Chapter 4: Worked Examples and Learning Path

Read the chapter, make predictions in the browser explorer, and then run the notebook. This sequence connects physical intuition to calculations that can be inspected and changed.

A suggested learning sequence¶

For a classroom session, use a repeated-symbol strip to identify the period, ask students to place equally spaced wavevector points between light-cone boundaries, and reveal the explorer’s answer. A second session can address finite arrays, power normalization, and solver setup. These are teaching suggestions rather than a fixed assessment schedule.

Example 1: six different elements¶

Let $p=0.5\,\mu\mathrm m$, $\Lambda=6p=3\,\mu\mathrm m$, and $\lambda_0=1.55\,\mu\mathrm m$. At normal incidence into air,

The strict open condition is $|q_m|<1$, giving $m=-1,0,+1$. Their angles are approximately $-31.11^\circ,0^\circ,+31.11^\circ$. The next orders have $|q_{\pm2}|\approx1.03333>1$, so they are evanescent in air.

If every element becomes identical, the smallest period can be $p=0.5\,\mu\mathrm m$. Then $\lambda_0/p=3.1$ and only the primitive zeroth order is open in air. A six-element computational cell still permits the old order labels, but smaller translation symmetry forces the extra coefficients to vanish.

Check your reasoning: the $3\,\mu\mathrm m$ channel map never established that either first order carries appreciable power. A scattering calculation is needed for that claim.

Example 2: change the output medium¶

Keep the six-element geometry and normal incidence, but set $n_o=1.5$. The open condition becomes

Orders $-2,-1,0,+1,+2$ are open. Their angles are approximately $-43.54^\circ,-20.15^\circ,0^\circ,+20.15^\circ,+43.54^\circ$. The same tangential momentum is divided by a larger exterior wavenumber, so the first-order angles shrink while additional orders become possible.

Try next: let $n_i\ne n_o$ and use oblique incidence. Explain why changing the incident index changes $q_i$, while changing only the output index changes the light-cone boundary.

Example 3: oblique microwave incidence¶

Use $\lambda_0=30\,\mathrm{mm}$, $\Lambda=25\,\mathrm{mm}$, $n_i=n_o=1$, and $\theta_i=20^\circ$. The wavelength corresponds to 10 GHz using the rounded teaching value $c=3.00\times10^8\,\mathrm{m/s}$.

Thus $m=0$ has $q_0=0.342020$ and $\theta_0=20^\circ$. Order $m=-1$ has $q_{-1}\approx-0.857980$ and $\theta_{-1}\approx-59.09^\circ$. Order $m=+1$ has $q_{+1}\approx1.542020$ and is closed. All remaining integer orders are also closed.

The incident tangential wavevector is

The field phase advance over one positive period is $k_{x,i}\Lambda\approx1.791\,\mathrm{rad}\approx102.61^\circ$. The complex multiplier is $e^{+i1.791}$ under our convention. Mapping in the reverse direction uses its inverse.

Diagnostic: reporting $m=+1$ as the open first order while keeping our stated equation would be a sign error. A solver using the opposite labeling convention can still describe the same physical beam if its wavevectors and labels are mapped explicitly.

Example 4: opening a channel at cutoff¶

Take $\Lambda=0.8\,\mu\mathrm m$, $\lambda_0=1.55\,\mu\mathrm m$, $n_i=1$, and increase positive incidence. For $m=-1$,

The order first reaches the negative output light-cone boundary when

For $n_o=1.5$, $\theta_c\approx25.94^\circ$. For air, $\theta_c\approx69.64^\circ$. At the exact threshold the order is grazing. Just below it the order is closed, and just above it the order is open. At $\theta_i=45^\circ$, the glass output supports $m=-1$ and $m=0$, whereas air supports only $m=0$.

The notebook tests points on both sides of the threshold using offsets larger than its numerical classification tolerance. A coarse angle grid can miss this change; an analytical cutoff is a useful independent check.

Example 5: amplitude, power, and the flat-interface test¶

At normal incidence from $n_i=1$ to $n_t=1.5$, the TE electric-field amplitudes are $r=-0.2$ and $t=0.8$. The power fractions are

The power sum is one. The calculation $|r|^2+|t|^2=0.68$ mixes unnormalized field amplitudes with power fractions.

Now assign that flat interface an artificial $3\,\mu\mathrm m$ period at $\lambda_0=1.55\,\mu\mathrm m$. The air and glass channel maps list additional open orders, but a flat interface scatters only into the specular order. Continuous translational symmetry makes every other amplitude zero.

Interpretation: the channel map and the amplitude calculation answer different parts of the same physical problem. A correct channel map is necessary for power accounting, but does not supply its entries.

Example 6: a scan-range spacing bound¶

Suppose $n_i=n_o=1$, $\lambda_0=1.55\,\mu\mathrm m$, and the full scan is $|\theta_i|\leq60^\circ$. A sufficient bound for keeping nonzero orders closed throughout the scan is

The bound is stricter than the normal-incidence-only condition $\Lambda<1.55\,\mu\mathrm m$. Equality permits a grazing order at the edge of the scan. If the output index or wavelength changes, recalculate the bound rather than treating half-wavelength spacing as a universal rule.

Report checklist¶

Retain the conventions and units, primitive-period argument, parameters, order-enumeration bounds, cutoff tolerance, open-order sets, signed angles, and the notebook’s analytical assertions. Discuss one case where an open order has zero amplitude and one reason why a closed order still matters in a solver.

For finite arrays, label the plotted quantity as a normalized scalar array factor. For a future full-wave report, add complex modal coefficients, incident-power normalization, all-channel power balance, and independent numerical refinements. An analytical angle alone does not validate a claimed efficiency.

Connection to the rest of the course¶

Chapter 3 uses these channels to compute and validate a dielectric TE unit-cell library. Chapter 5 uses them to verify full-wave beam steering and develop wavefront models. Chapter 6 develops numerical convergence in more detail. Chapter 10 needs a consistent channel basis when a learned model predicts a scattering response; adding or removing channels at cutoff can change both data representation and validation requirements.