 Solution

In a homogeneous, source-free, nonmagnetic dielectric the electric wave equation isSet $\theta =\omega t-kx$ and $E_y=f(z)\cos \theta$. Substitution gives $f^{\prime \prime }+({ϵ}_r{\omega }^2/c^2-k^2)f=0$. For the labelled field form (a), $f=E_0\cos Kz$, so $K^2={ϵ}_r{\omega }^2/c^2-k^2>0$. For form (b), $f=E_0^{\prime }{e}^{s{K}^{\prime }z}$ with $s=\pm 1$, so $K^{\prime 2}=k^2-{ϵ}_r{\omega }^2/c^2>0$. These give the two stated inequalities, with equality a zero-transverse-wavenumber limiting case.

Faraday's law gives the associated wave magnetic field, without an added static field,Thus in case (a), $H_x=-K{E}_0\sin (Kz)\sin \theta /({\mu }_0\omega )$ and $H_z=k{E}_0\cos (Kz)\cos \theta /({\mu }_0\omega )$. In case (b), $H_x=s{K}^{\prime }{E}_0^{\prime }{e}^{s{K}^{\prime }z}\sin \theta /({\mu }_0\omega )$ and $H_z=k{E}_0^{\prime }{e}^{s{K}^{\prime }z}\cos \theta /({\mu }_0\omega )$. In both cases $H_y=0$.

For the planar dielectric waveguide the core is dielectric but the exterior is vacuum. Consequently the exterior decay constant has $K^{\prime 2}=k^2-{\omega }^2/c^2$, while the interior oscillation has $K^2={ϵ}_r{\omega }^2/c^2-k^2$. Adding gives the relation labelled (c):For an even bound mode the complete profile can be writtenContinuity of tangential $E_y$ fixes the exterior amplitude; continuity of tangential $H_x$ makes $f^{\prime }$ continuous because both media have magnetic permeability ${\mu }_0$. At $z=d$,which is the relation labelled (d). The same even profile also matches at $z=-d$; continuity of $H_z$ follows from that of $E_y$.

In the requested $K,K^{\prime }$ plot, (c) is a first-quadrant circle and (d) rises from the origin to its first pole at $Kd=\pi /2$. Equivalently put $u=Kd$, $w=K^{\prime }d$, and $\mathcal{V}=(\omega d/c)\sqrt{{ϵ}_r-1}$. At $u=0$ the circle has $w=\mathcal{V}>0$ whereas $u\tan u=0$. At the other end of $0<u<\min (\mathcal{V},\pi /2)$, the rising tangent curve exceeds the falling circle. Monotonicity therefore gives one intersection on this first branch for every $\mathcal{V}>0$.

The fundamental even mode of a symmetric dielectric slab has no positive cutoff in index contrast or thickness. Its electric profile is a nodeless cosine maximum at the mid-plane, smoothly joined to decaying exterior tails. The qualifier ${ϵ}_r>1$ is essential: for the printed limiting value ${ϵ}_r=1$, the circle collapses to $K=K^{\prime }=0$ and there is no transversely confined bound mode. A uniform unconstrained wave is not a guided mode. This even-mode argument does not assert that the specified cosine describes every possible odd guide mode.

The original diagram uses the illustrative normalized radius $\mathcal{V}=1.4$; it is a scaled $K{K}^{\prime }$ plot with $u=Kd,w=K^{\prime }d$, followed by the corresponding field profile.