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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.

Laser pulse energy is the radiant energy carried by one specified laser pulse across a measurement plane. For an isolated pulse whose total radiant power is $P(t)\ge 0$ and whose background is zero,Here $P$ is already integrated over the transverse beam area; it is not local irradiance. The unit of $U$ is the joule. Radiant exposure instead divides received energy by a receiving area, so a beam's $U$ alone does not specify the exposure at its centre. The radiant exposure of a Gaussian beam makes that spatial distinction quantitative.

The Gaussian pulse energy and peak power relation gives $P_{\mathrm{pk}}\simeq 0.939437U/\tau$ when $\tau$ is the power full width at half maximum. The pulse energy shape factor records pulse-shape dependence, while the Gaussian pulse field-envelope width is $\sqrt{2}$ times the power width.

For an original example, stipulate $U=0.50\mu \mathrm{J}$ and a Gaussian pulse power full width at half maximum $\tau =10\mathrm{ns}$. Then $P_{\mathrm{pk}}=46.97\mathrm{W}$. Stretching the same model pulse to $40\mathrm{ns}$ while keeping $U$ unchanged reduces its peak power to $11.74\mathrm{W}$. Both curves below integrate to the same energy; their peak heights and widths differ.

A finite detection time gate collects the Gaussian pulse time-gate energy fraction, which depends on its width and alignment. A centred gate as wide as the power full width at half maximum receives only $76.10\%$ of $U$, because Gaussian tails extend beyond it. For a $\tau$-wide gate, a Gaussian gate centring error of $\tau /2$ reduces capture to $49.07\%$.

The average power of a laser pulse train follows by counting complete pulses. If a long observation of duration $T$ contains $N$ pulses of energies $U_1,\dots ,U_N$, with negligible edge contributions and no continuous background,Here $f_{\mathrm{rep}}$ is the pulse repetition frequency, not the inverse pulse width. An average power readout constrains $\overline{U}$ when $f_{\mathrm{rep}}$ is known; it does not resolve pulse-to-pulse energy fluctuation. Identical, well-separated $0.50\mu \mathrm{J}$ pulses at $1.0\mathrm{MHz}$ give $0.50\mathrm{W}$ average power, even though the stipulated $10\mathrm{ns}$ pulses peak near $47\mathrm{W}$.

If a steady optical background is present, state whether the reported energy includes it. For pulse energy above a measured background $P_{\mathrm{bg}}$, the pulse energy baseline correction uses ${\int }_{\mathrm{gate}}[P(t)-P_{\mathrm{bg}}]dt$. Failing to subtract a constant background adds $P_{\mathrm{bg}}{T}_g$ to the inferred energy. This optical-background model is distinct from an offset introduced later by detector electronics.

Historical NIST equipment illustrates what the readouts measure. Its TC-24 reference absorbed the beam in a glass plate, sensed a plate temperature difference with thermocouples, and integrated the thermopile's electric potential difference during beam injection. Before/after readings enabled background subtraction and compensated thermal drift. Counting pulses converted total injected energy to mean pulse energy. The semiconductor pulse photodetector used an integrating amplifier; calibration related the peak output electric potential difference to pulse energy. That electrical peak did not by itself establish optical peak power.^[ref] Pulse energy measurement with a photodiode connects this kind of measurement to collected electric charge; validating the detector response under the actual pulse conditions remains part of that interpretation.

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]

Spatially averaged radiant exposure is the received radiant energy divided by the specified receiving area. For a patch $A$ and local radiant exposure $H(\mathbf{x})$,Both local and averaged values have units $\mathrm{J}{\mathrm{m}}^{-2}$, but the average need not equal the local maximum. The radiant exposure definition uses incident energy and the receiving surface's area.^[ref]

For an original unequal-area example, let one $3\mathrm{m}{\mathrm{m}}^2$ region receive a uniform $2\mathrm{J}{\mathrm{m}}^{-2}$, and a neighbouring $1\mathrm{m}{\mathrm{m}}^2$ region receive $10\mathrm{J}{\mathrm{m}}^{-2}$. They receive $6$ and $10\mu \mathrm{J}$ respectively. The average over the combined patch is thereforeThe unweighted average of the two listed exposure values would be $6\mathrm{J}{\mathrm{m}}^{-2}$, which is wrong for these unequal areas. The area weighting of radiant exposure gives the larger region three times the influence. The actual peak remains $10\mathrm{J}{\mathrm{m}}^{-2}$ in the stipulated piecewise-uniform model.

A receiver integrating all incident energy over the complete patch can determine $Q_A$; division by its known collecting area then gives the average. That measurement alone cannot identify the peak or reconstruct the spatial distribution. Subdivision with a characterized spatial response supplies additional information. For a finite aperture intercepting a Gaussian beam, use energy actually entering the aperture, as calculated in radiant exposure of a Gaussian beam, rather than the total energy of the untruncated beam.