= Solution
Let the two tangential <aperture> fields be $F_y(y,z)$ and $F_z(y,z)$, extended by zero off the <aperture>. The normal component does not enter $\mathbf n\times\mathbf E_{\rm inc}$, which equals $-F_z\mathbf e_y+F_y\mathbf e_z$. Apply the <Weyl plane-wave representation> and <curl> to both components. With the same kernel normalization as part (a),
$$
\boxed{\mathbf E(\mathbf r)=\frac1{2(2\pi)^2}\int_{\mathbb R^2}
\left[\widehat F_y\left(\mathbf e_y-\frac\nu\kappa\mathbf e_x\right)
+\widehat F_z\left(\mathbf e_z-\frac\eta\kappa\mathbf e_x\right)\right]
e^{i\mathbf q\cdot\mathbf r}\,d\nu\,d\eta}.
$$
This is the <electromagnetic aperture angular spectrum>. The two <electromagnetic polarization> vectors are linearly independent and both orthogonal, bilinearly, to $\mathbf q$; they need not be orthogonal to each other. Arbitrary tangential <electromagnetic polarization> is therefore represented, with its outgoing longitudinal component fixed by $\kappa E_x+\nu E_y+\eta E_z=0$ rather than independently prescribed. Away from grazing components one could equally express the spectrum in a transverse-electric and transverse-magnetic basis. The same limiting outgoing branch handles grazing and <evanescent> components.
For each component the magnetic amplitude follows from <Faraday's law> as $\mathbf B=\mathbf q\times\mathbf E/\omega$. Together with $\mathbf q^2=k^2$ and <transversality of an electromagnetic plane wave>, this verifies the source-free <Maxwell equations>. If the doubled-kernel <aperture> convention is used, multiply the entire expression by two, exactly as in part (a).
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