A thin pure-phase object has amplitude transmittance with

PHYSICS · Subjective · Fermi University Physics · Optics & Photonics · Fourier optics and image formation · Fourier optics and image formation

What an Instrument Can See, Ch Seeing what the transfer function forbids: phase contrast and super-resolution

Problem

A thin pure-phase object has amplitude transmittance t(x)=eiϕ(x)t(x) = e^{i\phi(x)} with ∣ϕ(x)∣≪1|\phi(x)| \ll 1, illuminated by a unit-amplitude plane wave. Measure ϕ\phi from its spatial average, so that ⟨ϕ⟩=0\langle\phi\rangle = 0 (a uniform phase changes nothing measurable), and write ⟨ϕ2⟩\langle\phi^2\rangle for its mean square over the field. To second order the field just after the object is 1+iϕ(x)−ϕ(x)2/21 + i\phi(x) - \phi(x)^2/2. Its zero-frequency (unscattered) part is the average of that field, 1−⟨ϕ2⟩/21 - \langle\phi^2\rangle/2; the rest, iϕ(x)−[ϕ(x)2−⟨ϕ2⟩]/2i\phi(x) - [\phi(x)^2 - \langle\phi^2\rangle]/2, is the scattered part. The imaging system that follows is aberration-free (it introduces no extra phase anywhere in the pupil) but its pupil is not perfectly flat: it passes the zero-frequency component with amplitude 11 unchanged, and passes every nonzero-frequency component with a uniform real amplitude fraction η\eta, where 0≤η≤10 \le \eta \le 1. (a) Show that the image irradiance is I(x)=1−(1−η)⟨ϕ2⟩−η(1−η)ϕ(x)2I(x) = 1 - (1-\eta)\langle\phi^2\rangle - \eta(1-\eta)\phi(x)^2 to second order in ϕ\phi, that no term linear in ϕ\phi appears for any value of η\eta, and that the image is uniform at both η=0\eta = 0 and η=1\eta = 1, with I=1I = 1 at η=1\eta = 1. (b) At one point of the field ϕ=0.10 rad\phi = 0.10\ \mathrm{rad}, and at another ϕ=0\phi = 0. Find the irradiance difference I(0.10)−I(0)I(0.10) - I(0) between them, in units of the incident irradiance, when η=0.70\eta = 0.70.

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