= Fundamental even mode of a symmetric dielectric slab
{title2=$u\tan u=w,\quad u^2+w^2=\mathcal V^2$}
Set $u=Kd$, $w=K'd$, and $\mathcal V=(\omega d/c)\sqrt{\epsilon_r-1}$. The <circle> in the first quadrant falls from $w=\mathcal V$, while $u\tan u$ rises from zero to its first pole. A unique first-branch intersection exists for every $\mathcal V>0$. It yields a nodeless even bound field. At exactly zero index contrast there is no confined mode; as contrast tends to zero from above its tail extends without bound.
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