Correct Answer - Option 1 : 15 mm
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:
TE10 mode means m = 1, n = 0
The cut - off frequency of the dominant mode TE10 of the rectangular waveguide is:
\({f_c} = \frac{c}{{2a}}\)
Where a is the dimension of the inner broad wall
\(\begin{array}{l} a = 3 \times \frac{{{{10}^{10}}}}{{2 \times 10 \times {{10}^9}}}\\ a = 1.5\;cm \end{array}\)
= 15 mm