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Peak normalization $P(0)=1$ does not make the integral of a point spread function equal to one. In the paraxial approximation to the focal plane, define $b=\pi D/(\lambda f)$ and $u=br$. The unit-integral point spread function normalization isHere $p$ has units of inverse area; multiplying by the detected radiant flux gives image-plane irradiance. Its encircled energy, meaning the fraction inside radius $R=U/b$, isThis is an integral over area, so its radial weight is $u$. An unweighted integral of a line through the centre is a different quantity.

A short proof uses the Bessel function of the first kind identities $J_0^{\prime }=-J_1$ and $J_1^{\prime }=J_0-J_1/u$. They giveBoth expressions for $E$ vanish at zero. Since these Bessel functions of the first kind tend to zero as $U\to \infty$, $E(\infty )=1$, establishing the normalization. At the first dark ring,A circular detector region ending at the first zero therefore misses about $16.22\%$ of the ideal radiant flux, even though the surrounding rings have low peak light intensity. This matters when converting measured counts into total source brightness or choosing an image aperture for a photodetector.

An aperture is an opening that admits light into or through an optical instrument. Its shape and size select the transmitted light rays; an aperture alone does not bend them as an optical lens does. The aperture stop is the opening that most restricts the bundle from an on-axis object point. The entrance pupil is the image of that aperture stop formed by the preceding optical lens elements, viewed from the object side. Consequently the effective collecting diameter $D$ is the entrance pupil diameter, which can differ from the physical hole's diameter. For example, preceding optical lens elements that enlarge a $10\mathrm{mm}$ aperture stop by a factor of two give $D=20\mathrm{mm}$; measuring the physical hole alone would misidentify the focal ratio. ^[ref].

For an unobscured circular entrance pupil under a uniform beam normal to its plane, the collecting area and admitted radiant flux arewhere $E_{\mathrm{pupil}}$ is incident irradiance. Doubling $D$ quadruples admitted radiant flux. Obstructions, nonuniform illumination and losses after the entrance pupil require the actual illuminated area and transmission to be included. The same $D$ also sets an angular diffraction-limited resolution scale $\lambda /D$ at wavelength $\lambda$. Collection and diffraction therefore answer different questions from field of view: an aperture selects the bundle from each object point, while the receiving plane's extent helps select which object directions appear in the image.

NASA/JPL's published pinhole demonstration uses two white cards, foil taped over a window in one card, and a small hole pierced in the foil. Sunlight passes through the hole onto the second card; the projected image is viewed on that card. The activity describes a larger image when the card separation increases, improved visibility when the receiving card is shaded, and a separate image from each of several holes. These are qualitative observations of image formation, rather than calibrated measurements of radiant flux or diffraction. ^[ref].

A proposed quantitative extension replaces sunlight by a steadily illuminated asymmetric target and puts the foil and a translucent receiving screen in a dark box. Record target distance $u$, hole-to-screen distance $b$, hole diameter $d$, image height, orientation and edge blur. Measure screen brightness with a fixed photodetector, or record the screen with an optical camera whose integration time and gain stay fixed. Exchange centred foils of measured $d$ at fixed $u,b$; then vary $b$ with the same foil. Geometrical optics predicts an inverted image with transverse magnification $m=-b/u$ and a point's geometric blur diameter $d(1+b/u)$. Thus a larger hole admits more radiant flux without changing the central image scale, whereas increasing $b$ enlarges the image. These are predictions to test, not reported data. Keep target illumination and detector position fixed, subtract a hole-covered dark reading, repeat the original foil to check drift, and avoid clipping the detector response. Foil thickness and ragged edges can block oblique light rays; at sufficiently small $d$, diffraction invalidates the claim that ever smaller holes always sharpen the image. A lensless optical camera has no optical lens focal length: $b$ is a propagation distance, not that focal length.

Basic radiance is radiance divided by the square of the refractive index. For radiance $L$ at a fixed wavelength,Lossless propagation conserves reduced radiance along each light ray; $L$ itself is conserved only when the endpoint refractive indices agree. A larger irradiance at a smaller image can result from admitting more directions without increasing reduced radiance. Uniform wave absorption can lower radiant flux while leaving the occupied geometric etendue unchanged; an opaque aperture removes part of that occupied set. ^[ref]

Consider an ideal parallel circular beam with uniform irradiance inside radius $b$, zero outside, and total radiant flux $P=2\mathrm{W}$. A centred circular collector of radius $a$, perpendicular to the light rays, interceptsFor $b=1\mathrm{cm}$ and $a=0.5\mathrm{cm}$, $E_b=20000/\pi \mathrm{W}{\mathrm{m}}^{-2}\simeq 6366.20\mathrm{W}{\mathrm{m}}^{-2}$ and the collected radiant flux is $0.50\mathrm{W}$. Expanding the beam to $b=2\mathrm{cm}$ while keeping $P$ fixed divides irradiance and the same collector's radiant flux by four: $E_b\simeq 1591.55\mathrm{W}{\mathrm{m}}^{-2}$ and ${\Phi }_{\mathrm{collected}}=0.125\mathrm{W}$. A collector covering either entire beam still receives $2\mathrm{W}$.

Tilting a flat collector of actual area $A$ by an angle $\theta$ to the beam direction gives ${\Phi }_{\mathrm{collected}}=E_{b}A\cos \theta$ if its entire projected area remains inside the uniform beam. At $60^{\circ }$ the small collector's radiant flux halves. A collector large enough to intercept the whole beam after tilting still receives $P$: the footprint on its surface grows as $1/\cos \theta$ and cancels the irradiance factor. This model ignores diffraction, beam-edge structure, and changes in absorption with incidence angle; it counts incident rather than absorbed power.

Take a stipulated pair of effective rods with common length $L$, linear thermal expansion coefficient ${\alpha }_L$, heat capacity $C$, projected illuminated area $A$, and thermal conductance $G$ to an environment at constant temperature $T_a$. One reference represents the symmetric polished-rod assembly. Write ${\theta }_b=T_b-T_a$ and ${\theta }_r=T_r-T_a$. Let their effective radiometric absorptances be $a_b>a_r$ for a fixed incident optical spectrum, and let $E(t)$ be the common irradiance. A spatially uniform, linear heat-balance model givesHere $G$ includes a linear approximation to the relevant losses; changing the surface treatment can also change those losses in a real device. We stipulate equal $C$ and $G$ to isolate the response to absorbed illumination, with thermal coupling from the burner neglected. Small, mechanically unconstrained extensions then give a differential displacementThe time constant $\tau$ describes thermal memory. A common imposed temperature shift cancels from the extension difference under the matched-coefficient assumptions; mismatched materials, geometry or thermal coupling can spoil that cancellation.

For a step from darkness to constant irradiance $E_0$, with initially equal rod temperatures, the first-order linear differential equation givesChoose $L=0.20\mathrm{m}$, ${\alpha }_L=20\times 10^{-6}{\mathrm{K}}^{-1}$, $A=0.0020{\mathrm{m}}^2$, $a_b=0.9$, $a_r=0.1$, $C=5.0\mathrm{J}{\mathrm{K}}^{-1}$, $G=0.20\mathrm{W}{\mathrm{K}}^{-1}$ and $E_0=600\mathrm{W}{\mathrm{m}}^{-2}$. The calculated difference of absorbed powers is $0.96\mathrm{W}$, givingFor an illustrative ideal valve that closes at $\delta =g=10\mu \mathrm{m}$, the closing time isThis threshold is crossed only when $0<g<{\delta }_{\infty }$. These are chosen model parameters, not a fit to a historical valve.

After illumination stops at $t=t_1$, the displacement decays as $\delta (t)=\delta (t_1)e^{-(t-t_1)/\tau }$. With $t_1=180\mathrm{s}$, the model reopens about $16.3\mathrm{s}$ later. Thus a sun valve responds to the thermal history of illumination; it does not read a clock or directly measure the beacon's brightness.

The logarithmic rate at which downwelling spectral irradiance decreases with depth. It depends on absorption, scattering and the angular light field, rather than equaling the absorption coefficient.

Suppose an extended source fills every position of the entrance aperture with a cone of half-angle ${\theta }_{\mathrm{in}}$, and all these light rays must reach a one-sided planar receiver. If the receiver accepts at most a cone of half-angle ${\theta }_{\mathrm{out}}$, its available etendue bounds the admitted etendue:ConsequentlyThis original derivation combines conservation of etendue with the exact cone integral; the second inequality allows the complete output hemisphere. It bounds capacity and does not prove that a chosen shape attains the bound.

For the $30^{\circ }$ input cone with $n_{\mathrm{in}}=n_{\mathrm{out}}=1$, $C\le 4$: a $10\mathrm{mm}$ entrance cannot deliver the entire filled bundle to a circular receiver smaller than $5\mathrm{mm}$ in diameter. If the output half-angle is restricted to $60^{\circ }$, $C\le 3$ and the receiver diameter must be at least $10/\sqrt{3}\approx 5.7735\mathrm{mm}$. For a receiver medium with model refractive index $n_{\mathrm{out}}=1.5$, the full-hemisphere capacity bound instead gives $C\le 9$; that larger value also depends on transmitting the bundle into that medium.

Attempting $C=10$ in the air-to-air example forces rejection of some input. For the uniform radiance already specified, even filling the entire output hemisphere givesThe maximum mean irradiance gain remains $\eta C\le 4$. With nonuniform radiance, rejecting a given fraction of geometric etendue need not reject the same fraction of radiant flux.

These bounds use a filled extended-source bundle in passive reversible geometrical optics, comparing the same wavelength and the same radiation channels. An ideal optical point source does not fill that four-dimensional set. Diffraction, optical coherence, or coupling a Gaussian beam into an optical fiber require a physical optics calculation; the cone bound is not a universal guarantee for a single wave mode. Likewise, optical gain or fluorescence changes the propagation assumptions.

The focal ratio, or f-number, of a converging optical instrument focused at infinity is the dimensionless quantitywhere $f>0$ is effective focal length and $D$ is the on-axis entrance pupil diameter. The notation f/4 means $N=4$, not $f=4$ in a particular unit. A $100\mathrm{mm}$ focal length with $D=25\mathrm{mm}$ gives f/4. Closing the aperture stop until $D=12.5\mathrm{mm}$ gives f/8 while retaining the same focal length. The mechanical opening need not equal $D$. ^[ref].

For a thin lens in air with a circular aperture stop at the thin lens, the converging bundle from a distant on-axis point has half-angle $\alpha$ satisfyingThe last step uses the small-angle approximation. For a compound optical lens, the cone is set by the image-side optical pupil and its distance from the receiving plane; the simple triangle is not an exact high-angle rule. At finite object distance a thin lens instead gives $v=f(1+|m|)$, by its imaging equation and transverse magnification, so its working cone corresponds to $v/D=N(1+|m|)$. Nominal focal ratio therefore does not alone specify close-range image irradiance. ^[ref].

Two useful scalings follow under stated conditions. For a fixed uniform incident beam and unchanged transmission, collected radiant flux scales as $D^2$. For an extended source with fixed radiance, the central image irradiance scales approximately as $N^{-2}$ in the paraxial approximation: the solid angle of the cone scales as $(D/f{)}^2$. For an ideal uniformly illuminated unobscured circular entrance pupil, the first dark ring of the point spread function has angular radius and focal plane radiusHere $\lambda$ is wavelength. These are first-zero radii, not diameters or a universal rule for resolving arbitrary objects. A proof outline evaluates the circular aperture's Fraunhofer diffraction using its Fourier transform, then uses the first zero of the resulting Bessel function. ^[ref].

At $\lambda =0.55\mu \mathrm{m}$, the example's f/4 and f/8 first-ring radii are $2.684$ and $5.368\mu \mathrm{m}$. Closing the aperture stop halves $D$, quarters collected radiant flux and doubles this diffraction radius. It can also reduce optical aberrations by excluding outer light rays; actual sharpness reflects both effects. This explains why a smaller aperture need not produce a sharper image. ^[ref].

A proposed optical bench check puts a fixed collimator before a converging thin lens and a photodetector large enough to collect the entire image behind it. Compare two centred aperture stops of known diameters while fixing source output, detector gain and integration time. The integrated signal tests the $D^2$ collection law; a spatially resolved optical camera at the focal plane can separately measure the point spread function. Repeat a dark reading and the first aperture stop, and check that neither the incoming beam nor detector clips either bundle. A resolved pinhole source, imperfect collimation, optical aberrations and detector pixel size can dominate the apparent spot width; a single peak reading does not measure total collected radiant flux.

Consider an ideal coherent monochromatic wave from an unresolved source on the optical axis, a uniform circular entrance pupil of diameter $D$, and ideal relay optics without wavefront errors. In the scalar Fraunhofer diffraction approximation, a Fourier transform carries the pupil's complex amplitude into the image plane. Choose dimensionless pupil coordinates $\mathbf{x}$ in units of $D$ and angular coordinates $\mathbf{u}=D\mathbf{\theta }/\lambda$, where $\lambda$ is the wavelength. DefineThe coronagraphic focal plane mask has transmission $M=1-W$, with $W=1$ inside the occulting disk and $W=0$ outside. Thus the field immediately after the mask is $MA$. Outside the disk $M=1$, so the local wave intensity is still $|A{|}^2$: hiding the core has left those rings intact. Reimaging an unlimited pupil giveswhere $\ast$ is convolution, $L$ is the Lyot stop transmission, and the proportionality constant in the irradiance is omitted. Coordinate reversals from physical relay optics are absorbed into the coordinate choice. These equations use wave amplitudes throughout propagation; applying a Fourier transform to wave intensity would give a different and incorrect model. This is the pupil/image calculation described by Sivaramakrishnan et al.; the circular disk and numerical choices below are our illustrative specialization. ^[ref].

The subtraction explains the beauty of the arrangement: in much of the interior, the original pupil field and the convolution term nearly cancel. Near a boundary, that cancellation is incomplete. The Lyot stop can therefore block a region containing disproportionately much unwanted light. A hard-edged coronagraphic focal plane mask has a transform with extended tails, so the interior need not be perfectly dark. With an unlimited relay, $L=1$ gives $A_{\mathrm{final}}=MA$ exactly: no further ring suppression occurs. A stop covering the full original optical pupil instead uses $L=P$ and already clips field outside the optical pupil; an undersized Lyot stop clips more of the remaining rim.

The toy uses a coronagraphic focal plane mask of angular radius $3\lambda /D$ and a Lyot stop of diameter $0.8D$. The discrete Fourier transform grid has $1024\times 1024$ samples, with 128 across $D$ and angular sampling $0.125\lambda /D$; it approximates the continuous equations on a finite periodic domain. The plot averages wave intensity in annuli of width $0.25\lambda /D$. Over the annulus $4\le |\mathbf{u}|\le 10$, the mean wave intensity divided by the direct stellar peak is approximately $1.40\times 10^{-4}$ without a coronagraph, the same with only the focal mask and an unlimited relay, $4.44\times 10^{-5}$ with the mask and full pupil, and $1.66\times 10^{-6}$ with the mask and $0.8D$ Lyot stop. Thus the undersized opening reduces this particular annular mean about 27-fold relative to the full-pupil comparison. It does not suppress every direction by that factor, and these values do not predict a real instrument's detection limit.

The ideal toy excludes a central obstruction, support vanes, scattering of light, wavefront errors, atmospheric effects, a spread of wavelengths, pointing drift and detector noise. It also contains only the star: no simulated planet or solar corona. For a far off-axis source whose image misses the coronagraphic focal plane mask, the continuous geometric coronagraph throughput of the $0.8D$ opening tends to $0.8^2=0.64$ relative to the full disk. Rejecting stellar light therefore costs some desired light as well. Enlarging the coronagraphic focal plane mask narrows the characteristic scale of its pupil-plane transform by the scaling property of the Fourier transform, but also hides a larger neighborhood of the source.

Consider a uniform plane wave of unit complex amplitude transmitted by an ideal optical slit of full width $D$. Treat the field as invariant along the optical slit height: this is one-dimensional transverse Fresnel propagation, so the prefactor is $1/\sqrt{i\lambda z}$, rather than the two-dimensional prefactor $1/(i\lambda z)$. Removing the carrier $e^{ikz}$ givesHere $I_{\mathrm{inc}}$ is incident irradiance. This infinitely long slit model describes a transverse cut away from the ends of a sufficiently long physical slit; its finite conserved quantity is transmitted power per unit optical slit height.

Put $t=\xi /D$ and $q=xD/(\lambda z)$. Substitution reduces the integral toThis form separates the pattern's shape from its physical scale. Reflection $t\mapsto -t$ proves $A(-q,N_D)=A(q,N_D)$, so the light intensity is an even function of $x$. At $N_D=0$, direct integration gives $A(q,0)=\sin (\pi q)/(\pi q)$, with limit one at $q=0$. Hence the Fraunhofer diffraction comparison is the squared sinc function. In this paragraph the unnormalized sinc function takes argument $\pi q$.

For an original example, choose wavelength $\lambda =633\mathrm{nm}$ and full width $D=0.400\mathrm{mm}$. At $z=0.0200\mathrm{m}$, $N_D\simeq 12.6382$, $N_a\simeq 3.15956$ and the edge-to-centre wave phase change is $9.92604$ radians. At $z=2.00\mathrm{m}$, $N_D\simeq 0.126382$, $N_a\simeq 0.0315956$ and that wave phase change is only $0.0992604$ radians. Even the nearer screen is fifty slit widths away; that geometric comparison alone misses its strong Fresnel diffraction.

These are numerical predictions from the displayed scalar integral, using the composite trapezoidal rule, rather than measured profiles. Each curve is divided by its own peak light intensity; the two horizontal scales differ. At 2 cm the computed central irradiance is about $1.14812I_{\mathrm{inc}}$, with brighter off-axis maxima. At 2 m it is about $0.126272I_{\mathrm{inc}}$ and is the peak. The dashed Fraunhofer diffraction curve has its own central normalization; its first zeros are at $x=\pm \lambda z/D=\pm 3.165\mathrm{mm}$. The finite-distance Fresnel diffraction curve remains slightly nonzero there.

Normalization hides the brightness change, but not the change in shape. The Fourier transform propagation multiplier has unit complex modulus, so Plancherel theorem givesThis is conservation of transmitted power per unit height in the ideal scalar model. An integral over only the displayed screen interval misses the tails; a peak-normalized profile is not a unit-integral probability density. Spreading and decreasing brightness together preserve the transmitted power.

A fundamental Gaussian beam is an ideal beam of light whose transverse wave amplitude has a Gaussian function profile. Here consider a circularly symmetric beam in vacuum, with propagation along $z$, two transverse coordinates $x,y$, and radius $r=\sqrt{x^2+y^2}$ from the optical axis. With the beam waist at $z=0$, total power $P>0$, and vacuum wavelength $\lambda$, its time-averaged light intensity, meaning irradiance in $\mathrm{W}{\mathrm{m}}^{-2}$, isHere $w_0$ is the minimum beam radius, and $z_R$ is the Rayleigh range. Far from the beam waist, $w(z)\sim \theta |z|$ with beam divergence half-angle $\theta =\lambda /(\pi w_0)$. These are the fundamental circular Gaussian beam laws within the scalar paraxial approximation, appropriate when $w_0\gg \lambda$ and the beam divergence is small. The underlying envelope solves the parabolic wave equation exactly; this does not make it an exact vector solution of Maxwell equations. ^[ref].

The radius convention matters. Writing $U$ for the complex wave amplitude envelope, the same $w(z)$ is the wave amplitude $1/e$ radius and the light intensity $1/e^2$ radius:At $r=w(z)$ these ratios are $e^{-1}\approx 0.367879$ and $e^{-2}\approx 0.135335$, respectively. The light intensity $1/e$ radius is instead $w(z)/\sqrt{2}$. None of these radii is a sharp edge: the ideal Gaussian function has nonzero tails at every finite radius. Kogelnik and Li explicitly use the wave amplitude $1/e$ radius in their fundamental-mode treatment. ^[ref].

For a numerical example, take vacuum wavelength $\lambda =632.8\mathrm{nm}$ and beam waist radius $w_0=0.500\mathrm{mm}$. Substitution gives Rayleigh range $z_R\approx 1.24114754\mathrm{m}$ and far-field beam divergence half-angle $\theta \approx 0.402852992\mathrm{mrad}$. At $z=z_R$, the beam radius is $w=\sqrt{2}{w}_0\approx 0.707107\mathrm{mm}$ and the on-axis irradiance is half its waist value. At the same wavelength, halving the beam waist radius to $0.250\mathrm{mm}$ quarters the Rayleigh range to $0.310286885\mathrm{m}$ and doubles the beam divergence half-angle to $0.805705984\mathrm{mrad}$. At fixed power, that smaller beam waist also quadruples the peak irradiance.

The contours in the left panel are $r=\pm w(z)$ in a plane through the optical axis; they mark an light intensity level, not an outer boundary. The right panel compares the exact model radii with the far-field asymptotes $w=\theta z$. The horizontal and vertical axes use different length units, so the drawn slopes are not physical angles.

The normalization expresses conserved power during free propagation. In transverse polar coordinates, an annulus has area $2\pi rdr$, soA proof outline is the substitution $s=2r^2/w(z{)}^2$ in this integral. Spreading reduces peak irradiance as $w^{-2}$ while increasing transverse area as $w^2$, leaving power unchanged. In particular, the disk of radius $w(z)$ contains $1-e^{-2}\approx 0.8646647168$, or $86.4665\%$, of the total power; the local light intensity at its rim is only $13.5335\%$ of the on-axis light intensity. These are different quantities.

The left panel compares normalized wave amplitude magnitude and irradiance; the right panel integrates the transverse area to show enclosed power. For an ideal absorbing circular aperture of radius $a$ centered on the optical axis, the same fraction gives the power passing the aperture at that plane. An aperture of radius $w$ clips about $13.5335\%$ of the incident power; radius $2w$ clips only $e^{-8}\approx 0.000335463$, or $0.0335463\%$. After clipping, extra diffraction means the transmitted field no longer obeys the unmodified ideal Gaussian beam law. Every finite aperture cuts some ideal tail. Kogelnik and Li distinguish their large-aperture beam analysis from their finite-aperture diffraction analysis. ^[ref].

The useful tradeoff is that a small bright spot cannot remain narrow indefinitely. An optical lens can create a smaller beam waist, but diffraction then makes the beam spread faster. The particularly simple relation $w_0\theta =\lambda /\pi$ ties localization directly to beam divergence. A proof outline for the propagation law starts with $U(x,y,0)=U_0{e}^{-(x^2+y^2)/w_0^2}$. The parabolic wave equation separates into two transverse factors, and a Gaussian integral in each coordinate givesHere $q$ is dimensionless and $k$ is the wavenumber; the carrier convention is $e^{i(kz-\omega t)}$, with angular frequency $\omega$. Taking the squared complex modulus gives the stated $w(z)$ and its $w^{-2}$ light intensity scaling. Each transverse factor contributes $q^{-1/2}$, explaining the $q^{-1}$ prefactor here and the different prefactor in the preserved Gaussian beam with one transverse coordinate.

Historically, Gaussian beam theory developed alongside early laser resonator theory and lens-guided transmission. In their 1966 review, Kogelnik and Li credit Boyd and Gordon, and Goubau and Schwering, with early beam treatments; they summarize earlier work rather than claim the whole construction as a new discovery. Their paper brings beam propagation, focusing and finite-aperture diffraction into a common practical account. ^[ref].

Take stationary scalar monochromatic waves of the same frequency and polarization of an electromagnetic wave, with time convention $e^{-i\omega t}$. Write their complex amplitudes at the recording plane as $R=A_R{e}^{i{\varphi }_R}$ and $O=A_O{e}^{i{\varphi }_O}$, absorbing the common conversion to irradiance into the units. An exposure short enough to avoid fringe motion but long compared with an optical period recordsThe two cross terms are where relative wave phase enters the measured light intensity. Blocking the holographic reference wave leaves $|O{|}^2$ and removes them. Loss of mutual optical coherence, incompatible polarization of an electromagnetic wave or motion that averages the relative wave phase similarly reduces fringe contrast. These are useful controls for distinguishing a hologram from an ordinary brightness record. One local cosine value does not uniquely determine a wave phase; known reference structure and separation of the cross terms supply the additional information in the example below.

Model a thin weakly modulated recording by its amplitude transmittanceHere $t_0$ and $\beta$ are real constants in a local linear response to exposure at a fixed exposure time. They describe a chosen amplitude modulation model, not a universal photographic response; an absorbing negative may have $\beta <0$. Keep $0\le t\le 1$ for this passive real-transmission model. Ignore material thickness, saturation, shrinking, clipping and changes of wavelength. Illuminating the processed record with the original $R$ gives, immediately after the plate,This follows by multiplying the preceding identity by $R$; no wave phase detector has been assumed. If $|R|$ is uniform, the desired term reproduces $O$ up to a constant. A nonuniform $|R{|}^2$ weights the recovered field. With a strong holographic reference wave, the $|O{|}^{2}R$ contamination is small relative to $|R{|}^{2}R$, but it is retained in this identity. Subsequent diffraction propagates each term: reproduction at the plate is the starting boundary field, not yet a focused image.

The schematic uses a plane wave reference and one optical point source to separate the functions. The top row records $R+O$; the bottom removes the object and illuminates the developed hologram with $R$. Dashed extensions identify the desired virtual image; the conjugate component converges toward a real image. All three outgoing components coexist. The coloured paths distinguish terms, not different wavelengths; distances and strengths are schematic. This functional drawing is not a reconstruction of the historical apparatus.

Ice-ocean albedo feedback is the ocean-surface case of ice-albedo feedback: loss of bright sea ice, often covered by snow, exposes darker water, lowers surface albedo and increases absorbed sunlight. The extra stored ocean heat content can promote further melting and delay refreezing. Reversing the sequence can amplify cooling. This is a positive climate feedback, meaning that it reinforces the initial change; the word positive does not mean beneficial, and it does not guarantee runaway warming.

For unchanged downward solar irradiance $S_{\downarrow }$ on a newly exposed opaque patch, the extra absorbed heat flux isFor illustrative local surface albedos $0.60$ and $0.10$, and $S_{\downarrow }=200\mathrm{W}{\mathrm{m}}^{-2}$, the patch absorbs an additional $100\mathrm{W}{\mathrm{m}}^{-2}$. This is a local sunlight calculation, not a whole-planet radiative forcing: newly exposed area, illumination, clouds and season determine its global contribution. With no incident sunlight this particular heating pathway vanishes. NASA's notes describe the contrast between reflective ice and snow and darker water. ^[ref]

To isolate how a positive climate feedback competes with Planck feedback, retain the assumptions of the zero-dimensional planetary energy balance and prescribe the smooth broadband Bond albedoThe exponential function makes a smooth transition from a bright cold state near $0.6$ to a darker warm state near $0.3$. The midpoint $240\mathrm{K}$ and width $3\mathrm{K}$ are chosen to illustrate the dynamics. They are not a seawater freezing law, an observed global relation, or a calibration of modern Earth. No evolving ice thickness or latent heat is represented.

An equilibrium satisfies $F(T_{\ast })=0$. For a small perturbation $\delta T$, linear stability follows fromSince $C>0$, $F^{\prime }<0$ gives an asymptotically stable equilibrium, while $F^{\prime }>0$ gives an unstable equilibrium. With $x=(T-240\mathrm{K})/(3\mathrm{K})$, the derivative isThus the optical term $-S{\alpha }^{\prime }/4$ is destabilizing, while the thermal radiation term $-4\sigma T^3$ is stabilizing. Their sum, not the optical term alone, decides the linear stability.

For the same $S=1360\mathrm{W}{\mathrm{m}}^{-2}$ and Stefan-Boltzmann constant used above, the bisection method on the intervals $(220,223)$, $(238,241)$ and $(253,255)$, with temperature in kelvins, gives:

Unrounded numerical zeros of a function have $|F(T_{\ast })|<10^{-10}\mathrm{W}{\mathrm{m}}^{-2}$; the table rounds them for readability. In the positive-temperature domain the sign of $F$ alternates between heating, cooling, heating and cooling across these three equilibria. The middle equilibrium separates the basins of attraction of the two stable equilibria. For example, a starting temperature of $235\mathrm{K}$ cools toward the cold state, whereas $245\mathrm{K}$ warms toward the warm state: this is bistability at identical external illumination.

The same model illustrates hysteresis. Its equilibrium curve can be written as $S(T)=4\sigma T^4/[1-\alpha (T)]$. The two turning points, where $4[1-\alpha (T)]+T{\alpha }^{\prime }(T)=0$, are saddle-node bifurcations: approximately $(T,S)=(232.637\mathrm{K},1567.801\mathrm{W}{\mathrm{m}}^{-2})$ and $(245.893\mathrm{K},1250.481\mathrm{W}{\mathrm{m}}^{-2})$. For illumination strictly between these thresholds there are two stable equilibria and one unstable equilibrium. Under sufficiently slow increase of $S$, the cold branch ends at the upper threshold; under sufficiently slow decrease, the warm branch ends at the lower threshold. The ensuing warming and cooling transitions differ, so the state depends on its history. Finite sweep rates can delay the transitions; these thresholds describe this model's static equilibrium curve, not predicted tipping points for modern Earth.

The striking feature is that a smooth albedo law can produce distinct long-time states and abrupt transitions, although no discontinuity was inserted into the optical law. It is useful for understanding why positive climate feedback can amplify a response while stabilizing thermal radiation still limits it. Here $0.3<\alpha (T)<0.6$ bounds the incoming heating, whereas outgoing thermal radiation grows as $T^4$; high enough temperatures therefore cool rather than run away indefinitely.

Historically, M. I. Budyko's 1969 paper coupled changing albedo from expanding ice cover to a latitude-dependent temperature and heat transfer calculation, showing how that coupling amplified the effect of changed sunlight. ^[ref] William D. Sellers's 1969 paper constructed an annual energy-budget model in 10-degree latitude belts, including planetary albedo, atmospheric infrared transmission and atmospheric and oceanic exchange. ^[ref] These primary papers provide historical examples of ice-sensitive energy-budget modeling; the smooth single-temperature example above is a separate illustration, not a reproduction of either paper's equations or predictions.

Illuminance is incident luminous flux per receiving area, measured in lux. It is the visually weighted counterpart of irradiance. Uniform illumination of an aperture of area $A$ gives collected luminous flux $A{E}_v$. An aperture measurement alone does not give a source's total emitted luminous flux.^[ref]

Wave interference is the reinforcement or cancellation that occurs when overlapping wave amplitudes add by the superposition principle. For two monochromatic waves of the same frequency, mutual optical coherence means that their relative wave phase is stable during the observation. Equal wave phases give constructive interference; a half-cycle phase difference gives the destructive interference intensity minimum. Wave amplitudes add before their squares determine wave intensity. ^[ref].

Consider two overlapping beams in a homogeneous medium, with the same polarization of an electromagnetic wave and individual local time-averaged light intensities $I_1,I_2$. Their phase difference $\delta$ is measured in radians. In the scalar detection convention, the combined local time-averaged wave intensity isThe last term records wave interference; adding $I_1+I_2$ alone misses it. The formula describes irradiance when both fields propagate along the same direction, or when the same detector geometry factor is included in all three intensities. For crossing beams it describes the squared electric field response, proportional to scalar wave intensity; a specified component of the Poynting vector also requires the propagation geometry. The ideal crossing plane waves below make that distinction explicit.

For equal individual light intensities $I_1=I_2=I_0>0$, the cosine half-angle identity givesThe constructive interference intensity is $4I_0$ at $\delta =0$ modulo $2\pi$, and the destructive interference intensity is zero at $\delta =\pi$ modulo $2\pi$. At phase quadrature, $\delta =\pi /2$, it is $2I_0$. With $I_1=1$ and $I_2=0.25$ in the same arbitrary wave intensity unit, the maximum is $2.25$, the minimum is $0.25$, and the phase quadrature value is $1.25$: unequal wave amplitudes cannot cancel completely.

Both curves use the same wave intensity unit and a horizontal angle in radians. The dashed levels are the averages over a full phase difference cycle. Equal beams have interferometric visibility one; the unequal example has interferometric visibility $0.8$.

A spatially varying phase difference turns the same law into interference fringes. Take two ideal plane electromagnetic waves with common wavelength $\lambda$, equal individual light intensities $I_0$, and propagation directions at angles $+\theta$ and $-\theta$ from the $z$ axis in the $xz$ plane, with $0<\theta <\pi /2$. Both have linear polarization along $y$, perpendicular to both propagation directions. Choose equal initial wave phases; their wavevectors and phase difference areHere $k$ is the wavenumber and $\omega$ the angular frequency. The common $z$ and time terms cancel, so the interference fringes stay at fixed $x$ for every $z$. Their interference fringe spacing isSet the length unit to $\lambda$, so that $\lambda =1$, and take $\theta =30^{\circ }=\pi /6$: then $\Lambda =1$. The constructive interference intensity bands lie at integer $x/\lambda$; the destructive interference intensity bands lie at half-integer $x/\lambda$.

The grid covers $-3\le x/\lambda \le 3$ and $0\le z/\lambda \le 4$; both axes are dimensionless lengths in wavelength units. The colour scale is $I/I_0$ from zero to four, using the same single-beam normalization as the equal-beam curve above. This is an ideal plane wave superposition, with no slits or point sources and no near-field approximation. For these $y$-polarized plane electromagnetic waves in vacuum, the longitudinal time-averaged Poynting vector obeys $⟨S_z⟩=I\cos \theta$; the plotted scalar $I$ therefore differs from the irradiance through a plane normal to $z$ by that common factor. This relation follows by adding the magnetic fields as well as the electric fields.

The bright peaks do not create energy. Averaging over one interference fringe spacing, or equivalently over one full phase difference cycle, givesbecause the integral of the cosine is zero, with sine as its antiderivative. Equal beams thus average to $2I_0$, despite bright bands at $4I_0$. For the crossing-wave geometry, the average longitudinal energy flux is correspondingly $2I_0\cos \theta$. Bright and dark regions redistribute the incoming energy; conservation of energy concerns the full energy flux, rather than the height of one peak.

Wave interference is useful because phase differences convert small changes in optical path length into measurable changes of wave intensity, as in a Mach-Zehnder interferometer. Its beauty is that the same superposition principle produces both complete cancellation and fourfold brightening; adding more coherent contributions also produces the narrow peaks of a diffraction grating. Feynman's treatment connects this many-source pattern directly to addition of complex amplitudes. ^[ref].

A short proof outline uses local real fields $E_j(t)=A_j\cos (\omega t+{\varphi }_j)$ with parallel linear polarizations. Write $I=C⟨(E_1+E_2{)}^2{⟩}_t$ and $I_j=C{A}_j^2/2$ for the same positive conversion factor $C$. Averaging over many wave periods while the phase difference stays fixed uses $⟨{\cos }^2(\omega t+{\varphi }_j){⟩}_t=1/2$ and $⟨\cos (\omega t+{\varphi }_1)\cos (\omega t+{\varphi }_2){⟩}_t=\cos \delta /2$, giving the stated law. Equivalently, add the complex amplitudes and take the squared complex modulus. ^[ref].

For different fixed linear polarizations in a common transverse plane, the cross term acquires the dot product of their real unit vectors, equal to $\cos \alpha$ for relative angle $\alpha$. Orthogonal linear polarizations remove the cross term in polarization-insensitive scalar detection; different orientations generally only reduce it. A common polarizer can select overlapping components and recover wave interference. Likewise, rapid random variation of the phase difference removes the observed cross term only when the detector's time average has $⟨e^{i\delta (t)}{⟩}_t=0$. Slowly varying wave phases can instead move resolved interference fringes, and biased or incompletely sampled wave phases can leave residual contrast. Finite optical coherence therefore depends on the observation time as well as the fields.

Irradiance measures incident radiant flux per actual receiving area. Its units $\mathrm{W}{\mathrm{m}}^{-2}$ match radiant exitance, but it counts the incoming side of the radiation exchange. Measure incoming angles from the unit normal pointing toward the incident hemisphere. ThenFor $2\mathrm{W}$ distributed uniformly over $0.010{\mathrm{m}}^2$, $E=200\mathrm{W}{\mathrm{m}}^{-2}$. If the receiving plane is tilted $60^{\circ }$ in the same broad uniform parallel beam, the incident power per actual area is halved to $100\mathrm{W}{\mathrm{m}}^{-2}$. This receiving-surface cosine factor is geometric and does not assume Lambertian reflectance. Together, irradiance and radiant exitance describe how radiation supplies and removes energy at a surface, while radiance preserves the directional information.

Lambertian reflectance is the ideal diffuse-reflection model with direction-independent outgoing radiance, the frequency-integrated specific intensity. If a fraction $0\le \rho \le 1$ of incident radiant flux is reflected, a surface receiving irradiance $E$ has reflected radiance $L=\rho E/\pi$. The reflected outward power per area is $\pi L=\rho E$ by the same hemispherical integral used for radiant exitance. Its radiant intensity still acquires the projected area factor in Lambert's cosine law.

For $E=100\mathrm{W}{\mathrm{m}}^{-2}$ and $\rho =0.80$, the reflected radiance is $80/\pi \approx 25.4648\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}$. A patch of area $0.020{\mathrm{m}}^2$ reflects $1.6\mathrm{W}$ in total, and its radiant intensities at $0^{\circ }$ and $60^{\circ }$ are respectively $1.6/\pi$ and $0.8/\pi \mathrm{W}\mathrm{s}{\mathrm{r}}^{-1}$. The radiance stays fixed as the view changes, provided the illumination stays fixed. Changing the incident angle can change irradiance and therefore the reflected radiance.

A spherical body of radius R observed from distance $d\gg R$ is illuminated and viewed from the same direction, with incident irradiance $F_{\star }$. For uniform Lambertian reflectance A, a patch at angle $\theta$ has outgoing radiance $A{F}_{\star }\cos \theta /\pi$. The observer receives this radiance times projected area divided by $d^2$. Since ${\int }_{\mathrm{hemisphere}}{\cos }^2\theta dS=2\pi R^2/3$, integration proves the displayed full-phase irradiance. Its geometric albedo is $2A/3$. Total reflected power alone does not specify the brightness in one direction for a non-Lambertian surface.

For an ideal diffuse emitting or reflecting planar surface with uniform direction-independent radiance, the radiant intensity in a direction at angle $\theta$ to its unit normal is $I(\theta )=I(0)\cos \theta$. This follows immediately from $I=LA\cos \theta$ and the projected area formula. The worked radiance example and diagram show the $0^{\circ }$ and $60^{\circ }$ cases. Lambertian reflectance supplies one physical idealization realizing the law; an ideal blackbody provides an emitting example.

The historical setting is Johann Heinrich Lambert's Photometria sive de mensura et gradibus luminis, colorum et umbrae, published in Augsburg in 1760. Its Part I, §139, explicitly compares normal and oblique illumination using the sine of the source's elevation, equivalently the cosine of its angle from the receiving normal vector. This is an early receiving-surface projection calculation; the modern outgoing-patch statement above keeps radiance, radiant intensity and irradiance distinct. ^{[ref] [ref]}.

For a circular stellar disk whose specific intensity is independent of azimuth, its mean specific intensity over projected area isThe polar coordinates element is $d{A}_{\mathrm{proj}}=2\pi rdr$ after angular integration, and $rdr=-R_{\star }^2\mu d\mu$. Substitution of the linear limb-darkening law givesFor a distant observer at distance $D\gg R_{\star }$ receiving light normally to the detecting plane, the observed radiative flux isIt is spectral irradiance, with units $\mathrm{W}{\mathrm{m}}^{-2}\mathrm{H}{\mathrm{z}}^{-1}$ when $I_{\nu }$ is per hertz. The local outward radiative flux per actual surface area is instead $F_{\nu }=2\pi {\int }_0^1{I}_{\nu }(\mu )\mu d\mu =\pi {\overline{I}}_{\nu }$, a spectral radiant exitance. In a spherical model with the same stellar atmosphere everywhere, the luminosity is $L_{\nu }=4\pi R_{\star }^2{F}_{\nu }$ and $f_{\nu }=L_{\nu }/(4\pi D^2)$, consistent with the inverse-square law. Compare disks at fixed $I_c$ or fixed total observed radiative flux explicitly: increasing $u$ lowers the total in the first comparison, while the second requires a larger $I_c$.