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A fluorescence microscope maps photon emission to an image. Excitation light passes through a fluorescence excitation filter and reflects from a wavelength-selective beam splitter into the objective optical lens. Returning fluorescence passes through that beam splitter and a fluorescence emission filter to the photodetector. Rejecting excitation makes faint emission visible.^[ref]

An ordinary image records emission-band counts at each pixel. Adding pulse-delay measurements gives a fluorescence lifetime map; scanning emission wavelength gives spectral information instead. Confocal microscopy adds spatial rejection and confocal optical sectioning, as developed in its own branch. An unlabeled specimen under the same illumination reveals intrinsic fluorescence and excitation leakage; excitation-blocked frames reveal detector background. A blank cannot perfectly predict background in a differently prepared specimen.

A useful motivating experiment moves a bead emitting fluorescence through the selected optical focus and records the detector signal against position. Measurement of point-spread function (PSF) for confocal fluorescence microscopy by InCheon Song and others (2005, §2 and Fig. 1) describes a $543.8\mathrm{nm}$ laser delivered through a single-mode optical fiber, collimation, beam-size adjustment and a dichroic beam splitter. A $100\times$, numerical aperture $1.3$ oil-immersion objective optical lens excites slide-mounted fluorescent beads and collects their emission. Emission passes a filter and collecting optical lens, then a pinhole, before an avalanche photodiode. A three-dimensional positioning stage moves the specimen. The apparatus also contains a phase plate for deliberately modified responses.^[ref]

The published comparison uses nominal bead diameters $0.17,0.5,1\mu \mathrm{m}$: larger beads give broader measured spots. This is a control on the finite emitter, because the image is the convolution of bead emission with the confocal point spread function. The smallest bead has weaker signal relative to noise; incorrect bead size, defocus, bleaching and optical aberrations limit inference from a restored response. The reported bead-size comparison concerns lateral responses; it is not a measured pinhole-size series or a measured sheet-sectioning curve.^[ref]

To test axial pinhole selection with this kind of apparatus, the following is a proposed extension, not an additional result from that paper. Choose an isolated bead small relative to the response, centre it laterally, then step $z$ through optical focus while recording background-subtracted signal at several pinhole settings. Hold excitation power, dwell interval, detector gain, emission band and sampling fixed. Preserve raw peaks before also plotting peak-normalized traces. A lower peak and faster relative decay with axial displacement would support the paired-selection mechanism; fitting only normalized widths would miss the throughput cost.

Measure blocked-excitation background, include a bead-free region, avoid detector saturation, and repeat scans in reversed order to reveal drift or bleaching. Check pinhole centring separately, since misalignment can mimic signal loss. Record bead diameter, immersion medium, refractive index, numerical aperture, wavelengths and axial step. A finite bead is an approximation to an optical point source; a point trace does not certify confocal optical sectioning of a whole fluorescent sheet. Depth-dependent spherical aberration, scattering and sample motion can broaden or distort the response. Increasing magnification or making a display brighter does not correct those effects.

The focal ratio, or f-number, of a converging optical instrument focused at infinity is the dimensionless quantitywhere $f>0$ is effective focal length and $D$ is the on-axis entrance pupil diameter. The notation f/4 means $N=4$, not $f=4$ in a particular unit. A $100\mathrm{mm}$ focal length with $D=25\mathrm{mm}$ gives f/4. Closing the aperture stop until $D=12.5\mathrm{mm}$ gives f/8 while retaining the same focal length. The mechanical opening need not equal $D$. ^[ref].

For a thin lens in air with a circular aperture stop at the thin lens, the converging bundle from a distant on-axis point has half-angle $\alpha$ satisfyingThe last step uses the small-angle approximation. For a compound optical lens, the cone is set by the image-side optical pupil and its distance from the receiving plane; the simple triangle is not an exact high-angle rule. At finite object distance a thin lens instead gives $v=f(1+|m|)$, by its imaging equation and transverse magnification, so its working cone corresponds to $v/D=N(1+|m|)$. Nominal focal ratio therefore does not alone specify close-range image irradiance. ^[ref].

Two useful scalings follow under stated conditions. For a fixed uniform incident beam and unchanged transmission, collected radiant flux scales as $D^2$. For an extended source with fixed radiance, the central image irradiance scales approximately as $N^{-2}$ in the paraxial approximation: the solid angle of the cone scales as $(D/f{)}^2$. For an ideal uniformly illuminated unobscured circular entrance pupil, the first dark ring of the point spread function has angular radius and focal plane radiusHere $\lambda$ is wavelength. These are first-zero radii, not diameters or a universal rule for resolving arbitrary objects. A proof outline evaluates the circular aperture's Fraunhofer diffraction using its Fourier transform, then uses the first zero of the resulting Bessel function. ^[ref].

At $\lambda =0.55\mu \mathrm{m}$, the example's f/4 and f/8 first-ring radii are $2.684$ and $5.368\mu \mathrm{m}$. Closing the aperture stop halves $D$, quarters collected radiant flux and doubles this diffraction radius. It can also reduce optical aberrations by excluding outer light rays; actual sharpness reflects both effects. This explains why a smaller aperture need not produce a sharper image. ^[ref].

A proposed optical bench check puts a fixed collimator before a converging thin lens and a photodetector large enough to collect the entire image behind it. Compare two centred aperture stops of known diameters while fixing source output, detector gain and integration time. The integrated signal tests the $D^2$ collection law; a spatially resolved optical camera at the focal plane can separately measure the point spread function. Repeat a dark reading and the first aperture stop, and check that neither the incoming beam nor detector clips either bundle. A resolved pinhole source, imperfect collimation, optical aberrations and detector pixel size can dominate the apparent spot width; a single peak reading does not measure total collected radiant flux.

Consider an ideal coherent monochromatic wave from an unresolved source on the optical axis, a uniform circular entrance pupil of diameter $D$, and ideal relay optics without wavefront errors. In the scalar Fraunhofer diffraction approximation, a Fourier transform carries the pupil's complex amplitude into the image plane. Choose dimensionless pupil coordinates $\mathbf{x}$ in units of $D$ and angular coordinates $\mathbf{u}=D\mathbf{\theta }/\lambda$, where $\lambda$ is the wavelength. DefineThe coronagraphic focal plane mask has transmission $M=1-W$, with $W=1$ inside the occulting disk and $W=0$ outside. Thus the field immediately after the mask is $MA$. Outside the disk $M=1$, so the local wave intensity is still $|A{|}^2$: hiding the core has left those rings intact. Reimaging an unlimited pupil giveswhere $\ast$ is convolution, $L$ is the Lyot stop transmission, and the proportionality constant in the irradiance is omitted. Coordinate reversals from physical relay optics are absorbed into the coordinate choice. These equations use wave amplitudes throughout propagation; applying a Fourier transform to wave intensity would give a different and incorrect model. This is the pupil/image calculation described by Sivaramakrishnan et al.; the circular disk and numerical choices below are our illustrative specialization. ^[ref].

The subtraction explains the beauty of the arrangement: in much of the interior, the original pupil field and the convolution term nearly cancel. Near a boundary, that cancellation is incomplete. The Lyot stop can therefore block a region containing disproportionately much unwanted light. A hard-edged coronagraphic focal plane mask has a transform with extended tails, so the interior need not be perfectly dark. With an unlimited relay, $L=1$ gives $A_{\mathrm{final}}=MA$ exactly: no further ring suppression occurs. A stop covering the full original optical pupil instead uses $L=P$ and already clips field outside the optical pupil; an undersized Lyot stop clips more of the remaining rim.

The toy uses a coronagraphic focal plane mask of angular radius $3\lambda /D$ and a Lyot stop of diameter $0.8D$. The discrete Fourier transform grid has $1024\times 1024$ samples, with 128 across $D$ and angular sampling $0.125\lambda /D$; it approximates the continuous equations on a finite periodic domain. The plot averages wave intensity in annuli of width $0.25\lambda /D$. Over the annulus $4\le |\mathbf{u}|\le 10$, the mean wave intensity divided by the direct stellar peak is approximately $1.40\times 10^{-4}$ without a coronagraph, the same with only the focal mask and an unlimited relay, $4.44\times 10^{-5}$ with the mask and full pupil, and $1.66\times 10^{-6}$ with the mask and $0.8D$ Lyot stop. Thus the undersized opening reduces this particular annular mean about 27-fold relative to the full-pupil comparison. It does not suppress every direction by that factor, and these values do not predict a real instrument's detection limit.

The ideal toy excludes a central obstruction, support vanes, scattering of light, wavefront errors, atmospheric effects, a spread of wavelengths, pointing drift and detector noise. It also contains only the star: no simulated planet or solar corona. For a far off-axis source whose image misses the coronagraphic focal plane mask, the continuous geometric coronagraph throughput of the $0.8D$ opening tends to $0.8^2=0.64$ relative to the full disk. Rejecting stellar light therefore costs some desired light as well. Enlarging the coronagraphic focal plane mask narrows the characteristic scale of its pupil-plane transform by the scaling property of the Fourier transform, but also hides a larger neighborhood of the source.

The Franck-Hertz experiment connects a reproducible loss of Electron kinetic energy with excitation of a chemical atom. The original mercury-vapor experiments were performed during 1912–1914 while investigating ionization; they were not designed as a test of the Bohr model. Their later interpretation distinguished internal excitation from removal of an Electron. ^[ref].

In a modern mercury tube, a heated cathode emits Electrons and a positive perforated grid accelerates them through Hg vapor. A collector slightly below the grid's electric potential rejects slow Electrons. The oven controls vapor density and hence the mean free path. Keep heating and retarding settings fixed, sweep the accelerating electric potential difference $V$, and record collector electric current $I(V)$. These measured quantities indirectly probe atomic excitation. ^[ref].

The curve shows repeated current minima, typically about $4.9\mathrm{V}$ apart. In the collision model, an Electron gains approximately $eV$ of kinetic energy, with $e$ the positive elementary charge. An inelastic collision transfers about $4.9\mathrm{eV}$ to internal excitation. Near the grid, this can prevent the Electron crossing the retarding region, lowering collected electric current. Further acceleration restores collection; repeated excitation losses produce further minima. These represent multiple collisions, rather than equally spaced atomic energy eigenvalues. Contact potentials shift the first minimum, so compare several successive spacings instead of identifying the first dip's voltage with the excitation cost. ^[ref].

The optical comparison supplies a second observable: mercury radiation at a wavelength near $253.6\mathrm{nm}$ was detected through a window in the tube. Its inferred photon energy $hc/\lambda \approx 4.89\mathrm{eV}$ agrees with the characteristic collision loss, with $c$ the speed of light. This agreement supports the interpretation as a transition between internal atomic energies. It does not make the collector trace a complete spectrum. In particular, the lowest excited mercury level, labeled $6{}^3{P}_0$, is metastable because an electric-dipole selection rule forbids its direct transition to the ground state; the nearby $6{}^3{P}_1$ level emits the quoted ultraviolet line. Collision excitation and optical detection therefore have different sensitivities to the available excited states. ^[ref].

Initial Electron kinetic energy spread and varying collision positions broaden real minima. Repeat traces at different retarding settings and oven temperatures to distinguish excitation from collection effects. This probes a discrete energy-transfer channel, rather than every energy eigenspace, its dimension, or an absolute ground-state energy. ^[ref].

James Franck and Gustav Ludwig Hertz presented their paper on 24 April 1914. Its heated platinum wire sat inside a cylindrical platinum mesh, with a collector 1–2 mm beyond it. A heated bath and pumping prepared the vapor; guard rings suppressed leakage. They differentiated retarding-electric current scans to study outgoing energies.^[ref]

Their 1925 Nobel Prize preceded their lectures of 11 December 1926. James Franck described choosing pressure for many collisions before the mesh but few afterward.^[ref] Gustav Ludwig Hertz explained how a field-free collision region narrowed excitation thresholds.^[ref]

In 1961, Peter Alden Franken, Alan Hill, C. Wilbur Peters and Gabriel Weinreich focused a pulsed ruby laser through a red optical filter into quartz. The filter suppressed xenon-flash background. A quartz-prism spectrograph used red-insensitive Eastman Type 103 plates. A single shot recorded a feature near $347.2\mathrm{nm}$ from $694.3\mathrm{nm}$ input. Removing the quartz or substituting glass removed the feature; changing polarization of waves and orientation followed their predicted pattern. These controls supported second-harmonic generation; photographic density gave an estimated yield.^[ref]

A spectrograph turns output frequency into a recording position. Its spectral separation makes a faint new component distinguishable from bright input light. The reported rounded wavelengths do not constitute an independent precision measurement of the ratio two. Nor do these observations demonstrate perfect phase matching or complete conversion. The experiment and Nicolaas S later theoretical account are distinct contributions to nonlinear optics.

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.

Fraunhofer diffraction is the far-field approximation for diffraction: contributions from an aperture reach a selected observation direction with approximately linear path-dependent wave phases. Adding their complex amplitudes gives a Fourier transform of the field across the aperture. It also describes the appropriate rear focal plane of an ideal optical lens under coherent plane-wave illumination. The diffraction grating examples use this regime. ^[ref]; ^[ref].

For a slit of width $a$, wavelength $\lambda$ and screen distance $L$, sufficient conditions near the forward direction areThe last two statements are equivalent: the important comparison is with $a^2/\lambda$, rather than just with the slit width. For a finite rectangular aperture, check the corresponding condition for both transverse dimensions. The single-slit diffraction calculation treats the transverse cut associated with width $a$. ^[ref].

A short explanation is that the distance from slit coordinate $x$ to an observation point at distance $R$ and angle $\theta$ isThe linear term supplies the phase difference; the quadratic term supplies a wave phase error of order $a^2/(\lambda R)$, since the wavenumber is $2\pi /\lambda$. Near the forward direction $R\approx L$, so the stated condition makes this error small. This is why a distant screen permits the simple single-slit diffraction formula. ^[ref].

The free spectral range is the wavelength or frequency spacing between adjacent orders or resonances of an optical instrument. It identifies when different orders can overlap. Always specify the coordinate: a uniform spacing in frequency need not be uniform in wavelength. For an ideal Fabry–Pérot interferometer at normal incidence, with mirror separation $d$ and constant refractive index $n$, Fabry–Pérot resonance spacing gives $\Delta {\nu }_{\mathrm{FSR}}=c/(2nd)$, where $c$ is the speed of light in vacuum. The free spectral range of an echelle grating follows a different order condition.

At a fixed temperature, the frequency maximizing $B_{\nu }$ and the wavelength maximizing $B_{\lambda }$ in Planck's law are different ways of describing the same black-body radiation. They are maxima of different densities. The inverse relation between frequency and wavelength does not preserve those maxima because its interval-conversion factor varies with wavelength.

For a short extremum calculation, substitute $x=h\nu /(k_{B}T)$ into $B_{\nu }$, leaving a positive constant times $x^3/(e^x-1)$. Substitute $x=hc/(\lambda k_{B}T)$ into $B_{\lambda }$, leaving a positive constant times $x^5/(e^x-1)$. Setting each derivative to zero givesSelect the positive roots: $x_{\nu }=2.821439372\dots$ and $x_{\lambda }=4.965114232\dots$. Although $x=0$ solves the rearranged equations, it is an endpoint, not either peak. For $m=3$ or $5$, the function $m(1-e^{-x})-x$ initially increases, its derivative $m{e}^{-x}-1$ crosses zero once, and it eventually becomes negative. Hence it has exactly one positive root. Each density tends to zero at both ends, so this root gives its unique maximum. The bisection method brackets these roots in $(1,3)$ and $(1,5)$.

Consequently Wien's displacement law givesThese constants agree with NIST's tabulation. ^[ref] The wavelength peaks at $3000$, $4500$, and $6000\mathrm{K}$ are $965.924$, $643.949$, and $482.962\mathrm{nm}$. Doubling temperature halves this peak wavelength.

At $6000\mathrm{K}$, the frequency peak is ${\nu }_{\max }\approx 352.736\mathrm{THz}$, whose converted wavelength is $c/{\nu }_{\max }\approx 0.849907\mu \mathrm{m}$. The $B_{\lambda }$ peak is instead $0.482962\mu \mathrm{m}$, corresponding to about $620.737\mathrm{THz}$. These numbers are compatible: equal-width frequency bins and equal-width wavelength bins partition the same radiation differently. Specifying which density was maximized is essential when using a peak to infer temperature.