= Solution
For the shown sequence of free space, <thin lens>, and free space, multiply the <ray transfer matrices> in traversal order:
$$
M=P(L_2)F(f)P(L_1)=\begin{pmatrix}
1-L_2/f&L_1+L_2-L_1L_2/f\\-1/f&1-L_1/f
\end{pmatrix}.
$$
To obtain an ordinary <Fourier transform> of an arbitrary input field, neither an input quadratic <phase> nor an output quadratic <phase> may remain. Thus $A=D=0$, requiring
$$
\boxed{L_1=L_2=f,\qquad M=\begin{pmatrix}0&f\\-1/f&0\end{pmatrix}.}
$$
Using the physically consistent kernel derived in part b gives
$$
\boxed{E(x')=\frac{e^{ikL_0}}{\sqrt{i\lambda f}}\int E(x)e^{-i2\pi xx'/(\lambda f)}\,dx
=\frac{e^{ikL_0}}{\sqrt{i\lambda f}}\widehat E\!\left(\frac{x'}{\lambda f}\right).}
$$
The spatial Fourier coordinate is $x'/(\lambda f)$. If the source's schematic integral normalization is retained instead, set its slowly varying denominator to $L_0$ to obtain the same <Fourier transform> with overall prefactor $ik e^{ikL_0}/(2\pi L_0)$. Keeping the source's incorrect quadratic sign would reverse the Fourier coordinate. These overall-normalization conventions do not alter the focal-plane placement, but an unapproximated transverse-dependent denominator would not give an exact <Fourier transform>.
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