Rectangular waveguide with cross-section 2a × a: which statement about cutoff (critical) wavelengths is correct? (Use standard TE/TM cutoff: λc = 2 / sqrt((m/width)^2 + (n/height)^2), noting TE and TM have the same λc for the same (m,n) when they exist.)

Introduction:
Cutoff (critical) wavelength in rectangular waveguides depends on cross-sectional dimensions and the integers (m, n) that define the field variations. This question checks whether you know that TE and TM modes with the same (m, n) share the same cutoff condition (when the TM mode exists) and how different index combinations compare.

Given Data / Assumptions:

• Rectangular guide size: width = 2a, height = a.
• Cutoff formula: λc = 2 / sqrt((m/width)^2 + (n/height)^2).
• TM00 does not exist; otherwise TE and TM with identical (m, n) have identical λc.

Concept / Approach:

Because cutoff arises from boundary conditions on the axial field component and transverse field separability, TE and TM families obey the same eigenvalue relation for the same (m, n). Numerical comparisons between different (m, n) show how λc changes with indices and aspect ratio.

Step-by-Step Solution:

Verification / Alternative check:

General identity: for a given (m, n) with m, n ≥ 0 and not both zero, λc depends only on the geometry and indices, not on whether the mode is TE or TM (except that TM00 is forbidden). Hence TE10 and TM10 share λc.

Why Other Options Are Wrong:

• TE10 = TE02 or TE10 half of TE01, etc.: direct substitutions give 4a vs a or 2a; they are not equal/half as claimed.
• TE01 = TM00: TM00 does not exist.
• TE10 half of TM10: they are equal, not half.

Common Pitfalls:

Assuming TE and TM always differ in cutoff for the same (m, n). They do not; the difference is in field composition, not the eigenvalue.

Final Answer:

Critical wavelength for TE10 is the same as that for TM10