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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.

Phase-contrast microscopy converts specimen-dependent phase differences into visible light intensity differences. It can reveal structures in living, unstained biological cells, allowing their changes to be followed without first making an absorbing preparation.^[ref] The key question is how an optical microscope can read a phase when an ordinary detector records light intensity.

Consider a transparent specimen with complex amplitude transmission $t(x,y)=e^{i\varphi (x,y)}$. In ideal in-focus coherent imaging with all relevant spatial frequencies transmitted, its image has $I=|t{|}^2=1$: varying $\varphi$ alone produces no brightness detail. This is a particular model, not a claim that all transparent specimens are invisible. Diffraction, defocus, absorption and a finite optical pupil can already change a real image.

A phase-contrast microscope images an illuminating annulus onto a phase plate ring near the objective's rear focal plane. The aligned ring attenuates and shifts direct light relative to most diffracted light; their interference changes image brightness.^[ref]

The method is beautiful because the bright direct beam becomes its own reference: a small specimen-dependent complex amplitude can interfere with it and change the image at first order. Merely magnifying an invisible phase pattern would leave $|t{|}^2$ unchanged.