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

It is more natural to refer the reflection to the front face $y = d$ , where the reflected and the incident waves meet:

Γ_{d} = \frac{E ^{r} ( d )}{E ^{i} ( d )} = \frac{E _{0}^{r} e ^{- j β_{1} d}}{E _{0}^{i} e ^{j β_{1} d}} = Γ e^{- 2 j β_{1} d} . (6.3.38)

The transmission-line picture now gives $Γ_{d}$ in one line. Look from medium 1 into the layer and ask for the ratio of the total $E_{x}$ to the total $H_{z}$ at the front face, the input impedance $η_{in}$ (both fields taken with the reference directions of the incident wave). With Eq. (6.3.30),

E_{x} (d) = \frac{E _{0}^{t}}{2 η _{3}} [(η_{3} + η_{2}) e^{j β_{2} d} + (η_{3} - η_{2}) e^{- j β_{2} d}] = \frac{E _{0}^{t}}{η _{3}} [η_{3} cos β_{2} d + j η_{2} sin β_{2} d],

H_{z} (d) = \frac{E _{0}^{t}}{2 η _{2} η _{3}} [(η_{3} + η_{2}) e^{j β_{2} d} - (η_{3} - η_{2}) e^{- j β_{2} d}] = \frac{E _{0}^{t}}{η _{2} η _{3}} [η_{2} cos β_{2} d + j η_{3} sin β_{2} d] .

Their ratio is the input impedance. On the medium-1 side of the front face the same ratio is $η_{1} (1 + Γ_{d}) / (1 - Γ_{d})$ , and the continuity of $E_{x}$ and $H_{z}$ makes the two equal.

Input impedance of a layer

A layer of impedance $η_{2}$ and thickness $d$ in front of medium 3 presents to medium 1 the input impedance

η_{in} and Γ_{d} = η_{2} \frac{η _{3} + j η _{2} tan β _{2} d}{η _{2} + j η _{3} tan β _{2} d}, = \frac{η _{in} - η _{1}}{η _{in} + η _{1}} (6.3.39)

at its front face. This is the input impedance of a transmission line of length $d$ and characteristic impedance $η_{2}$ terminated by the load $η_{3}$ .

In plain words

Seen from the front, the layer and everything behind it act like one medium of impedance $η_{in}$ . The layer transforms the impedance $η_{3}$ into $η_{in}$ , and the transformation depends on how many wavelengths thick the layer is.

Quarter-wave layer

For $d = λ_{2} /4$ , $β_{2} d = π /2$ and $tan β_{2} d \to \infty$ . Dividing numerator and denominator of Eq. (6.3.39) by $tan β_{2} d$ gives $η_{in} = η_{2}^{2} / η_{3}$ . The reflection vanishes, $η_{in} = η_{1}$ , if

Quarter-wave matching layer

d = \frac{λ _{2}}{4}, η_{2} = η_{1} η_{3} (n_{2} = n_{1} n_{3} for nonmagnetic media) . (6.3.40)

In plain words

A quarter-wave layer whose impedance is the geometric mean of its neighbours removes the reflection completely. Two equal echoes, one from each face, come back half a wavelength apart and cancel.

Half-wave layer

For $d = λ_{2} /2$ (or any multiple), $tan β_{2} d = 0$ and $η_{in} = η_{3}$ : the layer disappears, whatever $η_{2}$ is. With $η_{3} = η_{1}$ there is no reflection at all, which is the half-wave window of case (i) above.

Reflected power of two layers designed for the frequency f_0. Solid: a quarter-wave layer n_2=2 between air and n_3=4 (dotted: the same interface without the layer, 36\,\%). Dashed: a half-wave slab n_2=3 in air. The quarter-wave match repeats at 3f_0,5f_0,\dots, the half-wave window at 2f_0,3f_0,\dots. In between the layers reflect, up to 64\,\% for the n_2=3 slab. — **Figure 6.3.7.** Reflected power of two layers designed for the frequency $f_{0}$ . Solid: a quarter-wave layer $n_{2} = 2$ between air and $n_{3} = 4$ (dotted: the same interface without the layer, $36 %$ ). Dashed: a half-wave slab $n_{2} = 3$ in air. The quarter-wave match repeats at $3 f_{0}, 5 f_{0}, \dots$ , the half-wave window at $2 f_{0}, 3 f_{0}, \dots$ . In between the layers reflect, up to $64 %$ for the $n_{2} = 3$ slab.

Both work at one frequency

$f_{0}$ is the design frequency. Away from it the electrical thickness $β_{2} d = \frac{π}{2} f / f_{0}$ (or $π f / f_{0}$ for the half-wave slab) changes, $η_{in}$ moves away from its design value and the reflection grows (Figure 6.3.7). The bandwidth over which a single layer works is the subject of Problem 6.3.5, Home exercise 6.3.4 and Exercise 6.3.4.

The coating of Problem 6.3.4 below ( $ε_{r 2} = 2$ , $d = 133 nm$ on glass with $ε_{r} = 4$ ) is designed for $750 nm$ . Drag along the lower plot from $750 nm$ to $420 nm$ and see the reflection rise from $0$ to $10 %$ .

InteractiveLoading simulation…

Quick check

Light of $λ_{0} = 600$ nm falls normally from air on a substrate with $n_{3} = 2.25$ . Which single layer removes the reflection?

$n_{2} = 1 \times 2.25 = 1.5$ and $d = λ_{2} /4 = λ_{0} / (4 n_{2}) = 600/6 = 100$ nm.

Common mistake · A quarter of the wrong wavelength

Making the layer $λ_{0} /4$ thick. The quarter wave is measured in the layer: $d = λ_{2} /4 = λ_{0} / (4 n_{2})$ , shorter by the factor $n_{2}$ . Likewise $β_{2} d$ in Eq. (6.3.39) uses $β_{2} = n_{2} β_{0}$ , never $β_{0}$ .

Application 6.3.1 · Antireflection coatings and radome windows

Reflection is a nuisance (glare, lens flare, lost solar power, mismatched radomes) and a resource (mirrors without metal, light guides, polarisers). The layer results of this section are the engineering answer to the nuisance, at optical and at microwave frequencies alike.

The quarter-wave coating. Bare glass ( $n = 1.52$ ) reflects $R = (\frac{1.52 - 1}{2.52})^{2} = 4.3 %$ per surface. A camera lens with ten glass surfaces in air would lose a third of the light ( $1 - 0.957 4^{10} = 35 %$ ) and fill the image with ghost reflections. A layer of thickness $d = λ_{0} / (4 n_{2})$ on the glass sends back two waves (1 and 2 in Figure 6.3.8a): wave 2 has travelled an extra round trip $2 β_{2} d = π$ through the layer, so it comes back half a wavelength behind wave 1, and the two cancel when their amplitudes match. The two faces reflect equally, $(n_{1} - n_{2}) / (n_{1} + n_{2}) = (n_{2} - n_{3}) / (n_{2} + n_{3})$ , exactly when $n_{2}^{2} = n_{1} n_{3}$ , the condition Eq. (6.3.40):

n_{2} = n_{1} n_{3}, d = \frac{λ _{0}}{4 n _{2}} . (6.3.41)

(a) The two echoes (rays tilted for clarity).

(b) Reflectance of coated glass, design at 550 nm. — (a) The two echoes (rays tilted for clarity).

The ideal $n_{2} = 1.52 = 1.233$ has no durable material. Magnesium fluoride, MgF $_{2}$ ( $n_{2} = 1.38$ , $d = 550/ (4 \times 1.38) = 99.6$ nm for $550$ nm), leaves

n_{in} R = \frac{n _{2}^{2}}{n _{3}} = \frac{1.3 8 ^{2}}{1.52} = 1.253, = (\frac{1 - 1.253}{1 + 1.253})^{2} = 1.3 %,

nearly flat over the visible (Figure 6.3.8b). Multilayer stacks reach below $0.3 %$ . On solar cells the coating is silicon nitride, Si $_{3}$ N $_{4}$ , and the gain is far larger because bare silicon reflects a third of the light (Worked example 6.3.1).

The half-wave radome. An antenna cover, a radome, must protect the antenna from the weather without reflecting its radiation. A wall of thickness

d = \frac{λ _{2}}{2} = \frac{λ _{0}}{2 ε _{r}} (or a multiple) (6.3.42)

has $η_{in} = η_{0}$ and is transparent at its design frequency: for $ε_{r} = 4$ at $12$ GHz ( $λ_{0} = 25$ mm) $d = 6.25$ mm. Thin walls ( $d ≪ λ_{2}$ ) are the broadband option, since $∣Γ ∣^{2} \propto sin^{2} β_{2} d$ in Eq. (6.3.36) stays small while $β_{2} d$ is small. Problem 6.3.5 designs a radome wall that must also be mechanically strong, and Worked example 6.3.2 treats an ordinary window pane as a (non-designed) radome for Wi-Fi signals.

Quick check

Why does a radome wall work for any permittivity when it is $λ_{2} /2$ thick, whereas a coating needs the right index?

At a half-wave wall the two echoes have opposite signs automatically (air to dielectric gives $Γ < 0$ , dielectric to air $Γ > 0$ , with equal magnitudes) and a full $2 π$ round trip, so they cancel whatever the impedance step. In a coating the two faces see different impedance steps, and only $n_{2} = n_{1} n_{3}$ makes them equal.

Table 6.3.1. The application of this section and where it is worked out.

Application	Physics used	Worked out in
Antireflection coatings	$η_{in} = η_{2}^{2} / η_{3}$ , $n_{2} = n_{1} n_{3}$	Worked example 6.3.1, Problem 6.3.4, Home exercise 6.3.4
Radome windows	$d = m λ_{2} /2$ , $∣ Γ ∣^{2}$ of a slab	Worked example 6.3.2, Problem 6.3.5

6.3.6 Quarter-wave and half-wave layers

Quarter-wave layer

Half-wave layer