Correct Answer - Option 1 : EM wave of frequency 4.5 GHz
Concept:
Dominant Mode:
1. The dominant mode in a particular waveguide is the mode having the lowest cutoff frequency.
2. For rectangular waveguide, this is the TE10 mode.
3. The critical frequency for the dominant mode is given as:
\({f_c} = \frac{c}{2}\sqrt {{{\left( {\frac{m}{a}} \right)}^2} + {{\left( {\frac{n}{b}} \right)}^2}} \;\)
Where a and b are sides of the rectangle in cm,
m and n describe the mode.
For TE10 mode:
\(f_c=\frac{c}{2a}\)
Calculation:
Given:
a = 30 mm
b = 15 mm
For TE10 mode,
\(f_c \ = \ \frac{3 \ \times \ 10^8}{2 \ \times \ 30 \ \times \ 10^-3}\)
\(f_c \ = \ 5 \ GHz\)
So, the waves having the frequency above cutoff frequency (fc) will only pass.
Hence from the options given 4.5 GHz will not pass.
Hence the solution is a option (1).