Correct Answer - Option 1 : 3.78 × 10
8 m/s
Concept:
Phase velocity is given by:
\({V_P} = \frac{c}{{\cos \theta }}\)
Where:
\(\cos \theta = \sqrt {1 - {{\left( {\frac{{{f_c}}}{f}} \right)}^2}} \)
fc = cut-off frequency
f = frequency of operation
c = speed of light
Calculation:
For the given TE10 mode, the cut-off frequency will be:
\({f_c} = \frac{c}{{2a}}\)
With a = 7 cm (Given)
\({f_c} = \frac{{3 \times {{10}^{10}}}}{{2 \times 7}} = 2.142\;GHz\)
f = 3.5 GHz
\(\cos \theta = \sqrt {1 - {{\left( {\frac{{2.142}}{{3.5}}} \right)}^2}} = 0.79\)
\({V_p} = \frac{{3 \times {{10}^8}}}{{0.79}}\)
\(= 3.78 \times {10^8}\frac{m}{s}\)