= Solution
Take transverse length units in which the <wave phase> modulation has <wavenumber> one. Set $E_0(z)=e^{i\cos z}$. A <Taylor expansion> of the free parabolic propagator gives
$$
E(x,z)=E_0+\frac{ix}{2k}E_0''-\frac{x^2}{8k^2}E_0''''+O(x^3).
$$
For a general real <wave phase> $\phi$,
$$
\frac{E_0''}{E_0}=i\phi''-(\phi')^2,\qquad
\frac{E_0''''}{E_0}=i\phi''''-4\phi'\phi'''-3(\phi'')^2-6i(\phi')^2\phi''+(\phi')^4.
$$
Multiplying the envelope expansion by its <complex conjugate> and collecting powers of $x$ gives the <phase-to-intensity conversion near a deterministic phase screen>,
$$
I=1-\frac{x}{k}\phi''+\frac{x^2}{k^2}\left[(\phi'')^2+\phi'\phi'''\right]+O(x^3).
$$
For $\phi=\cos z$, $\phi'=-\sin z$, $\phi''=-\cos z$ and $\phi'''=\sin z$, hence
$$
\boxed{I(x,z)=1+\frac{x}{k}\cos z+\frac{x^2}{k^2}\cos2z+O(x^3)}.
$$
The first-order brightening occurs where the <wave phase> curvature is negative. Both correction terms average to zero over a transverse period, in agreement with power conservation. If a dimensional modulation $\cos(qz)$ is used instead, the two terms are $(q^2x/k)\cos(qz)$ and $(q^4x^2/k^2)\cos(2qz)$.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2007/iii/paper-80-phase-screen.png]
{title=Phase curvature converts a unit-amplitude sinusoidal phase screen into intensity variations; exact paraxial propagation is compared with its quadratic near-screen approximation}
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