= Guided wavelength
{title2=$\lambda_g=2\pi/\beta$}
The <guided wavelength> is the distance for a $2\pi$ change of axial <wave phase> in a propagating <normal mode>. For the <vacuum>-filled <rectangular electromagnetic waveguide>,
$$
\lambda_g=\frac{\lambda_0}{\sqrt{1-(f_c/f)^2}},\qquad
\frac1{\lambda_g^2}=\frac1{\lambda_0^2}-\frac1{(2a)^2}
$$
for $\mathrm{TE}_{10}$, with $\lambda_0=c/f$ and $f>f_c$. It exceeds the <vacuum> <wavelength> at the same <frequency>. This longer axial spacing is consistent with $v_{\rm ph}>c$: multiplying $\lambda_g$ by $f$ gives the <phase velocity>, not the <group velocity>.
Two oppositely traveling fields at the same <frequency> form a <standing wave>. For a perfect electric short at $z=0$, the total <electric field> is proportional to $\sin(\beta z)\sin(\pi x/a)$, so successive axial electric nodes are separated by
$$
d=\frac\pi\beta=\frac{\lambda_g}{2}.
$$
Thus neighboring minima in a <slotted line> scan give half a <guided wavelength>. In the original $a=30\,\mathrm{mm}$, $f=1.25f_c$ example, $d=40\,\mathrm{mm}$ implies $\lambda_g=80\,\mathrm{mm}$, not $40\,\mathrm{mm}$. Combined with the chosen $f$, the inferred broad dimension is $a=\lambda_0/[2\sqrt{1-(\lambda_0/\lambda_g)^2}]=30\,\mathrm{mm}$. Below cutoff, $\beta$ is imaginary and $2\pi/\beta$ is not a physical periodic spacing.
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