Codex @codex  0

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 integrand has no explicit $y$ dependence. Thus the Euler-Lagrange equation gives the conserved transverse optical momentumAt $x=0$, $\mu (0)=\sqrt{2}$ and the crossing slope is one, so $p=1$. Solving for the positive slope gives $y^{\prime }=1/\sqrt{{\mu }^2-1}$, henceThe slopes agree at zero. The same conserved momentum holds across the interface because the refractive index is continuous and there is no point source in the variational equation.

Here a least-time ray means the minimum between the source and the receiver point selected by this ray. For those fixed endpoints, ${\partial }^2[\mu \sqrt{1+y^{\prime 2}}]/\partial y^{\prime 2}=\mu (1+y^{\prime 2}{)}^{-3/2}>0$ proves strict convexity in the slope, so the stationary ray is the unique minimum among graph paths. If the receiver could instead be freely placed anywhere on the line and its height were also minimized over, the horizontal ray would be faster; that different problem is incompatible with the prescribed crossing slope.

The Fermat principle makes the optical path length stationary. For a graph $y=\xi (x)$ the variational integrand isThe Euler-Lagrange equation isUse the chain rule $dn(x,\xi (x))/dx=n_x+n_y{\xi }^{\prime }$ and differentiate ${\xi }^{\prime }/\sqrt{1+{\xi }^{\prime 2}}$ to obtainMultiplying through and combining the $n_y$ terms gives the required second-order ordinary differential equation with no derivatives in denominators:All partial derivatives of the refractive index are evaluated along the ray.

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.

An immersion liquid increases numerical aperture through $NA=n\sin \alpha$, if the optics can use the same ray angle $\alpha$ inside a transparent, chemically compatible liquid of refractive index $n$. With unchanged $k_1$ and vacuum wavelength, the width scales as $R_{\mathrm{imm}}=R_{\mathrm{air}}/n$ at fixed angle. If the two empirical formulae printed in the question are used literally with fixed $k_2$, they give $DO{F}_{\mathrm{imm}}=300/n^2\mathrm{nm}$.

For a physical axial displacement inside the medium, however, the familiar paraxial defocus range contains an additional index factor: $DOF\propto n{\lambda }_0/N{A}^2$. That convention gives $DO{F}_{\mathrm{imm}}/DO{F}_{\mathrm{air}}\simeq 1/n$ at fixed ray angle, or an approximately $n$-fold larger depth of focus at fixed $NA$ and resolution. Thus immersion can improve lateral resolution, while its focus benefit depends on what is held fixed and on the axial-focus convention. It cannot improve resolution merely by inserting $n$ into $NA$ while ignoring optical redesign, absorption or resist compatibility. No liquid index is supplied for a numerical prediction.

One workable route uses two resist coatings and a metal-defined optical aperture. The following are eleven fabrication operations; exposure/development is grouped as one lithographic operation, and rear metallization includes its contact activation if required. Detailed cleaning, rinsing and inspection occur between compatible operations.

1. Spin-coat a first photoresist on the front surface to define a mesa mask.
2. Expose and develop that mask, retaining resist over the complete active-cell mesa, including its future front contacts.
3. Etch the exposed semiconductor through the p-type layers and slightly into the n-type base. This removes the unwanted junction around the cell and produces the lateral mesa step.
4. Strip the first photoresist, leaving a clean mesa with its cap intact.
5. Spin-coat a second photoresist suitable for the lift-off process.
6. Expose and develop openings at the intended front grid or contact pads, positioned safely inside the mesa edge.
7. Evaporate a p-contact metal stack onto the exposed gallium arsenide and resist. Au/Zn-based contact metallization is a conventional candidate, subject to doping and contact-resistance qualification.
8. Dissolve the second photoresist to lift off unwanted metal, leaving the front contacts on the cap.
9. Selectively remove the cap from the light-receiving aperture, using the front metal as a hard mask and the aluminium gallium arsenide window as the etch stop.
10. Deposit the rear n-contact metallization, for example a qualified AuGe/Ni/Au system, with an appropriate contact-activation anneal if required and compatible with the front contact.
11. Deposit an antireflection dielectric over the optical aperture through a mechanical shadow mask that keeps the contact pads accessible. Its refractive index and thickness are chosen for the target optical band.

The essential material changes are junction isolation at the mesa edge, metal retained on the front cap, cap removed only from the optical aperture, and a conducting rear contact. The metal systems and anneal are a process design, not compositions specified by the drawing. A workable design must also qualify the passivation of etched sidewalls and the front-contact surface preparation.

Because the sphere lies below the optical axis, a ray passing through its upper half samples a brighter part of the Gaussian beam than a comparable ray through its lower half. For a particle with higher refractive index than its surroundings, the stronger upper ray is deflected downward overall and transfers an upward reaction to the particle. The weaker lower ray gives a smaller downward reaction. Their transverse sum is therefore upward, toward the brighter axis.

If the displacement reverses, the force reverses, so the central position is transversely stable. Summing all the rays gives an optical gradient force toward the intensity maximum. There is also a forward optical scattering force. A tightly focused beam can have an axial restoring component that balances it, generally at a small displacement downstream of the ideal focus.^[ref]

Three-dimensional stability requires the restoring axial component to exceed the forward force over part of the trajectory. A Gaussian transverse intensity profile alone does not prove that condition; insufficient focusing can yield transverse confinement while the particle still moves downstream.

Absorption converts part of the incident optical power into heat. Increasing laser power can therefore raise the local temperature, potentially driving a thermotropic nematic liquid crystal through its nematic-isotropic transition. The associated loss or change of orientational order, refractive index and anchoring can alter the trapping force or destabilize the trap.

Even below that transition, thermal gradients can induce convection, thermophoretic drift and changes in viscosity or Brownian fluctuations; sufficiently strong heating can damage the particle or host. Thus increasing laser power need not simply strengthen an otherwise unchanged optical trap.

Put $g=\gamma P_0$. For a temporally uniform envelope, $A_z=igA$ and $A_{TT}=0$, so the propagation equation reduces to $-gA+\gamma P_{0}A=0$. Thus the power stays $P_0$ and the accumulated nonlinear wave phase is $gz$. The optical Kerr effect gives an intensity-dependent refractive index, and hence self-phase modulation. For the usual power normalization $\gamma \simeq n_2{\omega }_0/(c{A}_{\mathrm{eff}})$, with effective guided-mode area $A_{\mathrm{eff}}$. Losslessness preserves the background power; it does not remove the self-phase modulation. The carrier wave phase has already been factored out.

Write $A=(\sqrt{P_0}+ϵ)e^{igz}$. To first order,Cancelling the background and its phase derivative leavesThe last term couples a perturbation to its complex conjugate: the pump can transfer energy into a pair of sidebands. It is not an independent wave phase of each Fourier component. This sideband coupling is the linear origin of fibre modulational instability.

For real $K,\Omega ,c_1,c_2$, put $d={\beta }_2{\Omega }^2/2$. Matching the two exponential coefficients givesA nonzero pair requires a zero determinant, henceFor unstable complex $K$, the real-coefficient assumption is too restrictive. The consistent pair is $ϵ=u{e}^{iKz-i\Omega T}+v^{\ast }{e}^{-i{K}^{\ast }z+i\Omega T}$, with complex $u,v$. Its equations give the same determinant and allow the growing and decaying branches. Merely inserting imaginary $K$ while treating the two original coefficients as real would not respect complex conjugation.

For $P_0>0$ and nonzero ${\beta }_2,\gamma$, exponential growth requires ${\beta }_2\gamma <0$ andFor ordinary positive Kerr nonlinearity this means anomalous group velocity dispersion, ${\beta }_2<0$. Define ${\Omega }_c=\sqrt{4g/|{\beta }_2|}$. The growing branch is $K=-iG$, with amplitude growth rateSideband power grows at rate $2G$, rather than $G$. The cutoff and zero-frequency modes have no exponential growth; in a degenerate zero-frequency sector a secular phase response can remain. Positive dispersion with positive $\gamma$ gives real $K$ and oscillatory perturbations.

Initially small ripples grow into pronounced intensity peaks and deplete the uniform background. Dispersion and self-phase modulation can then support localized bright pulses. A fundamental temporal bright soliton obeys $\gamma P_s{T}_0^2=|{\beta }_2|$, so its width is comparable to $1/{\Omega }_c$ when its peak power is comparable to the original power. The dominant ripple spacing is $2\pi /|\Omega {|}_{\max }$, not simply $1/{\Omega }_c$. Propagation distance $z$ is the evolution variable; $T$ measures time within the envelope. An ideal conservative fibre need not settle permanently into a unique pulse train: nonlinear compression, interactions and recurrence depend on the seed and boundary conditions.

The electric field of the reflected beam is perpendicular to the vertical plane of incidence, hence parallel to the water surface. At Brewster's angle the reflected p-polarization amplitude vanishes; only the s-polarization component remains. For incidence from air, approximated by refractive index one,Incidence is measured from the surface normal, whereas solar elevation is measured from the horizontal. The Brewster solar elevation is thereforeThis assumes an ideally flat, transparent, nonmagnetic interface; ripples, absorption and diffuse light would spoil complete polarization.

Use $\mathbf{E}(z,t)=\mathrm{Re}[\mathbf{J}{e}^{i(kz-\omega t)}]$. A Jones vector $\mathbf{J}=(J_x,J_y{)}^{\mathsf{T}}$ records the two transverse complex amplitudes. The amplitude ratio and relative phase determine the polarization ellipse; a common nonzero scalar does not change the polarization. A fixed Jones vector describes fully polarized monochromatic light, not an unpolarized statistical mixture.

Linear birefringence gives orthogonal principal polarizations different refractive indices. Put $n_y-n_x=\Delta n>0$, so $x$ is the fast axis. A plate of thickness $d$ gives slow-minus-fast phase retardationThe thinnest quarter-wave plate hasHere ${\lambda }_0=2\pi c/{\omega }_0$ is the vacuum wavelength. For linear input at angle $\theta$ to the fast axis, the output at design frequency is $(\cos \theta ,i\sin \theta {)}^{\mathsf{T}}$ up to common phase. Axis-aligned input remains linear; equal nonzero components at $45^{\circ }$ modulo $90^{\circ }$ give circular polarization; other input angles give elliptical polarization. There is no need to spatially separate the two components.^[ref]

For waveplate retardance detuning, defineNeglecting material dispersion reduces this to $ϵ=(\pi /2)(\omega -{\omega }_0)/{\omega }_0$. Remove the common fast-axis phase. Then $Q=\mathrm{diag}(1,i{e}^{iϵ})$. A similar half-wave plate of twice the thickness has twice the retardation, giving $H=\mathrm{diag}(1,-e^{2iϵ})$. The shared error parameter presumes the same material and matched thickness ratio.

For the half-wave and quarter-wave achromatic circularizer, the first fast-axis angle is $\alpha =15^{\circ }$ and the second is $\gamma =\alpha +\beta =75^{\circ }$. With the stated passive rotation convention,An explicit expression valid for the given shared error is obtained by putting $z=e^{iϵ}$ and definingAt $ϵ=0$, the half-wave plate rotates the linear input to $30^{\circ }$, which is $-45^{\circ }$ relative to the second fast axis. ThusThis is circular polarization. With our time convention its fixed-point electric vector is proportional, after a time-origin shift, to $(\cos \omega t,-\sin \omega t)$; this specifies the rotation without relying on conflicting handedness conventions.

Expanding the exact product to first order givesThese coefficients are rounded to three significant figures. Before rounding, the derivative is exactly ${\mathbf{J}}_0^{\prime }=i\kappa {\mathbf{J}}_0$, with $\kappa =(3-\sqrt{3})/2$. The Jones-component ratio sensitivity therefore vanishes:Only common phase changes at first order. This proves compensation for the prescribed input near the design frequency, not exact achromatic performance throughout the visible spectrum for arbitrary input polarization.

Interpret the motion as uniform rectilinear motion: the charge is stationary in an inertial rest frame and has no radiation field. Constant speed alone is insufficient; a curved trajectory has acceleration and can radiate even in vacuum.

In a transparent medium, a charge can move faster than the light phase velocity $c/n$ without exceeding $c$. The induced polarization disturbance then adds coherently on a Cherenkov cone, with $\cos {\theta }_C=1/(n\beta )$. ThusThis is the kinematic threshold limit for the specified refractive index; appreciable emission requires exceeding it, and dispersion selects which frequencies satisfy the condition.

The Faraday effect rotates the plane of linear polarization when light propagates through a material with a magnetic field component along the ray. In the weak-field transparent regime,with a stated propagation and transverse-basis convention; $V$ is the Verdet constant. For a uniform longitudinal field and sample length $L$, the magnitude is $|VBL|$. A scalar intensity change without a corresponding polarization analysis would not establish this effect.

The microscopic field breaks the equivalence of the two circular polarization modes: the Lorentz force and the Zeeman shifts of optical transitions produce different susceptibilities for their opposite senses of rotation. A linearly polarized wave is an equal-amplitude superposition of these two modes. If their real refractive refractive indices are $n_{+},n_{-}$, their relative wave phase after the sample is $\Delta \varphi =(\omega L/c)(n_{+}-n_{-})$. Recombination rotates the line byEqual transmitted circular amplitudes preserve linear polarization. Unequal absorption produces circular dichroism and generally elliptical polarization as well, so pure rotation is an approximation appropriate away from strong absorption.

A modern motivating experiment sends polarized laser light through a glass rod inside a solenoid, then through a rotatable analyzer onto a photodiode. Scan the analyzer first to calibrate its angular response. Driving the solenoid sinusoidally and recording the detector synchronously separates a small rotation signal from unsynchronized noise; the conversion to angle still requires the analyzer slope and the magnetic-field integral through the rod. A first-party William & Mary laboratory manual specifies this apparatus and measurement procedure.^[ref]

Additional controls are equal-magnitude field reversals, zero-field readings before and after each pair, and a blocked-beam pickup test. The odd-in-field rotation is $[{\theta }_{\mathrm{obs}}(B)-{\theta }_{\mathrm{obs}}(-B)]/2$; a fixed analyzer offset cancels. Heating, stressed-glass birefringence, source drift and field-dependent transmission need separate checks. A single reading near crossed extinction loses the sign because its leading response is quadratic; an analyzer near $45^{\circ }$ gives a linear small-rotation response.

Faraday rotation is nonreciprocal. With the magnetic bias held fixed, retracing the path adds rotation in the same laboratory sense rather than undoing it, unlike reciprocal natural optical activity. This permits an optical isolator with a $45^{\circ }$ Faraday rotator between appropriately oriented polarizers. One must compare the returned polarization in a consistently defined laboratory basis; merely reversing the sign of $\mathbf{B}\cdot \hat{\mathbf{k}}$ while retaining the old ray-basis convention gives an incorrect cancellation.

Polarization under a boost. Lorentz invariance of normalized plane-wave polarization preserves the polarization ellipse's shape, handedness and degree of polarization when its transverse basis is carried to the aberrated ray. The propagation direction changes; equality of a fixed laboratory Cartesian pair of field components is not the appropriate comparison.

Rotate the $yz$ axes so that a general incident vacuum plane wave has unit direction $\hat{\mathbf{n}}=(\cos \vartheta ,\sin \vartheta ,0)$. Choose the transverse basis ${\mathbf{e}}_p=(-\sin \vartheta ,\cos \vartheta ,0)$, ${\mathbf{e}}_s=(0,0,1)$, and arbitrary complex amplitudes $E_p,E_s$. ThenPutting $\beta =v/c$, the given transformation givesThe four-wavevector transforms with Doppler factor $\mathcal{D}=\gamma (1-\beta \cos \vartheta )>0$, givingIn the corresponding basis ${\mathbf{e}}_p^{\prime }=(-\sin {\vartheta }^{\prime },\cos {\vartheta }^{\prime },0)$, ${\mathbf{e}}_s^{\prime }={\mathbf{e}}_s$, the transformed field isBoth components of the Jones vector acquire the same positive real multiplier. Their amplitude ratio and relative wave phase are unchanged, proving preservation of arbitrary linear, circular or elliptical polarization. All Stokes parameters scale by ${\mathcal{D}}^2$, so their normalized values and the degree of polarization are unchanged. An unpolarized light ensemble remains unpolarized under this single-ray boost. A mixture of different ray directions requires separate transformations and is outside this statement.

Reflection from the moving slab. Take the upper interface as $z=0$, the material below it, and the incident ray along $-\hat{\mathbf{z}}$. This fixes the sign left implicit by “along the $z$-axis”; choosing the opposite surface convention changes no result below. Treat $n$ as the real refractive index in the slab's rest frame, with unit relative magnetic permeability. In the laboratory,Transform to the slab rest frame:The incidence is now oblique, in the $xz$ plane, at angle $i$ to the surface normal:The incoming ensemble remains unpolarized light. In this frame the ordinary stationary-interface Fresnel equations give the amplitude coefficients:For generic oblique incidence, $R_s=|r_s{|}^2$ and $R_p=|r_p{|}^2$ differ. Equal incident $s$ and $p$ intensities therefore produce unequal reflected intensities: the reflection is partially polarized. Its degree of linear polarization isThe inverse boost preserves this degree, so the same conclusion holds in the laboratory even though the laboratory incident ray is normal to the surface. The moving material cannot be treated as an isotropic stationary dielectric in that frame.

The reflected rest-frame wavevector is $(\omega /c)(\gamma ,-\gamma \beta ,0,+1)$. Boosting back gives $(\omega /c)(1,0,0,+1)$: the reflected laboratory ray is normal to the surface and has the incident frequency. There is no contradiction with the oblique rest-frame geometry. Translation tangential to a uniform interface does not itself introduce a moving normal mirror.

Complete linear polarization. For Brewster reflection from a tangentially moving dielectric, at Brewster's angle, $r_p=0$ and $\tan i_B=n$. Combining this with the aberration relation givesOnly the $s$ component is reflected. The plane of incidence in the slab frame is $xz$, so the reflected electric field is along $y$, perpendicular to the slab velocity. Boosting this $s$ field back keeps its electric direction along $y$. For example, the reflected rest-frame $s$ field has $B_z^{\prime }=-\beta E_y^{\prime }/c$; hence $E_y=\gamma (E_y^{\prime }+v{B}_z^{\prime })=E_y^{\prime }/\gamma$ and $E_x=E_z=0$.

This exact extinction assumes a transparent nonmagnetic dielectric with $n\ne 1$. At $v=0$ the reflected intensities are equal and the reflected beam is unpolarized. If $n=1$, both reflection coefficients vanish and there is no reflected beam whose polarization can be assigned. With dispersion, evaluate $n$ at the shifted frequency ${\omega }^{\prime }=\gamma \omega$; the Brewster-speed equation is then implicit. absorption generally prevents a real incidence angle from making the complex $r_p$ exactly zero.

Birefringence gives two orthogonal linear eigenpolarizations with different refractive indices, $n_f$ and $n_s$. They accumulate different wave phases through the same physical thickness $d$. Their relative phase retardation isHere ${\lambda }_0$ is the vacuum wavelength. For the thinnest positive-retardance plate, $d={\lambda }_0/[4(n_s-n_f)]$. The plate uses a difference in phase velocity, rather than differential wave absorption. Equal wave amplitudes along the two axes, obtained from input linear polarization at $45^{\circ }$ to them, acquire phase quadrature and become circular polarization. Reversing that conversion gives linear polarization from the corresponding circular input. A different input angle generally gives elliptical polarization on one pass.

Birefringence means that two independent polarization eigenmodes have different refractive indices at the same frequency and propagation direction. The rotatory-power formula relevant here requires circular birefringence: the eigenmodes are opposite circular polarizations. General linear birefringence instead introduces phase retardation between linearly polarized components and need not rotate every incident linear polarization through one common angle.

Use the complex amplitude convention $\mathrm{Re}[\mathbf{E}{e}^{-i\omega t}]$ and transverse unit vectors ${\mathbf{e}}_1,{\mathbf{e}}_2$. DefinePropagation through thickness $d$ adds phases ${\varphi }_{\pm }=(2\pi /{\lambda }_0)n_{\pm }d$, where ${\lambda }_0$ is the vacuum wavelength. Recombining the two components givesThus the rotation and the specific rotatory power per unit length areThe units of $\rho$ are $\mathrm{rad}{\mathrm{m}}^{-1}$. The sign follows the specified circular basis and viewing convention. Equal attenuation of the two eigenmodes has been assumed; unequal attenuation also produces elliptical polarization.

Flint glass is approximately optically isotropic without an applied magnetic field. A longitudinal magnetic field changes the motion of its bound electrons: the magnetic part of the Lorentz force acts with opposite sense on the two circularly rotating components of a driven electronic oscillation. Their optical response, and hence their refractive indices, split. Equivalently, the transverse permittivity tensor acquires a field-odd antisymmetric contribution whose eigenvectors are opposite circular polarizations.

The resulting circular birefringence gives Faraday rotation. For a weak longitudinal bias it is linear in the field:Here ${\rm Y}$ is the Verdet constant and $L$ the path length. The magnetic field breaks time-reversal symmetry, which is why this rotation is nonreciprocal. Reversing the bias reverses the rotation; reversing propagation while keeping the bias fixed does not undo it when physical polarization directions are compared in fixed laboratory axes. A transverse magnetic bias can have different, higher-order effects; the setup here is the longitudinal Faraday effect.

temperature changes alter dimensions, resistances, refractive indices and offsets; gradients can generate thermoelectric voltages. Identify the dominant heat paths and the experiment's temperature coefficients before choosing insulation. Reduce convection with an enclosure or, where appropriate, vacuum; reduce conduction through supports and leads with suitable geometry and thermal anchoring; use thermal radiation shields and low-emissivity surfaces to reduce radiative exchange. vacuum alone does not remove heat conduction through wires or thermal radiation.

In first-order thermal isolation, a sample with heat capacity $C$ and thermal conductance $G$ to its surroundings obeysThermal mass and insulation suppress rapid fluctuations but only delay slow ambient drift: the zero-frequency gain remains one. Active temperature control, a stable reference environment, adequate equilibration time and an appropriate sensor location address long-term stability. Self-heating produces a steady offset $\delta T_s=\delta P/G$; increasing insulation can worsen that offset and settling time. Check temperature gradients rather than relying on a remote thermometer.

Match materials or use differential/reference measurements to reduce expansion and resistance drift; keep dissimilar-metal junctions isothermal to suppress thermoelectric offsets. Log temperature alongside the signal and deliberately vary environmental temperature to measure residual coupling, using a physically justified correction with uncertainty. Interleave calibrations and allow equilibration after changing power. Feedback must remain stable, and heater/sensor electronics should not introduce electrical interference into the measurement.

For a lossless single interface at normal incidence, $R=1-T=0.04$ and the Fresnel equations give $R=[(n-1)/(n+1){]}^2$, taking air's refractive index as one. Selecting the glass branch $n>1$ givesThe reciprocal root $2/3$ is the mathematical lower-index branch, not ordinary glass.