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

The branch of mode $m$ starts on the $u$ -axis, at $u = mπ /2$ and $w = 0$ . There $α = 0$ : the field outside no longer decays, the wave is no longer bound to the slab, and the mode ceases to be guided. The circle reaches the start of branch $m$ when $V = mπ /2$ , and since $V = π df NA / c$ by Eq. (7.6.19):

Cutoff of the slab modes

V = \frac{mπ}{2} ⟺ f_{c, m} = \frac{m c}{2 d n _{1}^{2} - n _{2}^{2}} = \frac{m c}{2 d NA}, m = 0, 1, 2, \dots (7.6.21)

TE $_{m}$ and TM $_{m}$ have the same cutoff. A slab at the normalised frequency $V$ guides

M = 1 + ⌊ \frac{2 V}{π} ⌋ TE modes, and as many TM modes. (7.6.22)

TE $_{0}$ and TM $_{0}$ have no cutoff: any slab, however thin, guides them at any frequency.
Only TE $_{0}$ and TM $_{0}$ propagate when $V < π /2$ , that is for

$d < \frac{λ _{0}}{2 NA} (single TE_{0} /TM_{0} pair). (7.6.23)$
Cheng's cutoff frequencies (Table 10-4 of Field and Wave Electromagnetics) are the same numbers, $f_{c} = (n - 1) / (d μ ε_{d} - μ_{0} ε_{0})$ and $(n - \frac{1}{2}) / (d μ ε_{d} - μ_{0} ε_{0})$ for his two families, which interleave to give Eq. (7.6.21).

The b–V diagram. How far between $n_{2}$ and $n_{1}$ does $n_{eff}$ lie? The normalised propagation constant

b = \frac{n _{eff}^{2} - n _{2}^{2}}{n _{1}^{2} - n _{2}^{2}} = \frac{w ^{2}}{V ^{2}} (7.6.24)

answers this. (The second form follows from $τ^{2} - k_{0}^{2} n_{2}^{2} = α^{2}$ : divide by $k_{0}^{2} (n_{1}^{2} - n_{2}^{2})$ and use $w^{2} / V^{2} = α^{2} / [k_{0}^{2} (n_{1}^{2} - n_{2}^{2})]$ .) $b$ grows from $0$ at cutoff, where $n_{eff} = n_{2}$ , towards $1$ far above cutoff, where $n_{eff} \to n_{1}$ (Figure 7.6.5). Because $n_{eff}$ changes with frequency, the guide is dispersive even though both media are not.

The normalised propagation constant b=(n_{\rm eff}^2-n_2^2)/(n_1^2-n_2^2) of the slab modes against V (computed). Solid: TE_0 to TE_4, labelled by m. Dashed: TM_0 and TM_1 for a slab of \varepsilon_r=5 in air, slightly below their TE partners. Mode m starts at V=m\pi/2 with b=0 (n_{\rm eff}=n_2) and approaches b=1 (n_{\rm eff}=n_1) far above cutoff. — **Figure 7.6.5.** The normalised propagation constant $b = (n_{eff}^{2} - n_{2}^{2}) / (n_{1}^{2} - n_{2}^{2})$ of the slab modes against $V$ (computed). Solid: TE $_{0}$ to TE $_{4}$ , labelled by $m$ . Dashed: TM $_{0}$ and TM $_{1}$ for a slab of $ε_{r} = 5$ in air, slightly below their TE partners. Mode $m$ starts at $V = mπ /2$ with $b = 0$ ( $n_{eff} = n_{2}$ ) and approaches $b = 1$ ( $n_{eff} = n_{1}$ ) far above cutoff.

Not like a metal guide

A hollow metal guide (Section 7.3) has infinitely many modes, and even its lowest mode is cut off at low frequencies, where it becomes evanescent along $z$ . A dielectric slab has a finite number of guided modes at any frequency, and TE $_{0}$ and TM $_{0}$ are guided down to d.c. Below its cutoff a dielectric mode is not evanescent along $z$ : its field outside the slab no longer decays ( $α$ would have to become imaginary), so it leaks sideways and radiates.

For the slab of Problem 7.6.1, $ε_{r} = 5$ and $d = 2 cm$ in air, $NA = 5 - 1 = 2$ and $f_{c, m} = m \times 3 \times 1 0^{8} / (2 \times 0.02 \times 2) = 3.75 m GHz$ .

Quick check

A polymer film of index $1.6$ is surrounded by air and used at $λ_{0} = 1 μ m$ . Up to which thickness does it guide only TE $_{0}$ and TM $_{0}$ ?

$NA = 1. 6^{2} - 1 = 1.249$ and, by Eq. (7.6.23), $d < λ_{0} / (2 NA) = 1/2.498 = 0.400 μ m$ . A film thicker than $0.4 μ m$ also guides TE $_{1}$ and TM $_{1}$ .

7.6.6 Cutoff frequencies and the number of modes