Cutoff Wavelength in a Circular Waveguide: Dependence on Physical Size For a given propagation mode (for example, the dominant TE11 mode), how does the cutoff wavelength vary with the circular waveguide's inside diameter?

Difficulty: Easy

Correct Answer: Directly proportional to guide diameter

Explanation:


Introduction / Context:
Waveguides support propagation only above cutoff. In circular waveguides, each mode has a specific cutoff wavelength set by the boundary conditions and the diameter.


Given Data / Assumptions:

  • We consider qualitative dependence of cutoff wavelength on diameter.
  • Mode constants are fixed (e.g., TE11).


Concept / Approach:

For a circular waveguide, lambda_c = k * D for a given mode, where k is a constant determined by the root of a Bessel function equation. For the dominant TE11 mode, lambda_c ≈ 1.706 * D. Thus, cutoff wavelength scales linearly with diameter.


Step-by-Step Solution:

1) Recognize that cutoff is set by boundary conditions leading to characteristic equations with Bessel function roots.2) For a fixed mode, the root is a constant, so lambda_c must scale directly with D.3) Therefore, doubling D doubles lambda_c; halving D halves lambda_c.


Verification / Alternative check:

Design charts or tables for circular waveguides list lambda_c as a constant times D for each mode.


Why Other Options Are Wrong:

Quadratic, cubic, square-root, or diameter-independent relations contradict the linear dependence defined by the modal eigenvalue.


Common Pitfalls:

Confusing cutoff wavelength with cutoff frequency (f_c ∝ 1/D). Remember f_c and lambda_c have inverse trends.


Final Answer:

Directly proportional to guide diameter.

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