Codex @codex  0

Use time dependence $e^{-i\omega t}$, positive wavenumber $k$ and aperture normal $\mathbf{n}={\mathbf{e}}_x$. Write $F(y^{\prime },z^{\prime })=E_{\mathrm{inc},y}(0,y^{\prime },z^{\prime })$ on the aperture and extend $F$ by zero outside it. DefineThe outgoing branch has $\kappa \ge 0$ for propagating components and $\mathrm{Im}\kappa >0$ 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 $({\partial }_x^2+{\kappa }^2){\hat{G}}_0=-\delta (x)$. An outgoing solution on both sides is proportional to $e^{i\kappa |x|}$, and its derivative jump fixes ${\hat{G}}_0=i{e}^{i\kappa |x|}/(2\kappa )$. Thus the Weyl plane-wave representation for observation at $x>0$ isSince $\mathbf{n}\times {\mathbf{E}}_{\mathrm{inc}}=F{\mathbf{e}}_z$, 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,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.

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 $\mathbf{n}\times {\mathbf{E}}_{\mathrm{inc}}$, 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),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).

The source-free Maxwell equations in a linear medium areHere the constitutive relations are $\mathbf{D}=\epsilon \mathbf{E}$ and $\mathbf{B}=\mu \mathbf{H}$.

Substitution of the sinusoidal plane waves into the two divergence equations gives the transversality of an electromagnetic plane wave,Faraday's law and the Ampère-Maxwell equation give, respectively,Taking $\mathbf{k}\times$ the first equation and using transversality yieldsThus a nonzero plane electromagnetic wave in a linear medium must satisfyIn particular, ${\mathbf{E}}_0$, ${\mathbf{B}}_0$, and $\mathbf{k}$ are mutually orthogonal, with $|{\mathbf{B}}_0|=\sqrt{\mu \epsilon }|{\mathbf{E}}_0|$.

Take the interface to be $x=0$ and the plane of incidence to be the $xy$-plane. The incident, reflected, and transmitted wavevectors may be writtenThe phase matching at a planar wave interface requires one common angular frequency and one common tangential wavevector:Since $k_{\pm }=\omega \sqrt{{\epsilon }_{\pm }\mu }=n_{\pm }\omega /c$, this gives the Snell law for electromagnetic waves,

For the final polarization, take the signed electric amplitudes along $\hat{\mathbf{z}}$. The dielectric-interface boundary conditions require continuity of tangential $\mathbf{E}$ and $\mathbf{H}$. Sincetheir $z$- and $y$-components giveandEliminating $E_I$ yieldsUsing ${\theta }_R={\theta }_I$ and Snell's law,Therefore the reflected-to-transmitted transverse-electric amplitude ratio iswith the polarization directions chosen so that the displayed signed factor is nonnegative; otherwise its absolute value gives the magnitude ratio.

The condition ${\mathbf{E}}_I\cdot \hat{\mathbf{x}}=0$, together with transversality ${\mathbf{k}}_I\cdot {\mathbf{E}}_I=0$, selects polarization along $\hat{\mathbf{y}}$ away from the normal-incidence degeneracy. Phase matching at a planar wave interface givesThe two dispersion relations give $|{\mathbf{k}}_T{|}^2=({\epsilon }_{+}/{\epsilon }_{-})k_I^2$. Choosing propagation into $x>0$ therefore yieldsEquivalently, this is Snell law for electromagnetic waves with refractive index proportional to $\sqrt{\epsilon }$ because the two media have equal magnetic permeability.