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

About the further exercises of Section 6.4

Ten further exercises in the spirit of Problems 6.4.1–6.4.6 and the home exercises, in two groups each graded from easy to harder: Exercises 6.4.1–6.4.5 on the coefficients, the critical angle and the Brewster angle, Exercises 6.4.6–6.4.10 on applications in radio, fibre optics, microscopy, radomes and metal walls. They use only the equations listed at the head of this subsection and, where stated, the normal-incidence results of Section 6.3.

Exercise 6.4.9

A radome wall at an angle

A flat radome panel of a ceramic-filled composite ( $ε_{r} = 6.25$ , lossless, nonmagnetic) is designed to be a half wave thick at $15 GHz$ for normal incidence. At oblique incidence the layer formula of Section 6.3 still holds, provided each medium is described by its transverse impedance, $Z = η / cos θ$ for S and $Z = η cos θ$ for P polarisation, and $β_{2} d$ is replaced by $β_{2} d cos θ_{t}$ . (a) Find the thickness $d$ . (b) Find the S and P reflectances at $3 0^{\circ}$ , $4 5^{\circ}$ and $6 0^{\circ}$ . (c) Which thickness is transparent at $4 5^{\circ}$ ? Is it the same for both polarisations? (d) What does that panel reflect at normal incidence? (e) Show that at the Brewster angle a P wave crosses the wall without reflection whatever its thickness.

Try it yourself first.

Strategy

With the same medium on both sides, the slab result of Problem 6.3.3 carries over with $ρ = Z_{2} / Z_{1}$ and $ϕ = β_{2} d cos θ_{t}$ :

∣ Γ_{d} ∣^{2} = \frac{( ρ - 1/ ρ ) ^{2} sin ^{2} ϕ}{4 + ( ρ - 1/ ρ ) ^{2} sin ^{2} ϕ} .

The single-interface coefficients are $(Z_{2} - Z_{1}) / (Z_{2} + Z_{1})$ , which reproduce $Γ_{⊥}$ and $Γ_{∥}$ of Eq. (6.4.13) and Eq. (6.4.29).

$λ_{0} = 2 cm$ , $n = 2.5$ , $λ_{2} = 8 mm$ : $d = λ_{2} /2 = 4.0 mm$ .

At $4 5^{\circ}$ : $sin θ_{t} = 0.7071/2.5 = 0.2828$ , $cos θ_{t} = 0.9592$ , $ϕ = π \times 0.9592 = 172.6 5^{\circ}$ , $sin^{2} ϕ = 0.0163$ . In units of $η_{0}$ :

S : ρ = \frac{0.4/0.9592}{1/0.7071} = 0.295, (ρ - 1/ ρ)^{2} = 9.59, P : ρ = \frac{0.4 \times 0.9592}{0.7071} = 0.543, (ρ - 1/ ρ)^{2} = 1.69,

so $R_{⊥} = \frac{9.59 \times 0.0163}{4 + 0.157} = 3.77 %$ and $R_{∥} = 0.69 %$ . In the same way:

$θ_{i}$	$ϕ$	$R_{⊥}$	$R_{∥}$
$3 0^{\circ}$	$176. 4^{\circ}$	$0.61 %$	$0.31 %$
$4 5^{\circ}$	$172. 6^{\circ}$	$3.77 %$	$0.69 %$
$6 0^{\circ}$	$168. 9^{\circ}$	$15.8 %$	$0.32 %$

Transparency needs $ϕ = π$ : $d = λ_{2} / (2 cos θ_{t}) = 4.0/0.9592 = 4.17 mm$ . The condition does not involve $Z$ , so it is the same for both polarisations.

At normal incidence the $4.17 mm$ panel has $β_{2} d = 1.0425 π$ , $sin^{2} β_{2} d = 0.0177$ , and since $(ρ - 1/ ρ)^{2} = 4.41$ at normal incidence (see the check below), $∣ Γ_{d} ∣^{2} = \frac{4.41 \times 0.0177}{4 + 4.41 \times 0.0177} = 1.9 %$ .

At $θ_{B} = arctan 2.5 = 68. 2^{\circ}$ , $θ_{t} = 21. 8^{\circ}$ and $Z_{∥ 2} = 0.4 cos 21. 8^{\circ} = 0.371 = Z_{∥ 1} = cos 68. 2^{\circ}$ : $ρ = 1$ and $∣ Γ_{d} ∣ = 0$ for any $ϕ$ . Inside, the ray meets the back face at $21. 8^{\circ} = arctan (1/2.5)$ , the Brewster angle seen from the panel, so neither face reflects.

Answer

d = 4.0 mm, 4 5^{\circ} : R_{⊥} = 3.8 %, R_{∥} = 0.7 %, d_{4 5^{\circ}} = 4.17 mm (both), θ_{B} = 68. 2^{\circ}

Check

At $θ_{i} = 0$ both $ρ$ equal $1/2.5$ and $(ρ - 1/ ρ)^{2} = 4.41 = (ε_{r} - 1)^{2} / ε_{r}$ , the coefficient of Problem 6.3.3. The single-face S coefficient at $4 5^{\circ}$ , $\frac{0.295 - 1}{0.295 + 1} = - 0.545$ , agrees with Eq. (6.4.13).

Exercise 6.4.9. Reflectance of the 4.0\,\mathrm{mm} panel at 15\,\mathrm{GHz} against the angle of incidence. The S reflection rises quickly beyond about 45^\circ. The P reflection vanishes at the Brewster angle 68.2^\circ and stays below 1\,\% up to about 76^\circ. — **Figure 6.4.23.** Exercise 6.4.9. Reflectance of the $4.0 mm$ panel at $15 GHz$ against the angle of incidence. The S reflection rises quickly beyond about $4 5^{\circ}$ . The P reflection vanishes at the Brewster angle $68. 2^{\circ}$ and stays below $1 %$ up to about $7 6^{\circ}$ .

Remark

A radome wall can be made transparent at one angle only. Nose radomes see angles of $6 0^{\circ}$ and more, where S reflection rises quickly. Designers therefore choose the thickness for the most important angles and, for P polarisation, exploit the Brewster angle, where every thickness works.

cf. Problems 6.3.3, 6.4.5

Exercise 6.4.9: A radome wall at an angle