Rectangular waveguide cutoff relationship In a rectangular waveguide, the cutoff wavelength for the dominant TE10 mode is measured as 8 cm. Determine the cutoff wavelength for the TE20 mode of the same waveguide, assuming standard air-filled operation and identical physical dimensions.

Introduction / Context:
The question checks understanding of how cutoff wavelengths scale with mode indices in rectangular waveguides. Knowing the geometry-to-cutoff relationship lets engineers quickly infer the behavior of higher-order modes once the dominant mode is known.

Given Data / Assumptions:

• Air-filled rectangular waveguide (μ_r ≈ 1, ε_r ≈ 1).
• Cutoff wavelength for TE10 is given as 8 cm.
• Standard rectangular guide dimensions: broad wall a, narrow wall b; TE_mn cutoff depends on a and b.

Concept / Approach:

For a rectangular waveguide, the cutoff wavelength for TE_mn modes is λ_c(mn) = 2 / √((m/a)^2 + (n/b)^2). For TE10, λ_c10 = 2a. Hence, if λ_c10 is known, we can compute a and then evaluate λ_c for other modes. For TE20, λ_c20 = a (since m = 2 gives 2 / (2/a) = a).

Step-by-Step Solution:

Verification / Alternative check:

Scaling logic: doubling the index along the a-dimension halves the cutoff wavelength (TE10 to TE20 halves λ_c). 8 cm → 4 cm confirms consistency.

Why Other Options Are Wrong:

8 cm assumes same mode; 6 cm and 2 cm do not match the exact TE_mn formula for m = 2; 12 cm would be larger than the dominant mode, which is impossible for higher-order modes.

Common Pitfalls:

Confusing TE20 with TE01 (which involves b), or thinking cutoff scales with physical length rather than modal indices.

Final Answer:

4 cm.