(a) Show that, in a planar optical waveguide, the electromagnetic energy per unit width and per unit length, in medium 1, comprises three terms:
$$
\mathscr{E}_{1}^{\prime}=\frac{\mu_{0} M^{2} \cos ^{2}(b-\alpha)}{8 k_{1 x}^{3}}\left(k_{1}^{2}+k_{1 x}^{2}+k_{x}^{2}\right)
$$
The first term is the electric energy; the second is the longitudinal magnetic energy, or the magnetic energy associated with the longitudinal component of $\boldsymbol{H} ;$ and the third term is the transverse magnetic energy. From Sec. $35.3$, $k_{1}^{2}=k_{z}^{2}-k_{1 x}^{2} .$ Thus there is more magnetic energy than electric energy in medium 1 .
(b) Show that $\mathscr{E}_{1}^{\prime}=\mu_{0} M^{2} k_{z}^{2} /\left[4 k_{1 x}(2,1)\right]$, where $(2,1)$ is defined in Sec. 36.6. In medium 3, by symmetry, $\mathscr{E}_{3}^{\prime}=\mu_{0} M^{2} k_{z}^{2} /\left[4 k_{3 x}(2,3)\right] .$ Here also there is more magnetic than electric energy.
(c) Show that
$$
\mathscr{E}_{2}^{\prime}=\frac{\mu_{0} M^{2}}{4 k_{2 x}^{2}}\left[2 a k_{2}^{2}+\frac{k_{1 x} k_{z}^{2}}{(2,1)}+\frac{k_{3 x} k_{z}^{2}}{(2,3)}\right]
$$
(d) Show that, for the three media together and for a symmetric guide,
$$
\mathscr{E}^{\prime}=\frac{\mu_{0} M^{2}}{2}\left[\left(\frac{k_{z}}{k_{2 x}}\right)^{2}\left(\frac{1}{k_{1 x}}+a\right)+a\right]
$$
As one would expect, $\ell_{1}^{\prime}=C_{m}^{\prime}$