Home exercise 7.6.1: A thin-film optical waveguide

$\mathrm{NA}=\sqrt{1.53^2-1.50^2}=\sqrt{0.0909}=0.3015$. Only TE${}_0$ and TM${}_0$ propagate for

$k_0=2\pi /(1.3\mu \mathrm{m})=4.833rad/\mu \mathrm{m}$ and $V=\frac{\pi \times 2}{1.3}\times 0.3015=1.457$. TE${}_0$: $u\tan u=\sqrt{V^2-u^2}$ gives $u=0.903$, $w=1.144$. Then $h=2u/d=0.903rad/\mu \mathrm{m}$ and

TM${}_0$: $w=(1.50/1.53{)}^{2}u\tan u=0.961u\tan u$ gives a slightly larger $u$ ($0.913$) and $n_{\mathrm{eff}}=1.5183$. The decay length of TE${}_0$ is

$\sin 2u/(2u)=0.539$ and ${\cos }^{2}u/w=0.335$:

TE${}_1$ appears at $V=\pi /2$, that is for ${\lambda }_0<2d\mathrm{NA}=2\times 2\times 0.3015=1.21\mu \mathrm{m}$.

$0.903\times \tan 0.903=1.144=w$ and $u^2+w^2=0.815+1.309=2.124=V^2$.