Light in Planar Stratified Media - part 1

ECE 5368/6358                                            han q le - copyrighted
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Part 2

5. The etalon effect

5.1 Thin soap or oil film introduction

Can we apply what we learn above to explain the soap or oily film effects?

Soap film is just a very thin layer of liquid that can be mostly water, perhaps with other chemicals such as glycerin and those in detergent.

We are not so much interested in its chemistry, but only its optical property which is essentially modeled as a thin layer of a dielectric of index of refraction n between air media.

If we shine a bright light behind a soap film and look into it, we hardly see anything else but blinded by the bright light. Clearly, what we see from the soap film is whatever reflected from it. It’s the reflected light behind us shining on the soap film that we see. The most beautiful color contrast we can see is when the background of the soap is completely dark. Thus, we want to calculate the reflection from a soap film. How do we solve the problem?

5.2 The etalon problem

This is also known as the etalon problem which can be very easily solved with the transfer matrix approach.

First, etalon is simply a parallel slab as shown below:

It may even have air in the center and dielectric on both side (which is usually called Fabry-Perot cavity). A practical etalon usually have thin film coatings on both side to enhance the reflectivity of the surface. But here, we will just solve the simple case as shown above.

For the transfer matrix approach, it is quite easy with just one layer. Let’s start with TE polarization.

From Eq. (3.3.9), the incident field at the first surface x=0 is:   
                                       (({{(Subscript[F, 1])[x=0]}, {(Subscript[F', 1])[x=0]}}))=(({{1, 1}, {i (Subscript[k, 1]), -i (Subscript[k, 1]) }})) (({{1}, {r}}))       (5.2.1)
The transfer matrix method tells us that (from (3.2.17)):
              (({{(Subscript[F, 2])[x=l]}, {(Subscript[F', 2])[x=l]}}))=(({{(Subscript[ψ, 1])[l], (Subscript[ψ, 2])[l]}, {(Subscript[ψ, 1])'[l], (Subscript[ψ, 2])'[l]}})) (((({{(Subscript[ψ, 1])[0], (Subscript[ψ, 2])[0]}, {(Subscript[ψ, 1])'[0], (Subscript[ψ, 2])'[0]}})))^(-1))(({{(Subscript[F, 2])[0]}, {(Subscript[F', 2])[0]}}))       (5.2.2)
Here is the transfer matrix:

FullSimplify[(({{(E^(I (Subscript[k, 2])l )), (E^(-I (Subscript[k, 2])l ))}, {I (Subscript[k, 2]) (E^(I (Subscript[k, 2])l )), -I (Subscript[k, 2]) (E^(-I (Subscript[k, 2])l ))}})).(Inverse[(({{1, 1}, {I (Subscript[k, 2]), -I (Subscript[k, 2])}}))])]

Hence:    
        S=(({{(Subscript[ψ, 1])[l], (Subscript[ψ, 2])[l]}, {(Subscript[ψ, 1])'[l], (Subscript[ψ, 2])'[l]}})) (((({{(Subscript[ψ, 1])[0], (Subscript[ψ, 2])[0]}, {(Subscript[ψ, 1])'[0], (Subscript[ψ, 2])'[0]}})))^(-1))=(({{cos(l (Subscript[k, 2])), ((sin(l (Subscript[k, 2])))/(Subscript[k, 2]))}, {-sin(l (Subscript[k, 2])) (Subscript[k, 2]), cos(l (Subscript[k, 2]))}}))      (5.2.3)

Lastly:
           (({{(Subscript[F, 3])[x=l]}, {(Subscript[F', 3])[x=l]}}))=(({{1, 1}, {i (Subscript[k, 1]) , -i (Subscript[k, 1])}})) (({{t}, {0}}))        (5.2.4)

Put all together:
    (({{1, 1}, {i (Subscript[k, 1]) , -i (Subscript[k, 1])}})) (({{t}, {0}}))=(({{cos(l (Subscript[k, 2])), ((sin(l (Subscript[k, 2])))/(Subscript[k, 2]))}, {-sin(l (Subscript[k, 2])) (Subscript[k, 2]), cos(l (Subscript[k, 2]))}})).(({{1, 1}, {i (Subscript[k, 1]), -i (Subscript[k, 1]) }})) (({{1}, {r}}))  (5.2.5a)
    
 Or:   (({{t}, {0}}))= (1/2)(({{ 1, -(i/( (Subscript[k, 1])))}, { 1, (i/(Subscript[k, 1]))}}))(({{cos(l (Subscript[k, 2])), ((sin(l (Subscript[k, 2])))/(Subscript[k, 2]))}, {-sin(l (Subscript[k, 2])) (Subscript[k, 2]), cos(l (Subscript[k, 2]))}})).(({{1, 1}, {i (Subscript[k, 1]), -i (Subscript[k, 1]) }})) (({{1}, {r}})) (5.2.5b)

r= .; t= .;  l=.;
FullSimplify[Solve[ (({{t}, {0}}))== (1/2)(({{ 1, -(I/( (Subscript[k, 1])))}, { 1, (I/(Subscript[k, 1]))}})) . (({{Cos[l (Subscript[k, 2])], ((Sin[l (Subscript[k, 2])])/(Subscript[k, 2]))}, {- (Subscript[k, 2]) Sin[l (Subscript[k, 2])], Cos[l (Subscript[k, 2])]}})) . (({{1, 1}, {I (Subscript[k, 1]), -I (Subscript[k, 1])}})) . (({{1}, {r}})),{r,t}]]

This Mathematica result however does not give much insight in the underlying physical process.

We can use slightly different algebraic expression (and a different but equivalent concept of transfer matrix involving forward-backward traveling wave), a much better looking formula for the transmission is:
                         t=((4 (Subscript[k, 1]) (Subscript[k, 2]) )/( ((((Subscript[k, 1])+(Subscript[k, 2])))^2)))(( (e^(i  l ((Subscript[k, 2])-(Subscript[k, 1])))))/( (1-(e^(2 i L (Subscript[k, 2]))) (((((Subscript[k, 1])-(Subscript[k, 2])))^2)/((((Subscript[k, 1])+(Subscript[k, 2])))^2)))))           (5.2.6a)
                        t=(( ((Subscript[t, 1])^((+))) ((Subscript[t, 2])^((+))) )/( (1-((Subscript[r, 1])^((+))) ((Subscript[r, 2])^((-))) (e^(2 i L (Subscript[k, 2]))) )))(e^(i  l ((Subscript[k, 2])-(Subscript[k, 1]))))              (5.2.6b)
where we define:      ((Subscript[r, 1])^((+)))=((((Subscript[k, 2])-(Subscript[k, 1])))/(((Subscript[k, 1])+(Subscript[k, 2])))); ((Subscript[t, 1])^((+)))=((2 (Subscript[k, 1]) )/( ((((Subscript[k, 1])+(Subscript[k, 2])))^2)))         (5.2.7) 
which are (see figure below for illustration), respectively, the back reflection of light going from soap back to air, and the transmission coefficient of light going from air into soap in the first surface. And: 
                                  ((Subscript[r, 2])^((-)))=((((Subscript[k, 2])-(Subscript[k, 1])))/(((Subscript[k, 1])+(Subscript[k, 2])))); ((Subscript[t, 2])^((+)))=((2 (Subscript[k, 2]) )/( ((((Subscript[k, 1])+(Subscript[k, 2])))^2)))        (5.2.8) 
are, respectively, the reflection and transmission coefficient of light from inside soap exiting into air at the second surface as shown below:

Note that ((Subscript[r, 1])^((+))) and ((Subscript[r, 2])^((-))) denote different quantities but are just coincidentally equal in this case because it is air on both sides. If we have oil on water for example, they would be different.

And for the reflection:
                   r=(((Subscript[k, 1])-(Subscript[k, 2]))/((Subscript[k, 1])+(Subscript[k, 2])))+((4  (Subscript[k, 1]) (Subscript[k, 2]))/((((Subscript[k, 1])+(Subscript[k, 2])))^2))(((Subscript[k, 2])-(Subscript[k, 1]))/((Subscript[k, 1])+(Subscript[k, 2])))((  (e^(2 i (Subscript[k, 2])l)))/(1-(((((Subscript[k, 1])-(Subscript[k, 2])))^2)/((((Subscript[k, 1])+(Subscript[k, 2])))^2))  (e^(2 i (Subscript[k, 2]) l))))           (5.2.9a)
 Or:             r=((Subscript[r, 1])^((-)))+((Subscript[t, 1])^((+)))((Subscript[r, 2])^((-)))((Subscript[t, 1])^((-))) (( (e^(2 i (Subscript[k, 2]) l)))/(1-((Subscript[r, 1])^((+))) ((Subscript[r, 2])^((-))) (e^(2 i l (Subscript[k, 2])))))            (5.2.9b)  
 where the various quantities are defined in the figure below.

Why do we go through the trouble to express r and t by Eqs. (5.2.6) and (5.2.9)? 
Let’s take a look at the term:            (1-((Subscript[r, 1])^((+))) ((Subscript[r, 2])^((-))) (e^(2 i L (Subscript[k, 2]))) )                  (5.2.10)  
in the denominator of   t=(( ((Subscript[t, 1])^((+))) ((Subscript[t, 2])^((+))) )/( (1-((Subscript[r, 1])^((+))) ((Subscript[r, 2])^((-))) (e^(2 i L (Subscript[k, 2]))) )))(e^(i  l ((Subscript[k, 2])-(Subscript[k, 1]))))              (5.2.6b)

We see that (5.2.10) has the general form:      (1-r (e^(i φ)) )                  (5.2.11)  
for which, (1/((1-r (e^(i φ)) )))=(1/(1-z))  is a “single-pole” (simple pole) term that appears over and over in many phenomena in nature when there is a resonance involved.

SinglePoleResonance2

Later, we will see that Eqs. (5.2.6) and (5.2.9) are generally valid expressions for a Fabry-Perot etalon with arbitrary relection/transmission properties at the facets (interfaces). This formulism exposes the underlying physics of cavity resonance and hence gives much more insight than the formula obtained by Mathematica brute force solutions.

We will learn a different but equivalent transfer matrix approach that is insightful for dealing with cavity resonance in other chapter.

5.3 Thin soap or oil film illustration

5.3.1 Soap film

Demo Soap film

Manipulate[Show[Graphics[{Hue[s,1,1],Rectangle[{0,0},{2,1}] }]],{s,0,1}]

5.3.2 Oil film

For oil film, it is slightly different because the exit medium is not air (or the same as the incident medium) but water or some other dielectric like glasses or plastics.
The equation is:

(({{t}, {0}}))= (1/2)(({{1, -(i/( (Subscript[k, 3])))}, {1, (i/(Subscript[k, 3]))}}))(({{cos(l (Subscript[k, 2])), ((sin(l (Subscript[k, 2])))/(Subscript[k, 2]))}, {-sin(l (Subscript[k, 2])) (Subscript[k, 2]), cos(l (Subscript[k, 2]))}})).(({{1, 1}, {i (Subscript[k, 1]), -i (Subscript[k, 1])}})) (({{1}, {r}}))   (5.3.1)

Solve[  (({{t}, {0}}))==  (1/2)(({{ 1, -(I/( (Subscript[k, 3])))}, {1, (I/(Subscript[k, 3]))}})).(({{Cos[l (Subscript[k, 2])], ((Sin[l (Subscript[k, 2])])/(Subscript[k, 2]))}, {-Sin[l (Subscript[k, 2])] (Subscript[k, 2]), Cos[l (Subscript[k, 2])]}})).(({{1, 1}, {I (Subscript[k, 1]), -I (Subscript[k, 1])}})). (({{1}, {r}})),{r,t}]

FullSimplify[%]

Demo Oil film

Notice:
We see strong reflectance at large angle, that is why we tend to look at very slant incident angle to see the oil film. If not, the colorful sheen can be swamped by the backscattered light (if there are bright objects in the water beneath the oil film or bright reflection from the bottom. That’s why oil sheen is best seen with very dark background).
The color variation we see does not come just from varying thickness, but also from the fact that up close, different part of the film is viewed at different angle.

Oil sheen is not just for “fun” to look at. People use it to detect oil leaks after an accident like a lost aircraft, boat,... or oil seeping from the sea floor.

Discuss the different viewing configuration and illumination for seeing color in soap and oil film

We see that there are differences in viewing them.

5.4 High dielectric etalon illustration

Below, we will see the transmission of a thin layer of a dielectric of high index of fraction.

What is most noticeable is that at certain frequencies, the transmission is 100%. It means that power-wise, the slab acts as if it were not there. These “resonance” frequencies are separated with regular interval.

How does this happen?

Simplify[Inverse[(({{1, 1}, {I (Subscript[k, 2]), -I (Subscript[k, 2]) }}))] . (({{1, 1}, {I (Subscript[k, 1]), -I (Subscript[k, 1]) }})) . (({{1}, {r}})) ]

(({{((Expand[(r+1) (Subscript[k, 2])-(r-1) (Subscript[k, 1])])/(2 (Subscript[k, 2])))}, {((Expand[(r-1) (Subscript[k, 1])+(r+1) (Subscript[k, 2])])/(2 (Subscript[k, 2])))}}))

(({{((-r (Subscript[k, 1])+r (Subscript[k, 2])+(Subscript[k, 1])+(Subscript[k, 2]))/(2 (Subscript[k, 2])))}, {((r (Subscript[k, 1])+r (Subscript[k, 2])+(Subscript[k, 2])-(Subscript[k, 1]))/(2 (Subscript[k, 2])))}}))

Notice the field at on-resonance vs. off-resonance

6. Anti-reflection coating and quarter-wave layer

6.1 Introduction

6.2 Concept

Let’s consider the thin film above again, but this time, we use a thinner film.

Notice what happens? the resonance is broader, much much broader and it seems we can use the case when there is very little or no reflection.
However, this is the case of a thin film where both sides are the same medium, such as air.

The more typical situation we have is this (remember, it is similar to the case of oil film above)

We know that from medium 0 such as air to medium 2 such as glass or some dielectric, reflection is inevitable given by (((Subscript[n, 0])-(Subscript[n, 2]))/((Subscript[n, 0])+(Subscript[n, 2]))). But we also see that a thinfilm can create the effect of resonance in which, the transmission is maximum as the reflected waves from the 2 interfaces destructively interfere with each other so that there is no reflection back to medium 0. We wonder if we can design something like that even for the case when exiting medium 2 is not the same as the incident medium. This is the concept of “anti-reflection” coating.
In fact, if we choose the thin film to be λ/4, two times reflection will be λ/2 which causes destructive interference. Let’s apply the concept and see what happens.
But didn’t we say it is similar to oil sheen problem? 
Well, not really oil has higher index than both air and water. Here, we have the medium as “high” index and the AR film, we will see, has lower index to be useful.

6.3 Single-layer λ/4 film

Let’s apply the S-matrix:

This time, we want to put in a slightly different algebraic form.

r= .; t= .;  l=.;
FullSimplify[Solve[ (({{t}, {0}}))== (1/2)(({{ 1, -(I/(Subscript[k, 3]))}, { 1, (I/(Subscript[k, 3]))}})). (({{(E^(I (Subscript[k, 2])l )), (E^(-I (Subscript[k, 2])l ))}, {I (Subscript[k, 2]) (E^(I (Subscript[k, 2])l )), -I (Subscript[k, 2]) (E^(-I (Subscript[k, 2])l ))}})) . (({{(1/2), -(I/(2 (Subscript[k, 2])))}, {(1/2), (I/(2 (Subscript[k, 2])))}})) . (({{1, 1}, {I (Subscript[k, 1]), -I (Subscript[k, 1])}})) . (({{1}, {r}})),{r,t}]]

We want to do a little reduction:

r->((((Subscript[k, 1])-(Subscript[k, 2])) ((Subscript[k, 2])+(Subscript[k, 3]))+((Subscript[k, 1])+(Subscript[k, 2])) ((Subscript[k, 2])-(Subscript[k, 3])) (e^(2 i (Subscript[k, 2]) l)))/(((Subscript[k, 1])+(Subscript[k, 2])) ((Subscript[k, 2])+(Subscript[k, 3]))+((Subscript[k, 1])-(Subscript[k, 2])) ((Subscript[k, 2])-(Subscript[k, 3])) (e^(2 i (Subscript[k, 2]) l))))=((((Subscript[k, 1])-(Subscript[k, 2])) ((Subscript[k, 2])+(Subscript[k, 3])))/(((Subscript[k, 1])+(Subscript[k, 2])) ((Subscript[k, 2])+(Subscript[k, 3])))) ((1+((((Subscript[k, 1])+(Subscript[k, 2])) ((Subscript[k, 2])-(Subscript[k, 3])) )/(((Subscript[k, 1])-(Subscript[k, 2])) ((Subscript[k, 2])+(Subscript[k, 3])))) (e^(2 i (Subscript[k, 2]) l)))/(1+((((Subscript[k, 1])-(Subscript[k, 2])) ((Subscript[k, 2])-(Subscript[k, 3])))/(((Subscript[k, 1])+(Subscript[k, 2])) ((Subscript[k, 2])+(Subscript[k, 3])))) (e^(2 i (Subscript[k, 2]) l)))) 
Or:   r=(( ((Subscript[k, 1])-(Subscript[k, 2])) )/(((Subscript[k, 1])+(Subscript[k, 2])))) ((1+((Subscript[r, 23])/(Subscript[r, 12])) (e^(2 i (Subscript[k, 2]) l)))/(1+(Subscript[r, 12]) (Subscript[r, 23]) (e^(2 i (Subscript[k, 2]) l))))= ((Subscript[r, 12])/(1+(Subscript[r, 12]) (Subscript[r, 23]) (e^(2 i (Subscript[k, 2]) l))))(1+ ((Subscript[r, 23])/(Subscript[r, 12])) (e^(2 i (Subscript[k, 2]) l)))                        (6.3.1a)
where we denote:    (Subscript[r, 12])≡((((Subscript[k, 1])-(Subscript[k, 2])))/(((Subscript[k, 1])+(Subscript[k, 2]))))    ;    (Subscript[r, 23])≡(( ((Subscript[k, 2])-(Subscript[k, 3])) )/(((Subscript[k, 2])+(Subscript[k, 3]))))                                       (6.3.1b)

Notice something here:
This term:                               (1/(1+(Subscript[r, 12]) (Subscript[r, 23]) (e^(2 i (Subscript[k, 2]) l))))                                 (6.3.2)
is a resonance term that is maximum when 1+(Subscript[r, 12]) (Subscript[r, 23]) (e^(2 i (Subscript[k, 2]) l)) is smallest. But if  (Subscript[r, 12]) (Subscript[r, 23])  is small because of small difference between the indices of refraction, it is NOT a strongly resonance term. It gives some small modulation but not important.

SinglePoleResonance2

However, if we look at this term:        (1+ ((Subscript[r, 23])/(Subscript[r, 12])) (e^(2 i (Subscript[k, 2]) l)))        (6.3.3)
we see that it can go to zero, or very small if        |((Subscript[r, 12])/(Subscript[r, 23]))|~ 1      (6.3.4a)
(side note: each reflectivity (Subscript[r, 12]), (Subscript[r, 23]) can be small by itself, but its ratio can be large)
and:                                                              (e^(2 i (Subscript[k, 2]) l-φ))=-1     (6.3.4b)
With real index of refraction for all layers, φ=0, and (6.3.4b) becomes:
                      2 (Subscript[k, 2]) l=π    or:    ((4 π)/(Subscript[λ, 2])) l=π  or:   l=((Subscript[λ, 2])/4)=(λ/(4 (Subscript[n, 2])))         (6.4.5)
This is known as the principle of quarter-wave (λ/4) layer AR coating.

What is the requirement of                |((Subscript[r, 12])/(Subscript[r, 23]))|~ 1      (6.3.4a)?

FullSimplify[(((((Subscript[k, 1])-(Subscript[k, 2])) )/(((Subscript[k, 1])+(Subscript[k, 2]))))/(( ((Subscript[k, 2])-(Subscript[k, 3])) )/(((Subscript[k, 2])+(Subscript[k, 3])))))]

Solve[((((Subscript[k, 1])-(Subscript[k, 2])) ((Subscript[k, 2])+(Subscript[k, 3])))/(((Subscript[k, 1])+(Subscript[k, 2])) ((Subscript[k, 2])-(Subscript[k, 3]))))==1, {(Subscript[k, 2])}]

What we need is that             (Subscript[k, 2])=(Sqrt[(Subscript[k, 3])(Subscript[k, 1])]) 
Or:                                          (Subscript[n, 2])=(Sqrt[(Subscript[n, 1])(Subscript[n, 3])]) =(Sqrt[(Subscript[n, 3])])    if          (Subscript[n, 1])=1        (6.4.6)
This is the well-known formula for λ/4 AR-coating.

Additional note about transmittance: 
 From the above:           t=((4 (Subscript[k, 1]) (Subscript[k, 2]) (e^(i (Subscript[k, 2]) l)))/(((Subscript[k, 1])+(Subscript[k, 2])) ((Subscript[k, 2])+(Subscript[k, 3]))-((Subscript[k, 2])-(Subscript[k, 1])) ((Subscript[k, 2])-(Subscript[k, 3])) (e^(2 i (Subscript[k, 2]) l))))
           t=((4 (Subscript[k, 1]) (Subscript[k, 2]) (e^(i (Subscript[k, 2]) l)))/(((Subscript[k, 1])+(Subscript[k, 2])) ((Subscript[k, 2])+(Subscript[k, 3]))-((Subscript[k, 2])-(Subscript[k, 1])) ((Subscript[k, 2])-(Subscript[k, 3])) (e^(2 i (Subscript[k, 2]) l))))=((4 (Subscript[k, 1]) (Subscript[k, 2]))/(((Subscript[k, 1])+(Subscript[k, 2])) ((Subscript[k, 2])+(Subscript[k, 3]))))(( (e^(i (Subscript[k, 2]) l)))/(1-((((Subscript[k, 2])-(Subscript[k, 1])) ((Subscript[k, 2])-(Subscript[k, 3])) )/(((Subscript[k, 1])+(Subscript[k, 2])) ((Subscript[k, 2])+(Subscript[k, 3]))))(e^(2 i (Subscript[k, 2]) l))))   (6.4.7)
          t=(Subscript[t, 12])(Subscript[t, 23])(( (e^(i (Subscript[k, 2]) l)))/(1-(Subscript[r, 21]) (Subscript[r, 23])(e^(2 i (Subscript[k, 2]) l))))                                                        (6.4.8)
  Note here that we write: (Subscript[r, 21])≡((((Subscript[k, 2])-(Subscript[k, 1])) )/(((Subscript[k, 1])+(Subscript[k, 2])) )) that is different from (Subscript[r, 12])≡((((Subscript[k, 1])-(Subscript[k, 2])))/(((Subscript[k, 1])+(Subscript[k, 2])))) 
  The formula (6.4.8) is the same as (5.2.6b) above:
                          t=(( ((Subscript[t, 1])^((+))) ((Subscript[t, 2])^((+))) )/( (1-((Subscript[r, 1])^((+))) ((Subscript[r, 2])^((-))) (e^(2 i L (Subscript[k, 2]))) )))(e^(i  l ((Subscript[k, 2])-(Subscript[k, 1]))))              (5.2.6b)
The phase term difference: (e^(-i (Subscript[k, 1]) l )) is simply because we choose a different axis origin for the output wave.

6.4 Advanced topic: multi-layer AR coating

This is based on a fundamental difference with resonance effect above: continuous or adiabatic index variation (graded index).