Understanding the Concept:
In a rectangular waveguide with internal dimensions $a$ (width) and $b$ (height) where $a > b$, electromagnetic waves propagate in distinct modes ($TE_{mn}$ or $TM_{mn}$). The cut-off frequency ($f_c$) is the minimum frequency required for a specific mode to propagate through the waveguide. The general formula for the cut-off frequency is:
\[
f_c = \frac{c}{2} \sqrt{\left(\frac{m}{a}\right)^2 + \left(\frac{n}{b}\right)^2}
\]
where $c$ is the speed of light in free space ($3 \times 10^8 \, \text{m/s}$), and $m, n$ are the mode integers.
Step 1: Identifying the Dominant Propagation Mode
The dominant mode of a waveguide is the specific mode that has the lowest cut-off frequency. For a standard rectangular waveguide where the width $a$ is greater than the height $b$ ($a > b$), the dominant mode is always the $\text{TE}_{10}$ mode ($m = 1, n = 0$).
Step 2: Simplifying the Cut-off Formula for the Dominant Mode
Substitute $m = 1$ and $n = 0$ into the general cut-off frequency equation:
\[
f_c(TE_{10}) = \frac{c}{2} \sqrt{\left(\frac{1}{a}\right)^2 + \left(\frac{0}{b}\right)^2} = \frac{c}{2} \sqrt{\frac{1}{a^2}} = \frac{c}{2a}
\]
Step 3: Substituting the Given Numerical Values
The problem provides the following dimensions for the waveguide:
• Width: $a = 2.5 \, \text{cm} = 2.5 \times 10^{-2} \, \text{m} = 0.025 \, \text{m}$
• Height: $b = 1 \, \text{cm} = 1.0 \times 10^{-2} \, \text{m}$ (Not needed for the dominant mode calculation)
• Speed of light: $c = 3 \times 10^8 \, \text{m/s}$
Substitute these values into the simplified dominant mode formula:
\[
f_c = \frac{3 \times 10^8}{2 \times (2.5 \times 10^{-2})}
\]
Step 4: Performing the Calculation
Simplify the terms in the denominator:
\[
2 \times 2.5 \times 10^{-2} = 5.0 \times 10^{-2} = 0.05
\]
Now perform the division to find the frequency:
\[
f_c = \frac{3 \times 10^8}{5 \times 10^{-2}} = \frac{3}{5} \times 10^{10} = 0.6 \times 10^{10} \, \text{Hz}
\]
Convert the result into Giga-Hertz ($\text{GHz}$), where $1 \, \text{GHz} = 10^9 \, \text{Hz}$:
\[
f_c = 6 \times 10^9 \, \text{Hz} = 6 \, \text{GHz}
\]
The cut-off frequency for the dominant mode is exactly 6 GHz.