 Solution

Represent the thin layer by its wave phase jump at the exit $x=0$ and take the incident envelope to have amplitude one. Then $E(0,z)=e^{i\varphi (z)}$. Stationarity makes the wave phase mean a constant $\mu$; it need not be zero, since the question specifies only the variance. For a normal distribution marginal, write $\varphi =\mu +\sigma Z$ with $Z\sim N(0,1)$. If $F(t)=\mathbb{E}[e^{itZ}]$, integration by parts in the Gaussian density gives $F^{\prime }(t)=-tF(t)$ and $F(0)=1$, hence $F(t)=e^{-t^2/2}$. Thus its characteristic function givesThe Fourier transform of this constant is $2\pi M\delta (\nu )$, interpreted distributionally for the infinite plane wave. Applying the propagator from part (a),If the wave phase has zero mean, omit $e^{i\mu }$. A nonunit incident amplitude simply multiplies the expression. Only the one-point Gaussian distribution is needed for this expected value; no assumption about joint Gaussian distributions at separated points is needed.