Home exercise 7.8.3: Modes of a circular cavity

Guide cutoffs $4.775x/2$: TE${}_{11}$ $4.396$, TM${}_{01}$ $5.74$, TE${}_{21}$ $7.29$, TM${}_{11}$ and TE${}_{01}$ $9.15\mathrm{GHz}$. Hence, in GHz,

The chart plots $y=(2af{)}^2$ in $(\mathrm{GHz}\mathrm{cm}{)}^2$ against $(2a/d{)}^2$. Here $(2a/d{)}^2=(4/3{)}^2=1.78$. Going up this vertical line the chart shows TM${}_{010}$, TE${}_{111}$, TM${}_{011}$, TE${}_{211}$, TM${}_{110}$, TE${}_{011}$/TM${}_{111}$, and $f=\sqrt{y}/(2a)$. For TE${}_{111}$, $y=309.1+225\times 1.78=709$ and $f=\sqrt{709}/4=6.66\mathrm{GHz}$, as in (a).

At $5.74\mathrm{GHz}$ $R_s=19.8\mathrm{m\Omega }$:

Yes: $d=3\mathrm{cm}<2.03a=4.06\mathrm{cm}$. With $d=5\mathrm{cm}$ TE${}_{111}$ moves down to $\sqrt{4.396^2+3^2}=5.32\mathrm{GHz}$, below TM${}_{010}$ ($5.74\mathrm{GHz}$, unchanged), and becomes the dominant mode.

TE${}_{011}$ and TM${}_{111}$ coincide because $x_{01}^{\prime }=x_{11}=3.832$ ($J_0^{\prime }=-J_1$), for every $a$ and $d$: on the chart they are the same line.