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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.

Fluorescence quantum yield counts emitted fluorescent photons per absorbed excitation photon. It differs from quantum efficiency, which describes the photodetector.^[ref]

For an original constant-rate model, let $k_{\mathrm{r}}$ be the radiative decay rate and $k_{\mathrm{nr}}$ the sum of competing rates without fluorescent photon emission. An initial population $N_0$ in the emitting excited state obeys the rate equationIntegrating $k_{\mathrm{r}}N(t)$ over all time givesHere every absorbed excitation is assumed to reach this emitting population; additional losses before that stage require another factor. Under steady weak excitation, absorbed photon rate $R_{\mathrm{abs}}$ and total detection fraction $\eta$ give $C=\eta \Phi R_{\mathrm{abs}}$. The count rate therefore mixes excitation, yield, and detection, whereas $\tau$ specifies a decay timescale.

The Förster distance is the separation giving FRET efficiency $E=1/2$ for the specified pair and conditions. It is a property of the optical pair and its environment, rather than a universal length. In the usual pair model,where ${\kappa }^2$ is the FRET dipole orientation factor, $Q_D$ the FRET donor quantum yield, $J$ the FRET spectral overlap integral, and $n$ the medium's refractive index. The omitted proportionality constant depends on the unit convention for $J$.^[ref]

If FRET efficiency is known and those assumptions hold, inversion givesThe derivative explains why errors become troublesome near $E=0$ or $E=1$. An illustrative pair with $R_0=4$ nanometres and another with $R_0=6$ nanometres both give $E=1/2$ at their different Förster distances: knowing FRET efficiency alone does not determine $r$.

A heterogeneous population adds another ambiguity. In a synthetic example with equal weights per FRET donor excitation, efficiencies $0.1$ and $0.9$ average to $0.5$, but their separations are $9^{1/6}{R}_0$ and $9^{-1/6}{R}_0$. Their average separation is approximately $1.068R_0$, whereas inverting the average FRET efficiency gives $R_0$. A measured emission ratio has additional detection weights; it need not equal even this excitation-weighted average. The transferable lesson is that an average readout does not uniquely specify a distribution of molecular separations.

Förster resonance energy transfer is a nonradiative transfer of energy from an excited state of a FRET donor to a nearby FRET acceptor, through coupling of their electric dipoles. The transfer step does not emit a photon that the FRET acceptor subsequently absorbs. It competes with the FRET donor's own fluorescence and other routes out of its excited state. A fluorescent FRET acceptor can then emit fluorescence; an absorbing but nonfluorescent FRET acceptor can still receive the transferred energy.^[ref]

The striking use of FRET is to turn a change in molecular proximity into an optical readout. Proteins carrying a FRET donor and a FRET acceptor may change the readout when they associate, as in the chemotaxis FRET assay; two labels on one protein may report a change in its shape. Interpretation depends on the FRET spectral overlap, FRET dipole orientation factor and labeling geometry, as well as separation. A change in FRET is therefore evidence about proximity under specified conditions, rather than a direct measurement of an arbitrary distance.^[ref]

For a single pair with fixed Förster distance $R_0$ and label separation $r$, the FRET efficiency isAt $r=R_0$, half the FRET donor excitations transfer energy. At $r=R_0/2$, the model gives $E=64/65\approx 98.46\%$; at $r=2R_0$, it gives $E=1/65\approx 1.54\%$. The sixth power makes FRET especially useful near the Förster distance, with little sensitivity to further separation once transfer is already negligible. These are model calculations, not measurements of a particular pair.

FRET efficiency is the probability that an excitation of the FRET donor ends in transfer to the FRET acceptor. Write $k_D=k_{\mathrm{r}}+k_{\mathrm{nr}}$ for the total decay rate without a FRET acceptor, including radiative fluorescence and other decay, and $k_T$ for the transfer rate. For constant competing rates,The integral counts transfer events among all excitations in a population with exponential decay. Substitution gives the separation law above. This short rate argument assumes one FRET donor, one FRET acceptor, weak coupling, an available FRET acceptor, and fixed geometry during the decay; it is not a model for arbitrary interacting labels.^[ref]

FRET optical crosstalk includes FRET donor emission leaking into the FRET acceptor detection band and direct excitation of the FRET acceptor by the nominal FRET donor excitation light. Measure donor-only, acceptor-only and unlabeled preparations with the same fluorescence microscope settings to separate these contributions and background.^[ref]

An explicit linear measurement model makes the controls' purpose clear. Let $D$ be background-subtracted donor-band counts under donor excitation, $A$ acceptor-band counts under acceptor excitation, and $F$ acceptor-band counts under donor excitation. Donor-only and acceptor-only preparations determine leakage coefficients $a$ and $b$. Thenis a corrected FRET sensitized emission signal under the assumed linear, fixed optical response. In a synthetic example $D=100$, $A=80$, $a=0.10$, $b=0.05$ and $F=34$ give $S=20$. Even a nontransferring sample would give $F=14$ from these two contaminants. Correcting that signal still does not convert $S$ into FRET efficiency without a calibration of excitation, yields and detection.

A biological comparison also needs a pairing-negative preparation and a known transferring positive preparation. Match label abundance and localization where possible; changing concentration can change encounter frequency as well as brightness. Check that attaching the labels preserves the protein function being investigated. Otherwise the measurement can faithfully report a perturbation introduced by the labels themselves.

In FRET photobleaching controls, selectively destroy the FRET acceptor's ability to absorb and test whether FRET donor fluorescence increases. Under unchanged excitation and detection, complete loss of transfer, and an intact FRET donor,This is a destructive comparison. Monitor acceptor loss, donor-only samples exposed to the bleaching illumination, and an unbleached reference region or time trace. Incomplete bleaching, donor damage, movement or label replenishment violate the comparison. During ordinary recordings, gradual unequal bleaching can instead produce an apparent change of FRET without a biological change.^[ref]

FRET sensitized emission is FRET acceptor fluorescence caused by transfer from an excited FRET donor. Excite predominantly the FRET donor and collect both emission bands. Reduced FRET donor emission and increased FRET acceptor emission are useful together, but an acceptor-to-donor emission ratio is not automatically the FRET efficiency: brightness, label abundance and channel collection also enter.^[ref]

FRET spectral overlap quantifies overlap between the FRET donor's emission and the FRET acceptor's absorption as functions of wavelength $\lambda$. With normalized donor emission $f_D$ and acceptor molar absorption coefficient ${\epsilon }_A$, one convention isMatching two emission colors is insufficient: the receiving absorption matters.^[ref]

A Jablonski diagram arranges molecular energy eigenstates vertically and draws arrows for changes between them. In the usual closed-shell molecular example, $S_0$ is the electronic ground state, $S_1$ and $S_2$ are electronic excited states with singlet state spin, and $T_1$ has triplet state spin. Each electronic state also has levels of molecular vibration. The diagram records possible processes, not molecular paths in space.^[ref]

Photon absorption can populate an upper level of molecular vibration in $S_1$. Relaxation removes excess energy without photon emission; fluorescence commonly starts from relaxed $S_1$.^[ref] The drawing also includes relaxation from a higher electronic excited state and a competing spin-changing route to a triplet state.

The Stokes shift compares the lowest absorption-band maximum with the fluorescence emission spectrum maximum. Relaxation often makes $\Delta \lambda >0$. This comparison differs from using an arbitrary excitation wavelength. A spectral-density maximum can move when changing from per unit wavelength to per unit spectroscopic wavenumber; specify the spectral convention.^[ref]

An original illustrative energy budget makes the distinction concrete. Excitation at $450\mathrm{nm}$ and emission at $520\mathrm{nm}$ correspond to about $2.76$ and $2.38\mathrm{eV}$ per photon, using $E=hc/\lambda$ with Planck constant $h$ and speed of light $c$ in vacuum. The difference, about $0.37\mathrm{eV}$, must go elsewhere in the complete energy balance. These chosen wavelengths are not measured band maxima for a particular fluorophore.

Anti-Stokes emission has shorter wavelength than excitation. Additional energy, for example from a thermally populated level of molecular vibration or multiple excitation photons, makes it possible. The one-photon, initially relaxed drawing does not cover these cases.^[ref]