Solution:
Air-filled rectangular waveguide,
$
\begin{aligned}
\therefore \mu_r & =1, \varepsilon_r=1 \text {. } \quad \mu=\mu_0, \quad \varepsilon=\varepsilon_0 \\
a \times b & =2.28 \mathrm{~cm} \times 1.01 \mathrm{~cm} \\
f & =9.2 \mathrm{c}_{\mathrm{Hz}}
\end{aligned}
$
mode of wave $=$ dominant le. $T E_{10}$
$
\begin{aligned}
& T E_{m n}=T E_{10} \
& m=1, n=0
\end{aligned}
$
a) Calculation of cut-off frequency (fe),
$$
f_c=\frac{1}{2 \pi \sqrt{\mu \varepsilon}}\left[\sqrt{\left(\frac{m \pi}{a}\right)^2+\left(\frac{n \pi}{b}\right)^2}\right]
$$
$=\frac{1}{2 \pi \sqrt{4 \pi \times 10^{-7} \times 8.854 \times 10^{-12}}}\left[\sqrt{\left(\frac{\pi}{2.28 \times 10^{-2}}\right)^2+(0)^2}\right]$
$=6.58 \mathrm{GHz}^2$
b) Calculation of guide wavelength $\left(\lambda_9\right)$
$$
\begin{aligned}
\lambda_c & =c / f_c=0.0455 \mathrm{~m} \\\\
\lambda & =c / f=0.0326 \mathrm{~m} \\\\
\lambda_g & =\frac{\lambda}{\sqrt{1-\left(\lambda / \lambda_c\right)^2}}=0.046 \mathrm{~m}\\
\end{aligned}\\
$$
c) $\quad \eta=\frac{\eta}{\sqrt{1-(/ \pi)^2}}=540,42 \Omega$