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At normal incidence, the incident light ray follows the surface normal vector. Snell's law gives ${\theta }_t=0$ for positive refractive indices, so the transmitted light ray does not bend. Nevertheless, its phase velocity and wavelength change if the refractive index changes. A single axial light ray therefore cannot measure apparent depth: neighboring light rays and their directions supply that information. A reflected component can still be present; no bending of the transmitted light ray does not imply complete transmission.

An optical aberration is a departure from ideal image formation: light rays from one object point may miss their intended common image point, or image positions may depart from the intended mapping. Spherical aberration, coma, and astigmatism describe different ways of spreading a point's light rays; optical distortion can change image shape without spreading each point. In physical optics, these departures can be described by wavefront error relative to a reference wavefront. Diffraction also gives a finite point spread function in a perfect optical instrument, so an extended image alone does not establish an optical aberration.

An optical fiber is a thin, flexible electromagnetic waveguide that carries light along its length. In a conventional step-index optical fiber, an optical fiber core with refractive index $n_1$ is surrounded by an optical fiber cladding with lower refractive index $n_2$. Suitable light rays remain confined by total internal reflection at this interface. The cladding is an optical part of the guide, distinct from its protective coating.^[ref]

The useful surprise is that a transparent strand can deliver light around a route that a free beam would leave. This makes optical fibers useful for communications and for carrying collected light to a fiber-fed spectrograph. Guidance, successful launch, and surviving optical power are separate questions: the optical fiber acceptance angle constrains launch geometry, while optical fiber attenuation describes loss along the route.

The optical fiber acceptance angle is the largest launch half-angle admitted by the specified guide model. For a flat entrance face perpendicular to the optical axis, external refractive index $n_0$, and a straight step-index optical fiber, Snell's law at the entrance givesAt the core–cladding interface, the critical angle for total internal reflection $i_c$ satisfiesFor a meridional ray, strict total internal reflection requires $i>i_c$. At the limiting geometry $i=i_c$, the refracted light ray is tangent to the interface, soThis identifies the numerical aperture with an entrance cone under these assumptions. With $n_0=1$ for air, the conventional inclusive boundary condition isfor a nonnegative launch angle $\theta$; a practical launch stays inside the boundary. If $\mathrm{NA}>n_0$, this ideal model imposes no further restriction within the forward hemisphere, and $\arcsin (\mathrm{NA}/n_0)$ must not be used. The light ray must also intersect the optical fiber core: a suitable angle alone does not correct a misplaced beam.

For an original example, take $n_1=1.50$, $n_2=1.48$ and air at the entrance. ThenThe internal limiting angle is only ${\varphi }_{\max }\simeq 9.37^{\circ }$, while the critical angle for total internal reflection is $i_c\simeq 80.63^{\circ }$. A $10^{\circ }$ launch has $\varphi \simeq 6.65^{\circ }$ and $i\simeq 83.35^{\circ }$, so it satisfies total internal reflection. A $20^{\circ }$ launch has $\varphi \simeq 13.18^{\circ }$ and $i\simeq 76.82^{\circ }$, so it fails this test.

This cone is a geometrical optics result for meridional rays. A single-mode optical fiber requires the spatial overlap of the incident electromagnetic wave with its guided normal mode; its numerical aperture does not impose an exact angular cutoff on arbitrary beams. Diffraction, finite cladding, and bends require a physical optics description.^[ref]

An angular readout using a reflected light ray: rotating a mirror by $\theta$ rotates the reflected light ray by $2\theta$, giving spot displacement $x\simeq 2D\theta$ at distance $D$ for small angles.

An optical microscope uses light and optical lens assemblies to reveal small structures, such as the boundaries of biological cells or defects in a manufactured surface. In a compound optical microscope, the microscope objective supplies an enlarged real image; the microscope eyepiece makes that intermediate microscope image subtend a larger viewing angle. The transverse magnification of the first stage and the angular magnification of the second answer different questions.^[ref]

Consider an original ideal finite-conjugate microscope in air, with both optical lens assemblies treated as converging thin lens elements in the paraxial approximation. Choose objective focal length $f_o=10\mathrm{mm}$ and object distance $u_o=12\mathrm{mm}$. The thin lens equation and signed transverse magnification givePlace a microscope eyepiece of focal length $f_e=25\mathrm{mm}$ a further $25\mathrm{mm}$ beyond the intermediate microscope image. The objective-to-eyepiece separation is therefore $85\mathrm{mm}$. Each object point then supplies a parallel outgoing bundle of light rays, appropriate to a relaxed eye focused at infinity. Define the unaided reference as the same object viewed at distance $D=250\mathrm{mm}$; this is a rating convention, not a universal property of eyesight. The eyepiece angular magnification and total signed angular magnification areFor instance, a hypothetical object height $h_o=0.020\mathrm{mm}$ becomes $h_i=-0.100\mathrm{mm}$ at the intermediate microscope image. The signed viewing angles are approximately $h_i/f_e=-0.004\mathrm{rad}$ with the instrument and $h_o/D=0.00008\mathrm{rad}$ unaided. Their ratio is $-50$: enlarged and inverted. These are stipulated model dimensions, not measurements or specifications of a modern commercial optical microscope.

The attraction of an optical microscope is that seeing can become measuring: a stage micrometer supplies a known object-space length, and microscope scale calibration converts an image readout into a specimen length. This also separates a large-looking image from a trustworthy size estimate. Meanwhile, diffraction-limited resolution depends on numerical aperture and wavelength; empty magnification enlarges an existing blur without revealing additional structure.

For a light ray parametrized by arc length $s$, with unit tangent vector $t$ in a smooth positive refractive index $\mu$, the optical momentum is $p=\mu t$. This is the momentum for optical path-length normalization. The Euler-Lagrange equation for Fermat principle, $\int \mu (r)|r^{\prime }(\tau )|d\tau$, gives $d(\mu t)/ds=\nabla \mu$. Therefore an index independent of $z$ conserves $p_z=\mu dz/ds$. In coordinates $r(z),\theta (z)$ this is $\mu /\sqrt{1+r^{\prime 2}+r^2{\theta }^{\prime 2}}$, the first integral supplied by the Beltrami identity for several dependent variables.

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.

An optical pinhole is a small aperture used to select a narrow part of a beam or approximate an optical point source. Its finite diameter and diffraction limit that approximation; an illuminated optical pinhole at a collimator's optical focus supplies a narrow bundle of nearly parallel light rays.

The optical power is reciprocal signed focal length, $\Phi =1/f$. With $f$ expressed in metres, optical power is expressed in dioptres: $f=1\mathrm{m}$ gives $\Phi =+1\mathrm{D}$, while $f=-0.5\mathrm{m}$ gives $\Phi =-2\mathrm{D}$. The sign distinguishes converging and diverging optical lens elements; the magnitude measures how strongly they change the direction of a light ray at a given height. ^[ref].

For two ideal thin lens elements in contact in air, optical powers add: ${\Phi }_{\mathrm{total}}={\Phi }_1+{\Phi }_2$. Thus $+2\mathrm{D}$ and $-1\mathrm{D}$ combine to give $+1\mathrm{D}$, or $f_{\mathrm{total}}=1\mathrm{m}$. Separation changes the rule: for two ideal thin lens elements separated by $d$ in air, their equivalent optical power is ${\Phi }_1+{\Phi }_2-d{\Phi }_1{\Phi }_2$. The contact rule is the $d=0$ case.^[ref]

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.

Optics studies light, its propagation, and instruments for forming images and measuring optical spectra.

The central question is how light carries information from a source to an observer. A useful description must explain both where the light goes and how its distribution changes. In geometrical optics, light rays locate paths and images; in physical optics, an electromagnetic wave has a wave amplitude and a wave phase, so overlapping contributions can produce wave interference and diffraction. These descriptions address different levels of detail within the same subject. ^[ref].

An image is especially interesting because a single object point emits light in many directions. An optical lens redirects part of that spreading bundle toward a corresponding image point; repeating this construction over the object preserves spatial information. Simply gathering more light is insufficient: light rays from distinct object points must arrive at distinct locations. This is why an optical camera needs both an optical lens and a correctly positioned receiving surface. The thin lens example below makes that correspondence quantitative. ^[ref].

A parabolic reflector has a paraboloidal reflecting surface. The law of reflection makes all incident light rays parallel to its optical axis pass through one focus in ideal geometrical optics. This removes on-axis spherical aberration and explains its use as a receiving antenna for distant sources.

The Euler-Lagrange equation is $f_y-d(f_{y^{\prime }})/dx=0$. Since $f$ has no explicit $x$ dependence,Thus a stationary curve satisfies the Beltrami identity, $f-y^{\prime }{f}_{y^{\prime }}=C$.

By Fermat principle, the light ray makes the travel time $\int \sqrt{1+y^{\prime 2}}/(y+c_0)dx$ stationary. Its Beltrami identity becomesWriting $R=1/C$ and integrating the resulting first-order equation gives $(x-x_0{)}^2+(y+c_0{)}^2=R^2$. Both endpoints have height zero, so subtracting their equations gives $x_0=0$, and then $R^2=a^2+c_0^2$. The arc lying in $y>0$ isIts maximum height is $\sqrt{a^2+c_0^2}-c_0$. The ray rises into the faster region before returning to the boundary; its circle centre is $(0,-c_0)$. The prescribed boundary endpoints are understood by continuity from the open half-plane. This is the circular light ray in a linear speed profile.

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.

For endpoint-preserving variations of a functional, the Euler-Lagrange equation isAlong an extremal, the chain rule gives $d(F-y^{\prime }{F}_{y^{\prime }})/dx=F_x+y^{\prime }[F_y-d{F}_{y^{\prime }}/dx]=F_x$. Thus when $F$ has no explicit $x$ dependence, the Beltrami identity isFor a positive speed profile, the elapsed-time integrand in Fermat principle is $F=\sqrt{1+(y^{\prime }{)}^2}/c(y)$. Its first integral givesIntegrating each monotone-height branch yieldsReal branches require $Ac(y)\le 1$. Turning points are joined in the original graph $y(x)$, even though the inverse derivative is singular there.

For the inverse-height speed profile, this becomes $dx/dy=\pm a/\sqrt{y^2-a^2}$, with $a=A>0$. Integration gives $x-b=\pm a\mathrm{arcosh}(y/a)$, so the light rays are catenariesEqual endpoint heights at opposite $x$ values force $b=0$, since the real hyperbolic cosine is even and strictly increases with the absolute value of its argument. The remaining endpoint condition is $a\cosh (1/a)=\cosh 1$. One exact solution is $a=1$, givingHere $ds/dx=\cosh x$ and $1/c(y)=y=\cosh x$ account for the squared hyperbolic cosine in the time integral.

The stated endpoints also allow a second stationary ray. Set $t=1/a$. The function $\cosh t/t$ decreases and then increases: its derivative has the sign of $t\sinh t-\cosh t$, whose derivative is $t\cosh t>0$. Its unique minimum occurs where $t\tanh t=1$, at $t\simeq 1.199679$. Since the known root $t=1$ lies below this minimum's location, there is exactly one further root, $t\simeq 1.4242930655$. Thus the two stationary catenary rays in an inverse-height speed profile also includeNo extra condition in the question selects uniquely between these rays. The minimum height is a minimum along the chosen ray, not a global travel-time assertion. Indeed, competitors descending vertically to height $\eta >0$, crossing horizontally, and rising again have time ${\cosh }^2(1)-{\eta }^2+2\eta$. As $\eta \downarrow 0$ this tends to ${\cosh }^2(1)\simeq 2.381098$, below both stationary times. Their corners can be smoothed, so Fermat principle here must be understood as stationarity rather than unrestricted global minimization.

For $c>1$, the catenary has a positive minimum height $x_0=(c-1)/\beta$ above the road. Its derivatives areThus the light ray is symmetric about $y=y_0$, descends towards a horizontal tangent at its minimum, and then rises. It never intersects the road. This smooth upward bending in the refractive index gradient produces an inferior mirage, without reflection at the road.

By Fermat principle, a light ray makes optical travel time stationary. Multiplication by the constant vacuum speed of light does not change the stationary paths, so the functional can be taken to be the refractive index integrated over arc length. The constant denoted $c$ below is an integration constant, not that vacuum speed.

The Fermat principle states that the actual path of a light ray makes its travel time stationary under small admissible variations with fixed endpoints. Equivalently, its optical path length is stationary:where $v$ is the local light speed, $n$ is the refractive index, and the two integrals differ by the constant vacuum light speed. In a uniform medium this reduces to stationarity of geometric path length. The principle is a stationarity statement; for the planar reflection considered next the stationary path is also the unique minimum.

Follow the branch of a light ray that undergoes one internal specular reflection in the spherical drop. At entry, refraction turns its direction by $\alpha -\beta$. The internal specular reflection turns it by $\pi -2\beta$, because the radius is the normal vector and the incidence and reflection angles are both $\beta$. At exit the second refraction contributes another $\alpha -\beta$. Thus the total deviation angle isSnell's law at entry gives $\sin \alpha =n\sin \beta$, taking the surrounding air's refractive index as one. The same internal angle at successive encounters follows from the isosceles triangles formed by each chord and the two radii. Internal specular reflection here need not be total internal reflection; the transmitted branch at that encounter does not belong to the one-reflection trajectory.