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Stereoscopy supplies depth because different cameras see different projections of the same track. A functional reconstruction first matches the same interaction vertex and track segments across views. A calibrated lens and window model maps each measured image coordinate back to a ray inside the liquid. Refraction at the windows must be included. Intersecting the rays, or finding their closest agreement when measurements have errors, estimates three-dimensional coordinates. Several points then determine a trajectory, rather than only a projected arc.

An original ideal-camera model makes the geometry explicit. Take two parallel pinhole cameras with centers $(0,0,0)$ and $(b,0,0)$, focal length $f$, and optical axes along $z$. Use virtual image planes in front of the pinholes so that image coordinates have the same orientation as the scene. A point $(X,Y,Z)$ then givesSubtracting the horizontal image coordinates gives disparity $d=u_L-u_R=fb/Z$, henceFor stipulated $b=0.20\mathrm{m}$, $f=0.050\mathrm{m}$, $u_L=0.010\mathrm{m}$, $u_R=0.005\mathrm{m}$ and $v_L=0.002\mathrm{m}$, the reconstructed point is $(0.40,0.08,2.0)\mathrm{m}$. These are ideal optical inputs, not dimensions of Alvarez's apparatus. This model omits refraction and lens distortion; actual calibration replaces its simple ray geometry. Since $|dZ/dd|=fb/d^2$, the same disparity error gives worse depth precision for distant points.

The Lorentz force then supplies a momentum scale. With magnetic field $B$ uniform and negligible stopping power over a fitted segment, bubble-chamber track readout gives $p_{\perp }=|q|Br$. The local three-dimensional helix tangent supplies the missing direction: if $\alpha$ is its angle to the field, $p=p_{\perp }/\sin \alpha$. In practical units,Here $e$ is the elementary charge and $c$ the speed of light. A short, almost straight arc poorly constrains its radius of curvature; a visually sharp track can therefore have poor momentum precision.

Stereoscopy enables three-dimensional track reconstruction.^[ref] Calibrated camera positions and optical geometry are needed before a photographed arc becomes a physical radius of curvature. Neutral particles do not directly leave the same continuous ionization trail as charged particles; an interaction or decay can instead reveal them through charged secondary tracks. A photon's inferred path between interactions is not a photographed luminous line.

In a uniform magnetic field, with negligible stopping power and elastic scattering over the fitted segment, the transverse projection is a circle. The Lorentz force turns transverse momentum at angular rate $v_{\perp }/r$, givingHere $r$ is measured in the plane perpendicular to the magnetic field. The relation remains valid with relativistic physical momentum; it does not require $p=mv$. Thus curvature determines magnetic rigidity, not mass or kinetic energy alone. For chosen $|q|=e$, $B=1.5\mathrm{T}$ and $r=0.40\mathrm{m}$, $p_{\perp }\simeq 0.180\mathrm{GeV}/c$. This is an invented example, not a fit to a historical photograph. Reconstructing a three-dimensional helix also determines the component parallel to the magnetic field; the transverse radius of curvature alone does not. Determining the sign of electric charge additionally requires the field direction and the direction of travel, as in the charge-sign ambiguity of a curved track.

An optical camera can focus a white object sharply in one colour while leaving coloured fringes around its edges in another. Chromatic aberration is the dependence of image position or image scale on wavelength, caused by optical dispersion in the optical lens materials. Axial chromatic aberration separates the optical foci along the optical axis; lateral chromatic aberration changes the transverse magnification with wavelength. Moving the receiving plane can favour one colour without bringing all colours into focus. ^[ref]; ^[ref].

For ordinary optical glass in the visible range, the blue refractive index exceeds the red refractive index. For a fixed positive thin lens in air, the lensmaker's equation then givesHere $\lambda$ is wavelength, $n$ is refractive index, $\Phi$ is optical power, $f$ is focal length, and $R_a,R_b$ are the signed surface radius of curvature values in the lensmaker's equation. The blue light rays therefore bend more strongly and meet sooner. This ordering is a statement about ordinary visible optical dispersion, not a universal rule for every material and every spectral range. ^[ref].

For a worked example, take a single positive thin lens with reference optical power ${\Phi }_d=5\mathrm{D}$ and Abbe number $V_d=60$, hence reference focal length $f_d=0.2\mathrm{m}$. Its F–C optical power difference is $\Delta \Phi ={\Phi }_F-{\Phi }_C={\Phi }_d/V_d=5/60\mathrm{D}\approx 0.08333\mathrm{D}$, by the achromatic doublet power balance relation below. This difference alone does not locate the reference optical power within the F–C interval. To draw a transparent example, introduce a synthetic colour coordinate $q$, a dimensionless quantity, with red $q=-1/2$, reference $q=0$, and blue $q=+1/2$, and choose the toy optical dispersion modelThe diagram's red and blue endpoints represent the C and F endpoints of this toy model. Its reference point is placed symmetrically in optical power by construction; $q$ is not a measured wavelength scale, and the model is not measured optical glass data. The endpoint focal lengths areThus the axial chromatic aberration in this example separates the endpoint optical foci by about $3.334\mathrm{mm}$. A single receiving plane cannot coincide with both optical foci.

The useful design idea is to cancel changes of optical power while retaining a positive total optical power. An achromatic lens does this with materials having different Abbe numbers. It greatly reduces chromatic aberration, but correcting two chosen wavelengths still leaves the possibility of a secondary spectrum.

A concave spherical mirror reflects at the inside of a spherical surface of radius of curvature $R$. For incident light rays near its optical axis, the law of reflection gives paraxial focal length $f=R/2$. Rays farther from the optical axis have different axial crossing positions, causing spherical aberration; one must specify a criterion to define the best full-aperture focus.

For a twice continuously differentiable regular curve $r(t)$ in Euclidean space, its unsigned curvature measures how quickly its unit tangent vector changes per unit arc length. WriteThe curvature iswhere the norm is the Euclidean norm. Differentiating with respect to arc length removes the effect of traversal speed; differentiating $T$ with respect to $t$ alone generally does not. See Ghomi's university lecture notes, section 1.7.

For a circle of radius $R>0$, use $r(t)=(R\cos t,R\sin t)$. Then $v=R$, $T=(-\sin t,\cos t)$ and $\Vert dT/dt\Vert =1$, giving $\kappa =1/R$. A straight line $r(t)=p+ta$ with $a\ne 0$ has constant unit tangent vector, so $\kappa =0$.

The same circle traversed as $q(u)=(R\cos (3u),R\sin (3u))$ has speed $3R$ and $\Vert dT/du\Vert =3$: its curvature remains $1/R$. By contrast, enlarging the circle to radius $2R$ changes its geometry and gives curvature $1/(2R)$. More generally, a dilation $r\mapsto ar$ with $a>0$ divides curvature by $a$, because it multiplies arc length by $a$ while leaving unit tangent vectors unchanged. Thus curvature has units of inverse length, and a reparametrization changes neither its value at a corresponding point nor the shape.

Differentiating $T\cdot T=1$, using the dot product, shows $T\cdot (dT/ds)=0$: the change is perpendicular to the unit tangent vector. Wherever $\kappa >0$, the principal normal vector $N_p=(dT/ds)/\kappa$ points toward the bend. For a plane curve, the osculating circle at such a point has radius of curvature $\rho =1/\kappa$ and center $r+\rho N_p$. It matches the point, tangent line, and curvature, providing a local circular approximation. At $\kappa =0$ there is no finite osculating circle defined by this formula. See Ghomi's university lecture notes, section 1.13.

This makes curvature useful for comparing bends without confusing them with speed. If $t$ denotes time, differentiating $r^{\prime }=vT$ gives acceleration $r^{\prime \prime }=v^{\prime }T+v^2(dT/ds)$, whose component perpendicular to the unit tangent vector has magnitude $v^2\kappa$. The curvature of a plane curve example below computes the bend and osculating circle of a parabola explicitly.

For smooth surfaces, bending has more than one direction. The principal curvatures $k_1,k_2$ instead define Gaussian curvature $K=k_1{k}_2$ and mean curvature $H=(k_1+k_2)/2$. These are surface quantities, distinct from the curvature of an individual curve. In particular, the circle value $1/R$ must not be confused with the Gaussian curvature $1/R^2$ of a sphere of radius $R$. See Ritter's Oxford lecture notes, section 12.2.

For a twice continuously differentiable regular curve $r(t)=(x(t),y(t))$ in the plane, unsigned curvature is the absolute value of its signed curvature:The nonvanishing velocity vector makes the denominator positive. For a graph of a function $(t,f(t))$, this reduces to $|f^{\prime \prime }(t)|/(1+f^{\prime }(t{)}^2{)}^{3/2}$; omitting the absolute value gives the signed curvature of a plane graph for increasing $t$.

For the parabola $r(t)=(t,t^2)$, the derivatives are $r^{\prime }=(1,2t)$ and $r^{\prime \prime }=(0,2)$. ThereforeThe curvature is greatest at the vertex $t=0$, where $\kappa =2$ and the radius of curvature is $1/2$. At $t=\pm 1/2$ the curvature is $1/\sqrt{2}$, so the parabola bends less sharply despite its larger speed $\sqrt{2}$.

Its unit tangent vector and left unit normal areSince $k>0$, $N_L$ is also its principal normal vector. At $t=0$ the osculating circle consequently has center $(0,0)+(1/2)(0,1)=(0,1/2)$ and equationThe lower branch $y=1/2-\sqrt{1/4-x^2}$ has value $0$, derivative $0$, and second derivative $2$ at $x=0$, matching the parabola $y=x^2$ through second order. Away from the vertex the two curves separate: an osculating circle describes a local bend, not the whole curve.

Vanishing curvature at one point does not imply a straight segment. The graph of a function $y=x^3$ has signed curvature $6x/(1+9x^4{)}^{3/2}$, which vanishes at its inflection point $x=0$ and changes sign across it. By contrast, if curvature vanishes throughout a connected regular interval, $dT/ds=0$ throughout that interval: the constant unit tangent vector forces the image to lie on a straight line.

The lensmaker's equation connects focal length to the refractive index and shape of a thin lens. For a thin lens in air with surfaces shaped as parts of spheres, refractive index $n$, and incident light travelling to the right,Each signed radius of curvature is positive when its centre of curvature lies to the right of that surface, and negative when it lies to the left. A symmetric biconvex thin lens with $n=1.5$, $R_a=1\mathrm{m}$, and $R_b=-1\mathrm{m}$ therefore has $1/f=0.5(1+1){\mathrm{m}}^{-1}=1{\mathrm{m}}^{-1}$, giving $f=1\mathrm{m}$ and optical power $+1\mathrm{D}$. This is an ideal shape example consistent with the thin lens diagrams, not a specification of their finite thickness.

In a uniform surrounding medium of refractive index $n_s$ on both sides, replace $n-1$ by $n/n_s-1$. Thus the same shape reverses its focusing sign if its refractive index is lower than its surroundings: for $n=1.5$ and $n_s=2$, the same radii give $1/f=-0.5{\mathrm{m}}^{-1}$, or $f=-2\mathrm{m}$. Positive focal length therefore cannot be inferred from convex shape in every environment. A proof outline applies Snell's law with the small-angle approximation at each surface, adds their direction changes, and neglects the distance between them. ^[ref].

Put $s=z-z_0$. In the source's complex beam parameter convention,Substitution separates the transverse wave amplitude and phase:Thus the Gaussian beam is circularly symmetric, has a Gaussian function transverse light intensity, and narrows to a beam waist at $z_0$ before spreading by diffraction. Its wavefront is planar at the beam waist, converging before it and diverging after it. The quadratic transverse phase identifies the signed radius of curvature:Define spot size as the wave amplitude $1/e$ radius, equivalently the light intensity $1/e^2$ radius. ThenThe Rayleigh range is where $w=\sqrt{2}{w}_0$; the far-field beam divergence half-angle is $\lambda /(\pi w_0)$. The factor $1/q$ also produces a longitudinal phase shift relative to a plane wave, changing by $\pi$ across the beam waist; its sign here follows the given $e^{i(kz-\omega t)}$ convention. The on-axis light intensity decreases as $1/(s^2+z_R^2)$ while transverse integration leaves the power constant. These are scalar paraxial approximation results.

A distant satellite supplies an approximately collimated incident beam. A concave spherical mirror of radius of curvature $R_c=1.5\mathrm{m}$ has paraxial focal length $f_m=R_c/2$. Put the receiving feed at the principal focus, on the optical axis in front of the dish:The radius of curvature relation is a paraxial approximation. A practical dish is usually a parabolic reflector, which focuses an on-axis collimated beam without spherical aberration; a literal spherical dish has spherical aberration, so an exact optimum over its full aperture would need a specified focusing criterion.

For a slender elastic beam undergoing small bending, assume plane cross-sections remain plane and normal to the neutral axis. Fibres a signed distance $z$ from that axis have longitudinal strain ${\epsilon }_{xx}=-z/R$, where $R$ is the local radius of curvature. Hooke's law gives stress ${\sigma }_{xx}=-Ez/R$, where $E$ is Young's modulus. Zero resultant axial force places the neutral axis through the section's centroid for a homogeneous beam.

The bending moment is the moment of this stress distribution. Its magnitude isThe second moment of area $I_A$ has units ${\mathrm{m}}^4$, and the bending stiffness $E{I}_A$ has units $\mathrm{N}{\mathrm{m}}^2$. A rectangular section of width $b$ and depth $h$ has $I_A=b{h}^3/12$, so placing more material far from the neutral axis strongly resists bending. This is the purpose of an I-section.

For small deflection $y(x)$, the signed curvature is approximately $y^{\prime \prime }$. A consistent variational sign convention follows from the elastic energy and a transverse load $q(x)$:For constant bending stiffness this is the Euler-Bernoulli beam equation $E{I}_A{y}^{\prime \prime \prime \prime }=q$. A clamped end fixes $y=y^{\prime }=0$; a simply supported end has $y=0$ and zero bending moment; an unloaded free end has zero moment and shear, $y^{\prime \prime }=y^{\prime \prime \prime }=0$ for uniform $E{I}_A$. Endpoint loads replace the corresponding zero-force boundary conditions.

For a cantilever of length $L$ carrying a tip force $F$, choose positive deflection along $F$. The internal bending gives $E{I}_A{y}^{\prime \prime }=F(L-x)$, with $y(0)=y^{\prime }(0)=0$. Two integrations giveThis approximation requires small slopes, elastic strains and a length large compared with section dimensions. Shear deformation, rotary inertia, plastic yielding and large rotations require extensions. Adding inertia gives flexural waves with ${\omega }^2=(E{I}_A/{\rho }_m{A}_s)k^4$ for a uniform beam, where ${\rho }_m{A}_s$ is mass per unit length.

Across a uniformly illuminated slit, the phase difference between its edges is $2\pi a\sin \theta /\lambda$. Integrating the complex amplitude across the slit gives the single-slit diffraction factorThus $\beta =\pi a/\lambda$ in the small-angle amplitude expression. The first noncentral zero has $q=\pi$, so ${\theta }_0\simeq \lambda /a$. At the stated approximate subsidiary maximum,The exact first side maximum solves $\tan q=q$, giving $q\simeq 4.493$ and intensity ratio $0.0472$; the requested approximation need not be mistaken for the exact maximum.

A virtual object is a point toward which incident rays are already converging and would meet if the lens were removed. With real-is-positive object/image distances it has $u<0$. The thin-lens equation gives $1/v=1/f-1/u$; as $u\to -\infty$, $v\to f$, so a distant virtual object images in the rear focal plane.

Without the lens, $y\simeq L\theta$ puts the central maximum at zero and the first dark fringes at $y=\pm L{\lambda }_0/a=\pm 0.100\mathrm{m}$. Subsequent zeros are integer multiples, and weaker subsidiary maxima lie between them.

For the finite-distance pattern compression, take the old screen as the prospective image plane at $L=6\mathrm{m}$, with $u=-L$, and the new screen at $d=0.10\mathrm{m}$. ThenThe plane surface has infinite radius of curvature. The transverse scale is $d/L=1/60$, yieldingThe finite-distance scaling can also be checked without treating an intensity pattern as a collection of geometrical point objects: the slit-to-screen Fresnel integral after a thin lens has quadratic coefficient $1/d-1/f=1/L$ and linear term $-y/d$. It therefore reproduces the old diffraction pattern at transverse coordinate $Y=Ly/d$, up to an overall amplitude and phase. If the old far-field plane is treated as infinitely distant, the equally useful leading approximation is $f=d$, $R=6.0\mathrm{cm}$; the finite-distance correction is only $1.7\%$.

The colour discussion below assumes a smooth white incident spectrum and neglects lens dispersion when comparing the ideal diffraction patterns.

A thin lens models an optical lens by collapsing its refracting surfaces into one plane, neglecting thickness relative to the relevant radius of curvature values and object and image distances. Here use an ideal converging thin lens in air and the paraxial approximation: light rays make small angles with the optical axis, so Snell's law can be linearized using the small-angle approximation. Equal surrounding refractive indices on both sides give the familiar symmetric equation.^[ref]

Put the thin lens at $x=0$, with incident light travelling to the right. An object at $x=-u$ has $u>0$. The image coordinate is $x=v$: $v>0$ for a real image and $v<0$ for a virtual image. The converging thin lens has focal length $f>0$, and heights $h_o,h_i$ are signed relative to the optical axis. The thin lens equation and transverse magnification areThe expression for $v$ requires $u\ne f$. Negative transverse magnification means inverted; positive transverse magnification means upright. Remember that the positive distance $u$ differs in sign from the object's coordinate $-u$. ^[ref].

For focal length $f=1\mathrm{m}$, object distance $u=3\mathrm{m}$, and object height $h_o=0.1\mathrm{m}$, calculation givesThe real image is inverted, half as tall, and located $1.5\mathrm{m}$ to the right of the thin lens. A screen at that position receives the image because the outgoing light rays actually meet there.

For the same focal length $f=1\mathrm{m}$ and object height $h_o=0.1\mathrm{m}$, move the object inside the front optical focus, to $u=0.5\mathrm{m}$. ThenThe virtual image is upright, twice as tall, and apparently $1\mathrm{m}$ to the left of the thin lens. Outgoing light rays still travel to the right; only their backward extensions intersect at the virtual image. A screen there cannot collect that image from the outgoing bundle. This geometry explains why a converging optical lens can act as a magnifier when held close to an object. ^[ref].

Both diagrams use two rules for ray tracing. A light ray from the object's tip parallel to the optical axis reaches the thin lens at $(0,h_o)$ and exits along the straight line through the rear optical focus $(f,0)$. Another light ray through the centre $(0,0)$ continues undeviated in the ideal thin lens model. Their intersection, or the intersection of their backward extensions, locates the image tip. ^[ref].

A short geometric proof outline makes the signs transparent. The two outgoing straight lines areFor nonzero $h_o$, equating their heights at $x=v$ gives $1-v/f=-v/u$, hence the thin lens equation; the centre light ray gives $h_i/h_o=-v/u$, hence the transverse magnification. This derives the image equations from the ideal ray tracing rules, whose physical justification comes from Snell's law in the paraxial approximation.

At $u=f$, the two outgoing straight lines are parallel: there is no finite image position. For $u>f$ the real image has $m<0$; for $0<u<f$ the virtual image has $m>0$. These cases let an optical camera designer choose a receiving position and let a user understand a magnifier. An actual optical lens has finite apertures, which cause diffraction and impose diffraction-limited resolution; optical aberrations can further blur the image. The diagrams are ideal geometrical optics, with exact coordinates within that model, rather than exact predictions of Maxwell's equations. ^[ref].