GATE 2018 EC – Question 58
The cutoff frequency of $TE_{01}$ mode of an air filled rectangular waveguide having inner dimensions $a$ cm $\times$ $b$ cm ($a>b$) is twice that of the dominant $TE_{10}$ mode. When the waveguide is operated at a frequency which is 25% higher than the cutoff frequency of the dominant mode, the guide wavelength is found to be 4 cm. The value of $b$ (in cm, correct to two decimal places) is ________.
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Correct answer: 0.74 to 0.76
Explanation
$f_{c,01}=\frac{c}{2b}=2f_{c,10}=\frac{c}{a}$, so $b=\frac a2$. At $f=1.25f_{c10}$, $\lambda=\frac{c}{f}=\frac{2a}{1.25}=1.6a$ and $\lambda_g=\frac{\lambda}{\sqrt{1-(1/1.25)^2}}=\frac{1.6a}{0.6}=2.667a=4$ cm. So $a=1.5$ cm and $b=0.75$ cm.