Exercise 7.3.9: Counting the modes of an oversized guide

$c/2a=6.56\mathrm{GHz}$ and $c/2b=14.76\mathrm{GHz}$. Below $30\mathrm{GHz}$: TE${}_{10}$ ($6.56$), TE${}_{20}$ ($13.12$), TE${}_{01}$ ($14.76$), TE${}_{11}$ and TM${}_{11}$ ($16.16$), TE${}_{30}$ ($19.69$), TE${}_{21}$ and TM${}_{21}$ ($19.75$), TE${}_{31}$ and TM${}_{31}$ ($24.61$), TE${}_{40}$ ($26.25$), TE${}_{02}$ ($29.53\mathrm{GHz}$): $9$ TE and $3$ TM, twelve modes.

$f>\frac{c}{2}\sqrt{(m/a{)}^2+(n/b{)}^2}$ is equivalent to $(\frac{m}{2af/c}{)}^2+(\frac{n}{2bf/c}{)}^2<1$. The quarter ellipse has the area $\frac{\pi }{4}AB=\pi ab{f}^2/c^2$, roughly the number of lattice points in it, and counting TE and TM gives

At $30\mathrm{GHz}$, $A=4.572$, $B=2.032$ and $N\approx 2\pi \times 0.02286\times 0.01016/0.01^2=14.6$.

The points on the axes ($m=0$ or $n=0$) are TE modes only, there are about $A+B$ of them, and the area estimate counts half of each. Hence

At $30\mathrm{GHz}$: $7.30\pm 3.30$, that is $10.6$ TE and $4.0$ TM, against $9$ and $3$. At these small numbers the count is still lumpy.

At $94\mathrm{GHz}$: $A=14.33$, $B=6.37$, $\frac{\pi }{4}AB=71.6$ and $\frac{A+B}{2}=10.3$: $N_{TE}\approx 82.0$, $N_{TM}\approx 61.3$ and $N\approx 143$, against the exact $82+62=144$.

Just below the TE${}_{20}$ cutoff ($13.12\mathrm{GHz}$) the formulas give $N_{TE}\approx 2.8$ and $N_{TM}\approx 0$, against the true $1$ and $0$: the estimate is meant for many modes.