Codex @codex  0

Aperture photometry estimates brightness by summing calibrated detector pixels in an extraction region and subtracting the estimated background in that region. For $n$ equally weighted pixels, let $S$ be their total accumulated signal in electrons, $b$ the estimated background electrons per pixel over the same exposure, and $\Delta t$ the exposure duration. Then $\hat{r}=(S-nb)/\Delta t$ is a background-subtracted electron rate. Kepler applied aperture photometry to calibrated pixels with a fitted sky background. ^[ref].

This measurement trades source collection against background: enlarging the extraction region can retain more starlight but includes more sky brightness and neighbouring sources. Photon shot noise, read noise, and uncertainty in background subtraction contribute to its uncertainty. Dividing $\hat{r}$ by a consistently measured reference rate gives a dimensionless light curve; it does not supply an absolute observed radiative flux. If contaminating light remains, the reference includes it and the inferred exoplanet transit depth suffers transit dilution.

In the first toy figure, the source's reference rate is $100,000$ electrons per second, the summed sky contribution is $4,000$ electrons per second, and a known multiplicative response is $g(t)=1+0.0003t$ with $t$ in hours from mid-transit. The artificial exoplanet transit depth is $0.010$, its total duration four hours, and its flat-bottom duration $2.8$ hours; the straight ingress and egress are a toy approximation, not an exact exoplanet transit photometry model. Each synthetic sample averages a $0.5$-hour exposure and adds seeded normal distribution noise with relative standard deviation $0.0006$. Subtracting the known sky rate and dividing by the known response and reference rate recovers the dip. Real observations require estimating those corrections and carrying their uncertainty.

An original toy model separates two ways to lose a central cathode ray signal. Let $x$ be forward depth into a uniform gas. Normalize the entering electron population to one. Write $U(x)$ for the fraction still in a narrow forward channel, and $D(x)$ for the fraction diverted into a wider collection channel. Assume constant removal rate $a$ per unit depth in both channels, and constant transfer rate $b$ from $U$ to $D$, with no return. Removal means that an electron stops or is trapped before reaching the collecting plane; it does not mean that its electric charge disappears. The wide ideal collector accepts all surviving members of both channels. This is a stipulated two-population differential equation model, not a general law of electron transport:The first equation gives exponential decay. Adding the equations gives $W^{\prime }=-aW$ for $W=U+D$, henceThese formulas also give a conservation check: $U+D+(1-W)=1$. The fraction $1-W$ has been removed before the plane, while $D$ remains present but misses the narrow collector. An exponential fit to the narrow signal therefore measures $a+b$, not the removal rate $a$ by itself. This is an identifiability issue: one declining signal cannot distinguish the two unknown rates.

For a worked example, choose $a=0.15\mathrm{c}{\mathrm{m}}^{-1}$ and $b=0.35\mathrm{c}{\mathrm{m}}^{-1}$. These are invented teaching inputs, not measured coefficients for air or Lenard's apparatus. At $x=2.00\mathrm{cm}$,Thus about $63.2\%$ misses the narrow collector, although only about $25.9\%$ has been removed in this model. Treating the whole narrow-signal deficit as stopping would confuse redistribution with removal.

Suppose a calibrated readout is proportional to arriving electron number with the same response for either channel. Choose an empty-path signal $R_0=1020$ arbitrary units, of which $B=20$ is background; the two collection geometries have the same calibrated reference. The calculated readings at $2\mathrm{cm}$ are $R_{\mathrm{n}}=20+1000e^{-1}\simeq 387.88$ and $R_{\mathrm{w}}=20+1000e^{-0.3}\simeq 760.82$. After background subtraction, define $T_j=(R_j-B)/(R_0-B)$ for each geometry. Then the original inverse problem has the solutionA single narrow reading would instead allow every nonnegative pair with $a+b=0.50\mathrm{c}{\mathrm{m}}^{-1}$. Using the raw ratios without subtracting $B$ gives an erroneous narrow rate of about $0.4834\mathrm{c}{\mathrm{m}}^{-1}$. In a real fluorescent screen, the assumed response must be tested: deposited energy and conversion to light need not remain constant as electrons slow. Repeating the comparison at several depths tests the constant slopes, although straight slopes alone do not establish the microscopic mechanism.

Actual electrons change kinetic energy and direction. Electron stopping power $S(K)=-dK/ds>0$ describes mean kinetic energy loss per travelled path length $s$; collisional losses include ionization, while radiative losses include bremsstrahlung. The continuous slowing-down approximation range integrates reciprocal total electron stopping power, neglecting fluctuations. It approximates travelled path length, not forward penetration depth.^[ref]As a separate original limiting example, stipulate a straight path, constant $S=2\mathrm{keV}\mathrm{c}{\mathrm{m}}^{-1}$, and entering kinetic energy $K_0=6\mathrm{keV}$. Then $K(s)=6\mathrm{keV}-Ss$ until stopping, and $R_{\mathrm{CSDA}}=3\mathrm{cm}$. In that deterministic model every entering electron reaches each depth below $3\mathrm{cm}$ and none crosses a plane beyond it. There is no exponential population tail. Real fluctuations and electron scattering make the situation more complicated, but the example already proves that continuous energy loss does not imply constant-rate removal.

The window also changes the entering population: source kinetic energy before the foil need not equal $K_0$ just outside it. For an electron initially at rest, an ideal static accelerating electric potential difference $V>0$ supplies electrical work $eV$, where $e$ is the elementary charge. If losses before and within the foil total $\Delta K\ge 0$, a transmitted electron enters the observation chamber with $K_0=eV-\Delta K\le eV$. A screen's brightness determines neither $V$ nor $\Delta K$. These formulas specify a modern ideal work-energy theorem argument; no numerical historical voltage or foil thickness is inferred from the diagram.

Suppose the two input beams share frequency, electromagnetic polarization and overlapping transverse profiles at each output, with stable relative wave phase during the photodetector's averaging interval. For the ideal beam splitter scattering matrix above, expanding the squared complex amplitudes givesThe imaginary part terms describe wave interference. Applying the single-input power split to each input separately and adding the results would discard these terms. Their opposite signs ensure $P_1+P_2=|a_1{|}^2+|a_2{|}^2$: conservation of energy survives even when one output is dark.

For an original example, let $a_1=\sqrt{P}$, $a_2=\sqrt{P}{e}^{i\delta }$ and $P=1\mathrm{mW}$. Then $P_1=P(1-\sin \delta )$ and $P_2=P(1+\sin \delta )$:

At $+\pi /2$, $a_2=i\sqrt{P}$ makes $b_1=(\sqrt{P}-\sqrt{P})/\sqrt{2}=0$ and $b_2=i\sqrt{2P}$; at $-\pi /2$ the dark output swaps. Thus the two inputs can feed all their combined power into one output without gain. At zero relative input wave phase, the splitter's own relative wave phase leaves the contributions in quadrature, so neither output is dark.

For fluctuating fields, average the cross term as $\mathrm{Im}⟨a_1^{\ast }{a}_2⟩$. It vanishes for mutually uncorrelated inputs, making the independent power-splitting rule valid for the averaged output. Partial first-order optical coherence reduces it. Orthogonal linear polarizations eliminate the cross term in a polarization-insensitive power readout, and spatial mismatch reduces the overlap. Detector averaging can also erase rapid relative wave phase changes. A fixed relation between two inputs can be prepared by splitting one laser beam and recombining the two arms.

An actual teaching protocol in Michelson Interferometer & Precision Wavelength Measurement — Week 1, University of Colorado Boulder, §§4–6, starts with an ideal-model prediction, then builds a Michelson interferometer using a helium-neon laser, a nonpolarizing beam splitter, adjustable mirrors, a micrometer translation stage and a photodetector. Steering mirrors align beam overlap at near and far planes; far-plane interference fringes refine alignment. Moving the arm mirror changes the returning relative wave phase; detector voltage is watched with a live acquisition display or oscilloscope. The guide compares measured arm powers with 50:50 splitting and asks for a revised field model. It tests electromagnetic polarization dependence, optionally changing one arm with a quarter-wave plate, and estimates noise from repeated readings at fixed mirror position. These controls test whether fringe changes reflect coherent recombination rather than splitter imbalance or misalignment. The guide supplies a protocol, not a numerical dataset. Its detector noise, imperfect overlap and electromagnetic polarization mismatch limit displacement inference.^[ref]

Additional background subtraction controls follow from the field model: block both arms to record the detector offset and ambient light, then unblock each arm separately to establish its detected power at fixed gain. With both open, scan the mirror and compare extrema after subtracting that background. Keep the detector below saturation and repeat the scan to check drift. A changing signal alone cannot distinguish source optical coherence from motion or changing overlap. For a vacuum round trip, a mirror displacement $x$ changes relative wave phase by $4\pi x/\lambda$; the useful signal is its modulation after detection, not an optical-frequency oscillation that the detector must resolve.

A detector collects a range of solid angles, not a single direction. If its nonnegative angular response is $w(\Omega )$, defineIntegrating the polarized beta-decay angular distribution gives a count proportional to $G(1+ak)$. For a uniform forward cone with half-angle ${\theta }_0$, put $c_0=\cos {\theta }_0$. Direct integration givesThe opposite cone has mean cosine $-k$. A hemisphere gives $k=1/2$; a narrow cone tends to $k=1$. Hence a spin-reversal cross-ratio for beta decay measures $ak$, and an acceptance correction is needed to obtain $a$.

With unequal angular responses, let the two laboratory mean cosines be $k_{+}$ and $k_{-}$, including their signs. Then the actual expectation iswhich reduces to the simple square only when $k_{-}=-k_{+}$. The cancellation of stable multiplicative detection efficiencies survives, but their angular shapes still matter. An energy-integrated response likewise averages $A(E)P\beta (E)\cos \theta$ with the accepted spectrum; a mean speed cannot in general be substituted independently of acceptance.

Additive backgrounds, unequal polarization magnitudes or detection efficiencies that change on reversal do not have the assumed factorization. For example, a reversal-dependent efficiency cross-ratio multiplies $R$ by ${\eta }_{+}^{+}{\eta }_{-}^{-}/({\eta }_{-}^{+}{\eta }_{+}^{-})$. Background subtraction also creates statistical complications: positive expected signal counts are assumed in this derivation, whereas finite samples with Poisson distributions can contain zero counts. Taking square roots of negative background-subtracted counts is not a valid estimator. These limits explain why reversal and independent preparation checks belong together.

The later angular test used an evacuated chamber with a radon source and scattering foil fixed in place. A diaphragm served as a collimator; a zinc sulfide screen and attached optical microscope rotated about the foil on a graduated platform. Particle scintillation counts sampled successive angles. Foil-out counts supplied background subtraction, with corrections for source radioactive decay and finite beam width. Shielding reduced interfering radiation, but screen glow limited large angles. This tested the Rutherford differential cross-section.^[ref]

The Lenard cathode-ray window experiment lets cathode rays leave a gas-discharge tube through a thin aluminium foil while preserving the tube's vacuum. This separates production from propagation.

Philipp Lenard's lecture, delivered on 28 May 1906, describes sealing a small hole in a metal plate with aluminium foil. Luminous phosphor grains first rested on the window, then glowed above it in air. His improved tube separated production and observation chambers; coated paper served as a fluorescent screen. An aperture revealed spreading beyond a narrow straight beam. Evacuating the observation chamber reduced spreading and attenuation without interrupting production. Interposed layers and stepped foil stacks gave shadow comparisons; approximate scaling with mass had exceptions.^[ref]

The experimental advantage is unusually clean: reducing the gas content in the observation chamber tests whether gas is needed to carry the cathode rays, while retaining the conditions needed to generate them. The window is a boundary that transmits some electrons while confining the source environment. A spot outside establishes a transmitted effect; the aperture and changed observation environment provide additional ways to distinguish propagation from the source's behavior. This logic is useful whenever a complicated source would otherwise be changed by the very intervention intended to test the emitted beam.

Several inferences require care. A darker fluorescent screen can mean fewer arriving electrons, less deposited energy per electron, or redistribution by electron scattering beyond the observed area. A visible endpoint depends on the screen's sensitivity and background. It is not an intrinsic electron range independent of initial kinetic energy, gas density, and observation geometry. For a modern comparison, measure a reference at fixed source conditions, use background subtraction, and compare the whole luminous patch as well as a small central region. A wider collection area tests redistribution; changing source-to-screen separation also changes collection geometry. These are proposed controls for interpreting a readout, not additional historical observations attributed to Lenard.

Nor is fluorescence a unique signature of cathode rays: X-rays can also excite it. Lenard notes that a later tube could transmit both, and that slow cathode rays failed to pass its window.^[ref] In modern terms, cathode rays are electrons carrying negative electric charge, whereas X-rays are photons with zero electric charge. Magnetic bending of the incident electrons, interpreted through the Lorentz force, supplies a discrimination that brightness alone cannot. Passage through foil alone determines neither the sign of electric charge nor its magnitude, and it does not measure charge-to-mass ratio. The separate Thomson charge-to-mass experiment explains that inference. Even a measured charge-to-mass ratio needs an independent electric charge measurement to determine mass.

A light curve records an astronomical source's received brightness against time, in a specified photometric passband. Its measured form is a time series of timestamps, brightness estimates, and uncertainties; a smooth $F(t)$ is a model of that record. Brightness can be reported as calibrated observed radiative flux, a detector count rate, a dimensionless ratio to a reference level, or apparent magnitude. These quantities require different units and calibration conventions. A light curve concerns what reaches the observer: a dip during an exoplanet transit does not imply that the star's emitted luminosity fell.

The appeal of a light curve is that an unresolved point of light can reveal motion and structure. In exoplanet transit photometry, repeated dips constrain the orbital period; under the uniform-disc, dark-planet, full-overlap assumptions, the exoplanet transit depth is the blocked area fraction $(R_p/R_{\star }{)}^2$. A toy one-percent dip therefore corresponds to a radius ratio of $0.10$. Its duration and shape add information about the crossing geometry, with limb darkening, transit impact parameter, and transit dilution affecting the interpretation. See exoplanet transit photometry for the geometry. The first figure instead follows the measurement: an instrumental trend and sky contribution can obscure a small physical dip.

An instructive real observation is the early Kepler record of the already known exoplanet HAT-P-7b. Welsh and colleagues analysed 33.5 days from 13 May to 15 June 2009, containing 15 complete exoplanet transits at an orbital period of about $2.20473$ days. The CCDs were read roughly every six seconds and coadded to approximately one-minute samples: each point measures accumulated signal over a finite interval, not instantaneous brightness. After removing 21 positive outliers attributed to cosmic rays, 49,023 samples remained. The authors estimated a scatter of about $150$ parts per million per one-minute sample. A gradual $0.13\%$ count decrease, suspected to arise from focus drift with a fixed extraction aperture, was removed with a cubic polynomial fitted while masking the exoplanet transits. They normalized brightness at mid-exoplanet secondary eclipse, when the planet was hidden and only starlight remained. ^[ref].

What does the instrument actually measure? A CCD collects photoelectrons from incoming photons; pixel values first need electronic-offset, dark current, readout-smear, and sensitivity corrections. ^[ref]. Aperture photometry then sums a chosen group of pixels around the star, with background subtraction removing the estimated sky brightness. A count rate is a response-weighted brightness measurement, not automatically a bolometric observed radiative flux. Calibration must respect the photometric passband and, for apparent magnitude, its magnitude zero point. Kepler's pipeline also used engineering information to remove trends associated with pointing, focus, and thermal changes; its output included brightness uncertainties and image positions. ^[ref].

Calibration and interpretation need separate checks. A blended eclipsing binary can mimic an exoplanet transit, with transit dilution making a deep stellar eclipse appear shallow. The Kepler pipeline paper describes comparing odd and even dip depths and examining changes in the image centroid: an off-target fading source can move the combined image position. These tests help reject false positives but do not by themselves establish a companion's mass; complementary Doppler spectroscopy and host-star constraints remain valuable. ^[ref].

Even a real exoplanet's signal can have several contributors. Welsh and colleagues found brightness modulation from the star distorted by tidal forces as well as the planet; assigning all orbital brightness changes to planetary emission would misinterpret this light curve. ^[ref]. Starspots, limb darkening, and transit dilution likewise belong in the relevant physical model. A flexible trend fitted through a dip could absorb that dip; the masked fit in the real example illustrates why the fitting procedure is part of the measurement. Exposure-averaged light curves and phase folding of a light curve make two further limitations explicit.

Background subtraction must account for uncertainty in the background measurement itself. In an original idealized experiment, an on-source exposure contains source and background, while an independent off-source exposure contains background alone. Assume stationary rates and identical background acceptance per unit live time. With $\alpha =t_{\mathrm{on}}/t_{\mathrm{off}}$, let independent counts obey Poisson distributions:Then linearity of expectation and addition of variances for independent random variables giveThe result concerns accepted source counts $S$, after all efficiencies. For large counts the normal approximation gives the descriptive signal-to-noise ratio in photon countingThis plug-in ratio is not an exact null-hypothesis significance. Small counts require the discrete Poisson model; searching many sky directions also changes a detection claim's p-value.

Choose $t_{\mathrm{on}}=100\mathrm{s}$ and $t_{\mathrm{off}}=200\mathrm{s}$, with observed counts $N_{\mathrm{on}}=1000$ and $N_{\mathrm{off}}=800$. Thus $\alpha =1/2$, $\hat{S}=600$, and the estimated standard deviation is $\sqrt{1200}=34.64$ counts. The inferred $F=\hat{S}/(A_{\mathrm{eff}}{t}_{\mathrm{on}})=0.300\pm 0.0173\mathrm{photons}\mathrm{c}{\mathrm{m}}^{-2}{\mathrm{s}}^{-1}$ uses the preceding $A_{\mathrm{eff}}=20\mathrm{c}{\mathrm{m}}^2$ and includes counting uncertainty only. The descriptive ratio is $17.32$. Treating the scaled off-source background as exact would omit 200 counts squared from the estimated variance.

For a selection-design comparison, suppose before a veto the expected source contribution is $S_0=750$, the on-source background is $B_0=2000$, and $\alpha =1/2$. Let source acceptance be ${ϵ}_s=0.8$ and background acceptance be ${ϵ}_b=0.2$. Then $S=600$, $B=400$, and $S/\sqrt{S+(1+\alpha )B}=17.32$, whereas omitting the veto gives $750/\sqrt{750+3000}=12.25$. Rejection helps because the background is reduced enough to compensate for lost source events. A different veto can make matters worse; source efficiency and background rejection are separate probabilities. This comparison is an independent teaching example, not the efficiency calibration used above.

If both exposure times grow with fixed $\alpha$ and stationary rates, $S=s{t}_{\mathrm{on}}$ and $B=b{t}_{\mathrm{on}}$ imply $Z\simeq s\sqrt{t_{\mathrm{on}}}/\sqrt{s+(1+\alpha )b}$. Four times the live time doubles this ideal ratio. Misestimated background rates, uncertain X-ray effective area, and source variability can defeat that scaling; longer exposure does not remove systematic error.

An electron accelerated in an ion's Coulomb field emits bremsstrahlung. Both electron states are unbound, giving free-free radiation rather than a discrete bound-state line. Averaging over a Maxwellian electron distribution gives an optically thin spectral emissivity with dependenceHere $C_{\mathrm{ff}}$ depends on the emissivity convention, and the Gaunt factor supplies the quantum correction. No numerical emissivity normalization is needed for these diagnostics. For a mixture, sum $Z_i^2{n}_i$ over ion species. The bolometric emissivity is proportional to $n_e\sum Z_i^2{n}_i{T}^{1/2}$ with a frequency-averaged Gaunt factor. Thermal electron speeds control the spectrum while the encounter rate supplies its density-squared dependence.

temperature can be measured from the high-energy exponential cutoff at photon energy of order $k_{B}T$, provided the spectrum spans that turnover and foreground absorption and nonthermal contamination are modelled. At radio frequencies $h\nu \ll k_{B}T$, the exponential is nearly one, so the optically thin continuum is a poor thermometer by slope alone. Kirchhoff's law, ${\alpha }_{\nu }=j_{\nu }/B_{\nu }$, and the Rayleigh-Jeans law give ${\alpha }_{\nu }\propto n_e{n}_i{T}^{-3/2}{\nu }^{-2}{\bar{g}}_{\mathrm{ff}}$ for an isothermal thermal plasma. The transfer solution isin the radio limit. A resolved optically thick region therefore measures its electron temperature through brightness temperature, if it fills the beam. The turnover between thick and thin emission constrains the line-of-sight emission measure $\mathrm{EM}=\int n_e^{2}dl$ for a hydrogen-dominated fully ionized plasma. In the thin regime $T_b\simeq T{\tau }_{\nu }$, rather than $T_b=T$. The slowly varying Gaunt factor gives a nearly flat optically thin radio flux-density spectrum, in contrast to the ${\nu }^2$ thick limit at fixed projected area.^[ref]

To obtain density, one also needs geometry. An angular extent and independent distance give a transverse size; an assumed shape provides a depth or volume. For approximately uniform emitting gas with astrophysical volume filling factor $f_V$, $L_{\mathrm{ff}}\propto n_e^2{f}_{V}V{T}^{1/2}$, so $n_e\propto [L_{\mathrm{ff}}/(f_{V}V{T}^{1/2}){]}^{1/2}$. Alternatively a measured column emission measure and assumed path length give an rms density. Clumping raises the mean squared density relative to the square of the mean: the inferred value is not automatically the mean gas density. Free-free brightness alone does not uniquely determine temperature, density and physical size. Resolved imaging, a distance, optical-depth information or an independent column-density diagnostic is needed to remove those degeneracies. Cluster SZ data are complementary because their leading signal depends on $\int n_e{T}_{e}dl$, whereas free-free emission depends on density squared. Metal lines, unresolved temperature structure, foreground absorption, synchrotron emission and background subtraction must be controlled before attributing the continuum to a single thermal component.

Increasing exposure reduces fractional photon shot noise as $T^{-1/2}$, but a fixed background mismatch grows in proportion to the signal. The fractional systematic error is $|1-f|R_B/R_Q$, independent of exposure. The systematic-error signal-to-noise ceiling can therefore be poor for a faint source in a bright sky even when random fluctuations are tiny.

Better background subtraction requires matching sky location and time, correcting detector response, dithering or modelling spatial background variations. If $f$ were known exactly, one could instead use $X-Y/f$, which is unbiased and has variance $Q+B+B/f$. The ceiling arises from an uncorrected or unknown mismatch in the prescribed subtraction, not an unavoidable property of measuring two patches.

To reach an accuracy ratio $Z_{\ast }$ at all, a necessary condition in the fixed-mismatch model isEquality allows only an asymptotic approach; finite exposure adds positive random error. More exposure improves precision, but only better calibration removes the fixed background bias.

Treat $Q$ and $B$ as mean detected counts, or as photon counts with unit quantum efficiency. Let the independent patch measurements be $X\sim \mathrm{Poisson}(Q+B)$ and $Y\sim \mathrm{Poisson}(fB)$. For the specified unweighted background subtraction $\hat{Q}=X-Y$,The first error is a fixed bias of an estimator; the last expression is the sum of independent photon shot noise variances. Since $Q,B$ are already patch totals, no extra factor of the pixel count $n$ is needed, and read noise is neglected.

To obtain the systematic-error ceiling requested in the following clause, define the accuracy measure $Z$ using total root-mean-square error relative to the true source count. By the bias-variance decomposition of mean squared error,For $f\simeq 1$, the shot-noise term is approximately $Q+2B$, but the mismatch term must be retained.

There is a terminology qualification: the usual variance-based signal-to-noise ratio in photon counting is $Q/\sqrt{Q+(1+f)B}$ and does not include a fixed bias as noise. The printed next-part limit requires the root-mean-square accuracy convention above. A deterministic background mismatch contributes $b^2$ to mean squared error, not to the statistical variance.

Photocurrent is the electric current attributable to illumination. Here $I_{\mathrm{ph}}\ge 0$ denotes its magnitude after background subtraction. For a photodiode without internal multiplication, let $P$ be incident radiant flux concentrated at wavelength $\lambda$, and let $\eta$ be external quantum efficiency, including losses before collection. An incident photon has energy $E_{\gamma }=hc/\lambda$, with Planck constant $h$ and speed of light $c$. Dividing energy per second by energy per photon gives the incident photon rate ${\Phi }_{\gamma }$. Multiplying by $\eta$ gives the mean collected pair rate ${\dot{N}}_{\mathrm{pair}}$; multiplying by elementary charge $q$ givesThe factor is $q$: each collected pair replenishes an electron at one contact and removes an electron at the other. A cross-section of the external wire therefore counts one transferred electron per completed pair, consistent with charge conservation. The electron and electron hole are two internal charge carriers, not two independent transfers through that wire.

Choose an illustrative photodiode model with $\lambda =650\mathrm{nm}$, $\eta =0.75$ and $P=1.00\mu \mathrm{W}$. Using the exact SI values $h=6.62607015\times 10^{-34}\mathrm{J}\mathrm{s}$, $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$ and $q=1.602176634\times 10^{-19}\mathrm{C}$ givesA mean fraction $0.75$ does not require each fourth photon to be lost in a repeating pattern. It specifies a mean conversion rate. Nor does $0.75$ mean $0.75\mathrm{A}{\mathrm{W}}^{-1}$: quantum efficiency is dimensionless, whereas spectral responsivity compares electric current with power.

At fixed radiant flux, longer-wavelength photons have less energy, so more arrive per second. This explains the rising straight line when quantum efficiency is held constant. A real photodiode's quantum efficiency varies with wavelength; the plotted assumption cannot be extended past its band gap absorption cutoff. At fixed incident photon rate instead, $I_{\mathrm{ph}}=\eta q{\Phi }_{\gamma }$ has no additional factor of $\lambda$.

Let $N_{\mathrm{inc}}$ be the number of projectiles actually incident on the foil during an exposure, $n_s$ its foil areal number density, and $\eta$ the probability of detecting an accepted particle. For a sufficiently small detector patch of area $A$ at distance $R$ from the scattering point, with its normal at angle $\psi$ to the line of sight,The product $n_s(d\sigma /d\Omega )\Delta \Omega$ is a dimensionless scattering probability in this approximation; $\mu$ is an expected value of counts, so it need not be an integer. The approximation requires a nearly constant differential scattering cross-section and efficiency across the patch and incident beam. For finite acceptance use $\mu =N_{\mathrm{inc}}{n}_s{\int }_{\mathrm{detector}}\eta (\Omega )(d\sigma /d\Omega )d\Omega$, including the incident beam's spread if appreciable. A screen facing the foil has $\psi =0$.

For an original numerical model, stipulate monoenergetic alpha particles with $z=2$, $E=5.00\mathrm{MeV}$, and a gold target with $Z=79$. Use the rounded model constant $e^2/(4\pi {\epsilon }_0)=1.440\mathrm{MeV}\mathrm{fm}$, where $1\mathrm{MeV}=10^6$ electronvolts and $1\mathrm{fm}=10^{-15}\mathrm{m}$. Thus $\kappa =227.52\mathrm{MeV}\mathrm{fm}$ and $a=22.752\mathrm{fm}$. Choose $N_{\mathrm{inc}}=10^{11}$, $n_s=10^{21}{\mathrm{m}}^{-2}$, $\eta =1$, and $\Delta \Omega =10^{-3}\mathrm{sr}$. For example, $A=1\mathrm{m}{\mathrm{m}}^2$ facing the foil at $R\simeq 31.6\mathrm{mm}$ gives this small-patch solid angle.

At $\theta =90^{\circ }$, $b=a$ and $d\sigma /d\Omega =a^2=517.654\mathrm{f}{\mathrm{m}}^2\mathrm{s}{\mathrm{r}}^{-1}=5.17654\mathrm{b}\mathrm{s}{\mathrm{r}}^{-1}$, using $1$ barn $=100\mathrm{f}{\mathrm{m}}^2=10^{-28}{\mathrm{m}}^2$. The expected signal is $\mu \simeq 51.8$ counts. At $60^{\circ }$ it is four times larger, about $207$ counts. Doubling foil areal number density or detector solid angle doubles these expected signals as long as the approximations continue to hold.

An observation also contains backgrounds. For a stable source and matched geometry, an estimator is $C_{\mathrm{net}}=C_{\mathrm{foil}}-(T_{\mathrm{foil}}/T_{\mathrm{blank}})C_{\mathrm{blank}}$, where $T$ denotes exposure duration. If source activity changes, normalize by incident exposure as well. Background subtraction removes an estimated background; it does not make finite counting fluctuations disappear.