= Solution
Place the input plane one <focal length> before a single positive <lens> and the output plane one <focal length> after it. The <lens> maps each incident transverse spatial <frequency> to its corresponding point in the back focal plane. Referencing the input to the front focal plane removes the otherwise present output <phase curvature>, giving an exact scaled <Fourier transform> within the same paraxial model.
A short <Ray transfer matrix>[ray-matrix] check makes the geometry explicit. With free propagation $P(f)$ and <lens> <matrix> $L(f)$,
$$
\boxed{P(f)L(f)P(f)=
\begin{pmatrix}1&f\\0&1\end{pmatrix}
\begin{pmatrix}1&0\\-1/f&1\end{pmatrix}
\begin{pmatrix}1&f\\0&1\end{pmatrix}
=\begin{pmatrix}0&f\\-1/f&0\end{pmatrix}.}
$$
The vanishing diagonal entries are the Fourier-plane condition; the output is proportional to $\widetilde E_{\mathrm{in}}(kx/f)$. The final panel of the original diagram in part a shows the two focal distances. A <lens> with input immediately on its surface and output at the back focal plane would generally leave a quadratic output <phase>; the front-focal-plane choice avoids it.
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