Correct Answer - Option 1 : 2.94 GHz
Concept:
The dominant mode in a particular waveguide is the mode having the lowest cut-off frequency.
The cut-off frequency for a rectangular waveguide with dimension ‘a (length)’ and ‘b (width)’ is given as:
\({{f}_{c\left( mn \right)}}=\frac{c}{2}\sqrt{{{\left( \frac{m}{a} \right)}^{2}}+{{\left( \frac{n}{b} \right)}^{2}}}\)
'm' and 'n' represents the possible modes.
c = speed of light = 3 × 1010 cm/s
Calculation:
For TE10 mode, m = 1, and n = 0.
The cut-off frequency will be:
\({{f}_{c\left( mn \right)}}=\frac{c}{2}\sqrt{{{\left( \frac{1}{a} \right)}^{2}}+{{\left( \frac{0}{b} \right)}^{2}}}\)
\({{f}_{c\left( mn \right)}}=\frac{c}{2a}\)
With a = 5.1 cm, the cut-off frequency will be:
\({{f}_{c\left( mn \right)}}_{TE_{(1,0)}}=\frac{3\times 10^{10}}{2\times 5.1}\)
fc(1,0) = 2.94 GHz