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The Airy disk is the central bright region, up to the first dark ring, in the scalar diffraction pattern of a uniformly illuminated, unobscured circular aperture. The whole pattern includes the weaker surrounding rings; its radius is distinct from its full width at half maximum.

For a circular optical pupil of diameter $D$ under normal, uniform monochromatic plane wave illumination, Fraunhofer diffraction givesHere $I_0$ is the on-axis light intensity, $\theta$ is the angle from the optical axis, $\lambda$ is the wavelength in the propagation medium, and $J_1$ is the order-one Bessel function of the first kind. The paraxial approximation replaces $\sin \theta$ by $\theta$, measured in radians. The first positive zero $j_{1,1}=3.831705970\dots$ givesThis is the angular radius of the Airy disk; the angular diameter is twice as large. The Airy disk derivation explains the origin of the Bessel function of the first kind and of the numerical coefficient.^[ref]

The image and both radial curves show the same calculated light intensity, divided by its peak. The image's logarithmic scale makes the weak rings visible; its colour is a display choice. The linear curve shows how small the ring peaks are compared with the centre. A dark ring marks a zero of wave amplitude, produced by cancellation across the optical pupil; it is not the edge of all transmitted light. The first Airy disk contains about $83.78\%$ of the total radiant flux, as calculated in Airy disk normalization and encircled energy.

For an original design example, take an ideal optical lens in air with focal length $f=100\mathrm{mm}$, circular optical pupil diameter $D=5.00\mathrm{mm}$ and wavelength $\lambda =550\mathrm{nm}$. In the focal plane, the radial coordinate $r\simeq f\theta$ givesThe first-zero diameter is $26.8327\mu \mathrm{m}$. Solving $P(u_{1/2})=1/2$ gives $u_{1/2}=1.616339948\dots$, so the diameter measured by full width at half maximum is onlyThus a quoted “spot diameter” is incomplete without its definition. Doubling $D$ halves both widths at fixed wavelength and focal length; increasing magnification alone changes the image scale without improving object-space discrimination.

The Airy disk supplies a calculable reference for a point spread function, allowing diffraction to be distinguished from defocus and optical aberrations. Its beauty is that one circular boundary determines an entire sequence of bright and dark rings through a Fourier transform. The ideal formula assumes an unobscured pupil, uniform illumination, ideal optical focus, a single wavelength, and a scalar description with small output angles. A blocked centre, nonuniform pupil transmission or a significant pupil wave phase error changes the response. High numerical aperture may require polarization of an electromagnetic wave and a vector field description. The circular-aperture diffraction laboratory experiment tests the far-field pattern directly; the Rayleigh criterion uses its first zero to define a two-source convention.

The antenna effective area converts incident energy flux into available received power for a specified direction and matched polarization of an electromagnetic wave. In a lossless free-space link, a transmitter of antenna gain $G$ at range $r$ and a receiver of effective area $A_{\mathrm{eff}}$ have $P_r/P_t=G{A}_{\mathrm{eff}}/(4\pi r^2)$. Impedance or polarization mismatch reduces the delivered power.

In his 1958 lecture, Pavel Cherenkov recalls the 1934 reports by himself and Sergey Vavilov: radium gamma rays illuminated solutions and their solvents. Weak light was compared using dark-adapted human vision and threshold photometry. Heating and added fluorescence quenching agents left the solvent component and its polarization of an electromagnetic wave unchanged; parallel aqueous fluorescent controls visibly quenched. Its electric vibration direction was parallel to the exciting beam, unlike the compared polarized fluorescence. Further experiments attributed the emission to fast electrons produced by Compton scattering, rather than directly to gamma rays. These comparisons supported a mechanism distinct from ordinary luminescence.^[ref]

The evidential step is the controlled comparison, rather than the mere presence of light. A weak glow could include fluorescence from impurities; suppressing a known fluorescent control tests whether the intervention actually works. Visual threshold comparisons also require a stable observer and viewing conditions. These early controls do not themselves calibrate absolute photon yield or supply a single-electron angular scan. Its polarization of an electromagnetic wave comparison also does not by itself measure the Cherenkov cone angle of an individual electron.

Ilya Frank and Igor Tamm supplied the theory in 1937; Ilya Frank also investigated nuclear physics experimentally and later directed neutron research.^[ref] Sergey Vavilov, an optics and luminescence researcher and collaborator in the discovery, died in 1951. The 1958 Nobel Prize in Physics went to Pavel Cherenkov, Ilya Frank and Igor Tamm; Sergey Vavilov was not a recipient.^[ref]

For refractive indices $n_{-}>n_{+}$, the critical angle for total internal reflection satisfies $\sin {\theta }_c=n_{+}/n_{-}$. Above it, the transmitted normal wavenumber is imaginary and the transmitted field is an evanescent wave.

The critical angle for total internal reflection is the boundary between outgoing refraction and an evanescent wave, rather than an angle at which the external electric field abruptly disappears. Write $n_1=n_{-}$ and $n_2=n_{+}$. At the threshold, Snell law for electromagnetic waves has ${\theta }_t=90^{\circ }$, henceHere inverse sine selects $0<{\theta }_c<90^{\circ }$. For $n_1=1.5$ and $n_2=1$, $\sin {\theta }_c=2/3$ and ${\theta }_c\approx 41.8103149^{\circ }$. The threshold is independent of polarization of an electromagnetic wave under the stated assumptions.

A short proof using phase matching at a planar wave interface reveals all three cases at once. Choose a Cartesian coordinate system with boundary $z=0$, incident medium $z<0$, and transmitted medium $z>0$. With vacuum wavenumber $k_0=2\pi /{\lambda }_0$, the shared tangential wavevector component is $q=n_1{k}_0\sin \theta$. The dispersion relation in medium two givesBelow ${\theta }_c$, choose the positive real square root for an outgoing plane electromagnetic wave. At ${\theta }_c$, $k_{2z}=0$: there is a grazing wave, with zero normal average power but no exponential confinement in $z$. Above ${\theta }_c$, choose $k_{2z}=i\kappa$ with $\kappa >0$, so the wave amplitude decays into the second half-space.

If $n_1\le n_2$, no nongrazing angle of incidence produces this threshold. As $n_2/n_1\to 1$ from below, ${\theta }_c\to 90^{\circ }$; as that ratio tends to zero, ${\theta }_c\to 0$. These limits explain why a larger refractive index contrast permits confinement over a wider range of internal angles.

Diffraction-limited resolution is a scale for distinguishing details with a specified ideal diffraction response and a specified criterion. For two equally bright, mutually incoherent optical point sources imaged through a uniformly transmitting, unobscured circular optical pupil, the Rayleigh criterion places the centre of one Airy disk at the first zero of the other. At small angles in air this separation isHere $D$ is pupil diameter, $\lambda$ is wavelength in air, and $j_{1,1}$ is the first positive zero of the order-one Bessel function of the first kind. This is the radius to the first zero, not the diameter of the central spot. Fraunhofer diffraction gives the radial peak-normalized light intensityThe removable value at zero follows from $J_1(q)\sim q/2$. The pupil Fourier transform reduces to a radial integral by circular symmetry, which produces the Bessel function of the first kind; its first zero supplies the factor $1.22$.^[ref]

In the usual scalar circular-pupil approximation for a fluorescence microscope, the corresponding object-plane Rayleigh criterion isHere ${\lambda }_0$ is the vacuum emission wavelength, and the object-side numerical aperture already includes the medium's refractive index. Do not divide ${\lambda }_0$ by $n$ again. For small cone angles in air, $\mathrm{NA}\simeq D/(2f)$ and image-plane radius $f{\theta }_{\mathrm{R}}$ agrees with this form. The assumptions include a single wavelength, ideal optical focus, and mutually incoherent emitters. High numerical aperture can require a vector description with polarization of an electromagnetic wave and emitter orientation; confocal microscopy has its own excitation-and-detection response. The same $0.61$ formula therefore does not describe every microscope or illumination method.^[ref]

In this original model, ${\lambda }_0=550\mathrm{nm}$ and numerical apertures $0.25$ and $0.50$ give first-zero radii $1.342\mu \mathrm{m}$ and $0.671\mu \mathrm{m}$. Each point spread function is divided by its own peak to compare shape. The pupil axes are transverse wave vector components multiplied by ${\lambda }_0/(2\pi )$, so pupil radius equals numerical aperture. A wider pupil accepts a wider range of transverse wave vectors, producing a narrower image. Increasing magnification alone enlarges the same spot and does not reduce its object-plane radius.

The Rayleigh criterion is a useful visual convention, not an absolute information barrier. For equal incoherent sources with total detected radiant flux $F$ and separation vector $\mathbf{s}$, the ideal image isBelow the Rayleigh criterion this still differs from $Fp(\mathbf{x})$; for small $\mathbf{s}$, symmetry cancels the linear change and the first change is quadratic in separation. Estimating two unknown source positions is then increasingly sensitive to noise, background, relative brightness and uncertainty in the point spread function. Localizing one isolated source and recovering an arbitrary extended object are different inference tasks. Measurements that separate optical modes can also extract separation information differently from direct image light intensity measurement.^[ref]

There remains a real limit for arbitrary images in this model: the optical transfer function is zero above the transmitted spatial frequency band. Its cutoff for scalar incoherent imaging is ${\nu }_c=2\mathrm{NA}/{\lambda }_0$ in object-plane cycles per unit length. The Nyquist–Shannon sampling theorem therefore suggests object-referred detector pixel pitch $p_{\mathrm{pix}}\le {\lambda }_0/(4\mathrm{NA})$, with finer sampling providing margin. Finite detector pixels also integrate over their area, suppressing contrast; coarse sampling can cause aliasing. Finer sampling cannot alone recover spatial frequencies removed by the optics, and Fourier deconvolution amplifies noise where the optical transfer function is small. Photon shot noise, read noise, optical aberrations and defocus can all worsen practical discrimination. A useful resolution claim therefore reports the object model, criterion, wavelength, numerical aperture, sampling and signal-to-noise ratio in photon counting.

Digital optical phase conjugation uses a wavefront sensor and a spatial light modulator to measure an incident field and launch a beam with the opposite spatial wave phase. A phase-only spatial light modulator illuminated uniformly generally does not reproduce the incident wave amplitude as well; exact $\gamma U^{\ast }$ requires the appropriate wave amplitude and polarization of an electromagnetic wave too. The experimentally useful question is how much of the original optical mode can be recovered with the measured and controlled field components.

In 2010, Meng Cui and Changhuei Yang demonstrated digital optical phase conjugation through a prepared multiple scattering sample. They dried mixed-size polystyrene microspheres on cover glass, added immersion oil, and sealed them with another cover glass. A 532 nm laser supplied signal and reference beams. The signal crossed the sample between two objectives; the reference interfered with it on a charge-coupled device. Phase-stepped wave interference recovered the spatial wave phase; a calibrated spatial light modulator imposed its negative on the reference-derived return beam. The reference supplied playback power, without nonlinear mixing pumps. An optical camera read out the recovered signal. With the same sample and a zero-phase spatial light modulator control, the return instead produced a scattering of light pattern. This comparison supports mode recovery by spatial wave phase control, not merely increased illumination. A separate ordinary-mirror control tested source displacements. Finite sampling and collection limited recovery; the system controlled wave phase rather than the complete vector field. These restrictions distinguish demonstrated reconstruction from exact reversal of every scattered component.^[ref]

In a thin rectangular electromagnetic cavity, the window $\hslash c\pi /L_x,\hslash c\pi /L_y\ll k_{B}T\ll \hslash c\pi /L_z$ freezes the nonzero thickness indices while retaining many in-plane normal modes. The zero-thickness-index family has one polarization of an electromagnetic wave and density of states $dN=Akdk/(2\pi )=2\pi A\nu d\nu /c^2$. Multiplying by $h\nu /(e^{h\nu /k_{B}T}-1)$ and integrating gives $U_{2D}=4\pi \zeta (3)A(k_{B}T{)}^3/(h^2{c}^2)$. Thus thermal energy scales as area times $T^3$. The Riemann zeta function appears through ${\int }_0^{\infty }{x}^2/(e^x-1)dx=2\zeta (3)$.

This is an intermediate low-temperature regime of an anisotropic finite cavity. When $k_{B}T$ also falls below the first in-plane excitation, thermal energy becomes exponentially small. If a different model excludes every zero thickness index, its first transverse mode has a finite gap and the proposed $T^3$ regime does not occur. In the three-dimensional continuum regime the corresponding density of states is proportional to volume times ${\nu }^2$ and thermal energy is proportional to volume times $T^4$.

Measuring how specular reflection changes the polarization of an electromagnetic wave to infer optical properties.

For an elliptically polarized wave, the tip of its transverse vector traces an ellipse at a fixed point during each wave period. For polarization of an electromagnetic wave, this is the electric field vector. With the real components in polarization of waves, assume $A_x,A_y>0$ and $\sin \delta \ne 0$. Eliminating time givesA short derivation uses $\varphi =\omega t$ and the cosine addition formula to obtain $E_y/A_y=(E_x/A_x)\cos \delta -\sin \varphi \sin \delta$; squaring and using the sine and cosine identity ${\sin }^2\varphi =1-{\cos }^2\varphi$ gives the displayed equation. Outside phase quadrature, the mixed term gives a tilted ellipse; $A_x,A_y$ are component wave amplitudes and need not be its semi-major and semi-minor axes.

In phase quadrature, $\delta =\pm \pi /2$, the mixed term vanishes. For $(A_x,A_y)=(2,1)V/m$, the ellipse is $E_x^2/4+E_y^2=1$, its semi-major and semi-minor axes have lengths $2$ and $1V/m$, and its electric field magnitude varies between $1$ and $2V/m$. At $\varphi =\pi /4$ with $\delta =\pi /2$, the electric field is $(\sqrt{2},-1/\sqrt{2})V/m$ and has magnitude $\sqrt{5/2}V/m$. Equal wave amplitudes instead give circular polarization, a special case of elliptical polarization with constant magnitude. As $\delta$ approaches $0$ or $\pi$, the ellipse collapses to a line segment and gives linear polarization; this limit is degenerate rather than a nondegenerate ellipse.

Changing $\delta$ from $\pi /2$ to $-\pi /2$ reverses traversal of the same ellipse. In the displayed $x,y$ plot, $\delta =\pi /2$ moves from $(A_x,0)$ toward negative $y$ as time increases. A label such as right- or left-handed circular polarization additionally requires a stated viewing direction and naming convention: viewing along propagation and viewing against propagation reverse the apparent rotation. ^[ref].

The ellipse records both the relative wave amplitudes and the phase difference of two transverse oscillations. This separates a change of polarization of waves from a change of overall wave amplitude: multiplying both components by the same positive factor enlarges the ellipse while preserving its shape and orientation.

A rotating ideal linear polarizer can probe this geometry. For the $2:1$ ellipse in phase quadrature, orthogonal projection onto a pass axis at angle $\theta$ gives $E_{\mathrm{out}}(t)=2\cos \theta \cos \omega t-\sin \theta \sin \omega t$, in $V/m$. Averaging its square over a wave period gives $(4{\cos }^2\theta +{\sin }^2\theta )/2$, while the incoming squared electric field magnitude averages to $5/2$. Thus the transmitted fraction of light intensity is $(4{\cos }^2\theta +{\sin }^2\theta )/5$: it is $4/5$ at $0^{\circ }$, $1/2$ at $45^{\circ }$ and $1/5$ at $90^{\circ }$. There is no single incident line of linear polarization to insert into Malus's law; projecting the electric field and averaging its square is the appropriate generalization. This measurement cannot distinguish $\delta =\pi /2$ from $\delta =-\pi /2$, because reversing the traversal leaves all these squared averages unchanged.

The evanescent field in total internal reflection oscillates and advances in wave phase along the boundary while its wave amplitude decays away from it. With the coordinates above and time dependence $e^{-i\omega t}$, a TE polarization has complex amplitudeThe physical electric field is $\mathrm{Re}[\mathbf{E}(x,z)e^{-i\omega t}]$. The decaying sign excludes a field that grows without bound at $z\to \infty$. Dielectric-interface boundary conditions determine $A$; for nongrazing incidence it is generally nonzero.^[ref]

Zero normal energy flux follows directly from Maxwell equations. The wavevector is $(q,0,i\kappa )$; the relation $\mathbf{H}=(\mathbf{k}\times \mathbf{E})/(\omega {\mu }_0)$ givesConsequently, the cycle-averaged Poynting vector isThe electric field and $H_x$ are a quarter-cycle apart, so their product has zero cycle average. Tangential energy flux survives. This is a local calculation for the single decaying evanescent wave in the infinite second half-space.

To check reflected power independently, put $a=n_1{k}_0\cos \theta >0$. The existing transverse-electric Fresnel reflection coefficient becomes $r=(a-i\kappa )/(a+i\kappa )$. Its numerator and denominator are complex conjugates, so $|r{|}^2=1$, while $1+r=2a/(a+i\kappa )\ne 0$ supplies the surface electric field relative to the incident wave amplitude. Thus a unit optical reflectance does not require a zero transmitted field. The same normal power balance holds for TM polarization; its reflected wave phase generally differs from the TE polarization phase. A mixture can therefore change its polarization of an electromagnetic wave on reflection even though all normal incident power returns.

Consider two identical, lossless parallel mirrors with intensity reflectivity $0\le R<1$, separation $d$, and a uniform transparent medium of refractive index $n$. Fix one polarization of an electromagnetic wave and an internal angle of incidence $\theta$ measured from the mirror normal vector. Illuminate with a monochromatic wave having planar wavefronts; the emerging contributions must overlap coherently in the detected direction. Define $T=I_{\mathrm{transmitted}}/I_{\mathrm{incident}}$ using the same incident and exit media. Each mirror transmits a fraction $1-R$ of incident power. Its reflected wave amplitude has magnitude $\sqrt{R}$, so two internal specular reflections multiply the circulating complex amplitude by magnitude $R$, not $R^2$.

Let $\lambda$ be the vacuum wavelength and choose the reflection wave phase convention in which the product of the two internal amplitude reflection coefficients is real and positive. The phase difference between successive transmitted contributions, evaluated on a common outgoing wavefront, isWith other mirror coatings or phase conventions, use the total $\delta =4\pi nd\cos \theta /\lambda +{\varphi }_{\mathrm{rt}}$, where ${\varphi }_{\mathrm{rt}}$ is the sum of the two internal reflection wave phases. The stated resonance positions presume ${\varphi }_{\mathrm{rt}}=0$ modulo $2\pi$; a constant nonzero ${\varphi }_{\mathrm{rt}}$ shifts them, while a frequency-dependent one can also change their spacing.

Why does the cosine appear rather than its reciprocal? Inside the gap, the extra two legs have optical path length $2nd/\cos \theta$. But the later emerging contribution starts $2d\tan \theta$ farther along the mirror. Comparing both on one outgoing wavefront subtracts $2nd\tan \theta \sin \theta$ from that path difference, leaving $2nd\cos \theta$. Equivalently, a plane wave's mirror-normal wavevector component is $k_z=(2\pi n/\lambda )\cos \theta$, so the round-trip wave phase is $2k_{z}d$. The diagram uses the same medium outside and inside to show the common-wavefront subtraction directly; at a refracting exit interface, tangential wavevector matching in Snell's law supplies the same result.

Absorb the common first-pass wave phase into the incident complex amplitude $E_{\mathrm{in}}$. Successive transmitted contributions are $(1-R)E_{\mathrm{in}}$, $(1-R)E_{\mathrm{in}}R{e}^{i\delta }$, and so on. The superposition principle and the convergent geometric series giveThe last step expands the squared complex modulus as $1+R^2-2R\cos \delta$. This transmission profile is often called the Airy transmission function; it is distinct from the Airy disk, a diffraction image of a circular aperture. The University of Washington's lab handout gives the same transmission profile and distinguishes mirror reflectivity from other sources of peak broadening.^[ref]

For an original example, set $R=3/4$. At $\delta =0$, the amplitudes add to $(1/4)/(1-3/4)=1$, so $T=1$. At $\delta =\pi$, they alternate in sign and add to $(1/4)/(1+3/4)=1/7$, so $T=1/49$. At $\delta =\pi /2$, the amplitude is $4/25+3i/25$ and $T=1/25$. Adding the individual intensities instead would give $(1-R{)}^2{\sum }_j{R}^{2j}=(1-R)/(1+R)=1/7$, independent of wave phase; that discards the wave interference responsible for the peaks.

In general the maximum is $T_{\max }=1$ at $\delta =2\pi m$, with integer $m$, and the minimum is $T_{\min }=[(1-R)/(1+R){]}^2$ halfway between peaks. Perfect resonant transmission requires identical lossless mirrors, coherent illumination, and sufficient time to establish the steady field. Absorption, unequal mirror transmission, finite beams that fail to overlap, and fluctuations of $d$ can reduce it. The exactly reflecting case $R=1$ is excluded: no incident power enters through a perfectly reflecting mirror, and the resonant limit $R\to 1^{-}$ is singular.

The Fresnel equations give reflection and transmission amplitudes for both transverse-electric and transverse-magnetic polarization of an electromagnetic wave at a planar interface. They follow by applying interface conditions for electromagnetic fields to incident, reflected and refracted plane electromagnetic waves.

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.

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.

Guiding light and retaining useful optical power are different achievements. Charles Kuen Kao and George Alfred Hockham argued that impurities need not impose the measured losses of existing glasses on every future optical fiber. Their 1966 paper treats wave absorption, scattering of light and guidance separately; reducing impurity absorption leaves intrinsic material losses and geometric losses to address. This was a route toward practical communication, building on existing optical fibers, rather than the invention of every use of guided light.^[ref]

The launch experiment used clad glass optical fibers fixed in capillary tubes with epoxy for end polishing. Laser illumination gave transverse normal mode patterns; input-spot displacement and polarization of an electromagnetic wave controlled excitation. These observations tested guidance, not kilometre-scale optical fiber attenuation.^[ref]

Charles Kuen Kao's lecture describes the complementary material experiment: short, polished glass samples, including plasma-deposited fused silicon dioxide, were examined with purpose-built single-beam and later double-beam spectrophotometers. A separate ellipsometry measurement assessed end-surface specular reflection, a check against confusing interface loss with bulk wave absorption. The readout was loss versus wavelength. Comparing glass samples supported impurity reduction as a route to lower loss. Surface contamination and the tiny bulk loss of short samples limited inference; the result did not itself demonstrate a long drawn optical fiber. The lecture also credits subsequent manufacturing work at Corning.^[ref]

An original scale comparison makes the material problem tangible. In the uniform optical fiber attenuation model, a stipulated $20dB/km$ leaves $10^{-2}=0.01$ of launched optical power after $1\mathrm{km}$; a stipulated $0.2dB/km$ leaves $10^{-0.02}\simeq 0.95499$. Both guides could satisfy the same total internal reflection condition. A successful spot pattern at the output therefore cannot by itself establish a useful communication distance.

Ignoring magnetization, electric polarization charge transport gives bound current ${\mathbf{J}}_{\mathrm{b}}={\partial }_t\mathbf{P}$. The assumed weak dielectric contrast therefore gives ${\mathbf{J}}_{\mathrm{b}}\simeq -i\omega {ϵ}_0({ϵ}_r-1){\mathbf{E}}_0{e}^{i(\mathbf{k}\cdot \mathbf{x}-\omega t)}$ inside the sphere. This is the first Born approximation: the incident field supplies the electric polarization to leading order in ${ϵ}_r-1$.

For $\mathbf{x}=r\mathbf{n}$, the far-distance expansions are $|\mathbf{x}-{\mathbf{x}}^{\prime }|=r-\mathbf{n}\cdot {\mathbf{x}}^{\prime }+O(a^2/r)$ and $|\mathbf{x}-{\mathbf{x}}^{\prime }{|}^{-1}\simeq r^{-1}$. Substitute the retarded time into the phase in the Lorenz gauge vector potential. The phase becomes $kr-\omega t+(\mathbf{k}-k\mathbf{n})\cdot {\mathbf{x}}^{\prime }$, so, using ${\mu }_0{ϵ}_0=c^{-2}$,The assumptions $r\gg a$ and $k{a}^2/r\ll 2\pi$ control amplitude and phase errors respectively. In the radiation zone the supplied electric and magnetic fields satisfy ${\mathbf{B}}_{\mathrm{scatt}}=c^{-1}\mathbf{n}\times {\mathbf{E}}_{\mathrm{scatt}}$. Complex phasors therefore give the time-averaged radial Poynting flux $|{\mathbf{E}}_{\mathrm{scatt}}{|}^2/(2{\mu }_{0}c)$. The incident flux is $|{\mathbf{E}}_0{|}^2/(2{\mu }_{0}c)$, soOrient spherical coordinates along $\mathbf{q}$ to evaluate the form factor:ConsequentlyAt $q=0$ the bracket is interpreted by its limit $a^3/3$, not a singularity. The transverse factor accounts for polarization of an electromagnetic wave; for complex polarization of an electromagnetic wave its squared norm means the Hermitian norm.

In a box of volume $V$, the electromagnetic mode density in a box, including the two transverse polarizations, isEach photon has energy $\hslash \omega =\hslash ck$, and the given mean occupation number is $(e^{\hslash ck/T}-1{)}^{-1}$. Omitting the temperature-independent zero-point energy, the total energy is thereforeWith the change of variables $x=\hslash ck/T$,

An ideal cross-Kerr interaction between two single modes is $H_{\mathrm{int}}=\hslash \chi n_1{n}_2$, with $n_j=a_j^{†}{a}_j$. The number operators commute with the interaction, so this coupling changes phases without transferring photons between the modes. To implement the specified positive conditional phase, usewhere $n_{2V}=|V⟩⟨V|$ within the one-photon polarization subspace. The horizontal branch has zero interacting occupation. Since $e^{i\Theta n}|\alpha ⟩=|e^{i\Theta }\alpha ⟩$, the specified conditional rotation follows. This is an ideal lossless single-mode model; an actual travelling-wave medium's bandwidth, loss and noise are not established by writing this Hamiltonian.

Use $X_1=(a+a^{†})/2$ and $X_2=(a-a^{†})/(2i)$. A coherent state has centre $(\mathrm{Re}\alpha ,\mathrm{Im}\alpha )$ and variance $1/4$ in either optical quadrature. Multiplication of $\alpha$ by $e^{i\Theta }$ rotates that centre anticlockwise through $\Theta$, keeping the coherent state noise disk unchanged. For real positive $\alpha$ the two centres are $(\alpha ,0)$ and $(\alpha \cos \Theta ,\alpha \sin \Theta )$.

Normalize the initial polarization of an electromagnetic wave qubit. The output of the medium isThe joint state is generally an entangled state. The overlap magnitude is $|⟨\alpha |\beta ⟩|=e^{-2|\alpha {|}^2{\sin }^2(\Theta /2)}$. The state factorizes when the amplitudes coincide, while nearly distinguishable amplitudes make the polarization nearly maximally entangled with the optical mode. Before any polarization conditioning, tracing out mode 2 gives the mixture $(|\alpha ⟩⟨\alpha |+|\beta ⟩⟨\beta |)/2$, with no interference between branches.

Include the normalization factors in the half-wave plate map: $|H⟩\mapsto (|H⟩+|V⟩)/\sqrt{2}$ and $|V⟩\mapsto (-|H⟩+|V⟩)/\sqrt{2}$. The transformed joint state isConditioning on a vertical result and setting $\Theta =\pi$ therefore preparesThis pure optical cat state is the conditional output; it is not the unconditioned reduced state immediately after the Kerr medium. The measurement erases the polarization label, allowing the two amplitudes to interfere. The calculated optical Wigner functions below display the resulting fringes; their negative regions are quasiprobability, not negative detection probabilities. These illustrative plots choose $\alpha =1.15$: the rotation panel uses $\Theta =2\pi /3$, while the cat panels use the conditional Kerr rotation $\Theta =\pi$ and three different relative phases $\phi$.

The coherent-state overlap follows by expanding in the Fock basis:Thus $s=⟨\alpha |-\alpha ⟩=e^{-2|\alpha {|}^2}$ andThe postselection probability for Kerr-cat preparation includes both the normalized input qubit and the wave plate. At $\alpha =0,\varphi =\pi$ the vertical branch has zero probability and no normalized conditional state. For the remaining calculation take $\alpha$ real, as required.

Using $a|\pm \alpha ⟩=\pm \alpha |\pm \alpha ⟩$, put $D=1+s\cos \varphi$. Directly evaluating diagonal and cross terms givesThe last quantity is the mean photon number in the conditional cat, which differs from the input coherent state mean ${\alpha }^2$. Since $a{a}^{†}=n+1$,However, its first moment iswhich is generally nonzero. The original PDF's assertion of a zero mean for arbitrary $\varphi$ is inconsistent with its stated pure state. Consequently its displayed expression is the second moment, not the general variance. The phase-dependent cat-state quadrature variance isThe printed result does become the variance for $\sin \varphi =0$, notably the even and odd cat states. It could also describe an ensemble mixing equal weights of $+\varphi$ and $-\varphi$: that different state's mean vanishes while its second moment is unchanged. Such phase averaging is not part of the specified pure-state preparation.

For nonzero finite $\alpha$, $0<s<1$ and $D>0$. The correct quadrature squeezing condition in this quadrature is thereforeFor a representative phase in $-\pi <\varphi \le \pi$, this is $|\varphi |<\arccos (-e^{-2{\alpha }^2})$. If instead one applies the printed second moment as a variance, the resulting condition is $\cos \varphi >0$, or $|\varphi |<\pi /2$ modulo $2\pi$. That is sufficient, but not necessary, for the actual pure state's squeezing. For example, at $\varphi =\pi /2$, $⟨X_2^2⟩=1/4$ while $\mathrm{Var}{X}_2=1/4-{\alpha }^2{e}^{-4{\alpha }^2}<1/4$. This explicitly tests the difference.

For a fixed phase other than $\pi$ modulo $2\pi$, expandAs the input mean number ${\alpha }^2\to 0$, the normalized state tends to vacuum, $⟨X_2⟩\sim -\alpha \tan (\varphi /2)$ andThis expansion is nonuniform near the odd-cat phase. At exactly $\varphi =\pi$ the vacuum terms cancel and the normalized odd coherent-state superposition tends to $|1⟩$, giving $\mathrm{Var}{X}_2\to 3/4$, while its postselection probability tends to zero. The even coherent-state superposition at $\varphi =0$ instead contains only even photon numbers and tends smoothly to vacuum.

At large input mean number, $s\to 0$, the two coherent state branches become nearly orthogonal, and $\mathrm{Var}{X}_2\to 1/4$ for every fixed phase. The mean tends to zero as well. This does not make the state an incoherent mixture: phase-space interference still distinguishes the ideal quantum superposition, although its contribution to this low-order moment is exponentially small. The loss of coherence in an optical cat state arises when ordinary loss reveals the branch to the environment and suppresses that interference, with factor $e^{-2(1-\eta ){\alpha }^2}$ for transmissivity $\eta$.

For the especially simple even cat,It starts at $1/4$, falls to about $0.1108$ at $|\alpha |\simeq 0.7995$, and returns towards $1/4$ from below. The odd-cat curve decreases from $3/4$ towards $1/4$ from above. Other fixed-phase curves have the small- and large-amplitude limits derived above; for negative $\cos \varphi$, squeezing ends when $e^{-2{\alpha }^2}=-\cos \varphi$. In balanced homodyne detection, centre each quadrature distribution on its measured mean before inferring its variance.

The transmitting antenna is matched to its input transmission line, so its real terminal electrical resistance is $Z_i$. The identical receiving antenna has the same electrical impedance at the operating frequency. Its available received power for a matched load is the incident energy flux times its antenna effective area:The real receiving load is $Z_t$, so its voltage-wave reflection coefficient relative to $Z_i$ is ${\Gamma }_L=(Z_t-Z_i)/(Z_t+Z_i)$. The fraction of available power delivered to the output line isConsequently, neglecting path losses, mismatch of polarization of an electromagnetic wave and line losses,This is the Friis transmission equation with a receiving impedance-mismatch factor. For fixed $P_i,G_a,d,Z_i,Z_t$, it scales as ${\lambda }^2$. That scaling is conditional on holding antenna gain fixed: a fixed physical aperture can instead have $G_a\propto {\lambda }^{-2}$, and any wavelength dependence of the impedances also enters the expression. The matched-output check $Z_t=Z_i$ gives $P_t=P_i\mathcal{F}$.

The Johnson-Nyquist noise formula gives the open-circuit mean-square voltage in a small frequency band $d\nu$. When a source electrical resistance $R$ drives a load $R$ with impedance matching, the load sees half the source voltage, so the available noise power isThe factor of four in the open-circuit formula is therefore not an extra factor in the delivered power.

The transmitter's antenna gain $G$ makes the axial spectral energy flux at range $d$Multiplication by the receiving antenna effective area $A$ gives $d{P}_r=k_B{T}_{t}GAd\nu /(4\pi d^2)$. Define the equivalent receiving antenna temperature $T_A$ by the available matched-load noise $d{P}_r=k_B{T}_{A}d\nu$. ThenThis is the transmitter-induced equivalent noise temperature of the receiver port. It does not determine the actual physical temperature of its internal electrical resistance, or any additional receiver and background noise. The calculation assumes the usual far-field link, negligible path loss beyond geometric spreading, matched polarization of an electromagnetic wave at the two antennas and the stated effective collecting area. No additional one-half polarization factor is needed for the matched polarization of radiation from one transmitting antenna.