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The encircled radiant energy of a Gaussian beam is the radiant energy inside a centred circular receiver at a specified transverse plane. For fixed Gaussian beam radius $w$ and total laser pulse energy $U$,The radiant exposure of a Gaussian beam contains the radial integration and aperture-capture plot. Its $w$ is the irradiance $e^{-2}$ radius, as specified in the primary beam account.^[ref]

An original annular example distinguishes energy near an axis from energy between two radii. Subtracting two encircled values givesChoose $U=5\mu \mathrm{J}$, $w=0.50\mathrm{mm}$, $R_1=0.25\mathrm{mm}$ and $R_2=0.50\mathrm{mm}$. The annulus receives $5(e^{-1/2}-e^{-2})\approx 2.356\mu \mathrm{J}$, or 47.12 percent of the full beam energy. A bright central point does not imply that most energy lies in an arbitrarily small disk around it.

Indeed, the radial radiant-energy density of a Gaussian beam is $dQ/dR=2\pi RH(R)$, where $H$ is local radiant exposure. For this profile it is largest at $R=w/2$, even though $H$ is largest at zero radius. Increasing a centred aperture and comparing integrated detector readings can test the cumulative prediction. Detector linearity, aperture alignment and the finite collection boundary must be controlled; a displaced aperture is not described by the centred formula.

Let $a=L/2$, so each mirror in the symmetric confocal optical resonator has radius $R=L=2a$. At mirror 1, an incident beam whose waist is at the centre has $q_{\mathrm{in}}=a+i{z}_R$, with distances measured along its incoming direction. Reflection givesAfter propagating the distance $a$ back to the centre, its parameter is purely imaginary. Thus the reflected waist is also at the cavity centre andThe incident and reflected waist radii need not agree. With axial position $z$ measured from the centre, their radii obeyAt either mirror, $z^2=a^2$ and the two spot sizes agree, as reflection requires. Between the mirrors they generally have different waists while still reproducing themselves after a full round trip.

Radiant exposure measures the radiant energy received per unit surface area during a specified interval. At a fixed surface point $\mathbf{x}$, it is the time integral of incident irradiance $E_{\mathrm{e}}$:The area belongs to the receiving surface: its orientation and location matter. An irradiance in $\mathrm{W}{\mathrm{m}}^{-2}$ specifies a rate, while radiant exposure in $\mathrm{J}{\mathrm{m}}^{-2}$ specifies the accumulated arrival of energy.^[ref]

For a uniform rectangular pulse, $H_{\mathrm{e}}=E_0\tau$, with constant irradiance $E_0$ and duration $\tau$. Stipulate one pulse at $20\mathrm{W}{\mathrm{m}}^{-2}$ for $1\mathrm{s}$, and another at $80\mathrm{W}{\mathrm{m}}^{-2}$ for $0.25\mathrm{s}$. Both deliver $20\mathrm{J}{\mathrm{m}}^{-2}$, but their peak irradiances differ by a factor of four. The two radiant exposures agree only after each complete pulse has arrived. A shorter observation window can intercept different fractions of them, and a nonlinear receiver can respond differently even to their complete pulses.

Integrating over a receiving patch gives its incident radiant energy:Here ${\overline{H}}_A$ is the spatially averaged radiant exposure, rather than necessarily the local maximum. Uniform illumination is the special case where this spatially averaged radiant exposure equals the local value everywhere on the patch. In the rectangular-pulse example, a uniformly illuminated $1\mathrm{c}{\mathrm{m}}^2=10^{-4}{\mathrm{m}}^2$ patch receives $20\times 10^{-4}=2\mathrm{mJ}$.

For an original spatial example, consider a circular Gaussian beam at a fixed transverse plane, with constant Gaussian beam radius $w$ throughout the pulse. Let $P(t)$ be its total radiant power and $U=\int P(t)dt$ its laser pulse energy. Integrating its transverse irradiance in time gives the radiant exposure of a Gaussian beam:The normalization follows by integrating $H(r)2\pi rdr$ from $0$ to $\infty$. The disk of radius $w$ contains only $U(1-e^{-2})$, so its actual spatially averaged radiant exposure is $U(1-e^{-2})/(\pi w^2)$, not $U/(\pi w^2)$. With stipulated $U=1\mu \mathrm{J}$ and $w=1\mathrm{mm}$, the peak is $0.6366\mathrm{J}{\mathrm{m}}^{-2}$, while the average inside that disk is $0.2752\mathrm{J}{\mathrm{m}}^{-2}$. State the Gaussian beam radius convention and the receiving area with any reported value.

An energy readout alone therefore does not specify a surface's radiant exposure. In an ideal linear model for pulse energy measurement with a photodiode, dark-subtracted photocurrent $i_{\mathrm{ph}}(t)=R{P}_{\det }(t)$ has constant spectral responsivity $R$ at the chosen wavelength. Integrating the electric current gives signal electric charge $Q_{\mathrm{sig}}=R{U}_{\det }$. If a known patch of area $A$ is uniformly illuminated and all of its incident energy is collected, thenAs a separate stipulated example, $Q_{\mathrm{sig}}=0.40\mathrm{nC}$ and $R=0.40\mathrm{A}{\mathrm{W}}^{-1}$ correspond to $U_{\det }=1.0\mathrm{nJ}$. Uniform illumination over $A=1.0\mathrm{m}{\mathrm{m}}^2$ gives $H_{\mathrm{e}}=1.0\mathrm{mJ}{\mathrm{m}}^{-2}$. Background subtraction, collection geometry, detector linearity and an integration window covering the full response are part of this measurement model. Continuous-wave responsivity alone does not establish linearity for a short pulse: silicon-photodiode experiments have observed pulse-dependent saturation.^[ref]

A historical NIST measurement calibration separates these operations experimentally. A modulator produces pulses from a continuous-wave laser; a calibrated beam splitter sends one beam to a slow thermal reference and another to the pulse detector. The reference records the accumulated energy, while the pulse repetition rate and injection duration determine the mean energy per pulse. The detector's integrating electronics provide a pulse-energy readout. Converting such a readout to radiant exposure additionally requires the illuminated area and its spatial distribution.^[ref]

The beam-use term fluence denotes the same time integral for a directed beam normal to the chosen surface. For light arriving from many directions, specify the angular geometry: planar irradiance and spherical irradiance are different quantities, so a value defined using one cannot simply be substituted for the other.^[ref]

The radiant exposure of a Gaussian beam is its time-integrated transverse irradiance. At a fixed plane normal to a circular Gaussian beam, suppose the Gaussian beam radius $w$ is unchanged during a pulse and denotes the radius where irradiance is $e^{-2}$ of the on-axis value. With total laser pulse energy $U$,The primary beam account specifies this radius convention and the corresponding irradiance normalization; the radiant exposure parent derives its time integral.^[ref]

An original aperture example asks how much of that energy a finite receiver collects. Integrating over a centred disk of radius $R$ gives the encircled radiant energy of a Gaussian beam:For stipulated $U=2\mu \mathrm{J}$, $w=0.80\mathrm{mm}$ and $R=0.40\mathrm{mm}$, the receiver collects only $39.35$percent, or $0.7869\mu \mathrm{J}$. Its spatially averaged radiant exposure is about $1.566\mathrm{J}{\mathrm{m}}^{-2}$, compared with on-axis $H(0)\approx 1.989\mathrm{J}{\mathrm{m}}^{-2}$. Dividing the total $U$ by this receiver's area would count energy that missed it.

To collect a chosen fraction $0<f<1$ under this ideal model, the centred aperture needs $R=w\sqrt{-\log (1-f)/2}$. For95percent, $R\approx 1.224w$. A possible experimental check measures integrated received energy through several known apertures and compares their ratios. Aperture centring, detector linearity and any profile change during the pulse affect that comparison. A time-varying radius must remain inside the time integral rather than being replaced by an unspecified average radius.