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Phase-contrast microscopy converts specimen-dependent phase differences into visible light intensity differences. It can reveal structures in living, unstained biological cells, allowing their changes to be followed without first making an absorbing preparation.^[ref] The key question is how an optical microscope can read a phase when an ordinary detector records light intensity.

Consider a transparent specimen with complex amplitude transmission $t(x,y)=e^{i\varphi (x,y)}$. In ideal in-focus coherent imaging with all relevant spatial frequencies transmitted, its image has $I=|t{|}^2=1$: varying $\varphi$ alone produces no brightness detail. This is a particular model, not a claim that all transparent specimens are invisible. Diffraction, defocus, absorption and a finite optical pupil can already change a real image.

A phase-contrast microscope images an illuminating annulus onto a phase plate ring near the objective's rear focal plane. The aligned ring attenuates and shifts direct light relative to most diffracted light; their interference changes image brightness.^[ref]

The method is beautiful because the bright direct beam becomes its own reference: a small specimen-dependent complex amplitude can interfere with it and change the image at first order. Merely magnifying an invisible phase pattern would leave $|t{|}^2$ unchanged.

For an oscillation written $A\cos (\omega t-\varphi )$ relative to $A\cos (\omega t)$, a positive $\varphi$ represents delayed maxima. The corresponding unwrapped delay is $\varphi /\omega$, while a single periodic observation determines the phase difference only modulo $2\pi$.

Two sinusoidal signals are in phase quadrature when their phase difference is a quarter cycle, for example $\cos \varphi$ and $\sin \varphi$. Their time averages have zero correlation over a cycle. In a plane wave, this separates fields such as displacement and its time derivative, velocity.

For the overlapping monochromatic wave components, let $d$ be the thickness and ${\lambda }_0$ the vacuum wavelength. Define $\Delta n=n_s-n_f>0$ using the slow and fast refractive indices. With complex amplitude convention $e^{i(kz-\omega t)}$, the accumulated slow-minus-fast phase difference isThis is the optical path difference $\Delta nd$ expressed as a wave phase. It is a signed accumulation convention; with the opposite time convention the sign reverses. The outgoing polarization of an electromagnetic wave depends on $\delta$ modulo $2\pi$. Here the positive slow-minus-fast $\Delta n$ differs from the signed principal-index $n_e-n_o$ used in the parent birefringence entry when $n_e<n_o$.

For an original worked example, stipulate $\Delta n=0.0100$ at ${\lambda }_0=600\mathrm{nm}$ and a uniform thickness $d=15.0\mu \mathrm{m}$. These are design assumptions, not measured material data. The optical path difference is $150\mathrm{nm}$, soIt is a quarter-wave plate. At the same wavelength and refractive indices, doubling $d$ to $30.0\mu \mathrm{m}$ gives $\delta =\pi$, a half-wave plate. A $1.0\mu \mathrm{m}$ thickness increase changes $\delta$ by $\pi /30$, or $6^{\circ }$; retaining the same birefringence at $500\mathrm{nm}$ would instead give $\delta =3\pi /5$ for the $15.0\mu \mathrm{m}$ plate. Real refractive indices also vary with wavelength, so this last calculation explicitly neglects optical dispersion.

For these equal transmitted wave amplitudes, orthogonal projection onto a polarizer crossed with the initial $45^{\circ }$ linear polarization gives a simple independent readout. After removing a common wave phase, the component complex amplitudes are proportional to $(1,e^{i\delta })/\sqrt{2}$; orthogonal projection onto $(1,-1)/\sqrt{2}$ gives $(1-e^{i\delta })/2$. Squaring its modulus yields $I_{\perp }/I_{\mathrm{in}}={\sin }^2(\delta /2)$: zero at $d=0$, one half at $15\mu \mathrm{m}$, and one at $30\mu \mathrm{m}$. This assumes ideal polarizers, negligible specular reflection and wave absorption, overlapping components and stable relative wave phase. It measures a periodic function of phase retardation, so one brightness measurement cannot determine the thickness uniquely.

Polarization of waves describes the direction and time-dependent geometry of a transverse wave's vector during its oscillation at a fixed position. In three spatial dimensions, a transverse wave can have two independent directions of oscillation, both perpendicular to its direction of propagation. For polarization of an electromagnetic wave, the relevant vector is the transverse electric field; for a mechanical transverse wave, it can be the transverse displacement. A purely longitudinal wave has no choice between these transverse directions, and the orientation of a static electric field is not polarization of waves.

Choose transverse axes $x,y$ in a Cartesian coordinate system. A monochromatic wave with a fixed phase difference has the real electric fieldHere $A_x,A_y$ are wave amplitudes, $\omega$ is the angular frequency, and $\delta$ is the phase difference of the $y$ oscillation relative to the $x$ oscillation. The vector's tip traces its path during the wave period $T=2\pi /\omega$. Both plotted coordinates are real electric field components; neither axis represents the imaginary part of a complex amplitude.

For nonzero wave amplitudes, $\delta =0$ or $\pi$ modulo $2\pi$ gives linear polarization: the vector moves back and forth along a fixed line. Equal nonzero wave amplitudes and $\delta =\pm \pi /2$ modulo $2\pi$ give circular polarization: the tip traces a circle of constant radius. Unequal nonzero wave amplitudes with $\delta =\pm \pi /2$ give a noncircular ellipse whose axes coincide with $x,y$. More generally, nonzero wave amplitudes and $\sin \delta \ne 0$ give elliptical polarization, with circular polarization as a special case. If exactly one wave amplitude vanishes, the nonzero wave has linear polarization for every $\delta$; if both vanish, there is no nonzero wave to assign a polarization of waves. ^[ref].

For example, take $\omega =2\pi {\mathrm{s}}^{-1}$, so the wave period is $T=1\mathrm{s}$. The choices $(A_x,A_y,\delta )=(2,1,0)$, $(1,1,\pi /2)$ and $(2,1,\pi /2)$, with electric field wave amplitudes in $V/m$, produce respectively the line segment $E_y=E_x/2$, the circle $E_x^2+E_y^2=1$ and the ellipse $E_x^2/4+E_y^2=1$, when coordinates are measured in $V/m$. At $t=T/4$, their tips are respectively $(0,0)$, $(0,-1)$ and $(0,-1)V/m$. The last two share a point without having the same path.

Resolving the electric field into two transverse components makes polarization of waves useful: the same frequency and wave amplitudes can describe different paths when their phase difference changes. An ideal linear polarizer selects the component along its pass axis, letting its orientation probe the direction of linear polarization through Malus's law. A fixed pair of wave amplitudes and a fixed phase difference describe one stable path; unpolarized light requires a statistical description rather than one fixed ellipse.

A Michelson interferometer turns displacement into a count of interference fringes: a moving mirror changes both the outward and return paths. This lets a wavelength act as a ruler whose divisions need no engraving. Assume normal incidence, a fixed reference arm, stable refractive index $n$ along the moving arm, and fixed vacuum wavelength ${\lambda }_0$. Write $\lambda ={\lambda }_0/n$ for the wavelength in that medium and take $\Delta L$ positive away from the beam splitter. The change in optical path difference, the accumulated phase difference, and the signed number of complete interference fringe cycles areThe factor two comes from propagation geometry: moving the mirror by $\Delta L$ adds $\Delta L$ before reflection and $\Delta L$ afterward. One bright-to-bright cycle means mirror travel $\lambda /2$; bright-to-dark means $\lambda /4$, half a cycle. All lengths must use compatible units, and $\Delta N$ is dimensionless. A constant reflection phase difference changes the initial fringe position but cancels from this change formula.

For an original hypothetical example, suppose $\lambda =600\mathrm{nm}$ in the actual surrounding medium and a continuous scan gives $40.25$ cycles, including the final fractional cycle. Then $\Delta L=(600\mathrm{nm})(40.25)/2=12075\mathrm{nm}=12.075\mu \mathrm{m}$. With the wavelength treated as exact, a fringe-reading measurement uncertainty of $0.02$ cycle contributes $6\mathrm{nm}$ to the displacement measurement uncertainty. An uncertainty of $0.01\mathrm{rad}$ in a local phase difference instead corresponds to $600(0.01)/(4\pi )\mathrm{nm}\approx 0.477\mathrm{nm}$. This illustrates measurement resolution, not a complete claim of accuracy: wavelength calibration, air changes, mirror tilt and missed cycles can dominate.

To separate the oscillating readout from its optical coherence envelope, use the local modelHere $I$ is detected light intensity, $\bar{I}$ is the mean level, ${\varphi }_b$ is a fixed bias wave phase, and $\mathcal{V}$ is interferometric visibility. A fine scan of $D$ measures the carrier phase difference; a coarse scan can map the temporal optical coherence envelope through $\mathcal{V}(D)$. The envelope maximum locates matched delays only under the source and apparatus assumptions; it is not a bright fringe identifying an arbitrary mirror position. Even with known $\mathcal{V}$, one intensity reading leaves the cosine sign and whole-cycle ambiguities. Continuous motion of known direction, or an independently obtained pair of signals in phase quadrature, is needed to recover signed travel through reversals.

The figure's original control example stipulates equal arm light intensities, matching linear polarizations, and a Lorentzian optical spectrum giving $\mathcal{V}(D)=\exp (-|D|/L_c)$ with $L_c=1\mathrm{mm}$. In vacuum, arm mismatches $|L_2-L_1|=0.60\mathrm{mm}$ and $0.10\mathrm{mm}$ give $\mathcal{V}=e^{-1.2}\approx 0.301$ and $e^{-0.2}\approx 0.819$. Matching the arms improves the phase difference readout without changing its ${\lambda }_0/2$ travel per cycle. A fine mirror scan of one cycle changes $D$ by only $600\mathrm{nm}$ for the stipulated $600\mathrm{nm}$ source, much less than $L_c$. These are calculated values, not historical data.

A proposed measurement sequence follows from this model. Overlap the returning beams, match their linear polarizations, find a high-contrast delay and use a small calibrated mirror scan to determine the sign and fringe period. Read a fixed spatial sample of the interference fringes with a photodiode; record each arm separately and the dark signal, avoid detector saturation, and retain enough time resolution to count every cycle. Approach mechanical settings consistently and repeat the return scan to expose lost motion or drift. For very small changes, bias near phase quadrature, where $|dI/d\varphi |=\bar{I}\mathcal{V}$ is largest; an intensity extremum has zero first-order sensitivity. A known drive voltage alone does not provide measurement calibration for mirror travel: compare the actual travel with the fringe count using an independently known wavelength.

Control temperature, pressure, vibration and air currents because the instrument measures optical path difference, not mirror motion alone. For a fixed reference arm and fixed ${\lambda }_0$, the first-order change of the moving arm's round-trip path is $\Delta D\approx 2n\Delta L+2L\Delta n$. Thus an unrecognized $\Delta n$ mimics $\Delta L_{\mathrm{app}}=L\Delta n/n$. A hypothetical $L=0.10\mathrm{m}$, $n\approx 1$, and $\Delta n=10^{-7}$ give $10\mathrm{nm}$ apparent travel. Holding the mirror still while varying or monitoring the air conditions tests this confounder. Separate checks of alignment, wavelength stability and thermal expansion are necessary before sub-fringe measurement resolution becomes a defensible measurement uncertainty.

Use the complex amplitude convention $v(t)=\mathrm{Re}(\hat{V}{e}^{\mathrm{i}\omega t})$, where $\omega =2\pi f$ is the angular frequency and $f$ is the frequency in cycles per second. Substitution into the RC low-pass filter equation replaces the time derivative with multiplication by $\mathrm{i}\omega$, giving its continuous-time frequency responseHere $\varphi$ is the output-minus-input phase difference; a negative value means the output lags. Equivalently, the voltage divider formed from the electrical impedances $R$ and $1/(\mathrm{i}\omega C)$ gives the same ratio. For an input $A\cos (\omega t)$, the sustained output is $A|H|\cos (\omega t+\varphi )$ after the initial exponential decay has died away.

Define the cutoff frequency by a voltage amplitude ratio $1/\sqrt{2}$ relative to the low-frequency value:On a Bode plot, the voltage gain in decibels isAt the cutoff frequency it is $-3.0103\mathrm{dB}$, and far above that frequency it approaches a decrease of $20\mathrm{dB}$ per factor of ten in $f$. There is a smooth transition rather than an abrupt wall. The squared voltage ratio is one half at the corner; converting that statement to a power ratio requires the same real electrical resistance for the compared voltages. It does not mean that an unloaded capacitor dissipates half the source power.

For $R=2.2\mathrm{k}\Omega$ and $C=47\mathrm{nF}$, the cutoff frequency is $f_c\approx 1.5392\mathrm{kHz}$. A $1\mathrm{V}$ peak input at $0.1f_c$, $f_c$, and $10f_c$ produces peak output amplitudes of approximately $0.9950\mathrm{V}$, $0.7071\mathrm{V}$, and $0.09950\mathrm{V}$, with phase differences $-5.711^{\circ }$, $-45^{\circ }$, and $-84.289^{\circ }$. Thus a wanted oscillation at $0.1f_c$ is almost preserved while an unwanted one at $10f_c$ is reduced by about a factor of ten in voltage. The superposition principle lets these two responses be added when both oscillations occur in the same signal.

The University of Colorado Boulder's Lab 3: Filters, §§7.1–7.4, specifies measuring the resistor and capacitor, then driving the assembled RC low-pass filter with a function generator. A two-channel oscilloscope reads actual input and output using high electrical impedance. Sweep the frequency, record the voltage amplitude ratio, and locate $1/\sqrt{2}$. A control compares a cable with a $10\times$ oscilloscope probe on a resistive voltage divider, isolating measurement capacitance before using the probe for the filter. A slow square wave then lets discharge timing estimate the time constant independently. Agreement supports the RC low-pass frequency response; source output impedance, probe loading, and instrument signal bandwidth limit it. These are prescribed measurements, with no measured dataset supplied here.^[ref]

The two-channel setup can also test the predicted phase difference directly: for positive frequency $f$, measure the output's time lag $\Delta t$ between corresponding features of the two sinusoids. For this single-stage model the phase lies between $-\pi /2$ and $0$, so choose the lag within a quarter cycle and calculate $\varphi =-2\pi f\Delta t$. At the cutoff frequency the predicted lag is one eighth of a cycle. This tests the timing part of the frequency response as well as the voltage ratio.

The step response describes a sudden persistent input change. Suppose the input becomes a constant $V_s$ at $t=0$, and the initial capacitor voltage is $v_0$. Finite electric current leaves that voltage continuous. Subtract $V_s$ from the RC low-pass filter equation: the remaining error satisfies $\tau de/dt+e=0$. Separating variables givesFor an initially uncharged capacitor and a $0$ to $3\mathrm{V}$ step in the constructed example,At one time constant, $t=103.4\mu \mathrm{s}$, the voltage is $1.8964\mathrm{V}$, about $63.2\%$ of its final value. The voltage error then continues its exponential decay. This is different from a pure delay: the output starts changing immediately, yet takes time to approach the target.

Define settling time relative to the total voltage change: require $|v_{\mathrm{out}}-V_s|\le ϵ|v_0-V_s|$ thereafter, with $0<ϵ<1$. The exponential decay is monotone, soFor $1\%$ settling, $t_{0.01}=4.6052\tau \approx 476.17\mu \mathrm{s}$; the $10\%$ to $90\%$ rise takes $\tau \ln 9\approx 227.19\mu \mathrm{s}$. Reducing the cutoff frequency by doubling $C$ doubles both times. This reciprocal relation is the useful connection between signal bandwidth and response speed: stronger suppression of rapid noise comes with slower tracking of wanted changes.

The rotatory power is the rotation of transmitted linear polarization per unit path length, in $\mathrm{rad}{\mathrm{m}}^{-1}$. With equally transmitted circular eigenmodes $({\mathbf{e}}_1\mp i{\mathbf{e}}_2)/\sqrt{2}$ and time convention $e^{-i\omega t}$, their accumulated phase difference gives $\theta =({\varphi }_{+}-{\varphi }_{-})/2$. For uniform circular birefringence, $\rho =\pi (n_{+}-n_{-})/{\lambda }_0$. Generic linear birefringence need not admit one rotation angle common to all incident polarization directions. In a weak longitudinal Faraday effect, $\rho =V{B}_{\parallel }$, with signed Verdet constant $V$.

A Sagnac interferometer uses a beam splitter to send light with optical coherence around a common closed optical path in opposite directions, then recombines the beams to read their phase difference. Mirrors or a guiding path define the loop; the source, beam splitter, loop and readout rotate together in the basic experiment. A small relative beam tilt makes spatial interference fringes whose displacement reveals a change of phase difference. At one fixed output, a photodetector can instead record the corresponding change in light intensity.

In 1913 Georges Sagnac mounted a lamp, collimator, beam splitter, four mirrors and photographic plate on a horizontal rotating platform. The mirrors defined a closed polygon; the two beams traversed it in opposite senses. A slight beam splitter adjustment produced visible interference fringes. A motor brought the platform to steady rotation, and switching the lamp exposed the plate. Comparing opposite senses of rotation revealed a reversing displacement. Changing the beam splitter adjustment supplied a further check against optical deformation. The report compared measured and calculated displacements. Georges Sagnac interpreted this as evidence for a luminiferous aether; the relativistic derivation below reproduces the delay without that hypothesis.^[ref]

The readout measures a relative wave phase, not two stopwatch readings separated by an optical transit time. Reversal isolates a contribution odd in angular velocity: if a signed interference fringe position is $q(\Omega )=q_0+k\Omega$ to leading order, then $[q(\Omega )-q(-\Omega )]/2=k\Omega$ cancels the fixed offset $q_0$. Changes of alignment or temperature correlated with reversal can survive this subtraction. Thus the experimental inference concerns a reproducible rotating-loop phase difference, with controls for disturbances; it is not by itself a unique test of an underlying propagation hypothesis.

Consider a stipulated series circuit containing an ideal resistor of electrical resistance $R=100\Omega$ and an ideal capacitor of capacitance $C=1\mu \mathrm{F}$. The same electric current passes through both. Their terminal electric potential differences add by Kirchhoff's voltage law, soA quick derivation uses $i=Cd{v}_C/dt$: taking the derivative of the complex exponential multiplies the capacitor voltage phasor by $\mathrm{i}\omega$, hence ${\hat{V}}_C=\hat{I}/(\mathrm{i}\omega C)$. Dimensionally, $1/(\omega C)$ has units $s/F=V/A=\Omega$, matching $R$.

At frequency $f=1000\mathrm{Hz}$, the angular frequency is $\omega =2\pi f$. The signed capacitive electrical reactance, modulus and complex argument of the total electrical impedance areApply $V_{\mathrm{rms}}=1.00\mathrm{V}$ across the complete series circuit. Its root mean square electric current isWith voltage as the zero-phase reference, the current phasor has complex argument $+57.858092^{\circ }$: electric current leads. The phasor drops have moduli $V_{R,\mathrm{rms}}=0.532018\mathrm{V}$ and $V_{C,\mathrm{rms}}=0.846733\mathrm{V}$; these add as complex numbers, not as positive magnitudes. Their squared magnitudes sum to $(1.00\mathrm{V}{)}^2$.

The mean absorbed power uses the root mean square values:The ideal capacitor has zero cycle-averaged absorbed power; all the mean absorbed power becomes Joule heating in the resistor.^[ref] To see why the phase factor appears, average $p(t)=v(t)i(t)$ over a full cycle with $v(t)=\sqrt{2}{V}_{\mathrm{rms}}\cos \omega t$ and $i(t)=\sqrt{2}{I}_{\mathrm{rms}}\cos (\omega t-\varphi )$. The cosine addition formula reduces the average to $V_{\mathrm{rms}}{I}_{\mathrm{rms}}\cos \varphi$; the term at $2\omega$ averages to zero. Independently, $p_C=d(C{v}_C^2/2)/dt$, and the capacitor energy returns to its starting value after each cycle.

At fixed applied $V_{\mathrm{rms}}$, the limit $\omega \to 0^{+}$ gives $|Z|\to \infty$, $\varphi \to -\pi /2$ and $I_{\mathrm{rms}}\to 0$: the ideal capacitor blocks steady electric current. The limit $\omega \to \infty$ gives $Z\to R$, $\varphi \to 0$ and $I_{\mathrm{rms}}\to V_{\mathrm{rms}}/R$. These are limits of the ideal lumped electrical circuit; unwanted lead inductance and component losses can invalidate that model at high frequency. At $\omega RC=1$, the equal legs of the impedance right triangle give $|Z|=\sqrt{2}R$ and $\varphi =-\pi /4$ without numerical evaluation. Measuring both the magnitude and the sign of the phase difference distinguishes this capacitor response from an equally large positive electrical reactance.

Single-slit diffraction is the spreading of light transmitted by one slit. Consider an ideal rectangular slit of width $a$, illuminated normally by a uniform plane wave with a single wavelength $\lambda$ in the surrounding medium. This monochromatic wave has uniform wave amplitude and equal incident wave phases across the slit, providing the required optical coherence. In the scalar Fraunhofer diffraction model, the normalized angular wave intensity isHere $\theta$ is measured from the forward normal in the plane across the slit width. The sinc function uses the unnormalized convention $\mathrm{sinc}u=\sin u/u$. The limit makes the central value one. This ideal envelope neglects additional direction-dependent propagation and detector factors; near the forward direction it describes the normalized screen irradiance. ^[ref].

The noncentral zeros satisfyThe restriction follows from the range of the sine of a real angle. The first minima have $\sin \theta =\pm \lambda /a$ when $\lambda /a\le 1$. If $a<\lambda$, this model has no noncentral zero in the forward hemisphere; equality puts the first zeros at grazing directions, outside the small-angle approximation and off any finite perpendicular screen. ^[ref].

For $0<\lambda /a<1$, define the central width as the separation between the two first minima. Its angular width and its width on a perpendicular screen areThe angular approximation uses radians and requires $\lambda /a\ll 1$; the screen approximation uses $y\approx L\theta$. The tangent relation is exact straight-line geometry, even though the underlying Fraunhofer diffraction field is approximate. This width is distinct from the full width at half maximum.

For $\lambda =500\mathrm{nm}$, $a=0.1\mathrm{mm}$ and $L=1\mathrm{m}$, the positive first minimum isThe dimensionless far-field parameter is $a^2/(\lambda L)=0.02$. Doubling the slit width to $a=0.2\mathrm{mm}$ gives $W\approx 5.00002\mathrm{mm}$ and parameter $0.08$; both curves below use the Fraunhofer diffraction approximation. Choosing $L=2\mathrm{m}$ halves those parameters and doubles both screen widths.

Each curve is normalized to its own $I_0$; the arrows measure between first minima. The calculation uses $\theta =\arctan (y/L)$ before evaluating the sinc function, rather than replacing $\sin \theta$ by $y/L$. The lower diagram shows the exact screen geometry with the vertical scale enlarged. Normalization removes the change in peak brightness when the slit width changes.

The wave interference explanation is especially simple: divide the slit into equal narrow strips and add their complex amplitudes, rather than their wave intensities. With wavenumber $k=2\pi /\lambda$, the continuous slit givesIntegrating the complex exponential gives the endpoint difference; its continuous value at zero is one. Taking the squared complex modulus gives the stated wave intensity. This is the Fourier transform of an interval indicator expressed in angular coordinates. Feynman's continuous-source argument obtains the same pattern by letting the number of sources tend to infinity. ^[ref].

At the positive first minimum, the wave phases span exactly $2\pi$ across the slit. Every strip in one half has a partner $a/2$ away, with equal wave amplitude and phase difference $\pi$, so the pairs cancel. At $\theta =0$, all the wave phases align instead. ^[ref].

The drawing uses $N=8$ equally weighted midpoint samples $x_j/a=-1/2+(j+1/2)/8$, with $j=0,\dots ,7$. At the first minimum their complex amplitudes are a common rotation of the eight roots of unity, so their head-to-tail polygon closes exactly. This finite sum illustrates cancellation; the continuous integral, rather than eight point sources, defines the slit pattern at general angles. The circle shows the continuous range of wave phases; coloured arrows show the eight samples.

A narrower aperture requires a larger angle to accumulate the cancelling phase difference, so narrowing the slit widens the beam. This is a useful limit on geometrical optics: confining light more tightly at an opening increases its subsequent angular spread. At fixed incident irradiance and slit height, reducing $a$ also reduces transmitted power, exposing a practical balance between diffraction-limited resolution and light throughput. In a diffraction grating, this single-slit diffraction envelope modulates the wave interference between different slits.

Spatial optical coherence compares field samples at different positions, usually across a transverse plane at equal time. Two small openings separated by $d$ sample this correlation: recombine their light with nearly matched travel times and measure interferometric visibility near the central interference fringes. A narrow opening in front of a broad-spectrum lamp can improve transverse spatial optical coherence while leaving its short temporal optical coherence. These are separate controls: narrow spectral selection by itself need not make an extended source spatially coherent. MIT's lecture notes illustrate the two-opening test and these contrasting source preparations. ^[ref].

To isolate source width, stipulate a distant, narrowband source with uniform angular brightness on $-\Theta /2\le \alpha \le \Theta /2$, and mutually uncorrelated emission from distinct source directions. Here $\Theta \ll 1$ is the full angular width in radians, $\lambda$ the common wavelength, and the two openings are separated by $d$ along the same transverse direction as the source extent. Neglect amplitude differences across the openings. Each source direction contributes a phase difference approximately $2\pi d\alpha /\lambda$. Averaging its complex cross term givesThe sinc function is extended continuously to one at $q=0$. The sign convention in the exponential depends on which opening is numbered first; the symmetric model gives the same real result either way. Equal detected single-opening light intensities make the magnitude equal to interferometric visibility. A displaced source center adds a deterministic wave phase, shifting the interference fringes without changing their contrast.

The strips show $I/(I_1+I_2)=1+g_{12}^{(1)}(0)\cos \varphi$ with equal single-opening light intensities, a common phase difference coordinate $\varphi$ and no single-slit diffraction envelope. At $q=1/2$, contrast is $2/\pi \approx 0.636620$; at $q=1$, it vanishes. At $q=3/2$, the correlation is $-2/(3\pi )\approx -0.212207$: weak fringes return with bright and dark positions exchanged. For $\lambda =500\mathrm{nm}$ and $d=0.5\mathrm{mm}$, the first zero occurs at $\Theta =1\mathrm{mrad}$. This is a calculated example, not an experimental measurement or a universal source-width law.

A proposed source-width control follows directly from this model: illuminate two fixed openings with a distant adjustable luminous slit of full width $w$ at distance $R$, so $\Theta \approx w/R$ under the small-angle approximation, and hold $d$ and the spectral selection fixed. Increasing $w$ overlays more mutually incoherent, displaced interference fringe patterns. Record profiles over several near-central fringes, measure each opening separately and correct intensity imbalance as in first-order optical coherence. Restrict the recorded region so the path difference stays small compared with the optical coherence length, and resolve the fringes with the detector. Real source nonuniformity, finite opening widths and spectral spread modify the stipulated sinc function law. This protocol is a model-based extension of the verified two-opening method, not a claim that the plotted curves were measured.

Thus an extended incoherent source can give locally excellent wave interference when the sampled separation is small, and almost none when it is larger. The useful scale $d\Theta \sim \lambda$ expresses how source extent limits the aperture over which coherent addition is available. The same angular averaging explains why narrowing a source opening can recover visible interference fringes, even without narrowing its optical spectrum.

A node of a standing wave must have $y(x,t)=0$ for every time, not merely at one snapshot. For $y=2a\sin (kx)\cos (\omega t)$, with $a,k,\omega >0$, set $t=0$ to see that this requires $\sin (kx)=0$. ThusAn antinode instead maximizes the oscillation amplitude $2a|\sin (kx)|$ and lies halfway between successive nodes of a standing wave, at $x=(2j+1)\lambda /4$. Neighbouring antinodes are also separated by $\lambda /2$; the distance from a node of a standing wave to its nearest antinode is $\lambda /4$. These distances refer to the string's longitudinal coordinate, not its curved arc length.

Within one interval between nodes of a standing wave, all moving points have the same oscillation phase. Adjacent intervals have opposite signs of $\sin (kx)$, so their phase difference is $\pi$: one loop rises while the next falls. The oscillation phase at an exact node of a standing wave is undefined because its amplitude is zero. A ruler measures the spacing of the fixed pattern; it does not track a moving wave crest.

A thermal diffusion wave is a periodically forced solution of the heat equation, with spatially decreasing amplitude and varying wave phase. In a semi-infinite rod $x\ge 0$, assume constant thermal diffusivity, negligible side losses, a nearly uniform temperature across each cross-section, and the boundary conditions $\theta (0,t)={\Theta }_0\cos \omega t$ and $\theta (x,t)\to 0$ as $x\to \infty$. A steady periodic thermal state then satisfiesChoose the square root in the complex numbers with $a>0$ so that the amplitude decays into the rod. Substitution of the phasor into the heat equation proves the formula: $i\omega \hat{\theta }=\alpha {\hat{\theta }}^{\prime \prime }$. Taking the real part gives the displayed cosine. In this convention $a$ is the spatial attenuation slope and $b$ is the spatial lag slope, both in ${\mathrm{m}}^{-1}$; $\gamma$ combines them into one complex number. If a complex wavenumber $K$ is defined by $e^{i\omega t-iKx}$ instead, $K=b-ia$.

With $\Delta x=x_2-x_1>0$, let $A_j=|\hat{\theta }(x_j)|$ and define a positive downstream lag $\Delta \varphi$. Thenand either measurement determines thermal diffusivity:The logarithm uses an amplitude ratio, not a ratio of squared amplitudes. The lag is in radians and must be tracked continuously across cycles: a measured phase difference modulo $2\pi$ leaves an integer-cycle ambiguity. Use closer sensor spacings or vary frequency continuously to resolve that ambiguity. Equal attenuation and lag slopes provide a useful diagnostic of the no-side-loss model.

The apparent phase velocity is $\omega /b=\sqrt{2\alpha \omega }$, which depends on frequency. A thermal diffusion wave therefore does not give a material-independent travel speed for heat; it is a forced pattern in a dissipative diffusion equation. The ideal heat equation still has infinite propagation speed, and the observed lag is not the arrival time of a sharp front.

A two-element radio interferometer coherently combines signals from two receiving antennas. With equal single-arm wave intensity $I_0$, projected baseline $b$, free-space wavelength $\lambda$ and imposed phase difference $\delta$, its directional response is the displayed formula. It follows by adding the two complex amplitudes and taking the squared modulus. Near the normal to the baseline, the first null is $\lambda /(2b)$ from a maximum. This narrow response repeats in interference fringes, so a long baseline alone does not give an unambiguous image.

Stipulate a distant circular source of uniform angular brightness, full angular diameter $\theta \ll 1$ in radians, and narrow optical bandwidth around wavelength $\lambda$. Distinct source patches emit independently. Assume each collector is small enough not to resolve the disk individually. For a sky offset $\mathbf{\alpha }$ from the disk centre, the baseline-dependent phase difference is $2\pi {\mathbf{b}}_{\perp }\cdot \mathbf{\alpha }/\lambda$, up to channel-sign convention. Summing independent patch contributions makes the equal-delay first-order optical coherence the normalized Fourier transform of angular brightness:Choose the horizontal coordinate along ${\mathbf{b}}_{\perp }$, put $k=2\pi B/\lambda$, and integrate along vertical chords of the disk:The imaginary part cancels between opposite chords. Here $J_1$ is the order-one Bessel function of the first kind. One way to obtain the last equality is to integrate the exponential's power series term by term on the finite disk. Equivalently the result is specified by the convergent seriesAt $B=0$ take the continuous limit $\gamma =1$; $V^2=|\gamma {|}^2$ starts as $1-x^2/4+O(x^4)$. This is why very short interferometric baselines give little leverage on angular diameter. Larger disks lose spatial optical coherence at shorter baselines.

The first positive zero of $J_1$ is $x_1\approx 3.831705970$, soThis is a full diameter convention. The angle must be converted before substitution: one arcsecond is $\pi /648000$ radians, and one milliarcsecond, abbreviated mas, is $10^{-3}$ arcseconds. For the stipulated $\theta =5\mathrm{mas}$ and $\lambda =500\mathrm{nm}$, $\theta =2.4240684\times 10^{-8}\mathrm{rad}$ and $B_0\approx 25.1575\mathrm{m}$. At $B=10\mathrm{m}$, the model gives $x\approx 1.52309$ and $V^2\approx 0.542783$.

Beyond $B_0$, $\gamma$ changes sign but $V^2$ stays nonnegative and develops weak further lobes. A single measured squared visibility can therefore match several angular diameters; sampling several interferometric baselines constrains the fit. Limb darkening, spectral averaging, or resolved collector apertures change the uniform-disk prediction. A uniform-disk diameter is a parameter of that brightness model; converting it to a physical stellar radius also requires distance and an appropriate brightness profile.