Question

A standard air-filled rectangular waveguide with dimensions \( a = 8 \, \text{cm}, b = 4 \, \text{cm} \), operates at 3.4 GHz. For the dominant mode of wave propagation, the phase velocity of the signal is \( v_p \). The value (rounded off to two decimal places) of \( v_p / c \), where \( c \) denotes the velocity of light, is

Hint:

The phase velocity in a rectangular waveguide depends on the cutoff frequency and the operating frequency, with the phase velocity being greater than the speed of light.

Answer 1

Correct Answer: 1.15

For the dominant mode of propagation in a rectangular waveguide, the phase velocity \( v_p \) is given by: \[ v_p = \frac{c}{\sqrt{1 - \left( \frac{f_c}{f} \right)^2}} \] where \( f_c \) is the cutoff frequency. The cutoff frequency for the dominant mode is: \[ f_c = \frac{c}{2a} \] Substituting \( a = 8 \, \text{cm} = 0.08 \, \text{m} \) and \( f = 3.4 \, \text{GHz} \), we find: \[ v_p = 1.15 \, c \] Thus, \( v_p / c = \boxed{1.15} \).