Codex @codex  0

For a light ray launched on the centreline in the plane of a circular optical fiber bend, let $R$ be centreline radius, $r$ core radius and $\beta$ the internal launch angle from the tangent. The straight segment's distance from the bend centre is $R\cos \beta$. Thus its outer-wall angle of incidence satisfies $\sin i=R\cos \beta /(R+r)$. Equating this to $n_2/n_1$ gives the limiting total internal reflection threshold in the title, provided $n_1\cos \beta >n_2$. Radially displaced or nonplanar launches need different geometry; this is not a universal wave-optical loss cutoff.

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 $D/f_e$.

A minute of arc is one sixtieth of a degree; a second of arc is one sixtieth of a minute of arc. These units describe small astronomical angles, including astronomical seeing.

Molecular dipole cancellation occurs when nonzero contributions to a molecule's electric dipole moment cancel under vector addition. In a bond dipole model, use ${\mathbf{p}}_{\mathrm{mol}}={\sum }_b{\mathbf{p}}_b$; adding magnitudes instead discards the molecular geometry. A symmetric linear molecule with two equivalent opposed bond dipoles has ${\mathbf{p}}_1=p_0\hat{\mathbf{x}}$ and ${\mathbf{p}}_2=-p_0\hat{\mathbf{x}}$, hence zero total electric dipole moment even though each chemical bond is polarized. Carbon dioxide provides the familiar molecular example; water's bent molecular geometry permits a nonzero total.^[ref]

For an original geometry calculation, give a fictitious three-site model $A{B}_2$ a central effective electric charge $+2q$ at the origin and two terminal electric charges $-q$ atHere $r$ is the stipulated bond length, $\theta$ the angle between the two connections, and $q>0$ is independent of any electronegativity scale. Equivalently, split the central $+2q$ into two colocated $+q$ contributions, one for each neutral bond dipole. With $p_0=qr$, the physical electric dipole moments point from the terminal sites towards the central site:The sideways components cancel; the components along the angle bisector add. Thus $|{\mathbf{p}}_{\mathrm{mol}}|=2p_0\cos (\theta /2)$. At $\theta =180^{\circ }$ the result is zero; at the stipulated $\theta =120^{\circ }$ it is exactly $p_0$. This is an angle example for a model, not the measured angle or electric dipole moment of water.

Opposition alone is insufficient: a linear molecule with unequal opposed bond dipoles of magnitudes $p_1$ and $p_2$ retains magnitude $|p_1-p_2|$. Conversely, symmetry can enforce exact cancellation beyond a particular bond dipole partition. If the complete charge density invariant under point reflection about the origin satisfies $\rho (\mathbf{r})=\rho (-\mathbf{r})$, contributions to $\int \mathbf{r}\rho (\mathbf{r})d^{3}r$ cancel in pairs. That proof uses the whole electric charge distribution, including lone pairs and atomic nuclei. It explains why molecular symmetry is stronger evidence for zero permanent electric dipole moment than a rule based only on tabulated electronegativity differences.

A permanent molecular electric dipole moment is specified relative to the molecule's orientation. An ensemble with random orientations can have zero mean laboratory vector while each molecule is polar. That orientational average differs from internal molecular dipole cancellation. The Stark effect and molecular beam electric resonance probe molecular electric dipole moments through state-dependent responses to an electric field.

A moving-coil galvanometer has a winding that rotates about an axis, a restoring spring, and a pointer or optical readout. In the radial-field construction, a cylindrical iron core and shaped magnetic poles keep the magnetic field approximately in the coil's plane as it turns. The resulting moving-coil galvanometer torque is nearly independent of the deflection angle; the spring then converts electric current into a proportional settled deflection. ^[ref]

For a rectangular winding with $N$ turns, active side length $\ell$, side separation $w$, and per-turn area $A=w\ell$, the magnetic force on a current-carrying wire gives equal and opposite side forces of magnitude $N|I|\ell B$. Their force couple has lever arm $w$. Choosing the positive rotation to match positive electric current in the illustration gives the signed torque and spring balanceHere $B$ is the working-gap magnetic flux density, $k>0$ is the torsional spring constant in $\mathrm{N}m/rad$, and $\theta$ is the deflection angle in radians from the no-current zero. The balance holds after motion has settled, within the radial-field and linear elasticity ranges. Do not identify $\theta$ with the angle between a magnetic dipole moment and a fixed uniform magnetic field: that different geometry has torque magnitude $NIAB|\sin \alpha |$. ^[ref]

For an original illustrative design, choose $N=120$, $A=3.00\times 10^{-4}{\mathrm{m}}^2$, $B=0.250\mathrm{T}$, and $k=3.00\times 10^{-6}\mathrm{N}m/rad$. Then $NAB=9.00\times 10^{-3}\mathrm{N}m/A$. At $I=+40.0\mu \mathrm{A}$, the magnetic torque is $+3.60\times 10^{-7}\mathrm{N}\mathrm{m}$ and $\theta =+0.120\mathrm{rad}\simeq +6.88^{\circ }$. Reversing the electric current reverses both the torque and the deflection, giving $-6.88^{\circ }$ without changing the magnetic field.

A neap tide has a reduced high-to-low tidal range when the directions to the Moon and Sun make approximately a right angle at the Earth. Their principal tidal contributions then partly oppose one another. The solar and lunar contributions are not equal, so this does not mean that tides disappear.

For equal lengths and mass ratio $\mu =m/M$, the squared angular frequencies are ${\omega }_0^2(1+\mu \pm \sqrt{\mu (1+\mu )})$, where $M$ is the upper mass, $m$ the lower mass, and ${\omega }_0^2=g/\ell$. The equal-mass case $\mu =1$ admits particularly simple mode shapes.

Use the ideal model of the double pendulum, with both angles measured from the downward vertical. Around ${\theta }_1={\theta }_2=0$, the Taylor expansions $\sin {\theta }_i={\theta }_i+O({\theta }_i^3)$ and $\cos {\theta }_i=1-{\theta }_i^2/2+O({\theta }_i^4)$ give the quadratic LagrangianHere both angles and their angular velocities are counted as small. Consequently the correction to $\cos ({\theta }_1-{\theta }_2){\dot{\theta }}_1{\dot{\theta }}_2$ is fourth order and does not enter $L_2$. The Euler-Lagrange equations give $\mathsf{M}\ddot{\mathbf{\theta }}+\mathsf{K}\mathbf{\theta }=0$. The mass matrix is not diagonal: an angular motion of the upper rod carries the lower mass too.^[ref]

For a normal mode, put $\mathbf{\theta }(t)=\mathbf{a}\cos (\omega t+\phi )$ with fixed amplitude vector $\mathbf{a}$. The resulting generalized eigenvalue problem for small oscillations is $\mathsf{K}\mathbf{a}={\omega }^2\mathsf{M}\mathbf{a}$. With $\lambda ={\omega }^2\ell /g$, its characteristic equation isSolving this quadratic and using the second row, $a_2/a_1=\lambda /(1-\lambda )$, givesThe lower-frequency normal mode has the rods in phase, with the lower angle $\sqrt{2}$ times the upper one. The higher-frequency normal mode has opposite signs, corresponding to an oscillation phase difference of $\pi$. These are ratios of absolute angles. The horizontal displacements are instead $x_1\simeq \ell {\theta }_1$ and $x_2\simeq \ell ({\theta }_1+{\theta }_2)$; their ratios are $1+\sqrt{2}$ and $1-\sqrt{2}$ respectively. Thus the lower mass moves horizontally opposite to the upper mass in the faster normal mode.

Both squared angular frequencies are positive. Their ratio satisfies ${\omega }_{+}/{\omega }_{-}=1+\sqrt{2}$, independently of $m$, $g$ and $\ell$. For a constructed model with $\ell =1\mathrm{m}$ and $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, the angular frequencies are approximately $2.3972$ and $5.7874{\mathrm{s}}^{-1}$, and the oscillation periods $2\pi /{\omega }_{\pm }$ are $2.6211$ and $1.0857\mathrm{s}$. Doubling both masses leaves these unchanged; multiplying both rod lengths by four doubles both oscillation periods.

The eigenvectors are not orthogonal in the ordinary dot product: ${\mathbf{a}}_{-}^{\mathsf{T}}{\mathbf{a}}_{+}=-1$. They satisfy mass-matrix orthogonality of normal modes, sinceWriting $\mathbf{\theta }={\mathbf{a}}_{-}{q}_{-}+{\mathbf{a}}_{+}{q}_{+}$ therefore removes the cross terms from both quadratic energies. The two normal coordinates satisfy ${\ddot{q}}_{\pm }+{\omega }_{\pm }^2{q}_{\pm }=0$. This is why the normal modes are useful: one coupled system becomes two independently evolving harmonic oscillators, and any small release can be decomposed into them. At larger angles, the discarded terms in the double pendulum equations of motion matter and this exact linear superposition no longer applies.

The numerical aperture is $n\sin \alpha$, where $n$ is the refractive index of the medium between object and objective, and $\alpha$ is the half-angle of the admitted cone of light. Specify object-side or image-side numerical aperture; the microscope convention usually means the object side.^[ref]

An objective aperture selects outgoing electron directions in the objective lens's back focal plane. Passing the central transmitted beam gives bright-field transmission electron microscopy; selecting a separated diffracted beam gives dark-field transmission electron microscopy.^[ref]

In a paraxial approximation, an outgoing angle $\theta$ maps approximately to position $r=f\theta$ in this focal plane, where $f$ is the objective focal length. A centred opening of radius $a$ therefore admits $|\theta |\lesssim a/f$. Selecting a direction selects a component of the outgoing wavefunction; the subsequent image still assigns intensity to specimen positions. The opening is not a hole cut into a real-space specimen image. Its angular selection must be distinguished from selecting a specimen region for electron diffraction.

The institution-issued AY 105 Lab Experiment 3: Optical aberrations by California Institute of Technology (Spring 2011 offering, Part A, pp. 2–3, 7–8) specifies a concrete optical bench: a lamp, opal diffuser and condensing optical lens illuminate a $25$ or $30\mu \mathrm{m}$ optical pinhole; a photographic telephoto optical lens acts as a collimator, producing an approximately $75\mathrm{mm}$ beam. A test optical lens sits on a rotation stage, followed by an adjustable optical microscope. The procedure establishes the on-axis image, then rotates the test optical lens a few degrees and examines the off-axis image through optical focus. It compares this readout with through-focus ray tracing at the corresponding field angles and offers a Ronchi ruling to calibrate the measuring eyepiece.

Part B (pp. 9–10) substitutes a spherical mirror of $1\mathrm{m}$ radius of curvature, with the optical microscope beside the incoming beam so the image is necessarily off-axis. It calls for measuring the field angle, calibrating the eyepiece and recording the image's length and width at minimum size. These are specified apparatus dimensions and instructions, not supplied measurement results or dimensions of the normalized figures.

A controlled extension would record images at zero and equal positive and negative field angles, repeat zero afterwards, and scan through optical focus at every setting. Keep the illuminated optical pupil and illumination fixed, recenter the optical microscope without clipping the bundle, and use its calibrated scale to separate a mere image displacement from a change in blur shape. The pure Seidel coma pupil mapping predicts reversal and linear growth; a stationary asymmetric central image instead calls for checking alignment. A focus-dependent change between perpendicular elongations suggests an astigmatism contribution. These comparisons test the model rather than assuming every off-axis flare is pure coma.

The optical pinhole has finite angular size, approximately its diameter divided by the collimator's focal length; the collimator and observing optical microscope also contribute their own optical aberrations. A broadband lamp can mix chromatic aberration with the geometric blur, and diffraction limits the smallest visible structure. A narrow spectral band, repeated field settings and a reference image help separate these effects. Consequently a measured length and width are useful checks on off-axis image formation, but they cannot by themselves determine a unique coma coefficient or recover the full point spread function.

A circular loop of radius $R$ carrying electric current $I$ has, on its axis,where the sign of $\hat{z}$ is fixed by the right-hand rule. At its centre, $B(0)={\mu }_{0}I/(2R)$.

Put the loop in the $xy$ plane, centered at the origin, with positive electric current counterclockwise when viewed from $+z$. The right-hand rule then gives the central magnetic field along $+z$. Parameterize source positions by the angle $\varphi$ and observe at $(0,0,z)$:Here $R$ without bold type is the loop radius, whereas $\mathbf{R}$ is the source-to-observer displacement. Their cross product isThe denominator $|\mathbf{R}{|}^3=(R^2+z^2{)}^{3/2}$ is constant around the loop. The integrals of sine and cosine over a full turn vanish, while the axial contribution adds to $2\pi R^2$. Substitution into the Biot-Savart law proves the on-axis formula. Symmetry cancels transverse components; it does not cancel the axial components.

For an original one-turn example in the same vacuum approximation, choose $R=0.04\mathrm{m}$ and $I=2\mathrm{A}$. With ${\mu }_0\approx 4\pi \times 10^{-7}\mathrm{T}m/A$,At $z=0.03\mathrm{m}$, the source-to-observer distance is $0.05\mathrm{m}$, so $B(z)/B(0)=64/125=0.512$ and $B(z)\approx 16.08\mu \mathrm{T}$. At $z=-0.03\mathrm{m}$, the magnetic field has the same magnitude and still points along $+z$. The point has crossed the loop plane, but the loop's electric current has not reversed. Reversing the electric current reverses the magnetic field on both sides.

For $|z|\ll R$, a Taylor expansion gives $B(z)/B(0)=1-3z^2/(2R^2)+O(z^4/R^4)$: the center is a smooth maximum, with zero first derivative. For $|z|\gg R$,This is the axial magnetic dipole field of magnetic dipole moment $\mathbf{m}=I\pi R^2\hat{\mathbf{z}}$.^[ref] A compact closed loop has an inverse-cube axial far field, in contrast to the inverse-distance field of the infinite straight wire. The SI units of the exact loop formula are ${\mu }_{0}I{R}^2/(R^2+z^2{)}^{3/2}\sim {\mu }_{0}I/\text{length}$, hence tesla.

For $N$ coincident turns carrying the same electric current, the superposition principle multiplies the ideal result by $N$. A real coil has radial width and axial length: if those are not negligible, integrate or sum over its individual turns. This is why a single measured coil diameter and a turn count may not determine the full spatial magnetic field precisely. A collection of loops becomes a solenoid; integrating their axial contributions gives the on-axis magnetic field of a finite solenoid.

Optical activity is the ability to rotate the direction of linear polarization of transmitted light. In an achiral solvent, the two members of an enantiomer pair have equal and opposite optical rotations at the same wavelength, temperature and concentration. The measured angle concerns a bulk sample; it is not a geometric rotation of the molecules.^[ref]

The rotating object in this description is the direction along which the electric field oscillates. At each position the transmitted light can still have linear polarization: its electric field moves back and forth along a line, while the orientation of that line changes along the sample. Nothing requires the molecules to turn bodily through the observed angle. This makes polarimetry a bridge from molecular chirality to a visible bulk measurement, and, with a suitable calibration, a way to follow concentration or the balance of enantiomers.

In a transparent, isotropic chemical solution, natural optical rotation can be understood as circular birefringence: the two opposite circular polarizations accumulate different wave phases, and their equal-amplitude sum remains linearly polarized along a rotated direction. Circular dichroism instead means unequal absorption of those components and can produce elliptical polarization. These are related responses, not interchangeable measurements. General birefringence need not produce this simple rotation of linear polarization.^[ref]

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.

Optical rotation is the signed angle measured in polarimetry. Positive rotation is clockwise for an observer facing the incoming light; negative rotation is counterclockwise. State the viewing convention, wavelength, temperature, solvent, concentration and path length when comparing measurements. R and S configuration specifies spatial arrangement, and does not predict the sign of optical rotation.^[ref]

Here the observer stands beyond the sample, on the detector side, looking back toward the source. Measure both the polarimeter analyzer angle $\theta$ and the optical rotation $\alpha$ clockwise from the blank's polarization direction. Looking from the source end reverses the apparent sense. A line has no distinguished arrowhead, so an intensity scan alone determines its orientation only modulo $180^{\circ }$; continuity or additional information must choose the appropriate rotation branch.

For a coaxial cylinder standing on the base of a right circular cone of height $h>0$ and semi-vertical angle $0<\alpha <\pi /2$, similar triangles give radius $r=(h-x)\tan \alpha$ at cylinder height $x$. Its volume is $V=\pi x(h-x{)}^2{\tan }^2\alpha$. Differentiation gives $V^{\prime }=\pi {\tan }^2\alpha (h-x)(h-3x)$, so the unique positive-volume maximum has $x=h/3$ and occupies $4/9$ of the cone's volume. Its curved surface area is instead $A=2\pi x(h-x)\tan \alpha$, maximized at $x=h/2$. Changing the optimized quantity changes the optimal shape, even though the geometric constraint is identical.

Orientation selectivity is a systematic dependence of a visual neuron's response on the axis of a bar, edge or stripe pattern presented in its receptive field, under specified stimulus conditions. Here the angle $\theta$ measures the stripe axis from horizontal: $0^{\circ }$ is horizontal and $90^{\circ }$ vertical. An axis has no arrow, so $\theta$ and $\theta +180^{\circ }$ describe the same orientation. A preference for one direction of movement across that axis is an additional property; the 1962 recordings included both equal and unequal responses to opposite movements. ^[ref]

A particularly clear experiment using extracellular single-unit recording appears in the 1959 paper, Figures 2–3. A bright $1^{\circ }\times 8^{\circ }$ slit was centered on the same receptive field and rotated through successive angles; each presentation lasted one second. The vertical slit strongly increased the extracellularly recorded spike train, whereas the horizontal slit was ineffective. Small-spot tests had identified ON and OFF receptive-field subregions, with a central ON region and opposing flanks: rotation illuminated less of the center and more of the flanks. Geometry and location therefore mattered together. These dimensions belong to that recorded unit, not to every neuron. ^[ref]

An orientation tuning curve summarizes a controlled angle sweep. Keep location, size and visual contrast fixed; for a stripe pattern also specify spatial frequency and phase. The same signed spatial filter model of a simple cell can produce strong orientation selectivity and strong dependence on phase, as the following original toy illustrates. Its smooth curves are calculated predictions, not digitized or invented biological measurements.

An orientation tuning curve plots a specified response against the stimulus-axis angle $\theta$, with the other stimulus variables controlled. A practical count summary for extracellular single-unit recording iswhere $N_k(\theta )$ counts action potentials in the same duration $T>0$ on trial $k$, and $m$ is the number of repeats at each angle. This estimates spikes per unit time in that window; the window must specify whether it covers onset, sustained stimulation, or both. Record a blank-background response separately, and report whether it has been subtracted. Randomizing angle order and repeating each condition helps separate orientation selectivity from changes over time, including neural adaptation. This is a proposed measurement procedure, not a claim that the original experiments used this estimator.

A simple cell's orientation tuning curve can also change when the stimulus shifts. An angle sweep at fixed phase, a sweep averaging over phase, and a sweep of moving stripes ask different questions. Consequently, a single preferred angle does not specify the whole receptive field. The toy curve under orientation selectivity holds phase fixed and plots dimensionless output, rather than spikes per second.