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An optical path length measures how much propagation wave phase a path accumulates, expressed as the distance that would accumulate the same wave phase in vacuum. A centimetre in a uniform medium with refractive index $3/2$ has optical path length $1.5\mathrm{cm}$, although its geometric length remains $1\mathrm{cm}$. This makes optical path length useful whenever two contributions must arrive with matching wave phases, from an optical lens to a Michelson interferometer. ^[ref]

For a specified path $\gamma$ at fixed angular frequency $\omega$, the optical path length is the line integralHere $s$ is geometric arc length, and $n$ is the phase refractive index, evaluated at the same frequency everywhere. The propagation interpretation assumes a stationary linear medium with isotropy, a real positive scalar refractive index and negligible loss, on a valid geometrical optics branch. The path can curve; at sharp interfaces apply the appropriate matching conditions to each segment. A prescribed curve has a mathematical line integral, but it predicts a propagation wave phase only when it represents the physical branch. ^[ref] Optical fibers can require a mode wavenumber rather than the bulk refractive index along a guessed light ray. ^[ref]

Fix the vacuum wavelength ${\lambda }_0=c/f=2\pi c/\omega$, with frequency $f=\omega /(2\pi )$ and vacuum wavenumber $k_0=2\pi /{\lambda }_0$. In a locally uniform medium the wavelength is $\lambda ={\lambda }_0/n$, while the frequency stays unchanged at a stationary interface. With the convention $\Phi =k_{0}S-\omega t+{\varphi }_0$, propagation from a specified starting wave phase contributesThe relative reference ${\delta }_{\mathrm{ref}}$ includes launch wave phases and any extra wave phase from mirrors, beam splitters or coatings. Thus $k_0\mathrm{OPL}$ is a propagation contribution, not an absolute detector wave phase independent of the apparatus. If the reference contributions stay fixed, a change in optical path difference $h$ changes the phase difference by $2\pi h/{\lambda }_0$. Using the material wavelength again in this formula would count the refractive index twice. ^[ref]

A motivating measurement turns invisible optical oscillations into a slowly changing electrical signal. MIT's teaching experiment uses a laser, a beam splitter, two returning mirrors, an amplified photodiode, an oscilloscope and a calibrated mirror actuator driven by a function generator. The returning beams overlap on the photodiode; one channel records detector voltage and another the actuator drive. The measured observable is the changing light intensity, including its maxima and minima, rather than direct timing of an optical cycle. A calibrated displacement per drive voltage and the spacing of the extrema relate geometric motion to the phase difference. The guide infers the laser wavelength from known actuator calibration; conversely, with independently known vacuum wavelength and ambient refractive index, the same relation calibrates mirror motion. These two unknowns cannot both be inferred from fringe spacing alone. ^[ref]

For a proposed change-of-medium measurement, keep the mirrors fixed and put a transparent cell of internal length $L$ in one arm. Compare its contents at refractive indices $n_a$ and $n_b$, leaving its windows in place. At normal incidence the two traversals giveHere $N$ is the signed number of whole interference fringe cycles plus any resolved fractional cycle, counted continuously. A mirror displacement $x$ instead gives $\Delta \mathrm{OPL}=2n_{\mathrm{amb}}x$; distinguish this double-pass factor from a single pass in a Mach-Zehnder interferometer. OpenStax's gas-cell worked example provides the same double-pass reasoning; it is a teaching calculation rather than a published measurement dataset.^[ref]

An exact toy calculation makes the sensitivity visible. Stipulate ${\lambda }_0=600\mathrm{nm}$, $L=10\mathrm{mm}$ and $n_b-n_a=3/100000=0.00003$. Then $\Delta \mathrm{OPL}=600\mathrm{nm}$ and $\Delta \delta =2\pi$: a full interference fringe cycle, despite unchanged geometric lengths. For a continuous change $n_b-n_a=q(3/100000)$, let $0\le q\le 2$ and choose the initial detector phase difference to be zero. Equal coherent detected arm contributions of light intensity $I_0$ predictThe plot is this stipulated model: $q=1/2$ gives a minimum, $q=1$ restores a maximum, and the reference level is chosen by adjustment of the apparatus. It contains no measured values or assigned properties of a real gas. The equivalent mirror travel in vacuum is $300\mathrm{nm}$ per cycle. A single endpoint light intensity cannot distinguish $q=0$ from $q=1$; the continuous fringe count carries the missing information.

For an interpretable cell measurement, establish the initial phase difference with a small calibrated mirror scan, then track slow changes through successive interference fringes. Measure the dark signal and each arm separately, keep photodiode gain within its linear range, align both beams on the same region and match their linear polarizations. Monitor interferometric visibility and keep arm delays within the useful optical coherence time. Record cell length, contents, pressure and temperature; return to the starting condition to check drift. Changes in window deformation, cell length, external air or mirror position also change optical path length, so a reference cell or an independently monitored mechanical path is needed to isolate the contents' refractive index. These controls are deductions from the model, not a claim that the illustrated apparatus was operated. Loss of fringe count leaves a whole-cycle ambiguity; at an intensity extremum the first derivative with respect to phase difference vanishes, so a scan around phase quadrature is more sensitive to small changes. Broadband illumination, unstable optical coherence, vibration and strong loss can invalidate a simple one-frequency inference.

For the refractive index $\mu =1/y$ on the upper half-plane, Fermat principle and the Beltrami identity give $1/[y\sqrt{1+(dy/dx{)}^2}]=1/R$. The nonvertical rays are arcs of circles centered on $y=0$; vertical lines are the remaining rays. These are also geodesics of the Poincaré half-plane model. A ray entering the strip $0<y\le 1$ toward decreasing $y$ at angle $\theta$ to the vertical has limiting horizontal displacement $\tan (\theta /2)$, at most one. The line $y=0$ is approached only after infinite optical path length because the refractive index diverges there.

Optical tweezers hold and move a small particle using the force exerted by a tightly focused beam of light, usually from a laser. They can position transparent beads in a liquid and manipulate suitable biological cells. Attaching a molecule to a trapped bead turns the bead into a handle: a measured bead displacement, together with a calibrated spring constant, reveals the force exerted on the attachment. This makes microscopic manipulation and measurement possible without a mechanical tool touching the bead. Living biological cells were manipulated with infrared beams by Ashkin, Dziedzic and Yamane in 1987; Svoboda and Block used trapped beads to measure forces from individual molecules in 1994. ^{[ref] [ref]}.

Why can light pull a bead towards a focus? Light carries momentum. A transparent bead with a higher refractive index than the surrounding liquid bends transmitted light rays by refraction, changing their momentum; conservation of momentum requires an opposite transfer to the bead. In the geometrical optics picture, unequal illumination of a displaced bead produces a net restoring force towards the brighter region. For a particle much smaller than the wavelength, a physical optics description replaces that ray picture. The detailed force depends on particle size, refractive index, beam shape and wavelength; the qualitative attraction does not supply a universal formula for every particle.

The optical gradient force draws a suitable transparent bead towards high illumination, while the optical scattering force generally pushes along beam propagation. A stable three-dimensional optical trap needs a balance with a restoring force in each direction: transverse confinement alone does not ensure axial confinement. Strong focusing can make the backward axial gradient contribution large enough to balance the forward scattering contribution, with the equilibrium displaced from the focus. Higher refractive index alone is insufficient; absorbing particles, different index contrasts and different beam geometries can behave differently. ^[ref].

Historically, Ashkin's 1970 experiments accelerated and trapped particles using radiation pressure. The 1986 paper by Ashkin, Dziedzic, Bjorkholm and Chu demonstrated the single-beam gradient optical trap for transparent particles. These are distinct steps: the earlier work established optical manipulation, and the later work supplied the tightly focused single-beam arrangement. ^{[ref] [ref]}.

Near a stable equilibrium, a harmonic optical trap behaves like a spring; thermal calibration of an optical trap uses its Brownian motion to determine the effective spring constant. The later passage through a translating optical trap sections instead consider a moving localized potential energy, which need not remain harmonic throughout a passage.

The symmetry direction of a uniaxial optical medium. For a wavevector along it, the two refractive indices coincide: there is no birefringence for that propagation direction, even though $n_e\ne n_o$.

The ordinary wave's electric field is perpendicular to the plane containing the wavevector and optic axis. Its refractive index $n_o$ is direction independent at fixed frequency.

In first Born order the unknown potential enters linearly. Let $M_i$ multiply by the known incident field, $\mathcal{G}$ be the outgoing Green integral over the known support $D$, and $\mathcal{R}$ restrict the resulting field to the measurement set. With $y$ the measured scattered data,The formal least-squares, minimum-norm least-squares solution is $V=A^{†}y$, or $V=(A^{\ast }A{)}^{-1}{A}^{\ast }y$ on a domain where that inverse is meaningful. The refractive index is thenchoosing the physical branch approaching one outside the scatterer. At weak contrast $n-1\simeq V/(2k^2)$.

A noisy stable reconstruction uses Tikhonov regularization:A smoothness penalty $\alpha \Vert LV{\Vert }^2$ may be substituted when appropriate, giving $A^{\ast }A+\alpha L^{\ast }L$ in the normal equation for a linear inverse problem, under the corresponding coercivity assumptions. The parameter balances noise amplification and smoothing bias.

The data geometry matters: knowing a scattered field on a measurement surface does not automatically mean knowing it throughout the support. A single incident wave and fixed-frequency far field sample only a shifted sphere of the potential's Fourier transform, and in general do not determine an arbitrary three-dimensional potential uniquely. Tikhonov regularization stabilizes a selected minimum-norm least-squares solution; it cannot create missing information or guarantee uniqueness from inadequate data. If the total field and its derivatives were known throughout $D$, the first-Born equation could instead be inverted formally as $V\simeq -(\Delta +k^2){\psi }_s/{\psi }_i$ where ${\psi }_i\ne 0$, but differentiation still amplifies measurement noise.

The zero-mean fluctuation is $W$, not the refractive index $n$: $⟨n⟩=1$. Also the full scattering potential has a nonzero mean and is not Gaussian:The usual Gaussian closure requires $W$ to be a jointly Gaussian random field, not merely to have a Gaussian distribution at each point. We first give the intended leading weak-fluctuation result with the linearized potential $V_1=2\mu k_0^{2}W$, then account for the quadratic term.

Keep the definitions $q,B$ from the preceding solution and let ${\chi }_1=B{\int }_D{V}_1(y)e^{-iq\cdot y}dy$. Because the incident field has unit modulus,The exponent is a real Gaussian linear functional $Z={\int }_{D}h{V}_1$. If $C_W(y-z)=⟨W(y)W(z)⟩$, the linearized potential has autocorrelation function of a random field $C_1(s)=4{\mu }^2{k}_0^4{C}_W(s)$. Complete the square in the one-dimensional Gaussian density of $Z$: $⟨e^Z⟩=\exp (⟨Z⟩+\mathrm{Var}Z/2)$. Here its mean is zero, soThis is Gaussian intensity in the first Rytov approximation. In a more explicit complex notation, putThenThe second term is the pseudo-covariance of a complex random variable contribution. It does not vanish for an arbitrary finite real medium. The real-kernel expression shows that the combined exponent is nonnegative, even though its two complex-form contributions need not be separately nonnegative.

For the full potential, Gaussian moment pairing theorem gives its centered covariance functionIts uncentered autocorrelation is $R_V(s)=C_V(s)+{\mu }^4{k}_0^4$. The first Rytov approximation is linear in $V$, but its intensity is an exponential average; applying a Gaussian formula exactly to this quadratic potential would be incorrect. A consistent expansion through order ${\mu }^2$ isThe first term is the coherent mean-potential correction. Odd powers vanish by the symmetry of the joint Gaussian field. The variance term can equally use $C_V=R_V-{\overline{V}}^2$ through this order. Thus the mean shift must not be discarded while retaining the order-${\mu }^2$ fluctuation intensity.

An exact expression within the far-field, first-Rytov approximation model is also available when the full $n^2$ is retained. Assume the restriction of $W$ to $D$ is a Gaussian element of the real Hilbert space $L^2(D)$, with covariance operator $C$ of kernel $C_W(y-z)$. Let $H$ denote multiplication by $h$, ${\ell }_h=2\mu k_0^{2}h$, $Q={\mu }^2{k}_0^2{C}^{1/2}H{C}^{1/2}$ and $b_h=C^{1/2}{\ell }_h$. The Gaussian quadratic exponential moment isHere the determinant is the Fredholm determinant; require $I-2Q$ strictly positive for finiteness. Diagonalize the self-adjoint trace-class operator $Q$ and complete the square in each Gaussian coordinate to derive the formula. On a bounded $D$, unit pointwise variance makes $\mathrm{tr}C=|D|$, so the covariance is trace class under the stated measurable $L^2$ hypothesis. Small enough $\mu$ satisfies the positivity condition. Expanding the determinant and inverse recovers the preceding mean and covariance terms. For small $\mu$ this is also determined by the potential autocorrelation: since $C_W\in [-1,1]$,using the branch near zero. This makes explicit the extra joint-Gaussian assumption behind an exact correlation-based answer.

Finally, Gaussian marginals alone are insufficient. Take $X\sim N(0,1)$ and independent random signs $S_j$. The stationary sequence $W_j=S_{j}X$ has Gaussian marginals and covariance $⟨W_i{W}_j⟩={\delta }_{ij}$, just like independent standard Gaussian variables. However $⟨e^{t(W_1+W_2)}⟩=(1+e^{2t^2})/2$, whereas for the independent Gaussian pair it is $e^{t^2}$. The different intensities show why the joint hypothesis is needed. The mean-zero hint is interpreted for $W$, and the linearized Gaussian formula is not claimed to be exact for $V=k_0^2(n^2-1)$.

Choose the time dependence $e^{-i\omega t}$ and measure distances in background-wavenumber units. Thus $L_0=\Delta +1$, the incident field satisfies $L_0{\psi }_i=0$, and the scattering potential convention is $V=n^2-1$, giving $L_0\psi =-V\psi$. To match the minus sign in the paper's integral, take the outgoing Green function to satisfy $L_{0}G=\delta$. This is the negative of the frequently used outgoing Helmholtz equation fundamental solution satisfying $L_{0}G=-\delta$. In physical coordinates one restores $k_0^2$ in the scattering potential, or absorbs it in the integral kernel; the sign convention must remain consistent.

On a region where the incident and total fields are nonzero, introduce the logarithmic wave perturbationChoose a continuous logarithm branch connected to the unperturbed field. Substitution into the Helmholtz equation givesThe quadratic term uses the complex bilinear dot product, not the squared modulus of the gradient. Write $V=\epsilon V_1$ and $\chi =\epsilon {\chi }^{[1]}+\cdots$. The first-order Rytov approximation discards that quadratic term. Since $L_0({\psi }_i\chi )={\psi }_i{\mathcal{L}}_i\chi$, the outgoing first correction isand henceThe validity of the first Rytov approximation concerns the omitted logarithmic correction. Its next contribution satisfies ${\mathcal{L}}_i{\chi }_2=-\nabla {\chi }_1\cdot \nabla {\chi }_1$, so a useful explicit criterion is that the outgoing solution ${\chi }_2$ be small in the region of interest. Weak refractive index contrast and small wave phase gradients on a wavelength scale provide the usual perturbative regime, with weak amplitude fluctuations and no strong focusing or zeros that destroy the logarithm. A sufficient local source comparison is $|\nabla {\chi }_1\cdot \nabla {\chi }_1|\ll |V|$ where the scattering potential is nonzero, together with control of propagation of that error. This is not a universal pointwise test at zeros of $V$.

Small accumulated wave phase is not required in the same way as in a linear field approximation: the exponential retains that wave phase accumulation. Large gradients, strong multiple-scattering amplitude effects or field zeros can still invalidate the Rytov approximation. The integrals also require a finite scattering region or appropriate convergence conditions.

Let the stationary scattering potential fluctuation have autocorrelation function of a random fieldBecause $W$ is centered, this is also its covariance function. Multiplying the two real wave phase integrals and taking expectations givesThe two minus signs cancel. This is the wave phase variance, since the wave phase mean is zero. More generally the phase covariance in the first Rytov approximation replaces the first kernel by $b({\mathbf{r}}_1,{\mathbf{r}}^{\prime })$ and the second by $b({\mathbf{r}}_2,{\mathbf{r}}^{\prime \prime })$. Finite observation/scattering windows, or suitable weighted-integrability hypotheses, make these double integrals well-defined in a stationary infinite-medium model.

The correlation needed here is that of the scattering potential fluctuation. With the printed $V=n^2-1$, the covariance of a squared random field isIt involves a fourth moment of $n$, so its value is not generally determined by the ordinary two-point correlation $C_n=⟨n({\mathbf{r}}^{\prime })n({\mathbf{r}}^{\prime }+\mathbf{\rho })⟩$ alone. If one additionally assumes a zero-mean Gaussian random field, Isserlis theorem yields $C_W=2C_n^2$. That assumption is not printed and must not be inserted silently. Alternatively, for a physical weak fluctuation $n_{\mathrm{phys}}=1+\eta$, $W\simeq 2\eta$ gives $C_W\simeq 4C_{\eta }$. The general answer uses $C_W$; either reduction to a refractive index two-point correlation requires an extra assumption.

Assume a real refractive index and write the scattering potential as $V=⟨V⟩+W$, so $⟨W⟩=0$ by construction. With the real kernel components in the question, the logarithmic Rytov approximation separates intoInclude the incident wave phase and the mean-potential wave phase in the deterministic reference ${\varphi }_0$: at each observation point it is $\arg {\psi }_i-\int b⟨V⟩$. It need not be spatially constant. The amplitude is $|{\psi }_i|\exp [-\int aV]$. The phase covariance in the first Rytov approximation therefore starts from the fluctuating wave phaseUnder the integrability assumptions needed to interchange the expectation and integral,For a complex absorbing potential, the corresponding phase fluctuation is $-\int [b\mathrm{Re}W+a\mathrm{Im}W]$; the displayed scalar formula is the real-index case. The zero mean comes from centering $V$, not from setting the mean of $V$ equal to zero. In particular the printed $⟨n⟩=0$ does not imply $⟨V⟩=0$. A physical positive refractive index usually has a nonzero background mean; a zero-mean assumption normally refers to its fluctuation. The algebra above remains meaningful for a signed real random field and explicitly retains its mean scattering potential.

Interpret the longitudinal Dirac delta function as the Markov approximation for a random medium. The two arguments on the left of the printed covariance must be $x_1$ and $x_2$. The transverse covariance kernel is $B(|{\mathbf{z}}_1-{\mathbf{z}}_2|)$. Here statistical isotropy means transverse isotropy: a medium with a distinguished longitudinal white noise direction and a smooth transverse covariance is not literally isotropic in all three directions. Also, ideal Gaussian white noise replaces the original finite-variance field; it cannot simultaneously satisfy a pointwise normalization $\mathbb{E}{W}^2=1$.

Write $Q(r)={\mu }^{2}B(r)$ for the covariance strength of the refractive index fluctuations. Define the transverse power spectrum using the unnormalized forward Fourier transform printed in equation 3:This is the spectrum of the fluctuations, excluding the deterministic mean index. Equivalently it is the zero-longitudinal-frequency slice of the three-dimensional fluctuation spectrum before the Markov approximation. The angular integral is $2\pi J_0(\nu r)$, where $J_0$ is the Bessel function of the first kind. Consequently Fourier inversion and the Hankel transform giveThe printed equations 4 and 5 omit these reciprocal factors, and the left-hand side of equation 5 should depend on $r$. They are instead a consistent order-zero Hankel transform pair if their $F$ denotes $S_H=S_F/(2\pi )$.

Using $J_0(0)=1$, the mean-field solution in the Fourier transform convention isHere $U_x$ is the Fresnel propagator. Finiteness of the integral ensures a finite screen wave phase variance. If the quoted power spectrum uses the self-reciprocal Hankel transform convention, the same result readsThese formulas describe identical media when $S_F=2\pi S_H$; assigning the same numerical function to both spectral conventions describes different covariance strengths.

Use the time-harmonic wave convention $\psi e^{-i\omega t}$ and write $\mathbf{z}=(y,z)$, $E(0,\mathbf{z})=E_0(\mathbf{z})$, with a deterministic incident wave envelope. The printed speed ratio is inconsistent with $k=k_{0}n$: the refractive index used below is $n=c_0/c$. Substituting $\psi =e^{i{k}_{0}x}E$ in the Helmholtz equation givesThe paraxial approximation discards ${\partial }_x^{2}E$. The usual weak-fluctuation model also linearizes $n^2-1=2\mu W+O({\mu }^2{W}^2)$, givingDropping the quadratic contrast is an additional weak-fluctuation assumption, not a consequence of small propagation angles. Although the linearized random potential generates attenuation of order ${\mu }^2$, it does not retain every effect of that order in the literal finite-correlation index: the discarded quadratic contrast can also produce a mean wave phase shift. This linearization must precede a Gaussian white noise limit: the square of ideal white noise has no ordinary pointwise meaning.

The split-step Fourier method alternates free-space diffraction, $U_h=e^{hD}$, with a random phase screen,For jointly Gaussian random fields, Gaussian phase averaging gives $\mathbb{E}{e}^{i{\varphi }_h}=e^{-\mathrm{Var}({\varphi }_h)/2}$. More generally, a product of $p$ fields and $q$ conjugate fields picks up $\exp (i{\sum }_a{ϵ}_a{\varphi }_h({\mathbf{z}}_a))$, with signs ${ϵ}_a=+1$ or $-1$. Its screen average is determined entirely by the wave phase covariance matrix. The deterministic diffraction step acts on each coordinate, with opposite signs on factors formed by complex conjugation. This is the basis of the field-moment equations.

There is an important closure qualification. A stationary Gaussian random field need not have independent longitudinal increments. For finite-correlation fluctuations, the unlinearized parabolic wave equation instead givesIn the weak-fluctuation model, the exact first-moment equation isThe last term cannot in general be replaced by a constant times $m$. An exact generic solution of the linearized model iswhere $\mathcal{T}$ denotes time ordering of the propagation operators. For the unlinearized model, add $i{k}_0{\mu }^2{W}^2/2$ inside the propagation generator. The covariance function of the medium is needed to evaluate this expression; One-point Gaussian distributions alone would not even determine the joint wave phase statistics.

For example, omit diffraction and take a longitudinal autocovariance function $R(s)=\mathbb{E}[W(s,\mathbf{z})W(0,\mathbf{z})]$. Direct Gaussian phase averaging givesEven here, unit-variance stationary Gaussian random fields with different $R$ give different answers. In the general problem the diffraction and multiplication operators do not commute, so this scalar attenuation cannot simply be multiplied by $U_x$ without an additional approximation.

The standard closed answer uses the Markov approximation for a random medium, made explicit in part ii. Let the longitudinally integrated covariance kernel beReplace the medium by longitudinal Gaussian white noise with this strength. A screen of thickness $h$ then has $\mathbb{E}[{\varphi }_h({\mathbf{z}}_a){\varphi }_h({\mathbf{z}}_b)]=k_0^2{\mu }^{2}hB({\mathbf{z}}_a-{\mathbf{z}}_b)$ and is independent of the incoming field. For the moment $M=\mathbb{E}{\prod }_a{E}_a^{({ϵ}_a)}$, expanding both steps to order $h$ givesIn particular, the coherent attenuation in a white-noise random medium and its Fresnel propagator solution areEquivalently, write $Wdx=d{\beta }_x$, where ${\beta }_x(\mathbf{z})$ is Brownian motion in $x$ with transverse covariance kernel $\mathbb{E}[d{\beta }_x(\mathbf{z})d{\beta }_x({\mathbf{z}}^{\prime })]=B(\mathbf{z}-{\mathbf{z}}^{\prime })dx$. The Stratonovich integral formulation is $dE=DEdx+i{k}_0\mu E\circ d{\beta }_x$. Its Itô integral form isThe mean of the Itô integral vanishes, independently confirming the attenuation drift. For $x>0$ in two transverse dimensions,Thus a unit plane wave has $m=e^{-\gamma x}$. For a general incident wave envelope, the Fresnel propagator supplies its spreading. The attenuation is redistribution between the coherent and diffuse wave fields, rather than wave absorption: for a field and its conjugate at the same point, the screen contribution in the second-moment equation cancels.

Use the physical refractive index convention $k=k_{0}n$, with $k_0$ the background wavenumber and time dependence $e^{-i\omega t}$. Treat $B$ as a bounded region in ${\mathbb{R}}^3$; the printed $B\in {\mathbb{R}}^3$ should be $B\subset {\mathbb{R}}^3$. Define the contrast $q=n^2-1=2n_{ϵ}+n_{ϵ}^2$ and the outgoing Green functionThe Sommerfeld radiation condition selects this sign of the outgoing wave phase. Since $(\Delta +k_0^2)\psi =-k_0^{2}q\psi$, the Lippmann-Schwinger equation isThe positive sign in this integral follows from the minus sign in the defining Green function equation. One may instead define the scattering potential with the opposite sign, provided both equations change consistently.

If $\Vert T_q\Vert <1$ on the chosen field space inside $B$, its Neumann series converges. The Born series for the scattered field isFor observation points outside $B$, restrict the intermediate factors to $B$ and use the same outgoing integral for the final factor. For example, the first two terms areThe $m$th term describes $m$ successive scattering interactions. The first Born approximation for scalar wave scattering keeps one interaction and replaces the field inside the medium by the incident field. Higher terms describe multiple scattering and the resulting feedback on the internal field.

A concrete sufficient condition follows on $L^{\infty }(B)$. If $d_B$ is the diameter of $B$, then for $\mathbf{r}\in B$ the region lies in the ball of radius $d_B$ about $\mathbf{r}$, andThus $\boxed{\rho \ll 1}$ is a conservative sufficient condition for the first Born approximation for scalar wave scattering. When $\rho <1$, the omitted internal-field terms satisfyThe bound ignores cancellation in the oscillatory Green function, so it is sufficient, not necessary. A common physical small-contrast criterion for an extended weak medium is small accumulated extra wave phase, $k_0\ell \Vert n_{ϵ}{\Vert }_{\infty }\ll 1$, together with weak scattering and no resonant internal enhancement. Small local contrast alone is not enough for an arbitrarily large or resonant object.

Atmospheric extinction changes the amplitude of the incoming light through absorption and scattering. Its wavelength dependence alters the measured optical spectrum and colors; clouds and aerosols introduce additional time dependence. Scattered moonlight and atmospheric emission increase sky brightness.

Atmospheric refraction changes the apparent position of the source. Its wavelength dependence, atmospheric dispersion, spreads a broadband image toward the zenith. The mean refractive index gradient therefore affects direction even in the absence of small-scale turbulence.

Atmospheric turbulence produces rapidly varying optical path lengths, distorting the phase and curvature of a nominally plane wavefront. Different pupil regions acquire different phase delays, producing astronomical seeing, image motion, and short-exposure speckle patterns. Different lines of sight sample different fluctuations, giving anisoplanatism.

Propagation through the fluctuating medium also changes the intensity through atmospheric scintillation, the familiar twinkling of stars. A phase-only adaptive optics correction can reduce atmospheric phase errors but does not remove absorption, all scintillation, or the atmospheric emission background.

$V_d$ is the Abbe number, a measure of inverse relative optical dispersion. Ordinary crown glasses have comparatively large $V_d$ and lower refractive index, while ordinary flint glasses have smaller $V_d$ and often higher index. The Abbe diagram below spans typical optical-glass values; the shaded family ranges are schematic, and the marked examples are catalogue data. Modern glass compositions broaden and overlap the simple crown/flint picture.

For a positive doublet with $V_{1d}>V_{2d}$, the preceding formulas require $f_{1d}>0$ and $f_{2d}<0$: a converging crown glass element and a diverging flint glass element. Their dispersion corrections cancel, while their net optical power is positive. A small difference in Abbe numbers requires large opposing component powers, which makes the design harder to correct for other optical aberrations.

At minimum deviation, ${\theta }_2={\theta }_3=\varphi /2$ and ${\theta }_1={\theta }_4=({\gamma }_{\min }+\varphi )/2$. Substituting in Snell's law givesFor a known prism apex angle, measuring the minimum deviation with monochromatic light determines its refractive index. Repeating the measurement at several wavelengths measures optical dispersion. If the surrounding medium is not air, the measured ratio is relative to that medium's refractive index.

The condition ${\mathbf{E}}_I\cdot \hat{\mathbf{x}}=0$, together with transversality ${\mathbf{k}}_I\cdot {\mathbf{E}}_I=0$, selects polarization along $\hat{\mathbf{y}}$ away from the normal-incidence degeneracy. Phase matching at a planar wave interface givesThe two dispersion relations give $|{\mathbf{k}}_T{|}^2=({\epsilon }_{+}/{\epsilon }_{-})k_I^2$. Choosing propagation into $x>0$ therefore yieldsEquivalently, this is Snell law for electromagnetic waves with refractive index proportional to $\sqrt{\epsilon }$ because the two media have equal magnetic permeability.

First integral. If $F$ has no explicit $x$ dependence, the chain rule givesThe Euler-Lagrange equation makes this zero without requiring division by $y^{\prime }$. Hence the Beltrami identity is

Optical paths. By Fermat principle, a ray makes the optical path length stationary. Within the strip $0<y\le 1$, its functional for a graph $y(x)$ isThe Beltrami identity givesFor a nonvertical segment $k>0$, write $R=1/k$. Then $(y^{\prime }{)}^2=(R^2-y^2)/y^2$, soThus the optical rays in a reciprocal-height refractive index satisfyrestricted to the part in the strip. They are arcs of circles centered on $y=0$. Vertical lines are the additional solutions, obtained as a limiting case or directly from the constant horizontal optical momentum. The same curves are geodesics of the Poincaré half-plane model, whose length element is $ds/y$.

Horizontal displacement. Let the ray enter at $(x_0,1)$, travelling to the right and downward at angle $0<\theta <\pi /2$ to the vertical. The refractive index is continuous at $y=1$, so there is no instantaneous angular jump at entry. The tangent has $(dx/ds,dy/ds)=(\sin \theta ,-\cos \theta )$; hence $y^{\prime }(x_0)=-\cot \theta$. The first integral gives $k=\sin \theta$, or $R=\csc \theta$. The circle derivative $y^{\prime }=-(x-c)/y$ then fixes $c=x_0-\cot \theta$.

As $y\downarrow 0$, the rightward ray approaches $x=c+R$. The total limiting horizontal distance is thereforeFor vertical incidence, $D=0$ by continuity. For a leftward ray the displacement has the opposite sign, with the same magnitude. Since $\tan (\theta /2)$ increases on $[0,\pi /2]$,If entry is required to have a strictly downward component, this is a supremum rather than an attained maximum. A horizontal tangent at entry gives the limiting unit-radius arc, which immediately bends downward in the medium.

The lower boundary is a limit, rather than a point reached in finite optical travel time. On a nonvertical circular ray, $ds/dy=R/\sqrt{R^2-y^2}\to 1$ as $y\downarrow 0$. Thus the optical path length contains a divergent integral proportional to $\int dy/y$. The horizontal distance stays finite even though propagation toward $y=0$ takes infinite time.

Vary $r$ by $r+\epsilon \eta$ with $\eta (a)=\eta (b)=0$. The first variation of the functional isThe boundary term from integration by parts is zero. The fundamental lemma of the calculus of variations therefore gives $\boxed{F_{1,r}-d{F}_{1,r^{\prime }}/dz=0}$. For $I_2$, independently vary $r$ and $\theta$ with zero endpoint variations; the two Euler-Lagrange equations areFor the optical path, Fermat principle makes $\int \mu ds$ stationary. In cylindrical coordinates its integrand is $L=\mu (r,\theta )s$, with $s=\sqrt{1+r^{\prime 2}+r^2{\theta }^{\prime 2}}$. There is no explicit $z$ dependence. The Beltrami identity for several dependent variables givesThe conserved quantity is the axial component of the optical momentum of a ray. As $\mu >0$, squaring and inverting yieldsNow suppose $\mu =\mu (r)$ and seek a path of radius $R>0$. The angle is a cyclic coordinate, so $L_{{\theta }^{\prime }}=\mu R^2{\theta }^{\prime }/\sqrt{1+R^2{\theta }^{\prime 2}}$ is conserved. At fixed radius this is strictly increasing in ${\theta }^{\prime }$, forcing ${\theta }^{\prime }=\omega$ constant. The radial Euler-Lagrange equation reduces toConsequently the helical rays in a radially graded refractive index haveA real nonzero helix exists at such a radius when $0<-R{\mu }^{\prime }(R)/\mu (R)<1$, and its helical pitch is $2\pi /|\omega |$. If ${\mu }^{\prime }(R)=0$, the solution is instead a straight axial light ray. Thus radial dependence permits constant-radius rays for suitable profiles; an arbitrary radial profile need not support a nonzero real helix. This qualification is necessary, for example when the refractive index increases with radius.