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

For lossless media the time-average Poynting vectors $\frac{1}{2} Re {E \times H^{*}}$ of the three waves point along $+ a_{x}$ , $- a_{x}$ and $+ a_{x}$ , with the magnitudes (Section 6.1)

P^{i} P^{r} P^{t} = \frac{∣ E _{0}^{i} ∣ ^{2}}{2 η _{1}}, = \frac{∣ E _{0}^{r} ∣ ^{2}}{2 η _{1}} = ∣Γ ∣^{2} P^{i}, = \frac{∣ E _{0}^{t} ∣ ^{2}}{2 η _{2}} = \frac{η _{1}}{η _{2}} ∣ τ ∣^{2} P^{i} . (6.3.13)

No energy is stored at the interface in the time average, so what arrives must leave: $P^{i} - P^{r} = P^{t}$ , which requires $1 - ∣Γ ∣^{2} = \frac{η _{1}}{η _{2}} ∣ τ ∣^{2}$ . Indeed, with Eq. (6.3.10),

\frac{η _{1}}{η _{2}} \frac{4 η _{2}^{2}}{( η _{1} + η _{2} ) ^{2}} = \frac{4 η _{1} η _{2}}{( η _{1} + η _{2} ) ^{2}} = \frac{( η _{1} + η _{2} ) ^{2} - ( η _{2} - η _{1} ) ^{2}}{( η _{1} + η _{2} ) ^{2}} = 1 - Γ^{2} . (6.3.14)

Reflectance and transmittance

The fractions of the incident power reflected and transmitted at a plane interface between lossless media are

R = ∣Γ ∣^{2}, T = \frac{η _{1}}{η _{2}} ∣ τ ∣^{2}, R + T = 1. (6.3.15)

For nonmagnetic media $η = η_{0} / n$ with the refractive index $n = ε_{r}$ , and

Γ = \frac{n _{1} - n _{2}}{n _{1} + n _{2}}, R = (\frac{n _{1} - n _{2}}{n _{1} + n _{2}})^{2} . (6.3.16)

In plain words

The power is shared, never created. A larger transmitted $E$ always comes with a smaller transmitted $H$ , and their product, the power density, is what counts. $R$ is the same whichever way the wave crosses the interface.

Air on a lossless nonmagnetic dielectric of relative permittivity \varepsilon_{r2}: reflected and transmitted fractions of the power. Glass (\varepsilon_r=2.25, n=1.5) reflects 4\,\%, \varepsilon_r=9 (n=3) a quarter, and water with \varepsilon_r=81 (its conductivity ignored) 64\,\%. — **Figure 6.3.3.** Air on a lossless nonmagnetic dielectric of relative permittivity $ε_{r 2}$ : reflected and transmitted fractions of the power. Glass ( $ε_{r} = 2.25$ , $n = 1.5$ ) reflects $4 %$ , $ε_{r} = 9$ ( $n = 3$ ) a quarter, and water with $ε_{r} = 81$ (its conductivity ignored) $64 %$ .

A lossy medium 2

If medium 2 is lossy (medium 1 still lossless), $P^{i}$ and $P^{r}$ are unchanged, but just inside the interface the transmitted wave has $H^{t} = (E_{0}^{t} / η_{2}) a_{z}$ with complex $η_{2}$ , so

P^{t} (0) T = \frac{1}{2} Re {E_{0}^{t} (\frac{E _{0}^{t}}{η _{2}})^{*}} = \frac{∣ E _{0}^{t} ∣ ^{2}}{2} Re {\frac{1}{η _{2}^{*}}}, = η_{1} ∣ τ ∣^{2} Re {\frac{1}{η _{2}^{*}}}, (6.3.17)

which reduces to Eq. (6.3.15) for real $η_{2}$ . This power does not travel to infinity: it decays as $e^{- 2 α_{2} x}$ and is turned into heat in the first few skin depths. $R + T = 1$ still holds, $T$ now being the fraction absorbed by medium 2 (Problems 6.3.6 and 6.3.7).

Quick check

Medium 2 has three times the impedance of medium 1. Find $Γ$ , $τ$ , $R$ and $T$ . Is energy created, since $τ > 1$ ?

$Γ = \frac{3 - 1}{3 + 1} = \frac{1}{2}$ , $τ = \frac{3}{2}$ , $R = \frac{1}{4}$ and $T = \frac{1}{3} \times \frac{9}{4} = \frac{3}{4}$ : $R + T = 1$ . The transmitted $E$ is $1.5$ times the incident one, but the transmitted $H$ is only $τ η_{1} / η_{2} = \frac{1}{2}$ of the incident $H$ , and the power density is $\frac{3}{2} \times \frac{1}{2} = \frac{3}{4}$ .

Common mistake · Writing

T = ∣ τ ∣^{2}

$∣ τ ∣^{2}$ is the ratio of the squared fields, and the two fields live in media of different impedance. Only when $η_{1} = η_{2}$ , or when the transmitted wave returns to the first medium (a slab with the same medium on both sides), is $T = ∣ τ ∣^{2}$ . Check every answer with $R + T = 1$ .

6.3.2 Where the power goes

A lossy medium 2