscript[ϵ, 2]))(Subscript[k, 2])) ≠0       (2.3.14b)
(except when β=0, or normal incident), hence:
                              (({{(Subscript[r, 1])}, {(Subscript[r, 3])}})) =(({{0}, {0}}))  ;        (({{(Subscript[t, 1])}, {(Subscript[t, 3])}})) =(({{ ((Subscript[ϵ, 1])/(Subscript[ϵ, 2]))(Subscript[r, 1])}, {(Subscript[r, 3])}})) =(({{0}, {0}}))  ;                       (2.3.15)

TE Polarization is conserved. (Similar proof can be obtained for TM).

In other words, all E fields in both media are transverse, polarized only along (OverHat[y]).
This intuitively makes sense: if the incident beam is polarized perpendicular to the incident plane (and also parallel to the planar boundary), all waves should remain TE since there is no mechanism for the field to suddenly give rise to polarized components in the incident plane.

2.4 H field condition

Since the only remaining coefficients are (Subscript[r, 2]), (Subscript[t, 2]), we can drop the subscript 2. We have only one equation:
                                     (1+r)  =t             (2.3.9a)

To solve for r and t, we need to complete the set up of the boundary conditions with H field. We can use the equation:
                                    ∇×E+(μ/c)((∂ H)/(∂ t)) = 0                     (2.4.1)

This is simple and straight forward for harmonic plane waves as seen from the lecture and HW. In fact, it is quite simple to obtain r from these equations: 
          ∇×(Subscript[E, 1])=i ((Subscript[k, inc])×(Subscript[E, inc])+(Subscript[k, refl])×(Subscript[E, refl]) )=-(μ/c)((∂ H)/(∂ t)) =i (((Subscript[μ, 1]) ω)/c)(Subscript[H, 1]) =i (Subscript[μ, 1]) (Subscript[k, 0])(Subscript[H, 1])      (2.4.2a)
          ∇×(Subscript[E, 2])=i (Subscript[k, refr])×(Subscript[E, 2]) =i (((Subscript[μ, 2]) ω)/c)(Subscript[H, 2]) =i (Subscript[μ, 2]) (Subscript[k, 0])(Subscript[H, 2])                                       (2.4.2b)
Note:        (Subscript[k, inc])×(Subscript[E, inc])=Det(({{(OverHat[x]), (OverHat[y]), (OverHat[z])}, {(Subscript[k, 1]), 0, β}, {0, 1, 0}}))  ; (Subscript[k, refl])×(Subscript[E, refl])=Det(({{(OverHat[x]), (OverHat[y]), (OverHat[z])}, {-(Subscript[k, 1]), 0, β}, {0, r, 0}})) ;
                  (Subscript[k, refr])×(Subscript[E, 2])=Det(({{(OverHat[x]), (OverHat[y]), (OverHat[z])}, {(Subscript[k, 2]), 0, β}, {0, t, 0}}))
Applying continuity to the z-component (tangential H field) yields:
                            (1/(Subscript[μ, 1]))(Subscript[k, 1]) (1- r)=(1/(Subscript[μ, 2]))(Subscript[k, 2]) t                  (2.4.4)
with:                             (1+r)  =t                         (2.3.9a)

Solve[(1/(Subscript[μ, 1]))(Subscript[k, 1]) (1- r)==(1/(Subscript[μ, 2])) (Subscript[k, 2]) t &&  (1+r)  == t, {r,t}]

FullSimplify[t->1-(((Subscript[k, 2]) (Subscript[μ, 1])-(Subscript[k, 1]) (Subscript[μ, 2]))/((Subscript[k, 2]) (Subscript[μ, 1])+(Subscript[k, 1]) (Subscript[μ, 2])))]

We obtain:        r=(((Subscript[μ, 2])(Subscript[k, 1])-(Subscript[μ, 1])(Subscript[k, 2]))/((Subscript[μ, 2])(Subscript[k, 1])+(Subscript[μ, 1])(Subscript[k, 2])))   ;   t=((2(Subscript[μ, 2])(Subscript[k, 1]))/((Subscript[μ, 2])(Subscript[k, 1])+(Subscript[μ, 1])(Subscript[k, 2])))    (2.4.5)
Since for most dielectric, (Subscript[μ, 1])=(Subscript[μ, 2])=1,
                        r=(((Subscript[k, 1])-(Subscript[k, 2]))/((Subscript[k, 1])+(Subscript[k, 2])))   ;   t=((2(Subscript[k, 1]))/((Subscript[k, 1])+(Subscript[k, 2])))                 (2.4.6)
The problem is solved.

2.5 Obtain unique solutions

Formulate general wave solutions in each medium: there will be unknown quantities (or coefficients)

Match the solutions at the boundary(ies): Obtain boundary condition equations that involve the unknown quantities

Solve for the unknown quantities.

Substitute the unknown quantities to obtain specific and unique solutions.

(Subscript[E, 1])= (Subscript[E, inc]) +(Subscript[E, refl])                                            (2.5.1a)
                          =  (OverHat[y]) (Subscript[E, inc]) ((e^(i  (Subscript[k, 1]) x ))+r (e^(-i  (Subscript[k, 1]) x )))(e^(i  ( β z -ω t )))       (2.5.1b) 
where:               r=(((Subscript[k, 1])-(Subscript[k, 2]))/((Subscript[k, 1])+(Subscript[k, 2])))                                 (2.4.6)

(Subscript[E, 2])=  (OverHat[y]) (Subscript[E, inc]) t (e^(i  (Subscript[k, 2]) x ))(e^(i  ( β z -ω t )))       (2.5.2) 
where:                t=((2(Subscript[k, 1]))/((Subscript[k, 1])+(Subscript[k, 2])))                                  (2.4.6)
For reference:              ((Subscript[k, 1])^2)+(β^2) =(Subscript[ϵ, 1])(Subscript[μ, 1])((Subscript[k, 0])^2)=((Subscript[n, 1])^2)((Subscript[k, 0])^2)     ;     (Subscript[k, 0])≡((2 π)/λ)              (2.2.2b) 
                               (Subscript[k, 1])=(Subscript[k, 0])(Subscript[n, 1]) Cos[θ]  ;   β=(Subscript[k, 0]) (Subscript[n, 1])Sin[θ]                              (2.2.2c)
And:                     (Subscript[k, 2])= (Sqrt[(Subscript[ϵ, 2]) (Subscript[μ, 2]) ((Subscript[k, 0])^2) -(β^2) ])                           (2.3.10b)
                   β=(Subscript[k, 0]) (Subscript[n, 2])Sin[(Subscript[θ, r])]   ;    (Subscript[k, 2])= (Subscript[n, 2])(Subscript[k, 0]) Cos[(Subscript[θ, r])]         (2.5.3)