Represent the thin layer by its wave phase jump at the exit and take the incident envelope to have amplitude one. Then . Stationarity makes the wave phase mean a constant ; it need not be zero, since the question specifies only the variance. For a normal distribution marginal, write with . If , integration by parts in the Gaussian density gives and , hence . Thus its characteristic function gives
The Fourier transform of this constant is , interpreted distributionally for the infinite plane wave. Applying the propagator from part (a),
If the wave phase has zero mean, omit . 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.

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