Rectangular Waveguide Cutoff Relation (Dominant TE10) A rectangular waveguide has broad-wall width a. At cutoff for the dominant TE10 mode, which relation between a and the cutoff wavelength λc is correct?

Introduction / Context:
Cutoff conditions in rectangular waveguides determine the lowest frequency that can propagate for a given mode. For the dominant TE10 mode, the broad dimension a controls the cutoff wavelength and hence the usable frequency band for single-mode operation.

Given Data / Assumptions:

• Uniform, air-filled rectangular waveguide with width a (broad wall) and height b (narrow wall).
• Dominant mode is TE10, with one half-sine variation across a and none across b.
• Perfect electric conductor approximation for walls.

Concept / Approach:

The cutoff wavelength for mode TEmn in a rectangular guide is λc = 2 / sqrt((m/a)^2 + (n/b)^2). For TE10, m = 1 and n = 0, giving λc = 2a. Rearranging yields a = λc / 2. This simple proportionality is a foundational design rule for waveguides and is frequently used for selecting a standard WR size given a desired band.

Step-by-Step Solution:

Verification / Alternative check:

The well-known dominant-mode cutoff frequency is fc10 = c / (2a). Converting to wavelength (λc = c / fc10) again gives λc = 2a, consistent with the above.

Why Other Options Are Wrong:

a = λc or a = 2λc contradict the derived relation; a = λc / 4 is also incorrect; “independent of λc” ignores the defining cutoff equation.

Common Pitfalls:

Confusing λ (operating wavelength) with λc (cutoff wavelength); λ must be shorter than λc for propagation in a given mode.

Final Answer:

a = λc / 2