Coherent zero-mode model of phase contrast

The following original model isolates the conversion mechanism. Use a periodic image window, unit illumination, ideal unit magnification, and a phase plate that changes only the zero spatial frequency. Write $c_0=⟨e^{i\varphi }⟩$ for the spatial mean. Multiplying that single Fourier coefficient by $a{e}^{i\theta }$, with $0<a\le 1$, gives the exact image field in this model:This formula follows by splitting $e^{i\varphi }=c_0+(e^{i\varphi }-c_0)$: the second term has zero mean and is left unchanged. A real annular, partly coherent instrument alters a finite band of directions, so this calculation is a mechanism model rather than its full image-formation law.

For a weak phase object with $⟨\varphi ⟩=0$, one has $c_0=1+O({\varphi }^2)$ and henceThe cross term is $2\mathrm{Re}(a{e}^{-i\theta }i\varphi )=2a\varphi \sin \theta$. Thus $\theta =\pi /2$ makes positive $\varphi$ brighter and $\theta =-\pi /2$ makes it darker under this stated phase convention for complex amplitudes. The sign comes from this convention and chosen plate; a manufacturer’s positive/negative contrast labels should not be inferred from the sign of $\theta$ alone.

For example, choose $a=0.5$ and a zero-mean periodic pattern whose peak phase is $0.084$. The first-order peak intensities are $0.25\pm 0.084$, compared with the uniform ideal bright-field microscopy value $1$. The relative change about the reduced background is approximately $2\varphi /a$ for a quarter-cycle shift. Attenuating the reference increases relative contrast while reducing the background photon count; it is not light amplification or a universal improvement of signal-to-noise ratio.

Two checks expose what the model can and cannot measure. A spatially constant phase gives $E_{\mathrm{out}}=a{e}^{i\theta }{e}^{i\varphi }$ and $I_{\mathrm{out}}=a^2$, so no absolute phase is recovered. Orthogonality of the constant and zero-mean parts also givesFor $a\le 1$ this cannot exceed the incident mean light intensity. The image redistributes brightness and loses some reference light rather than creating energy.