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Area weighting of radiant exposure combines patch-average exposures in proportion to the receiving areas they represent. For disjoint patches with areas $A_j$ and mean radiant exposures $H_j$, the coefficients ${\alpha }_j=A_j/{\sum }_k{A}_k$ satisfy ${\sum }_j{\alpha }_j=1$, and the combined spatially averaged radiant exposure is ${\sum }_j{\alpha }_j{H}_j$. This follows by adding the received energies $A_j{H}_j$, rather than averaging a list of values without its geometry.^[ref]

An original coordinate example explains a calibration issue in image-based measurements. Let a rectangular receiving strip have length $\ell$ and constant width. Stipulate $H(x)=H_{0}x/\ell$ for $0\le x\le \ell$. Suppose a calibrated image coordinate $u\in [0,1]$ maps to physical position $x=\ell u^2$. Uniform sampling in $u$ is therefore nonuniform in physical area.

The physical area average isAn unweighted average over increasingly dense, uniformly spaced $u$ samples would instead converge to ${\int }_0^1{H}_0{u}^{2}du=H_0/3$. The factor $2u$ is the change of variables formula's area correction for the chosen mapping; the constant strip width and $\ell$ cancel in the normalized average.

Thus a Jacobian area correction for radiant-exposure maps can matter even if every local exposure value is correct. An optical camera measurement needs its spatial calibration and the receiving-surface map, not just pixel values. Overlapping patches must not be counted as disjoint areas, and local point samples need an explicit quadrature model before they can stand for patch averages.

A Jacobian area correction for radiant-exposure maps converts a coordinate-space integral into one over the physical receiving surface. For a one-to-one planar calibration map $\mathbf{x}=f(u,v)$,The Jacobian determinant supplies physical area per coordinate-area increment. This change of variables formula preserves the radiant exposure definition as incident energy per receiving area.^[ref]

For an original two-dimensional example, let $0\le u,v\le 1$ and map $x=au$, $y=bv(1+u)$, with $a=2\mathrm{mm}$ and $b=3\mathrm{mm}$. Then $|\det Df|=ab(1+u)$. Stipulate $H(f(u,v))=H_0(1+u)$ with $H_0=2\mathrm{J}{\mathrm{m}}^{-2}$. The physical receiving area is $3ab/2=9\mathrm{m}{\mathrm{m}}^2$, andIts spatially averaged radiant exposure is $28/9\approx 3.111\mathrm{J}{\mathrm{m}}^{-2}$. Uniformly averaging the exposure values over $u,v$ without the determinant gives only $3\mathrm{J}{\mathrm{m}}^{-2}$.

The correction applies to exposure densities represented in distorted coordinates. If each detector sample already reports energy integrated over its physical patch, summing those calibrated energies requires no second multiplication by patch area. Otherwise the area is counted twice. An optical camera reduction must specify whether its values represent energy, irradiance, exposure or a further processed signal.

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.