Exercise 7.6.5: A silicon-on-insulator waveguide layer

$\mathrm{NA}=\sqrt{3.476^2-1.444^2}=\sqrt{12.083-2.085}=3.162$, $k_0=2\pi /(1.55\mu \mathrm{m})=4.054rad/\mu \mathrm{m}$ and

only TE${}_0$ and TM${}_0$. The layer is single-mode up to $d={\lambda }_0/(2\mathrm{NA})=245\mathrm{nm}$.

TE${}_0$, $w=u\tan u$: $u=0.8888$, $w=1.0945$. With $h=2u/d=8.080rad/\mu \mathrm{m}$,

TM${}_0$, $w=0.173u\tan u$: $u=1.2506$, $w=0.6509$, $h=11.37rad/\mu \mathrm{m}$, $n_{\mathrm{eff}}=\sqrt{12.083-2.805^2}=2.053$. The birefringence is $\Delta n_{\mathrm{eff}}=0.794$.

$\cos u=0.630$: the field at the faces is $63\%$ of its central value. $1/\alpha =d/(2w)=110/1.0945=100.5\mathrm{nm}$. With $\sin 2u/(2u)=0.5507$ and ${\cos }^{2}u/w=0.3631$,

${\mathrm{e}}^{-\alpha s}=0.01$: $s=\ln 100/\alpha =4.605\times 100.5=463\mathrm{nm}$ for TE${}_0$. TM${}_0$ has $1/\alpha =110/0.6509=169\mathrm{nm}$, so $s=778\mathrm{nm}$.

Both effective indices lie between $n_2=1.444$ and $n_1=3.476$. For TE${}_0$, $\sin \theta =2.848/3.476$ gives the ray angle $\theta =55.0^{\circ }$, beyond the critical angle $24.5^{\circ }$.