Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 338 / 3 / a / iii / Solution

The speckle contrast of a sinusoidal wavefront error gives a peak modal wave amplitude $h_0=(\lambda /\pi )\sqrt{C}$. If all $M$ residual modes have that wave amplitude, their coefficient norm is $H_{\mathrm{coeff}}=\sqrt{M}{h}_0$. Substituting the printed quadrant count gives the intended expressionThere is a normalization issue if this quantity is called physical spatial root mean square wavefront error. For $h(\mathbf{r})={\sum }_k{h}_{0k}\cos (2\pi {\mathbf{u}}_k\cdot \mathbf{r}/D+{\psi }_k)$, the actual pupil-averaged value isOn an orthogonal full-period basis, $⟨s_k^2⟩=1/2$ and cross terms vanish. Alternatively, independent uniform random wave phases give this result after wave phase averaging, even on a circular optical pupil. With the same $M$ and equal contrasts the physical RMS is thenA single full-period cosine already demonstrates the issue: its RMS is $h_0/\sqrt{2}$, not $h_0$. On the actual circular optical pupil, any nonconstant unit-peak cosine also has mean square strictly less than one; averaging over its wave phase gives exactly $1/2$. On a finite circular optical pupil, phase-dependent mode overlaps can additionally matter. If a constant piston is removed before computing RMS, replace the mode Gram matrix by its centred optical pupil covariance. The printed value can be recovered as a modal coefficient norm, or by counting $2M$ orthogonal peak-amplitude quadratures in the physical RMS; neither convention is specified by the quoted mode definition. This is the distinction between modal amplitude norm versus wavefront RMS, not a correction silently made to the PDF. Unequal residual contrasts require ${\mathbb{E}}_{\psi }{h}_{\mathrm{rms}}^2={\lambda }^2{\sum }_k{C}_k/(2{\pi }^2)$, so one speckle's $C$ alone cannot determine the total RMS. The weak-aberration approximation also requires the total wave phase error to stay small, not merely each coefficient separately.