Correct Answer - Option 1 : 6.25 GHz
Concept:
Minimum operating frequency or the cut off frequency for a rectangular waveguide is given by:
\({f_c} = \frac{c}{2}\sqrt {\frac{{{m^2}}}{{{a^2}}} + \frac{{{n^2}}}{{{b^2}}}} \)
a = length of the waveguide
b = height of the waveguide
m,n = modes of operation
Calculation:
Given, a = 4 cm
b = 3 cm
The minimum frequency in TE11 is nothing but the cut-off frequency calculated as:
\({f_c} = \frac{{3 \times {{10}^8}}}{2}\sqrt {\frac{1}{{{{\left( {4 \times {{10}^{ - 2}}} \right)}^2}}} + \frac{1}{{{{\left( {3 \times {{10}^{ - 2}}} \right)}^2}}}} \)
\(f_c= \frac{{3 \times {{10}^8}}}{{2 \times {{10}^{ - 2}}}} \times \frac{5}{{4 \times 3}} = 6.25 \times {10^9}Hz\)
fC = 6.25 GHz