Rectangular Waveguide Cutoff Wavelength for TM11 For a rectangular waveguide of width a and height b, what is the cutoff wavelength λc for the TM11 mode (air-filled, uniform guide)?

Difficulty: Medium

Correct Answer: λc = 2 / sqrt((1/a)^2 + (1/b)^2)

Explanation:


Introduction / Context:
Cutoff characteristics of rectangular waveguides depend on geometry and mode indices. Correct formulas prevent design errors in frequency planning and component selection.


Given Data / Assumptions:

  • Mode TMmn with m = 1, n = 1.
  • Uniform, air-filled rectangular guide with dimensions a (width) and b (height).
  • Standard boundary conditions for perfect electric conductors.


Concept / Approach:

The general cutoff relation for TE/TM modes in a rectangular guide is λc = 2 / sqrt((m/a)^2 + (n/b)^2). Substituting m = 1 and n = 1 yields λc = 2 / sqrt((1/a)^2 + (1/b)^2). This expression assumes a and b are in meters so that λc is in meters.


Step-by-Step Solution:

1) Start with λc(m,n) = 2 / sqrt((m/a)^2 + (n/b)^2).2) Plug in m = 1, n = 1 → λc = 2 / sqrt((1/a)^2 + (1/b)^2).3) Confirm dimensional consistency: a and b in meters give λc in meters.


Verification / Alternative check:

Equivalent forms can be derived from the cutoff frequency fc = (c/2) * sqrt((m/a)^2 + (n/b)^2); using λc = c/fc yields the same expression.


Why Other Options Are Wrong:

Algebraic variants including ab or π terms here are incorrect for the rectangular-guide cutoff formula; the correct relation depends on the quadratic sum of inverse dimensions.


Common Pitfalls:

Confusing cutoff wavelength with guide wavelength; mixing TE and TM presence/absence conditions on m and n (TM requires both m and n nonzero).


Final Answer:

λc = 2 / sqrt((1/a)^2 + (1/b)^2)

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