= Solution
In a homogeneous, source-free, nonmagnetic <dielectric> the electric <wave equation> is
$$
\boxed{\nabla^2\mathbf E-\frac{\epsilon_r}{c^2}\partial_t^2\mathbf E=0.}
$$
Set $\theta=\omega t-kx$ and $E_y=f(z)\cos\theta$. Substitution gives $f''+(\epsilon_r\omega^2/c^2-k^2)f=0$. For the labelled field form (a), $f=E_0\cos Kz$, so $K^2=\epsilon_r\omega^2/c^2-k^2>0$. For form (b), $f=E'_0e^{sK'z}$ with $s=\pm1$, so $K'^2=k^2-\epsilon_r\omega^2/c^2>0$. 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,
$$
\boxed{\mathbf H=\frac1{\mu_0\omega}\left(f'(z)\sin\theta,0,kf(z)\cos\theta\right).}
$$
Thus in case (a), $H_x=-KE_0\sin(Kz)\sin\theta/(\mu_0\omega)$ and $H_z=kE_0\cos(Kz)\cos\theta/(\mu_0\omega)$. In case (b), $H_x=sK'E'_0e^{sK'z}\sin\theta/(\mu_0\omega)$ and $H_z=kE'_0e^{sK'z}\cos\theta/(\mu_0\omega)$. In both cases $H_y=0$.
For the <planar dielectric waveguide> the core is <dielectric> but the exterior is vacuum. Consequently the \i[exterior] decay constant has $K'^2=k^2-\omega^2/c^2$, while the interior oscillation has $K^2=\epsilon_r\omega^2/c^2-k^2$. Adding gives the relation labelled (c):
$$
\boxed{K^2+K'^2=\frac{(\epsilon_r-1)\omega^2}{c^2}.}
$$
For an even bound mode the complete profile can be written
$$
f(z)=\begin{cases}E_0\cos(Kz),&|z|\leq d,\\E_0\cos(Kd)e^{-K'(|z|-d)},&|z|\geq d.\end{cases}
$$
<Continuity> of tangential $E_y$ fixes the exterior amplitude; <continuity> of tangential $H_x$ makes $f'$ continuous because both media have <magnetic permeability> $\mu_0$. At $z=d$,
$$
-KE_0\sin(Kd)=-K'E_0\cos(Kd),\qquad
\boxed{K\tan(Kd)=K',}
$$
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 $K,K'$ plot, (c) is a first-quadrant <circle> and (d) rises from the origin to its first pole at $Kd=\pi/2$. Equivalently put $u=Kd$, $w=K'd$, and $\mathcal V=(\omega d/c)\sqrt{\epsilon_r-1}$. 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 $\mathcal V>0$.
\b[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 $\epsilon_r>1$ is essential: for the printed limiting value $\epsilon_r=1$, the <circle> collapses to $K=K'=0$ 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.
\Image[/past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2011/paper-2-slab-mode.png]
{title=Fundamental slab mode}
The original diagram uses the illustrative normalized radius $\mathcal V=1.4$; it is a scaled $KK'$ plot with $u=Kd,w=K'd$, followed by the corresponding field profile.
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