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Angular magnification is the ratio of the signed viewing angle with an optical instrument to the signed viewing angle in a stated reference arrangement. Measure both as apparent directions from the eye, positive toward the same side of the optical axis; then an inverted view has negative $M_{\theta }$. If only positive angular sizes are compared, the result is $|M_{\theta }|$ and carries no orientation information. ^[ref].

For the upright virtual image of a converging thin lens, let the eye be at $x=e\ge 0$ and take the unaided reference to be the same small object at a distance $D$ from the eye. Within the small-angle approximation,Specify $D$: keeping the object in its original position means $D=e+u$, whereas the customary magnifier rating uses a near-point reference, often $D=25\mathrm{cm}$. A reference distance is a convention, not a claim that every eye has that near point. ^[ref].

For the virtual image diagram, $m=4$, $v=-30\mathrm{cm}$ and $e=5\mathrm{cm}$. With $D=25\mathrm{cm}$, $M_{\theta }\simeq 20/7$, whereas keeping the object at $D=e+u=12.5\mathrm{cm}$ gives $M_{\theta }\simeq 10/7$. Thus neither viewing comparison gives the transverse magnification four. A person may also be unable to focus on the unaided object at the shorter reference distance.

At $u=f$, each outgoing bundle is parallel with slope $-h_o/f$; its apparent viewing direction has angle ${\theta }_i\simeq h_o/f$. Consequently $M_{\theta }\simeq D/f$ remains finite even though $m$ is undefined at a finite image plane. Equivalently, in the limit $u\to f^{-}$, the factors $m$ and $e-v$ both grow without bound but their ratio tends to $1/f$. For $f=10\mathrm{cm}$ and $D=25\mathrm{cm}$, the relaxed-view prediction is $M_{\theta }\simeq 2.5$. ^[ref].

A submerged object looks shallower because its emerging light rays bend away from the surface normal vector. Their backward extensions give an approximate virtual image nearer the surface. For an object at perpendicular depth $d$ in a medium of refractive index $n_{\mathrm{obj}}$, viewed through a flat boundary from a medium of refractive index $n_{\mathrm{obs}}$, the near-normal paraxial approximation gives $d_{\mathrm{par}}=(n_{\mathrm{obs}}/n_{\mathrm{obj}})d$. Thus the model water depth $12\mathrm{cm}$ appears approximately $9\mathrm{cm}$ from air. This is a bundle-of-light rays statement, not a deflection of the axial light ray.^[ref]

For general transmitted light rays, specify what “depth” means. Put the object at $(0,-d)$ below a horizontal boundary $z=0$. A light ray reaches the boundary at $(x,0)$, with object-side angle ${\theta }_{\mathrm{obj}}$ and observer-side angle ${\theta }_{\mathrm{obs}}$, both measured from the upward normal vector. For $x>0$, its backward extension meets the object's normal line $x=0$ at $z=-d_{\mathrm{axis}}$, whereThis axis-intercept depth is exact for that one transmitted light ray in the stated model. It tends to $d_{\mathrm{par}}$ as the angles tend to zero; at exactly zero the quotient is interpreted as a limit. These elementary tangent, sine and cosine relations show precisely where the small-angle approximation enters.

The original calculations below use $d=12\mathrm{cm}$, $n_{\mathrm{obj}}=4/3$ and $n_{\mathrm{obs}}=1$:

At appreciable obliquity these depths differ, so the full outgoing bundle has no single exact virtual image on the object's normal line. An oblique eye samples a narrow bundle; its local apparent position also depends on that bundle and cannot be inferred by declaring $d_{\mathrm{axis}}$ to be a universal image depth. A plane refracting boundary gives approximate image formation near normal incidence, with departures at larger angles. The construction assumes a flat surface, uniform transparent media with isotropy, and a propagating transmitted light ray; it does not extend to total internal reflection.

Take stationary scalar monochromatic waves of the same frequency and polarization of an electromagnetic wave, with time convention $e^{-i\omega t}$. Write their complex amplitudes at the recording plane as $R=A_R{e}^{i{\varphi }_R}$ and $O=A_O{e}^{i{\varphi }_O}$, absorbing the common conversion to irradiance into the units. An exposure short enough to avoid fringe motion but long compared with an optical period recordsThe two cross terms are where relative wave phase enters the measured light intensity. Blocking the holographic reference wave leaves $|O{|}^2$ and removes them. Loss of mutual optical coherence, incompatible polarization of an electromagnetic wave or motion that averages the relative wave phase similarly reduces fringe contrast. These are useful controls for distinguishing a hologram from an ordinary brightness record. One local cosine value does not uniquely determine a wave phase; known reference structure and separation of the cross terms supply the additional information in the example below.

Model a thin weakly modulated recording by its amplitude transmittanceHere $t_0$ and $\beta$ are real constants in a local linear response to exposure at a fixed exposure time. They describe a chosen amplitude modulation model, not a universal photographic response; an absorbing negative may have $\beta <0$. Keep $0\le t\le 1$ for this passive real-transmission model. Ignore material thickness, saturation, shrinking, clipping and changes of wavelength. Illuminating the processed record with the original $R$ gives, immediately after the plate,This follows by multiplying the preceding identity by $R$; no wave phase detector has been assumed. If $|R|$ is uniform, the desired term reproduces $O$ up to a constant. A nonuniform $|R{|}^2$ weights the recovered field. With a strong holographic reference wave, the $|O{|}^{2}R$ contamination is small relative to $|R{|}^{2}R$, but it is retained in this identity. Subsequent diffraction propagates each term: reproduction at the plate is the starting boundary field, not yet a focused image.

The schematic uses a plane wave reference and one optical point source to separate the functions. The top row records $R+O$; the bottom removes the object and illuminates the developed hologram with $R$. Dashed extensions identify the desired virtual image; the conjugate component converges toward a real image. All three outgoing components coexist. The coloured paths distinguish terms, not different wavelengths; distances and strengths are schematic. This functional drawing is not a reconstruction of the historical apparatus.

Image formation assigns each object point a corresponding image point: outgoing light rays from that object point either converge there, giving a real image, or diverge as though they originated there, giving a virtual image. The requirement applies separately to each point. Bringing unrelated light rays together makes a bright patch, but does not preserve the object's pattern. This point-by-point correspondence explains how an optical camera records a scene. ^[ref].

Use the same convention as thin lens: place the optical lens at $x=0$, with incident light travelling to the right and a real object at $x=-u$, $u>0$. Write the image coordinate as $x=v$, so $v>0$ for a real image and $v<0$ for a virtual image. Heights $h_o,h_i$ are positive above the optical axis and negative below it. In air, within the paraxial approximation, a thin lens obeysHere $f$ is the signed focal length, positive for a converging optical lens and negative for a diverging optical lens; the expression for $v$ assumes $u\ne f$. Object distance $u$ is positive even though the object's coordinate is negative. These coordinate rules describe transmission through the optical lens; a mirror reverses the outgoing direction, so do not identify its image distance with this same rightward coordinate. ^[ref].

The beauty of geometrical optics is that two convenient light rays can locate a point, while a whole admitted bundle forms its image. For a converging thin lens, an incident light ray parallel to the optical axis exits through the rear optical focus; a light ray through the centre continues undeviated. Real optical lens elements do not focus every admitted light ray exactly: optical aberrations and diffraction limit the point correspondence. The diagrams below are original ideal-model constructions, not measured data. ^[ref].

Magnification compares an image's scale with a specified reference. Transverse magnification compares signed lengths in object and image planes; angular magnification compares viewing angles at an observer. Both are dimensionless quantities, but they answer different questions. A projected real image twice as tall has $|m|=2$; an optical telescope's advertised viewing enlargement normally concerns angular magnification. ^[ref].

Always specify the sign and the reference arrangement. For signed transverse magnification, $m<0$ means inverted and $m>0$ means upright; $|m|>1$ means enlarged and $0<|m|<1$ means reduced. A larger virtual image can subtend a modest viewing angle if it is farther away. At $u=f$, an ideal converging thin lens has no finite image height and hence no finite transverse magnification, while angular magnification can remain finite. Enlargement also does not guarantee extra detail: diffraction-limited resolution and optical aberrations limit the information available to enlarge. ^[ref].

A magnifying glass uses a converging optical lens to present an upright enlarged virtual image. For focal length $f$ and eye near point distance $D$, the near-eye angular magnification is $D/f$ with a relaxed eye, or $1+D/f$ with the virtual image at the near point. These formulas assume negligible eyepiece-to-eye distance. With eye-to-lens distance $e$, object distance $u$, and the virtual image at the eye's near point, the angular magnification is $(D-e)/u$.

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 $D/f_e$.

Parallax localization of a virtual image matches the apparent depth of a virtual image to a directly viewed physical marker. A sideways change of viewpoint changes their relative apparent positions if their depths differ. Adjust the marker until that relative motion vanishes, and use its coordinate to estimate image position. The marker must be viewed directly; viewing it through the same optical lens compares two images rather than an image and a known physical depth. ^[ref].

For two nearly aligned points at distances $L_i,L_m$ from an eye displaced sideways by $\delta$, the small-angle approximation gives a change of relative viewing angle proportional to $\delta (1/L_i-1/L_m)$. It vanishes when $L_i=L_m$, and changes sign across that match. When both distances are large, the difference is small, explaining the poor depth precision near the thin lens condition $u=f$.

A real image is a point where the actual outgoing rays converge; a screen placed there can intercept the focused light. A virtual image is a point from which the outgoing rays appear to diverge: only their backward extensions intersect, so a screen at that apparent point does not receive a focused real image from the same rays.

For a converging thin lens with $f>0$ and a real object a distance $u>0$ away, the lens relation gives $v=fu/(u-f)$. If $u>f$, then $v>0$: the image is real, on the opposite side, and inverted. If $0<u<f$, then $v<0$: the image is virtual, on the same side as the object, upright and enlarged, with $m=-v/u=f/(f-u)>1$.

At $u=f$, the emergent rays are parallel and the image is at infinity; there is no finite focused image. For an object very far away, $v\to f$ from above. These are conditions for a real object in the stated lens convention, rather than claims about all systems with virtual incident objects.

The requested source-image relation is the thin lens equation, using positive real-object distance $u$ and signed image distance $v$:Here $f>0$ for a converging lens, and $v<0$ denotes a virtual image. This is usually distinguished from Lensmaker's equation, $1/f=(n-1)(1/R_1-1/R_2)$ for a thin lens in air with the usual signed surface radii. The latter relates focal length to the lens material and curvature, rather than to object and image positions.^[ref]

For a magnifying glass held close to the eye, compare the angle $h/u$ with the unaided near-point angle $h/D_n$, where $D_n=25\mathrm{cm}$. Maximum usable angular magnification occurs with the virtual image at the near point, so $v=-D_n$. ThusThe object lies inside the front focal length. The emerging light rays diverge, while their backward extensions meet at an upright enlarged virtual image on the object's side. With the image at infinity instead, the relaxed-eye angular magnification would be $D_n/f=6.25$, which is smaller. A finite eye-to-lens distance $s$ changes the near-point result to $1+(D_n-s)/f$; the quoted maximum uses the conventional $s\simeq 0$ limit.^[ref]

Use a converging thin lens as a magnifying glass, with the screen at or just inside its front focal plane. It presents an upright virtual image at infinity or at the eye's near point, respectively. Let the near point distance be $D$; the conventional numerical assumption is $D=250\mathrm{mm}$, which is not specified in the paper.

Under the usual approximation that the eye is close to the eyepiece, interpreting $50\mathrm{mm}$ as the focal length $f$ gives the familiar angular magnificationsThe first is for a relaxed eye with the screen at the front focal plane. For the second, the virtual image is at signed distance $v=-D$ and the thin lens equation gives screen distance $u=fD/(D+f)=41.7\mathrm{mm}$. Thus the often intended maximum $6$ requires a screen-to-eye distance of approximately $41.7\mathrm{mm}$, rather than exactly $50\mathrm{mm}$.

If $50\mathrm{mm}$ instead fixes the physical screen-to-eye separation $s$, an eye adjacent to the optical lens sees the angular magnificationThis is achieved by $f=s$ for relaxed viewing, or by $f=sD/(D-s)=62.5\mathrm{mm}$ for a virtual image at the near point. The latter follows from $1/f=1/s-1/D$.

More generally, the source does not specify the eyepiece-to-eye distance either. If that distance is $e$, then $u=s-e$ and a virtual image at the near point has transverse magnification $(D-e)/u$. Its angular magnification relative to unaided viewing at $D$ isFor example, $s=50\mathrm{mm}$ and $e=25\mathrm{mm}$ give $M=9$, with $u=25\mathrm{mm}$ and $f=28.125\mathrm{mm}$. This is a concrete counterexample to a universal maximum of $5$ at fixed screen-to-eye separation. In ideal geometrical optics, $e\to s^{-}$ allows arbitrarily large $M$ with an arbitrarily short focal length; practical aperture and optical aberrations constrain this. A unique numerical maximum therefore needs the conventional near-eye eyepiece geometry and near point assumption. The conventional $f=50\mathrm{mm}$ answer is $6$, with those assumptions stated explicitly.

The phrase “image” does not specify whether a real image or a virtual image is required. The later mention of an image plane makes a projected real image a natural interpretation. Let $u>0$ be object distance and $v>0$ real-image distance from a converging thin lens. A transverse magnification of magnitude $15$ gives $v=15u$, and the object-to-image separation gives $u+v=200\mathrm{mm}$. HencePlace the optical lens $12.5\mathrm{mm}$ beyond the object along the illuminating beam. The real image is inverted and has transverse magnification $-15$.

For the alternative single-optical lens magnifying glass interpretation, the virtual image is upright and lies on the object's side, so signed $v=-15u$. Its separation from the object is $|v|-u=14u=200\mathrm{mm}$. The thin lens equation then givesThe source does not explicitly select between these two geometries; they should not be conflated with a $200\mathrm{mm}$ lens-to-image distance.

Use the principle of dark-field microscopy: block the unscattered beam where it is concentrated, while transmitting most light scattering at nonzero angles. All direct light rays of the collimated illumination are parallel, so the thin lens focuses them to one point in its back focal plane. Put a small opaque circular stop centred on that point, one focal length behind the optical lens. In the real image geometry above this is $11.719\mathrm{mm}$ behind the optical lens, well before the real image at $187.5\mathrm{mm}$.

A light ray leaving object coordinate $y$ with a small initial direction $\theta$ reaches the optical lens at $y+u\theta$. The thin lens changes its direction to $\theta -(y+u\theta )/f$. At the back focal plane its coordinate is thereforeThus direct illumination with $\theta =0$ is concentrated at the centre, whereas scattered light at different directions is distributed away from it, independently of $y$. A stop of radius $r_s$ removes approximately $|\theta |\lesssim r_s/f$; choose it just large enough to cover the direct beam's finite diffraction spot, illumination divergence and alignment errors. Scattered light inside that small angular cone is also lost, so negligible loss requires most useful scattered light to lie outside it. A stop in the object's real image plane instead removes object detail and does not generally separate scattered and direct light. For the virtual image interpretation, the same physical back focal plane filtering requires a downstream observation arrangement that actually passes through this plane.

For $u=8\mathrm{cm}$ and negative focal length $f=-8\mathrm{cm}$,The transverse magnification is $M=-v/u=+1/2$, soThus the virtual image is upright, half the object's height, and $4.0\mathrm{cm}$ from the lens on the object's side. A parallel incident light ray diverges after the lens as though it originated at the near optical focus; the central light ray continues undeviated in the ideal thin lens model. Their backward extensions meet at the image tip, $(-4,1)\mathrm{cm}$.

A plane mirror has a flat reflecting surface. The law of reflection sends incident unit direction $\mathbf{d}$ to $\mathbf{d}-2(\mathbf{d}\cdot \mathbf{n})\mathbf{n}$, where $\mathbf{n}$ is its unit normal. It forms a virtual image of an object the same perpendicular distance behind the reflecting plane as the object is in front. Two successive plane mirrors whose planes meet at angle $\gamma$ rotate the ray geometry through $2\gamma$ about their intersection.

A real image lies where the outgoing light rays from each object point actually intersect in geometrical optics. A screen or detector placed at that image plane receives the focused pattern. The screen reveals the real image; it does not create the intersection. A viewer can also see a real image without a screen if their eye receives the diverging light rays beyond the intersection. ^[ref].

For a converging thin lens and real object with $u>f>0$, the real image has $v>0$ and signed transverse magnification $m<0$. Inversion is a property of this particular setup, not the definition of a real image. With $f=10\mathrm{cm}$ and $h_o=0.50\mathrm{cm}$, the following are predictions:

To test both image position and transverse magnification, use an optical bench with an illuminated arrow or millimetre target, a converging optical lens of nominal focal length $10\mathrm{cm}$, a movable white screen and a ruler. Measure corresponding target and image intervals, rather than guessing the whole image size when it exceeds the screen. PASCO's Experiment 12 compares distance-based and size-based magnification; the following procedure uses several object distances, as in the Northern Virginia laboratory. ^[ref]; ^[ref].

Keep the optical lens, illumination and central target detail fixed between trials. Approach best focus from both sides and record the interval over which sharpness is difficult to distinguish; repeat focusing to estimate its reproducibility. A modest central aperture stop tests whether peripheral light rays change the result, but also reduces brightness. Spherical aberration, chromatic aberration and diffraction can broaden the optimum; a thick optical lens also makes the chosen reference plane imperfect. Treat nominal $f$ as an initial prediction, and report position and interval-reading limits instead of forcing exact agreement. ^[ref].

Moving the object toward the front optical focus from outside makes the real image larger and farther away. A failed screen search near $u=f$ can therefore mean that the real image is beyond the bench, rather than that a virtual image has formed. At exactly $u=f$, the outgoing bundle from each point is parallel and no finite screen position focuses it.

At a smooth mirror with unit normal vector $n$, an incoming unit direction $i$ has outgoing direction $i-2(i\cdot n)n$. Thus the incident and reflected rays make equal angles with the normal.

More precisely, specular reflection redirects each incident light ray into the direction prescribed by the law of reflection, rather than spreading it over many outgoing directions. The incident light ray, reflected light ray, and local normal vector lie in one plane. The surface must be sufficiently smooth on the scale of the wavelength for this model to describe its reflected beam; a visibly flat surface can still scatter light if its microscopic roughness is appreciable. ^[ref].

The displayed formula uses unit vectors $\mathbf{i},\mathbf{n},\mathbf{r}$ and the dot product. It preserves the component parallel to the mirror and reverses the normal component. For a mirror along $y=0$, take $\mathbf{n}=(0,1)$ and $\mathbf{i}=(3/5,-4/5)$. Then $\mathbf{i}\cdot \mathbf{n}=-4/5$ and $\mathbf{r}=(3/5,4/5)$: both directions have vector norm one, and both make an angle $\arccos (4/5)\approx 36.87^{\circ }$ with the relevant normal direction. Replacing $\mathbf{n}$ by $-\mathbf{n}$ leaves the result unchanged.

This same local rule explains a flat mirror's virtual image. For a source at $(0,a)$ above $y=0$, backward extensions of reflected light rays intersect at $(0,-a)$. The light remains above the mirror; its apparent origin lies below it. A curved mirror changes its normal vector from point to point and can instead make reflected light rays converge. The geometry of image formation therefore starts with one simple direction rule. ^[ref].

// Target: optics.bigb

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].

Transverse magnification is the ratio of corresponding signed lengths perpendicular to the optical axis. For an ideal image of a small arrow of nonzero height,Use the same upward-positive height convention in both planes. Thus an upright $0.50\mathrm{cm}$ arrow imaged as $-0.25\mathrm{cm}$ has $m=-1/2$: inverted and half as tall. The sign describes orientation; real image versus virtual image describes ray intersections. They are separate properties, even when a particular thin lens setup correlates them.^[ref]

For a thin lens with object coordinate $-u$ and image coordinate $v$, the undeviated centre light ray is the straight line through $(-u,h_o)$ and $(0,0)$, so its height at $x=v$ is $h_i=-h_{o}v/u$. This is the signed similar triangles argument. Combining it with the thin lens equation givesFor $f>0$, $u>2f$ gives $-1<m<0$; $u=2f$ gives $m=-1$; $f<u<2f$ gives $m<-1$; and $0<u<f$ gives $m>1$. At $u=f$, the image is conventionally “at infinity”: the divergent expressions for $v$ and $m$ describe the limiting behaviour, not a finite image of infinite height. Different object points produce parallel bundles in different directions, which is why angular magnification remains meaningful.

The real image experiment tests $m$ in two independent ways: signed ratios of measured corresponding intervals, and $-v/u$ from measured longitudinal distances. As a further prediction, moving from $u=30\mathrm{cm}$ to $u=15\mathrm{cm}$ for $f=10\mathrm{cm}$ swaps $u,v$, changing $m=-1/2$ to $m=-2$. The two image sizes differ by a factor of four, while both images remain inverted. This reversal follows from the symmetric thin lens equation and connects to the Bessel lens displacement method.

For a uniform on-axis plane wave reference, choose its constant wave phase to be zero, so $R=A_R>0$. The reconstructed cross terms are then proportional to $O$ and its complex conjugate $O^{\ast }$. This pair creates the twin image in holography. An outgoing curved wavefront and its conjugate have opposite curvature: under forward propagation, one can appear to diverge from a virtual image while the other converges to a real image. In in-line holography their depths differ, but a defocused member still overlaps the other. This is the same conjugation of spatial wave phase used in phase conjugation, although a transmission hologram here launches both components forward rather than automatically creating a returning phase-conjugate mirror.