For a slowly varying envelope $E=\psi e^{-ikx}$, the <paraxial approximation> to the <Helmholtz equation> is
$$
E_x=\frac{i}{2k}E_{zz}
+\frac{ik}{2}(n^2-1)E.
$$
Because $n=1+\mu W$ and $\mu^2\ll1$, passage through a sufficiently thin <phase screen> produces
$$
E(0,z)=e^{i\phi(z)},
\qquad
\phi(z)=\frac{k}{2}\int_{-\xi}^{0}(n^2-1)\,dx
\simeq k\mu\xi W(z).
$$
Its <modulus> is one at the screen exit. Beyond the screen, $n=1$, so the <parabolic wave equation> is $E_x=iE_{zz}/(2k)$. A <Taylor expansion> in propagation distance gives
$$
E(x,z)=E(0,z)+\frac{ix}{2k}E_{zz}(0,z)+O(x^2).
$$
Since
$$
\frac{(e^{i\phi})_{zz}}{e^{i\phi}}
=i\phi''-(\phi')^2,
$$
we find
$$
E(x,z)=e^{i\phi(z)}
\left[1-\frac{x}{2k}\phi''(z)
-\frac{ix}{2k}(\phi'(z))^2\right]+O(x^2).
$$
It follows that
$$
\boxed{|E(x,z)|=1-\frac{x}{2k}\phi''(z)+O(x^2)},
$$
or, equivalently, $|E|^2=1-x\phi''/k+O(x^2)$. Thus random <phase curvature> produces local focusing and defocusing: <free-space diffraction> converts phase fluctuations into amplitude fluctuations immediately after the screen.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 335 / 2 / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-335/2/i/solution