The p-adic digits are the unique coefficients in $x={\sum }_{j\ge 0}{b}_j{p}^j$ for a p-adic integer $x$. Taking least nonnegative representatives of its congruence classes,so each new residue determines one more p-adic digit. For $13$ in ${\mathbb{Z}}_2$, the low-to-high digits are $1,0,1,1,0,\dots$, giving $13=1+0\cdot 2+1\cdot 4+1\cdot 8$. A finite prefix describes a whole coset, not a unique infinite expansion.

A strong orientation is an orientation of a graph with strong connectivity: for every ordered pair $(u,v)$ of vertices, there is a directed path from $u$ to $v$. A length-zero directed path permits $u=v$. Ordinary connectivity of the underlying undirected graph is necessary, but does not ensure this two-way reachability after directions are chosen. The acyclic $D_1$ is not a strong orientation; the cyclic $D_2$ is, since a journey can continue around its directed cycle.

Attach a new vertex $D$ to that triangle in a graph by the single edge $CD$. This edge is a bridge in a graph. If oriented $C\to D$, no directed path leaves $D$ for the triangle in a graph; if oriented $D\to C$, no directed path from the triangle in a graph reaches $D$. Neither choice gives a strong orientation, even when the triangle in a graph itself is cyclically oriented. By contrast, the four-vertex cycle graph has the strong orientationFollowing this directed cycle reaches any requested vertex in at most three steps. Removing any one of its undirected edges leaves a graph path through all four vertices, so it has no bridge in a graph.

For more than one vertex, a strong orientation cannot also be an acyclic orientation: directed paths from $u$ to $v$ and back give a nonempty closed directed walk, from which removing repeated portions leaves a directed cycle. A directed graph containing one directed cycle need not have strong connectivity, as the attached $D$ shows.

An acyclic orientation is an orientation of a graph whose resulting directed graph has no directed cycle, hence is a directed acyclic graph. A directed cycle must follow every arrow around its closed route. A cycle in a graph in the skeleton of a directed graph only requires the corresponding undirected edges; their arrows need not agree around the route.

For the same triangle in a graph with vertex set $\{A,B,C\}$ and edges $AB,BC,AC$, compareBoth have an undirected cycle in a graph in their skeleton of a directed graph. Only $D_2$ has a directed cycle. In $D_1$, $C$ has no outgoing directed edge, so a route cannot continue from $C$ back to $A$.

Thus an acyclic orientation does not require its underlying undirected graph to be a forest. Every orientation of a graph of a forest is acyclic, because a directed cycle would give an undirected cycle in a graph. Conversely, every finite simple graph, including one with many cycles in a graph, admits at least one acyclic orientation by the following ordering construction. On this triangle in a graph, six of the eight orientations of a graph are acyclic; the other two follow the directed cycle in opposite senses.

For a chemical solution in a validated linear regime,where $l$ is path length in $\mathrm{dm}$ and $c$ is solute mass per final solution volume in $\mathrm{g}\mathrm{m}{\mathrm{L}}^{-1}$. These conventions give specific rotation units $\mathrm{deg}\mathrm{mL}{\mathrm{g}}^{-1}\mathrm{d}{\mathrm{m}}^{-1}$. Specify solvent, temperature $T$ and wavelength $\lambda$: specific rotation is conditional on the sample environment.^[ref]

For an original model, stipulate an unnamed enantiomer pair in the same achiral solvent at $T=20^{\circ }\mathrm{C}$ and wavelength $\lambda =589\mathrm{nm}$. Assign the positive member specific rotation $+40\mathrm{deg}\mathrm{mL}{\mathrm{g}}^{-1}\mathrm{d}{\mathrm{m}}^{-1}$, assumed constant over the modeled concentrations. With $l=1\mathrm{dm}$ and $c=0.200\mathrm{g}\mathrm{m}{\mathrm{L}}^{-1}$,These are chosen model parameters, not measured properties of a named material. Doubling only $l$ or only $c$ doubles $\alpha$ within this stipulated linear model; doubling both multiplies it by four. A real chemical solution must be checked for concentration dependence rather than assuming that $0.200\mathrm{g}\mathrm{m}{\mathrm{L}}^{-1}$ is dilute enough.

For an additive mixture containing only this enantiomer pair as solutes, put $k=[\alpha {]}_{+}$ and write $c_{+},c_{-}$ for their mass concentrations. ThenBecause enantiomers have the same molar mass, their relative mole fractions within the pair also equal their relative mass fractions: $c_{+}=x_{+}c$, $c_{-}=x_{-}c$. Thus $\alpha =klc(x_{+}-x_{-})$, which connects optical rotation to enantiomeric excess.

Steven Chu's Bell Laboratories team prepared sodium atoms by pulsed-laser evaporation of a pellet. A frequency-swept opposing beam slowed them before six crossed laser beams supplied optical molasses. The vacuum chamber used a shield coated with graphite and cooled with liquid nitrogen as a pump. Fluorescence provided the readout; the initial beams were aligned as closely counterpropagating as possible. ^[ref].

After equilibration, the team switched off the cooling light for a variable interval, then restored it. Faster atoms escaped during the dark interval; slower atoms were recaptured. The recaptured fraction versus delay constrained the velocity distribution and hence temperature. The inference depended on the initial cloud shape and recapture volume: Steven Chu later acknowledged adjusting a geometry correction towards the expected Doppler cooling limit. Delay was the scanned variable, and the cooling beams themselves supplied the release/recapture control; visible fluorescence alone could not establish temperature. ^[ref].

Köhler illumination arranges illumination so that the light source is imaged into an aperture plane while the field diaphragm is imaged onto the specimen. This illuminates the observed field without projecting the lamp's structure into the specimen image.

After focusing the specimen, close the field diaphragm, adjust the condenser height until its edge is sharp, center that image and open it just beyond the viewed field. Separately adjust the condenser aperture to set the illumination cone and the balance between image contrast and optical resolution. Confusing the two diaphragms makes brightness changes difficult to interpret.^[ref]

Differential interference contrast microscopy converts small sideways differences of optical path length into visible intensity differences. A polarizer and birefringent prisms produce slightly displaced beams with perpendicular polarizations; after they pass through neighboring specimen positions, the beams are recombined and analyzed so that their relative wave phase controls brightness.^[ref]

For beam displacement $\mathbf{s}$ small compared with the scale of specimen variation, the phase difference is approximately $\mathbf{s}\cdot \nabla \varphi$. Thus a transparent object can appear shaded along the shear direction. Rotating the specimen changes that shading: the relief appearance does not by itself measure three-dimensional shape.

Empty magnification enlarges an image without revealing additional specimen detail. Adequate angular magnification can make existing detail visible to the eye, but it cannot undo the objective's diffraction blur. For equally bright, mutually incoherent optical point sources and an ideal circular optical pupil, a commonly used lateral Rayleigh criterion in object space iswhere $\lambda$ is the vacuum wavelength, $n$ is the specimen-medium refractive index, and $\alpha$ is the admitted half-angle. For this model, $\lambda =550\mathrm{nm}$ and $\mathrm{NA}=0.25$ give $d_{\mathrm{R}}\simeq 1.342\mu \mathrm{m}$; changing only numerical aperture to $0.50$ gives $0.671\mu \mathrm{m}$. Changing only the eyepiece leaves this object-space scale unchanged. Illumination, optical coherence, optical aberrations and specimen visibility matter in real instruments; this criterion is not a universal threshold.^[ref]

The geometric reason is especially simple. If a fixed recorded image is enlarged by a factor $q>0$, a feature separation $s$ and a blur width $b$ become $qs$ and $qb$. Their ratio remainsThis is the lower-panel construction in the diagram. A change to the microscope objective can change numerical aperture and hence the specimen detail captured; a display zoom changes how large that captured detail looks. Microscope scale calibration, diffraction-limited resolution and angular magnification are three complementary properties to report.

Microscope scale calibration compares an image readout with a known specimen length. This documented practical uses an eyepiece graticule, stage micrometer and Köhler illumination. Focus the graticule; place the reference on the stage with the objective-appropriate cover glass. Focus both scales with the chosen microscope objective, then rotate and translate them into parallel alignment. Read widely separated coincidences using consistent line edges. Record the optical configuration and recalibrate for each objective.^[ref]

The stage micrometer is the reference; preserve the optical configuration when substituting the specimen. Line width, focus, edge selection and optical aberrations limit accuracy. Additional recommended checks are realignment and another known interval: agreement tests repeatability and scale consistency, not absolute reference accuracy.^[ref]

Here is an original toy calculation, not the source's calibration example or measured data. Suppose a reference span $L=100\mu \mathrm{m}$ coincides with $N=40$ eyepiece intervals. Then the calibration factor and the length corresponding to $n=18$ specimen intervals areThis works because both readouts use the same image-to-specimen scale: $\ell /L=n/N$. One must count intervals, not tick marks; forty intervals have forty-one bounding ticks. In an ideal linear scale model, if identifying the calibration endpoints introduces a total readout error of $\pm 1$ interval, the inferred factor ranges from $100/41$ to $100/39\mu \mathrm{m}$ per interval. Holding $n=18$ exact gives $43.90\le \ell /\mu \mathrm{m}\le 46.15$. This isolates one contribution to measurement uncertainty; it is not a complete uncertainty budget or a measured confidence interval.

For a recorded real image, image a stage micrometer at the same capture settings to calibrate length per detector pixel. Resizing changes length per pixel; update the scale bar accordingly. Eye-viewing angular magnification alone supplies no digital image scale.^[ref]

An optical microscope whose microscope objective, with the specimen at its designed front focal plane, sends out a parallel bundle of light rays from each specimen point. Different specimen points generally give different bundle directions. A tube lens then forms the finite intermediate microscope image; the microscope eyepiece views it. This parallel region permits suitable optical accessories to be inserted. Finite-conjugate and infinity-corrected components must be matched to their intended system.^[ref]

In the ideal paraxial approximation, an object height $h_o$ at objective focal length $f_o$ gives bundle slope $-h_o/f_o$. A tube lens of focal length $f_t$ maps that slope to $h_i=-f_t{h}_o/f_o$, soThis is a two-assembly result. Substituting infinity into the single-objective transverse magnification $-v_o/u_o$ would not describe the finite recorded image. Infinity correction concerns the region after the objective; relaxed-eye viewing concerns the region after the eyepiece. They are distinct conditions.

An optical microscope whose microscope objective is designed to form a real image at a specified finite distance. The intended image distance and optical aberration corrections belong together; mechanical tube length is a mounting specification, not automatically the image distance measured from a principal plane.^[ref]

Why can a small displacement produce a large image? For a converging thin lens, write $u_o=f_o+ϵ$ with $epsilon>0$. Substitution into the thin lens equation givesThe similar triangles construction supplies the minus sign. The example has $epsilon=2\mathrm{mm}$ and hence $m_o=-5$. Letting $epsilon$ decrease moves the real image farther away; it does not keep the microscope eyepiece in focus at a fixed position. Large transverse magnification therefore requires the correct conjugate geometry, not merely a short focal length.

The real image of the specimen supplied to the microscope eyepiece or a recording surface. In the example above, a bare detector placed at $x=60\mathrm{mm}$ records an image five times the specimen's length, inverted. It does not record a fifty-times image: that number compares viewing angles under the stated reference convention. An optical camera receiving the eyepiece's outgoing light needs its own optical lens to form a further real image, with a recording scale determined by that added image formation.

The viewing optical lens assembly that acts as a magnifier of the intermediate microscope image. With that image at its front focal plane, the outgoing light rays from each point are parallel. In the small-angle approximation, the viewed angle is $h_i/f_e$, versus $h_i/D$ for the same image-sized object at reference distance $D$, giving $M_e=D/f_e$.^[ref]

A finite-distance virtual image requires a different eyepiece conjugate. In the ideal thin lens model, with the eye close to the eyepiece and the virtual image at distance $D$, the angular magnification becomes $1+D/f_e$. State the viewing condition before multiplying by the objective's transverse magnification; the relaxed-eye example above uses $D/f_e$.

The specimen-facing optical lens assembly of an optical microscope. A microscope objective collects light from the specimen and sets the object-side numerical aperture. In a finite-conjugate microscope it forms the intermediate microscope image directly; in an infinity-corrected microscope a tube lens completes that image formation. Its label alone does not calibrate a specimen length.

The optical material surrounding the optical fiber core. In a conventional step-index optical fiber, its lower refractive index permits total internal reflection; it is distinct from the protective coating.

The central guiding region of a conventional optical fiber; in a step-index optical fiber, its refractive index exceeds that of the surrounding optical fiber cladding.

Optical fiber attenuation is the reduction of transmitted optical power along an optical fiber. Define a positive loss in decibels by comparing optical power $P(0)$ already launched into the guide with surviving optical power $P(z)$:The factor ten applies to a power ratio; the factor twenty applies to a wave amplitude ratio only when power is proportional to its square under the same conditions.^[ref]

For a uniform illustrative model with constant loss coefficient ${\alpha }_{\mathrm{dB}}$ in $dB/km$ and distance $z$ in kilometres,For ${\alpha }_{\mathrm{dB}}=0.2dB/km$ over $10\mathrm{km}$, the loss is $2\mathrm{dB}$ andThus about $63.1\%$ of the launched optical power survives and $36.9\%$ is lost. This is an illustrative coefficient, not a property of all optical fibers, and is independent of the earlier illustrative refractive indices.

The exponential decay follows from assuming each equal distance interval transmits the same fraction. For two such intervals, power ratios multiply; their logarithms add, so the losses in decibels add. Two $5\mathrm{km}$ sections each transmit $10^{-0.1}\simeq 0.794328$; together they transmit $(0.794328{)}^2\simeq 0.630957$. The loss curve is straight in decibels and curved on a linear power scale.

Wave absorption and scattering of light can remove guided optical power even in a straight fibre, while bends can let it escape. Entrance losses, joints and connectors contribute additional losses; they belong separately from a uniform distance coefficient. This is why a measured end-to-end power ratio need not equal the straight-fibre model.^[ref]

MIT's Optical Waveguides and Fibers, Laboratory Exercise No. 7 (Fall 2005), describes a teaching apparatus. A cleaned, flat-ended multimode optical fiber receives helium–neon laser light through a microscope objective optical lens; a micropositioner aligns the input. The launch deliberately overfills the fibre's cone. A beam splitter sends a sample to a reference photodetector to track laser fluctuations. The fixed output faces a screen or a photodetector on a rotation stage with angular measurement calibration. Readings every two degrees, divided by simultaneous reference readings, give an angular profile. Repeating with an underfilled launch is a control: too few excited normal modes underestimate numerical aperture. Optional index-matching fluid removes cladding-carried light. Bending the fibre reveals leakage and reduced output on an oscilloscope. The handout supplies no measured trace or absolute detector measurement calibration, and this angular scan does not determine loss per kilometre.^[ref]

A graded-index optical fiber has an optical fiber core whose refractive index varies smoothly with distance from the optical axis. Refraction bends light rays through the varying medium; the straight zigzag of a step-index optical fiber no longer describes their paths. The local launch cone can depend on entry position.^[ref]

A single-mode optical fiber supports only its fundamental spatial guided normal mode at the operating wavelength, with possible distinct states of polarization of an electromagnetic wave. The designation must include wavelength: MIT's fibre laboratory explicitly notes that a fibre sold for single-mode operation can support several normal modes at its helium–neon laser wavelength.^[ref]