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For a rectangular waveguide of internal dimensions $a\times b$ $(a>b)$, the cut-off frequency for the TE$_{11}$ mode is the arithmetic mean of the cut-off frequencies for the TE$_{10}$ mode and TE$_{20}$ mode. If $a=\sqrt{5}$ cm, the value of $b$ (in cm) is

1. 1
2. 2
3. 3
4. 4

Correct answer: 2

Solution

For a rectangular waveguide, $f_{cmn}=\frac{c}{2}\sqrt{(m/a)^2+(n/b)^2}$. Thus $f_{c10}=\frac{c}{2a}$, $f_{c20}=\frac{c}{a}$, and $f_{c11}=\frac{c}{2}\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}$. Using the arithmetic mean condition gives $\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}=\frac{3}{2a}$, which leads to $b=2$ cm when $a=\sqrt{5}$ cm.

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