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

The problems of this section use the phasor convention $e^{j ω t}$ , the reference directions of Figures 6.3.1 and 6.3.6, and the constants $ε_{0} = 1 0^{- 9} / (36 π)$ F/m, $μ_{0} = 4 π \times 1 0^{- 7}$ H/m, $c = 3 \times 1 0^{8}$ m/s and $η_{0} = 120 π = 377 Ω$ . Every formula they need is collected in the first recipe below, and the second recipe is the order in which to use them. Problems 6.3.1–6.3.3 come with the theory, Problems 6.3.4–6.3.7 are four further problems at normal incidence (their oblique-incidence companions are in Section 6.4).

Recipe · The normal-incidence toolbox

One interface. $Γ = \frac{η _{2} - η _{1}}{η _{2} + η _{1}}$ , $τ = 1 + Γ$ , $S = \frac{1 + ∣Γ∣}{1 - ∣Γ∣}$ , $R = ∣Γ ∣^{2}$ , $T = \frac{η _{1}}{η _{2}} ∣ τ ∣^{2}$ (lossless media), $η = η_{0} / n$ for nonmagnetic media.
A layer of thickness $d$ on medium 3: $η_{in} = η_{2} \frac{η _{3} + j η _{2} tan β _{2} d}{η _{2} + j η _{3} tan β _{2} d}$ , $Γ_{d} = \frac{η _{in} - η _{1}}{η _{in} + η _{1}}$ . For $d = λ_{2} /4$ , $η_{in} = η_{2}^{2} / η_{3}$ . For $d = m λ_{2} /2$ , $η_{in} = η_{3}$ and the layer disappears. A slab in air: $∣ Γ_{d} ∣^{2}$ of Eq. (6.3.36).
A lossy medium.

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

For a good conductor ( $σ ≫ ω ε$ ) $η \approx (1 + j) ω μ /2 σ$ and $α = β = π f μ σ = 1/ δ$ . The transmitted (absorbed) power is $T = η_{1} ∣ τ ∣^{2} Re {1/ η_{2}^{*}}$ .

Recipe · The method, one recipe for every interface

Geometry. Normal or oblique incidence? (For oblique incidence decide S or P and apply Snell's law first, Section 6.4.) One interface or a layer?
Impedances. $η = η_{0} / n$ for lossless nonmagnetic media, the complex $η$ of Eq. (6.3.11) if a medium is lossy, and $η_{in}$ for a layer.
Coefficients. $Γ$ and $τ$ , or $Γ_{d}$ at the front face of a layer.
Power and checks. $R + T = 1$ (or $R + T + A = 1$ with absorption), limits ( $η_{2} \to η_{1}$ , $η_{2} \to 0$ ), units.

Problem	Incidence	Key step	Result
6.3.1	one interface, dielectric	$Γ = (η_{2} - η_{1}) / (η_{2} + η_{1})$ , $S$	$Γ = - \frac{1}{2}$ , $S = 3$ , $R = 25 %$
6.3.2	perfect conductor	$Γ = - 1$ , add the two waves	nodes $λ /2$ apart, $J_{S} = 0.1$ A/m
6.3.3	three media, slab	$d = λ_{2} /4$ , $η_{in} = η_{2}^{2} / η_{3}$	$Γ_{d} = - 0.6$ , $S = 4$
6.3.4	three media	$d = λ_{2} /4$ , $n_{2} = n_{1} n_{3}$	$ε_{r 2} = 2$ , $d = 133$ nm, $10.0 %$ at $420$ nm
6.3.5	three media	$d = m λ_{2} /2$	$d = 2.03$ cm, $1 %$ band $240$ MHz
6.3.6	good conductor	$η_{2} \approx (1 + j) ω μ /2 σ$ , $∣ τ ∣ E_{0} e^{- α d}$	$d_{m a x} = 37.9$ m
6.3.7	lossy dielectric	exact $ε_{r} - j σ / ω ε_{0}$	$Γ = 0.754 e^{j 177. 7^{\circ}}$ , $43 %$ absorbed

6.3.8 Problems and solutions

In this part

Problems · 7

Home exercises · 4

Further exercises · 10

Worked examples · 2