= Solution
Use the forward-slope convention established in part a, with $x'=Ax+B\theta$ and $\theta'=Cx+D\theta$, where $AD-BC=1$. Assume $B\ne0$. Solve for the endpoint slopes:
$$
\theta=\frac{x'-Ax}{B},\qquad \theta'=\frac{Dx'-x}{B}.
$$
<Fermat's principle> makes the optical path stationary against variations at internal interfaces. Consequently only endpoint terms survive when its endpoints change. For optical path $l(x,x')$, these are
$$
\frac{\partial l}{\partial x}=-\theta=\frac{Ax-x'}B,\qquad
\frac{\partial l}{\partial x'}=\theta'=\frac{Dx'-x}B.
$$
Integrate the first equation with respect to $x$, then match the second and impose $l(0,0)=L_0$. The resulting <ABCD optical path generating function> is
$$
\boxed{l(x,x')=L_0+\frac{Ax^2+Dx'^2-2xx'}{2B}.}
$$
Thus the geometrical kernel defined in the source would have the form
$$
\boxed{K(x,x')=\frac{\exp\!\left\{ik\left[L_0+(Ax^2+Dx'^2-2xx')/(2B)\right]\right\}}{L_0+(Ax^2+Dx'^2-2xx')/(2B)}.}
$$
\b[The printed minus sign is inconsistent with forward propagation under these conventions.] A concrete counterexample is free space: $A=D=1$, $B=L>0$, $C=0$. Its true geometrical path is
$$
\sqrt{L^2+(x'-x)^2}=L+\frac{(x'-x)^2}{2L}+O\!\left((x'-x)^4/L^3\right),
$$
whereas the printed expression predicts a path shorter than $L$ for nonzero transverse separation. For example, $L=1\,\mathrm m$, $x=0$, $x'=1\,\mathrm{mm}$ gives $1\,\mathrm m+0.5\,\mu\mathrm m$, rather than the printed $1\,\mathrm m-0.5\,\mu\mathrm m$. Reversing the slope convention would also reverse the free-space $B$ entry and must be done consistently; it cannot justify the minus sign with $P(L)$ from part a.
The printed form can be recovered by explicitly using the opposite slope coordinate $\widetilde\theta=-dx/dz$. Then $\widetilde B=-B$, free space has $\widetilde B=-L$, and the same physical path is $l=L_0-(Ax^2+Dx'^2-2xx')/(2\widetilde B)$. Thus the sign can be a convention choice, but the printed minus sign must not be combined with the positive-$B$ free-space <ray transfer matrix> used here.
For physical field propagation, the paraxial <diffraction> kernel retains this quadratic optical-path <phase> while using the leading spreading <amplitude>. In one transverse dimension the normalized <linear canonical transform> is
$$
E(x')=\frac{e^{ikL_0}}{\sqrt{i\lambda B}}\int\exp\!\left[\frac{ik}{2B}(Ax^2+Dx'^2-2xx')\right]E(x)\,dx,
$$
with a consistent square-root branch and $\lambda=2\pi/k$. The source's prefactor and $1/l$ kernel are a schematic one-dimensional version of the <Fresnel-Kirchhoff diffraction integral>, not this normalized general-system propagator. The quadratic denominator is normally replaced by its leading value when making the Fresnel approximation. When $B=0$, propagation is an imaging limit rather than an ordinary integral with division by $B$.
Back to article page
