Discussion
A square wave guide with perfect conductor boundaries is of cross-sectional area \(l^{2}\). The wave guide is placed in vacuum. The minimum frequency of the electromagnetic waves that can propagate through this wave guide is ........ .
Answer: C
The key's answer is \(\frac{\pi c}{l}\), the cut-off expressed as an angular frequency.
In a rectangular guide of sides a and b, the cut-off angular frequency of the \(TE_{mn}\) mode is \(\omega_{c} = c\pi\sqrt{(m/a)^{2} + (n/b)^{2}}\). The lowest mode in a square guide of side l is \(TE_{10}\) (or \(TE_{01}\)):
\[ \begin{aligned} \omega_{c} &= \frac{\pi c}{l} \\ f_{c} &= \frac{\omega_{c}}{2\pi} = \frac{c}{2l} \end{aligned} \]
Waves below this cut-off cannot propagate. The cycle frequency c/2l is not among the options, so the paper's 'frequency' is the angular frequency. \(\sqrt{2}\pi c/l\) is the cut-off of the higher \(TE_{11}\) and \(TM_{11}\) modes, and \(c/l\) and \(2c/l\) correspond to no lowest mode.
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