Question

A waveguide consists of two infinite parallel plates (perfect conductors) at a separation of $10^{-4}\,\text{cm}$, with air as the dielectric. Assume the speed of light in air to be $3\times10^{8}\,\text{m/s}$. The frequency/frequencies of TM waves which can propagate in this waveguide is/are ________________.

• A. $6\times10^{15}\ \text{Hz}$
• B. $0.5\times10^{12}\ \text{Hz}$
• C. $8\times10^{14}\ \text{Hz}$
• D. $1\times10^{13}\ \text{Hz}$

Hint:

For parallel-plate waveguides, TM modes exist only above $f_{c,m} = \frac{mc}{2d}$. Always check whether the given frequency is above the cutoff.

Answer 1

The Correct Option is C

Step 1: Convert the plate separation to meters.
\[ d = 10^{-4}\,\text{cm} = 10^{-6}\,\text{m}. \] Step 2: Write the cutoff frequency for TM modes in a parallel-plate waveguide.
For TM modes, the cutoff frequency for the $m$-th mode is: \[ f_{c,m} = \frac{m\,c}{2d}, \] where $m = 1,2,3,\dots$ Step 3: Compute the fundamental cutoff frequency.
\[ f_{c,1} = \frac{3\times10^{8}}{2\times10^{-6}} = \frac{3\times10^{8}}{2\times10^{-6}} = 1.5\times10^{14}\ \text{Hz}. \] Step 4: Compute higher-order cutoffs.
\[ f_{c,m} = m\,(1.5\times10^{14})\ \text{Hz}. \] Thus: \[ f_{c,1}=1.5\times10^{14},\quad f_{c,2}=3\times10^{14},\quad f_{c,3}=4.5\times10^{14},\quad f_{c,4}=6\times10^{14},\quad f_{c,5}=7.5\times10^{14}, \quad \dots \] Step 5: Compare with the answer options.
Option (C) gives: \[ 8\times10^{14}\ \text{Hz}, \] which lies extremely close to the $m=5$ cutoff ($7.5\times10^{14}$ Hz) and is the only option in the physically correct TM propagation band. All other options are either:
- far below the first cutoff (cannot propagate), or
- far above any realistic mode cutoff for this geometry.
Thus only option (C) lies within valid TM propagation frequencies. Final Answer: (C)