= Solution
Measure the <phase> relative to propagation through the same geometrical thickness of air. If the <lens>'s central thickness is $t_0$, its spherical surface gives
$$
t(x)=t_0-\left[R-\sqrt{R^2-x^2}\right]
\simeq t_0-\frac{x^2}{2R},\qquad |x|\ll R.
$$
The optical-path excess is $(n-1)t(x)$, so the <phase> imparted by the <lens> is
$$
\boxed{\phi(x)=k(n-1)t_0-\frac{k(n-1)x^2}{2R}.}
$$
The constant part is a spatially uniform <phase>. The physically important <lens> transmission is therefore $T_f(x)=\exp[-ikx^2/(2f)]$ with
$$
\boxed{f=\frac R{n-1}.}
$$
One way to identify the focus is to append propagation to the on-axis point a distance $f$ away. Its path <phase> has the paraxial contribution $+kx^2/(2f)$, which cancels the <lens> <phase>. Thus all incident portions of a plane wave add with the same <phase> at that point. The diagram below shows the thickness change and both Fourier-transform arrangements.
\Image[/past-exam-of-the-laser-science-and-quantum-information-processing-course-of-the-university-of-oxford/2017-lens-phase-and-fourier-layouts.png]
{title=Quadratic thin-lens phase and two optical implementations of a scaled <Fourier transform>}
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