Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

Let medium 1 ( $ε_{1}, μ_{1}$ ) fill the half space $x < 0$ and medium 2 ( $ε_{2}, μ_{2}$ ) the half space $x > 0$ , and let a uniform plane wave, linearly polarised along $a_{y}$ , travel in medium 1 towards $+ a_{x}$ (Figure 6.3.1b). The incidence is normal: the direction of travel is perpendicular to the interface. We take both media lossless first, so that the waves are described by the phase constants and the intrinsic impedances of Section 6.1,

β_{k} η_{k} = ω μ_{k} ε_{k}, = \frac{μ _{k}}{ε _{k}} (k = 1, 2), (6.3.1)

which some texts write as $k_{k}$ and $Z_{k}$ .

Three waves are needed. The incident wave is given. A reflected wave travels back into medium 1 and a transmitted wave travels on into medium 2. Why should these be plane waves of the same frequency, travelling along $\pm a_{x}$ ? The boundary conditions must hold at every instant and at every point $(0, y, z)$ of the interface. The incident fields there oscillate as $cos ω t$ and do not depend on $y$ or $z$ , so the other two waves must do the same: same $ω$ , no dependence on $y$ and $z$ , which means travel along $\pm a_{x}$ . With the phasor convention $e^{j ω t}$ (time factor suppressed) the three waves are

incident: E^{i} (x) = E_{0}^{i} e^{- j β_{1} x} a_{y}, H^{i} (x) = \frac{E _{0}^{i}}{η _{1}} e^{- j β_{1} x} a_{z}, (6.3.2)

reflected: E^{r} (x) = E_{0}^{r} e^{j β_{1} x} a_{y}, H^{r} (x) = - \frac{E _{0}^{r}}{η _{1}} e^{j β_{1} x} a_{z}, (6.3.3)

transmitted: E^{t} (x) = E_{0}^{t} e^{- j β_{2} x} a_{y}, H^{t} (x) = \frac{E _{0}^{t}}{η _{2}} e^{- j β_{2} x} a_{z} . (6.3.4)

The amplitude $E_{0}^{i}$ is known, $E_{0}^{r}$ and $E_{0}^{t}$ are the two unknowns. Each magnetic field follows from its electric field by the rule of Section 6.1 for a uniform plane wave travelling along the unit vector $a_{k}$ ,

H = \frac{1}{η} a_{k} \times E . (6.3.5)

For the incident and the transmitted waves $a_{k} = a_{x}$ and $a_{x} \times a_{y} = a_{z}$ . For the reflected wave $a_{k} = - a_{x}$ , and $(- a_{x}) \times a_{y} = - a_{z}$ : hence the minus sign in Eq. (6.3.3). Equivalently, $E \times H$ points along the direction of travel, and $a_{y} \times (- a_{z}) = - a_{x}$ .

Common mistake · Forgetting the minus sign of

H^{r}

Writing $H^{r} = + (E_{0}^{r} / η_{1}) e^{j β_{1} x} a_{z}$ because " $H = E / η$ ". The magnitude is right, but the reflected wave would then carry power towards the interface, and the boundary conditions give $Γ = (η_{1} - η_{2}) / (η_{1} + η_{2})$ , the wrong sign. Always build $H$ from Eq. (6.3.5) with the direction of travel of that wave.

The boundary conditions

The general boundary conditions at an interface are those of the static chapters, $E_{1 t a n} = E_{2 t a n}$ (Section 3.4.6) and $a_{n} \times (H_{1} - H_{2}) = J_{S}$ with $a_{n}$ the unit normal pointing from medium 2 into medium 1 (Section 4.2.3 writes it as $H_{1 τ} - H_{2 τ} = J_{n S}$ ), and they remain valid for time-varying fields (see the box below). A surface current $J_{S}$ needs an infinite conductivity. For two media of finite conductivity $J_{S} = 0$ and the tangential $H$ is continuous as well. At normal incidence both $E$ and $H$ are entirely tangential, so these two conditions are all there is. Applied at $x = 0$ , they read

E_{y}^{i} (0) + E_{y}^{r} (0) = E_{y}^{t} (0) ⟹ E_{0}^{i} + E_{0}^{r} = E_{0}^{t}, (6.3.6)

H_{z}^{i} (0) + H_{z}^{r} (0) = H_{z}^{t} (0) ⟹ \frac{E _{0}^{i}}{η _{1}} - \frac{E _{0}^{r}}{η _{1}} = \frac{E _{0}^{t}}{η _{2}} . (6.3.7)

Divide both equations by $E_{0}^{i}$ and introduce the ratios of the amplitudes.

Definition · Reflection and transmission coefficients

At normal incidence on a plane interface the reflection coefficient and the transmission coefficient are the ratios of the electric-field amplitudes at the interface,

Γ = \frac{E _{0}^{r}}{E _{0}^{i}}, τ = \frac{E _{0}^{t}}{E _{0}^{i}}, (6.3.8)

also written $R$ and $T$ in some texts (in this book $R$ and $T$ denote the power fractions).

The boundary conditions Eq. (6.3.6) and Eq. (6.3.7) become two linear equations,

1 + Γ = τ, 1 - Γ = \frac{η _{1}}{η _{2}} τ . (6.3.9)

Adding them eliminates $Γ$ : $2 = τ (1 + η_{1} / η_{2})$ , so $τ = 2 η_{2} / (η_{2} + η_{1})$ , and then $Γ = τ - 1$ .

Reflection and transmission at normal incidence

A uniform plane wave incident normally from medium 1 on medium 2 is reflected and transmitted with

Γ τ = \frac{η _{2} - η _{1}}{η _{2} + η _{1}}, = \frac{2 η _{2}}{η _{2} + η _{1}} = 1 + Γ. (6.3.10)

In plain words

Only the impedances matter. If medium 2 has the same intrinsic impedance as medium 1 the wave does not notice the interface ( $Γ = 0$ , $τ = 1$ ), however different the speeds are. The larger the impedance step, the stronger the echo, and its sign tells which medium has the larger impedance.

The transmission-line picture

Equations (6.3.9) are exactly those of two transmission lines joined end to end (Figure 6.3.2a). With the correspondence $V \leftrightarrow E_{y}$ , $I \leftrightarrow H_{z}$ and $Z_{0} \leftrightarrow η$ , the interface is the junction, the incident and reflected waves are the forward and backward voltage waves $V^{+}$ and $V^{-}$ on line 1, and medium 2, carrying only a forward wave, acts on line 1 as a load $Z_{L} = η_{2}$ . The reflection coefficient Eq. (6.3.10) is the familiar $Γ = (Z_{L} - Z_{0}) / (Z_{L} + Z_{0})$ . The analogy will pay off with two interfaces, where it turns a four-equation problem into one line of algebra.

Lossy media

Nothing in the derivation used the reality of $η$ . If a medium is lossy, its intrinsic impedance and its propagation constant are complex (Section 6.2),

η γ = \frac{j ω μ}{σ + j ω ε}, = α + j β = j ω μ (σ + j ω ε), (6.3.11)

the transmitted wave becomes $E_{0}^{t} e^{- γ_{2} x} a_{y} = E_{0}^{t} e^{- α_{2} x} e^{- j β_{2} x} a_{y}$ , and Eq. (6.3.10) holds unchanged with complex $Γ$ and $τ$ . Some texts write the same wave as $e^{- j k x}$ with the complex wave number $k = β - j α$ , so that $j k = γ$ . The two notations describe the same field.

When medium 2 is a perfect electric conductor (PEC), $σ_{2} \to \infty$ and

η_{2} = \frac{μ _{2}}{ε _{2} ( 1 - j \frac{σ _{2}}{ω ε _{2}} )} ⟶ 0 ⟹ Γ = - 1, τ = 0 : (6.3.12)

nothing enters the conductor and the reflected field cancels the incident one on its surface.

(a) The interface as the junction of two lines.

(b) \Gamma and \tau for lossless media. — (a) The interface as the junction of two lines.

Reading the formulas

For lossless media $- 1 \leq Γ \leq 1$ . $Γ < 0$ when medium 2 has the lower impedance, which for nonmagnetic dielectrics is the denser medium (larger $ε_{r}$ ). The transmitted field then has the opposite sign to the reflected one, and $∣ τ ∣ < 1$ . When medium 2 has the larger impedance, $Γ > 0$ and $τ = 1 + Γ$ exceeds $1$ (Figure 6.3.2b): the transmitted field is larger than the incident one. A larger field is not a larger power, as the next subsection shows.

Quick check

A plane wave in air falls normally on still water, treated here as a lossless dielectric with $ε_{r} = 81$ , $μ_{r} = 1$ . Find $Γ$ and $τ$ .

$η_{2} = η_{0} / 81 = η_{0} /9$ , so $Γ = \frac{1/9 - 1}{1/9 + 1} = - 0.8$ and $τ = 1 + Γ = 0.2$ . The reflected field is $80 %$ of the incident one and inverted. (Real water is lossy at most frequencies, which changes $η_{2}$ but not the method.)

6.3.1 Normal incidence on a plane interface

The boundary conditions

The transmission-line picture

Lossy media

Reading the formulas