= Solution
Use time dependence $e^{-i\omega t}$, positive <wavenumber> $k$ and <aperture> normal $\mathbf n=\mathbf e_x$. Write $F(y',z')=E_{{\rm inc},y}(0,y',z')$ on the <aperture> and extend $F$ by zero outside it. Define
$$
\widehat F(\nu,\eta)=\int_A F(y',z')e^{-i(\nu y'+\eta z')}\,dy'\,dz',\qquad
\kappa=\sqrt{k^2-\nu^2-\eta^2},\qquad \mathbf q=(\kappa,\nu,\eta).
$$
The outgoing branch has $\kappa\ge0$ for propagating components and $\operatorname{Im}\kappa>0$ for <evanescent> components.
Fix the <Green function> normalization by $G_0(\mathbf r)=e^{ik|\mathbf r|}/(4\pi|\mathbf r|)$, so $(\Delta+k^2)G_0=-\delta$. Its transverse <Fourier transform> satisfies $(\partial_x^2+\kappa^2)\widehat G_0=-\delta(x)$. An outgoing solution on both sides is proportional to $e^{i\kappa|x|}$, and its <derivative> jump fixes $\widehat G_0=i e^{i\kappa|x|}/(2\kappa)$. Thus the <Weyl plane-wave representation> for observation at $x>0$ is
$$
G_0=\frac{i}{2(2\pi)^2}\int_{\mathbb R^2}
\frac{e^{i\kappa x+i\nu(y-y')+i\eta(z-z')}}\kappa\,d\nu\,d\eta.
$$
Since $\mathbf n\times\mathbf E_{\rm inc}=F\mathbf e_z$, the given <aperture> integral becomes a superposition of <curls> of $\mathbf e_z e^{i\mathbf q\cdot\mathbf r}$. Each <curl> supplies $i\mathbf q\times\mathbf e_z=i(\nu,-\kappa,0)$. Therefore, with the unit prefactor in that integral,
$$
\boxed{\mathbf E(x,y,z)=\frac1{2(2\pi)^2}\int_{\mathbb R^2}
\widehat F(\nu,\eta)\left(-\frac\nu\kappa,1,0\right)
e^{i\kappa x+i\nu y+i\eta z}\,d\nu\,d\eta}.
$$
Every vector amplitude is transverse to $\mathbf q$, as <Maxwell's equations> require. The <polarizer> specifies the incident <aperture> field; <diffraction> can regenerate an $x$ component outside the <aperture>. In contrast, its $z$ component remains zero in this representation. <Evanescent> <plane waves> are needed for the complete near field, not just the propagating disk $\nu^2+\eta^2\le k^2$.
The factor $1/2$ follows explicitly from the standard <outgoing Green function for the three-dimensional Helmholtz equation> and the printed unit <curl> prefactor. If the intended kernel is the doubled half-space kernel $2G_0$, or the <aperture> formula carries the usual compensating factor $2$, delete that $1/2$. The latter normalization makes the tangential output trace equal to $F$. Keeping the convention explicit avoids silently changing the supplied representation; the angular dependence and <transversality of an electromagnetic plane wave> are the same in either normalization. Reversing the specified normal also reverses the overall sign.
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