EE 3340 Lect. 21 21- 1

21. Reflection and Transmission of Plane EM Waves

We have relied on the analogy between waves on transmission lines and

free-space waves to explain concepts such as impedance and propagation

constants, and we will continue to use that analogy as we now discuss reflection

and transmission of uniform plane waves. Again, it is unusual to encounter ideal

plane waves in practice, but since they are so simple to deal with, you can learn

a lot about how real waves behave by studying the behavior of idealized uniform

plane waves.

Before we get into the math, Iʼll show you some pictures that should help

you make the connection between transmission-line and free-space wave

transmission and reflection.

Fig. 21-1(a) shows a transmission line excited by a voltage source and

connected to another (semi-infinite) line that matches its (real) characteristic

impedance Z0. As the graphs of voltage and current magnitude show, both

magnitudes are constant all along the line (though they are both oscillating in

time at the frequency of the driving source). If the second line really goes to

infinity, no power is ever reflected and the load acts like a resistance R = Z0.

For a uniform plane wave, the situation that is analogous to Fig. 21-1 (a) is

Fig. 21-1 (b), which shows a plane wave traveling to the right (the +z direction) in

free space. We have shown a boundary between two regions of free space, but

EE 3340 Lect. 21 21- 2

Fig. 21-1. Reflection and transmission analogies, part 1.

EE 3340 Lect. 21 21- 3

just as with the transmission lines, the wave impedance of both regions is the

same, namely η0. Consequently, there is no reflection at the boundary and the

wave simply travels on its way to the right. If we observe the magnitudes of the

electric and magnetic fields, we find that they are also uniform with respect to the

direction of propagation, just as in the transmission-line case.

Now suppose we terminate the transmission line in a perfect short circuit

(R = 0) as shown in Fig. 21-1 (c). This imposes a boundary condition at the end

of the line, namely, V = 0. Accordingly, we have a standing wave on the line and

the voltage standing-wave ratio (VSWR) is infinite, meaning that the voltage goes

to zero at the end and every half-wavelength back from the end as well. The

current shows the same peak-and-trough pattern, but offset by a quarter

wavelength.

The plane-wave equivalent of a short circuit is a sheet made of a perfect

conductor. (In reality, copper or another good conductor will do nearly as well.)

This imposes the boundary condition that Etan = 0 at the reflecting plane,

analogous to the V = 0 condition at the end of the transmission line. As you

might expect, the plane wave is perfectly reflected at the conducting wall, and all

space in front of the wall is filled with two traveling waves of equal amplitude, one

going toward the wall and one returning. The result is a standing wave in which

the electric and magnetic fields vary over distance as shown. In particular, there

is a plane a half-wavelength away from the wall where the electric field is exactly

EE 3340 Lect. 21 21- 4

Fig. 21-2. Reflection and transmission analogies, part 2.

EE 3340 Lect. 21 21- 5

zero. And between that plane and the wall, a quarter-wavelength away from the

wall, the magnetic field is zero where the electric field is at a maximum.

What if a plane wave encounters a medium whose wave impedance is

different than that of free space, but not infinity or zero? This situation is

analogous to a transmission line feeding another line with a different impedance:

some of the incident wave will be transmitted and some will be reflected. In Fig.

21-2 (a), a transmission line feeds a second semi-infinite line whose

characteristic impedance ZL is higher than the first lineʼs impedance Z0. The

result is that the VSWR is between 1 and infinity, depending on the ratio of

impedances, and both voltage and current magnitudes vary along the line,

although not as extremely as they do in the case of a short.

The plane-wave analog of a high-impedance transmission line is a region

whose relative permeability µr is greater than 1 and whose relative permittivity εr

equals 1 (I donʼt know of any such substance, but there may be one!). This is

shown in Fig. 21-2 (b). Since the wave impedance η of a medium is

€

η = η0µ r

ε r , (21-1)

this medium will have a wave impedance higher than that of air, so we will get the

same kind of standing waves we saw in the transmission-line case. The electric

field on the interface between air and the medium is higher than it would be in

free space, and the magnetic field is lower.

If we connect a transmission line to a semi-infinite line with a lower

characteristic impedance, we will again get standing waves, as Fig. 21-2 (c)

EE 3340 Lect. 21 21- 6

shows. The analogous medium to a low-impedance transmission line is one in

which the relative permeability is 1, but the relative permittivity εr is larger than 1.

Most dielectrics have this property, so the kind of reflections we will see in Fig.

21-2 (d) are typical of what you find at many dielectric interfaces: the electric

field is somewhat lower than in free space and the magnetic field is higher. Of

course, if the dielectric is not semi-infinite (none are!) you will have reflections

from the back side which will interact with the transmitted wave and make things

more complicated. But as long as the uniform-plane-wave model suffices, this

situation is exactly analogous to a terminated transmission line, and you can take

the same mathematical machinery we developed to find impedances in

transmission lines, and use it to find the ratio of electric to magnetic fields in a

plane wave.

Now that you see the analogy, weʼll do some of the math. For simplicity,

we will consider only the case of normal (perpendicular) incidence, meaning the

wave encounters a boundary at an angle of 90 degrees. Although it is

straightforward to analyze reflection and transmission at angles other than

perpendicular, the math is tedious and we donʼt have time to go into it. But itʼs

out there if you ever need it.

EE 3340 Lect. 21 21- 7

Reflection and Transmission at a Boundary Between Semi-Infinite Media

All that long heading means is that we are considering only one reflection,

at the boundary between a medium with wave impedance η1 and a medium with

a different wave impedance η2, as shown in Fig. 21-3.

Fig. 21-3. Incident wave (i), transmitted wave (t), and reflected wave (r) at

plane interface between medium 1 and medium 2.

Our goal is to find expressions for the transmitted and reflected wave

amplitudes as a function of the incident wave amplitude, which is given. The

subscript S indicates all these quantities are phasors at a radian frequency ω.

We have conveniently chosen the origin of the z-axis to be at the interface

between the two media, so expressions for the fields there are particularly

simple.

Since z = 0 is a boundary, letʼs apply boundary conditions to the tangential

electric and magnetic fields. There are no charges or currents at the boundary,

so these conditions amount to simply

EE 3340 Lect. 21 21- 8

€

ESi + ES

r = ESt (21-2)

and

€

HSi − HS

r = HSt . (21-3)

The minus sign in front of the

€

HSr term appears because that vector points into

the page (the other magnetic-field vectors point out of the page).

So far we have four unknowns (the E and H reflected and transmitted

fields) but only two equations. Here are two more equations:

€

ESr

HSr = η1 (21-4)

€

ESt

HSt = η2 . (21-5)

Assuming we know the incident electric and magnetic fields, we now have four

equations and four unknowns, and we proceed to solve for the unknowns.

Sparing you the details, here are the results for the electric fields at z = 0:

€

ESt =

2η2η1 + η2

ESi (21-6)

€

ESr =

η2 −η1η1 + η2

ESi (21-7)

The transmitted and reflected magnetic fields can be found using Eqns. 21-4 and

21- 5 once you know the electric fields. And so the problem is solved.

EE 3340 Lect. 21 21- 9

The term in front of

€

ESi should look familiar. If you substitute

transmission-line impedances (Z1 and Z2) for the wave impedances η1 and η2,

you get the old familiar reflection coefficient Γ:

€

Γ =η2 −η1η2 + η1

=Z2 − Z1Z2 + Z1

. (21-8)

We can also define a transmission coefficient T:

€

T =2η2

η2 + η1 (21-9)

In fact, the whole mathematical machinery of cascade-connected transmission

lines transfers over exactly to plane waves in layered uniform media. If you like,

you can “translate” a uniform-plane-wave problem into its equivalent

transmission-line problem using Table 1 below. You solve the problem in its

transmission-line form, and then translate it back into electromagnetic form. If

this seems like too much work, never mind, but it seems to help some people

who are more comfortable with transmission lines than with plane waves.

Table 1: Transmission-Line to Uniform-Plane-Wave Translation

Transmission Line Term Uniform Plane Wave Term

Line impedances

€

Z1,

€

Z2 , . . . Wave impedances

€

η1,

€

η2

Prop. constants

€

α1,

€

β1,

€

α 2 ,

€

β2 . . . Prop. constants

€

α1,

€

β1,

€

α 2 ,

€

β2 . . .

Line lengths x1, x2, . . . Layer thicknesses x1, x2, . . .

Currents I1, I2, . . . Transverse magnetic fields H1, H2,. . .

Voltages V1, V2, . . . Transverse electric fields E1, E2, . . .

EE 3340 Lect. 21 21- 10

We will use this approach to solve a problem that more properly belongs in

optics, but is relevant to plane electromagnetic waves as well.

EXAMPLE

Anti-reflection coatings on glass lenses such as camera and eyeglass

lenses are designed to reduce the reflection loss of light that occurs when the

waves move from air (εr = 1) to glass, which at optical wavelengths has an index

of refraction as high as 1.897 (so-called high-index glass). Many anti-reflective

coatings use the compound magnesium fluoride (MgF2), which has an optical

index of refraction of 1.378. For the wavelength of green light (λ=550 nm), find

the thickness t of magnesium-fluoride coating on a semi-infinite slab of high-index

glass that will minimize the amount of reflected light. (See Fig. 21-4) Also,

calculate the fraction of incident electromagnetic power at that wavelength which

is reflected back into the air.

Answers: The first thing we need to do is to relate the index of refraction

to the dielectric constant. As you may recall from physics, the index of refraction

n in a certain medium is the ratio of the speed u of light in a vacuum to the speed

of light in the medium:

€

n =u(vacuum)u(medium)

(21-10)

If the mediumʼs permeability µ is equal to that of vacuum (and it usually is), then

EE 3340 Lect. 21 21- 11

Fig. 21-4. Magnesium-fluoride anti-reflection coating on spectacle lens:

application of quarter-wave matching principle to plane waves.

we use the fact that the speed of EM waves in a medium depends on the relative

dielectric constant in a way that can be derived from Eqns. 17-24 and 17-25:

€

u(medium) =1µε

. (21-11)

So for a medium having the permeability of free space but a relative permittivity εr

>1, the

EE 3340 Lect. 21 21- 12

RELATIONSHIP BETWEEN RELATIVE PERMITTIVITY AND INDEX OF

REFRACTION

is

€

n = ε r . (21-12)

So if we transform the indices of refraction into relative dielectric constants, we

find that the relative dielectric constant of high-index glass is εG = (1.897)2 =

3.598, and for MgF2 it is εM = (1.378)2 =1.899.

Returning to the problem of the anti-reflective coating, the easiest way to

do this is with a quarter-wave transformer, which we learned about in Lecture 4.

The two important characteristics of this device are its length, which has to be λ/4

(taking into account the wave velocity in the medium), and its impedance Z0. The

matching sectionʼs impedance is related to the sourceʼs characteristic impedance

ZIN and the loadʼs impedance ZL by Eqn. 4-5, reproduced here as Eqn. 21-13:

€

Z0 = ZINZL . (21-13)

Translating that into wave-impedance terms and using Eqn. 21-1 to find wave

impedances in terms of relative permittivities (we assume the relative

permeabilities µr are all 1), we find that the required matching-layer relative

permittivity εM is related to the relative permittivity of high-index glass εG by

€

η0εM

= η0 ⋅η0εG

(21-14)

or

€

εM = εG (21-15)

EE 3340 Lect. 21 21- 13

for a perfect match. We are very close, since εM is 1.899 and

€

εG is 1.897.

So what thickness of MgF2 amounts to a quarter-wavelength at a

(vacuum) wavelength λ0 of 550 nm? The easiest thing is to figure out the

frequency and then find the wavelength in MgF2. The frequency f is simply

€

f =cλ0

=3 ⋅108 m s−1

550 ⋅10−9 m= 545.4 THz (21-16)

(THz = terahertz = 1012 Hz). The wave velocity uM in MgF2 is

€

uM =cεM

=3 ⋅108 m s−1

1.9= 217.6 ⋅106 m s−1 . (21-17)

And so if we make t a quarter-wavelength in the matching material, we obtain

€

t =λM4

=uM4 f

=217.6 ⋅106 m s−1

4 545.4 THz( )= 99.7 nm (21-18)

A 100-nm layer is easily and precisely made with vacuum deposition technology,

and this is in fact how most anti-reflection coatings are applied.

Assuming all our numbers are exact, what is the remaining reflection of

green light at normal incidence once weʼve applied the coating? Once again, we

can apply the transmission-line analogy to find the “input” wave impedance at the

front surface of the coating. By analogy with Eqn. 4-3, we have

€

η IN =ηM2

ηG. (21-19)

Using our knowledge of the permittivities, we find this gives a value for the wave

impedance looking into the lens (so to speak!) of ηIN = 376.18 Ω. This is very

close to the wave impedance of free space, but not exactly equal.

EE 3340 Lect. 21 21- 14

Also, reflection and transmission coefficients are all in terms of wave

amplitudes. To find power transmission or reflection, you have to square the

amplitudes. This might be easiest to see if we use the reflection coefficient

instead. The reflection coefficient itself is

€

Γ =η IN −η0η IN + η0

= −6.9 ⋅10−4 (21-20)

The fraction of power reflected will be the squared magnitude of the reflection

coefficient, namely |Γ|2 = 4.7 x 10-7, which is essentially perfect. Real coatings do

not work nearly this well, but the principle is the same.