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 6.4.6

Radio waves skimming the ground

A $100 MHz$ wave is reflected by flat, dry ground, which may be taken as lossless and nonmagnetic with $ε_{r} = 4$ . Radio engineers use the grazing angle $ψ = 9 0^{\circ} - θ_{i}$ . Horizontal polarisation is S and vertical polarisation is P. (a) Find $Γ_{⊥}$ and $Γ_{∥}$ at $ψ = 6 0^{\circ}$ , $4 5^{\circ}$ , $1 0^{\circ}$ , $5^{\circ}$ and $1^{\circ}$ . (b) At which grazing angle is a vertically polarised wave not reflected at all? (c) Show that for $ψ \to 0$ both $∣Γ∣$ tend to $1$ , find first-order approximations, and show that for both polarisations the incident and reflected fields cancel at the ground.

Try it yourself first.

Strategy

In units of $η_{0}$ : $η_{1} = 1$ , $η_{2} = \frac{1}{2}$ , with $cos θ_{i} = sin ψ$ and $sin θ_{t} = cos ψ /2$ . Near grazing $cos θ_{i}$ is small while $cos θ_{t} \to 1 - 1/ ε_{r}$ stays finite, so the terms with $cos θ_{i}$ drop out of both Fresnel formulas.

Snell's law and the Fresnel formulas give

$ψ$	$θ_{t}$	$Γ_{⊥}$	$Γ_{∥}$
$6 0^{\circ}$	$14.4 8^{\circ}$	$- 0.382$	$- 0.283$
$4 5^{\circ}$	$20.7 0^{\circ}$	$- 0.451$	$- 0.204$
$1 0^{\circ}$	$29.5 0^{\circ}$	$- 0.819$	$+ 0.430$
$5^{\circ}$	$29.8 7^{\circ}$	$- 0.904$	$+ 0.665$
$1^{\circ}$	$29.9 9^{\circ}$	$- 0.980$	$+ 0.923$

For example at $ψ = 1 0^{\circ}$ : $cos θ_{i} = 0.1736$ , $sin θ_{t} = 0.4924$ , $cos θ_{t} = 0.8704$ and

Γ_{⊥} Γ_{∥} = \frac{0.5 \times 0.1736 - 0.8704}{0.5 \times 0.1736 + 0.8704} = - 0.819, = \frac{0.5 \times 0.8704 - 0.1736}{0.5 \times 0.8704 + 0.1736} = + 0.430.

$Γ_{∥} = 0$ at the Brewster angle $θ_{B} = arctan 2 = 63.4 3^{\circ}$ , that is at the grazing angle $ψ_{B} = 26.5 7^{\circ}$ . There $Γ_{⊥} = \frac{1 - ε _{r}}{1 + ε _{r}} = - 0.6$ .

With $cos θ_{i} = sin ψ \to 0$ and $cos θ_{t} \to (ε_{r} - 1) / ε_{r}$ , to first order in $sin ψ$

Γ_{⊥} Γ_{∥} \approx - 1 + \frac{2 sin ψ}{ε _{r} - 1}, \approx 1 - \frac{2 ε _{r} sin ψ}{ε _{r} - 1},

that is $- 0.980$ and $+ 0.919$ at $ψ = 1^{\circ}$ , against the exact $- 0.980$ and $+ 0.923$ . For S, $Γ_{⊥} \to - 1$ : the horizontal $E$ of the reflected wave cancels the incident one at the ground. For P, the reference directions of Figure 6.4.10 give the incident and reflected fields the same tangential reference but opposite normal references, so $Γ_{∥} \to + 1$ means that the vertical components, which carry nearly all of $E$ at grazing incidence, cancel.

Answer

ψ_{B} = 26.5 7^{\circ}, ψ \to 0 : Γ_{⊥} \to - 1, Γ_{∥} \to + 1 (the field cancels at the ground for both)

Check

At $ψ = 9 0^{\circ}$ both formulas give $Γ = \frac{1 - 2}{1 + 2} = - \frac{1}{3}$ . $Γ_{∥}$ changes sign at $ψ_{B}$ , as it must on passing through the Brewster angle (Figure 6.4.22).

Exercise 6.4.6. The two coefficients of dry ground (\varepsilon_r=4) against the grazing angle. Towards grazing incidence \Gamma_\perp\to-1 and \Gamma_\parallel\to+1, and \Gamma_\parallel passes through zero at \psi_B=26.57^\circ. — **Figure 6.4.22.** Exercise 6.4.6. The two coefficients of dry ground ( $ε_{r} = 4$ ) against the grazing angle. Towards grazing incidence $Γ_{⊥} \to - 1$ and $Γ_{∥} \to + 1$ , and $Γ_{∥}$ passes through zero at $ψ_{B} = 26.5 7^{\circ}$ .

Remark

Radio texts often define a vertical reflection coefficient with respect to the vertical field, equal to $- Γ_{∥}$ here, and then both coefficients tend to $- 1$ at grazing incidence. This is the basis of the plane-earth (two-ray) model of a link between two low antennas: the direct and the ground-reflected waves travel almost the same path and nearly cancel, and the received field grows in proportion to the antenna heights. That is why VHF and UHF antennas are put on masts.

cf. Problem 6.4.5

Exercise 6.4.6: Radio waves skimming the ground