The supplied <speckle contrast of a sinusoidal wavefront error> gives $h_0=\lambda\sqrt C/\pi$. Each sinusoid has <root mean square> error $h_0/\sqrt2$. For orthogonal modes, or for uncorrelated phases so that cross terms average to zero, the total mean square adds:
$$
h_{\rm rms}^2=M\frac{h_0^2}2=\frac{\pi N^2}8\frac{\lambda^2C}{2\pi^2}.
$$
Therefore the assumed equal-contrast modal model gives
$$
\boxed{h_{\rm rms}=\frac{N\lambda\sqrt C}{4\sqrt\pi}.}
$$
If the individual contrasts are not equal, the corresponding formula is $h_{\rm rms}^2=\lambda^2\sum_iC_i/(2\pi^2)$. The error here is optical-path <wavefront error>; for a near-normal reflecting mirror, physical surface error is half the optical-path error.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 338 / 2 / b / iii / Solution

{title2=$C=(\pi h_0/\lambda)^2$}

For a small sinusoidal <optical path length> error $h_0\cos(2\pi x/\Lambda)$, expand its phase factor to first order. The two exponential harmonics each have field amplitude $i\pi h_0/\lambda$ relative to the unperturbed field, so each speckle has contrast $C=(\pi h_0/\lambda)^2$. Its angular displacement is $\lambda/\Lambda$. Here $h_0$ is optical path error, not physical mirror displacement; a reflecting mirror introduces twice its surface displacement in the near-normal limit.

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