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Choose an original transparent-band model, defined only for $400\le \lambda \le 700\mathrm{nm}$:This chosen refractive index is not a fit to water, glass or the historical liquids. It decreases from $1.3425$ to approximately $1.300408$, so the threshold $1/n$ increases across the band. For $\beta =0.75$, solving $n({\lambda }_{\ast })=1/\beta$ givesOnly $400\le \lambda <{\lambda }_{\ast }$ emits. Faster particles with $\beta =0.78$ and $0.90$ emit throughout this chosen band. The Cherenkov cone angle and the Frank-Tamm formula are both evaluated with the same $n(\lambda )$.

For a $1\mathrm{cm}$ track the chosen model predicts about $0.589$, $27.40$ and $142.85$ emitted photons for $\beta =0.75$, $0.78$ and $0.90$, respectively. An expected count below one is meaningful over repeated equivalent tracks; it is not a fractional photon observed on one track. The near-threshold example illustrates why extending a constant-index approximation across an entire optical band can give the wrong emitting interval.

Finite bandwidth is essential. The $1/{\lambda }^2$ factor cannot be extrapolated down to arbitrarily small wavelength: the chosen material model stops at its stated band. Real optical dispersion changes the threshold and direction, and appreciable absorption requires a complex material response beyond the simple transparent-medium expression. A detected count must also include transport, collection geometry and the detector's wavelength-dependent response. Even a correct produced spectrum therefore does not, on its own, predict perceived colour or observed glow intensity.

For constant particle speed and electric charge $q=ze$, in a transparent, linear, nonmagnetic medium with isotropy, the Frank-Tamm formula gives the produced photon density:Set it to zero below threshold. Here $e$ is the positive elementary charge, $x$ is track length, $\lambda =2\pi c/\omega$ is vacuum wavelength, and $\alpha$ is the fine-structure constant. With SI lengths in metres, the density has units ${\mathrm{m}}^{-2}$; multiplying by $dxd\lambda$ gives a dimensionless expected count. The explicit SI convention isThese conventions agree.^[ref]

To check the convention, start from $E=\hslash \omega =hc/\lambda$ and $d^{2}N/(dxdE)=\alpha z^2{\sin }^2\theta /(\hslash c)$. Changing variables uses the positive density Jacobian $|dE/d\lambda |=hc/{\lambda }^2$; since $h/\hslash =2\pi$, the coefficient becomes $2\pi \alpha z^2$. No factor of the material refractive index belongs in this Jacobian: $\lambda$ is not the in-medium wavelength. Multiplication by $\hslash \omega$ gives the corresponding energy loss density in this SI nonmagnetic convention:

For an original constant-index example, stipulate $n=1.50$, $\beta =0.90$, $z=1$ and $\alpha =1/137.036$. Integration from $400$ to $700\mathrm{nm}$ givesThus a stipulated $1\mathrm{cm}$ track produces about $222$ photons in this band in the ideal model. This is an emitted count before absorption, collection losses and detector response, not a brightness measurement or a measured water yield.

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.

Phase matching makes wave amplitudes emitted at different depths add coherently. Consider collinear plane waves in a uniform transparent material of length $L$, with fixed polarization of waves and nonzero selected nonlinear optical susceptibility. Assume negligible pump depletion, negligible absorption, an initially absent harmonic, and a wide beam for which diffraction is negligible. The pump's wave amplitude is then constant throughout the interaction.

The pump electric field has spatial wave phase $k_{\omega }z$; its quadratic source has spatial wave phase $2k_{\omega }z$. The freely propagating harmonic has wave number $k_{2\omega }$. Define the second-harmonic phase mismatch byFor the chosen complex-field convention, the outgoing harmonic wave amplitude, up to an overall constant and propagation phase, is proportional toChoosing the conjugate complex-field convention reverses the exponent's sign without changing the wave intensity. Taking the squared complex modulus givesHere $C$ is fixed for the chosen modes and material; its dimensions accommodate the stated wave intensity convention. The quadratic pump dependence follows because the source wave amplitude is proportional to $E_0^2$, whereas $I_{\omega }$ is proportional to $E_0^2$. The sinc function comes from summing contributions along the whole material, just as a finite aperture produces diffraction through addition of complex amplitudes.

At $\Delta k=0$, the integral is $L$ and the harmonic wave intensity grows as $L^2$. At $\Delta kL=2\pi$, all those contributions cancel. This cancellation needs no absorption: their wave phases wind through a full turn. At high conversion, the constant-pump assumption fails, so the $L^2$ law cannot continue indefinitely. Tight focusing, finite optical pulses, absorption, and transverse separation of modes also require a more complete model.

Optical dispersion generally makes $n_{2\omega }\ne n_{\omega }$. In a suitable crystal, birefringence allows modes with different polarization of waves to experience different refractive indices, making their selected values equal. The appropriate crystal direction and allowed nonlinear optical susceptibility component must both work.^[ref]

For a worked calculation, choose $L_0=1\mathrm{mm}$ and $\Delta k=\pi \mathrm{m}{\mathrm{m}}^{-1}$. At $L=L_0$, the harmonic wave intensity is $4/{\pi }^2\simeq 0.4053$ of its perfectly matched value. At $L=2L_0$ it vanishes in this model. More material can therefore reduce the output. If instead $\Delta k=0$, doubling $L$ multiplies output by four. The two curves retain that common normalization, so their different peak heights are meaningful.

A specimen whose leading optical effect is a spatially varying phase, rather than absorption. A thin specimen with refractive index $n_s$ in a medium of refractive index $n_m$ has, in a straight-path approximation,Here $\lambda$ is the vacuum wavelength; the integral is an optical path length difference. For the chosen values $\lambda =550\mathrm{nm}$, thickness $1\mu \mathrm{m}$ and $n_s-n_m=0.005$, the phase is $\pi /55\simeq 0.0571$ radians. This calculation does not identify thickness separately from refractive index contrast.^[ref]