Correct Answer - Option 3 : 15 cm
Concept:
The dominant mode of a rectangular waveguide is TE10
The cutoff frequency is given as:
\({f_c} = \frac{c}{2}\sqrt {{{\left( {\frac{m}{a}} \right)}^2} + {{\left( {\frac{n}{b}} \right)}^2}} \)
Calculations:
\((f_c)_{TE10}= \frac{c}{2a}\)
= \(\frac{3\times 10^{10}}{20}=1.5 ~GHz\)
Operating frequency:
f = 2.5 GHz
Operating wavelength:
\(\lambda=\frac{c}{f}=\frac{3\times 10^{10}}{2.5 \times 10^{9}}\)
= 12 cm
Guide wavelength is given by:
\({\lambda _g} = \frac{\lambda }{{\sqrt {1 - {{\left( {\frac{{{f_c}}}{f}} \right)}^2}} }}\)
\(\lambda_g = \frac{{12}}{{\sqrt {1 - {{\left( {\frac{1.5}{{2.5}}} \right)}^2}} }}\)
\({\lambda_g} = \textbf{15}~\textbf{cm}\)
∴ The correct option is 3.
