Question

A rectangular waveguide of width w and height h has cut-off frequencies for $T{E}_{10}andT{E}_{11}$ modes in the ratio 1 : 2, The aspect ratio w/h, rounded off to two decimal places, is

• A. 1.732

Answer 1

The correct option is A 1.732
$f_{cmn}=\frac{c}{2}\sqrt{{\left(\frac{m}{a}\right)}^2+{\left(\frac{n}{b}\right)}^2}$

$ForT{E}_{10}mode,$

$f_{c10}=\frac{c}{2w}...\left(i\right)$

and For $T{E}_{11}$ mode,

$f_{c11}=\frac{c}{2}\sqrt{{\left(\frac{1}{w}\right)}^2+{\left(\frac{1}{h}\right)}^2}$

$=\frac{c}{2w}\sqrt{1+{\left(\frac{w}{h}\right)}^2}...\left(ii\right)$

given, $\frac{f_{c10}}{f_{c11}}=\frac{1}{2}...\left(iii\right)$

put (i), (ii) in (iii)

$\Rightarrow \frac{\frac{c}{2w}}{\frac{c}{2w}\sqrt{1+{\left(\frac{w}{h}\right)}^2}}=\frac{1}{2}\Rightarrow \sqrt{1+{\left(\frac{W}{h}\right)}^2}=2$

On solving above euation, we get,

$\frac{w}{h}=\sqrt{3}=1.732$