= Solution
For $n>0$, $k_0h\sin\alpha=n\pi$. Hence the numerator and denominator of the combined <independent-edge radiation from an open planar duct> both vanish at $\theta=\pm\alpha$. At the positive angle, the ratio has the removable limit
$$
\lim_{\theta\to\alpha}\frac{\sin(k_0h\sin\theta)}{\cos\theta-\cos\alpha}
=-(-1)^n k_0h\cot\alpha.
$$
At the negative angle the ratio changes sign, as does $\sin(\theta/2)$, so both directions give the same total <velocity potential>. Since $2\sin(\alpha/2)\cos(\alpha/2)=\sin\alpha$, the result is
$$
\boxed{\phi_d(\pm\alpha)\sim\frac{ikh}{2}\sqrt{\frac{2}{\pi k_0r}}e^{-ik_0r-i\pi/4}\qquad(n>0).}
$$
For the zero mode, use $\alpha=0$ directly: $\sin(\theta/2)\sin(k_0h\sin\theta)/(\cos\theta-1)\to-k_0h$. Its two directions coincide and
$$
\boxed{\phi_d(0)\sim ik_0hD(r)\qquad(n=0).}
$$
\b[The two-edge field is finite in the constituent propagation directions.] Its magnitude there is enhanced by a factor of order $kh$ over an ordinary isolated-edge diffracted amplitude away from a shadow boundary. An isolated-edge angular formula has a pole in these directions, because its <geometrical optics> and edge-wave split is nonuniform; the exact single-edge field is not infinite. In the combined field the equal and opposite pole residues cancel, and the difference of endpoint propagation phases produces the finite limit above. Taking this combined limit describes the leading far-field beam; it does not justify evaluating either nonuniform single-edge expression separately at its pole.
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