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For small measurement errors, write $f=L/4-d^2/(4L)$. Its partial derivatives and first-order propagation of uncertainty areHere the $\sigma$ values are standard deviations used to describe uncertainty, and $\mathrm{Cov}(L,d)$ is the covariance. Independent carrier-reading errors give ${\sigma }_d^2={\sigma }_{r_{-}}^2+{\sigma }_{r_{+}}^2$; correlated readings require the term $-2\mathrm{Cov}(r_{-},r_{+})$. Averaging repeated independent pairs reduces the random component of the uncertainty of the mean, but does not remove calibration or model errors.

As a stipulated uncertainty calculation, let $L=45\mathrm{cm}$, $d=15\mathrm{cm}$, ${\sigma }_L=0.10\mathrm{cm}$ and ${\sigma }_{r_{-}}={\sigma }_{r_{+}}=0.05\mathrm{cm}$, with all three input errors independent. Then ${\sigma }_d=0.0707\mathrm{cm}$, $A=5/18$, $B=-1/6$ and ${\sigma }_f\simeq 0.0302\mathrm{cm}$. The result would be $f=10.000\mathrm{cm}$ with this standard uncertainty in the stipulated thin lens model; these numbers are not a measured uncertainty budget. A small neglected principal-plane separation gives an additional first-order bias $f_{\mathrm{thin}}-f\simeq Ah$, distinct from that random budget.

The two kinds of sensitivity must be distinguished. At fixed focal length,These derivatives diverge as $L$ approaches $4f$ from above: a small change in object-screen spacing strongly changes the separation of the two settings. Yet the inferred focal length's displacement sensitivity $|B|=d/(2L)$ tends to zero. That does not make coalescing settings easy to measure. For fixed $L$, define the axial mismatch between the predicted real image and the screen as $g(x)=fx/(x-f)-(L-x)$, for $x>f$. At a focused setting,Near $u=v=2f$, moving the optical lens changes this mismatch only weakly, making visual discrimination of the two settings difficult. At the example's settings $g^{\prime }(15\mathrm{cm})=-3$ and $g^{\prime }(30\mathrm{cm})=3/4$, so equal carrier motion produces different axial defocus responses. This is an axial mismatch calculation, not a full prediction of perceived image sharpness.

Choose $L$ sufficiently above $4f$ to separate the two accepted focusing ranges, while retaining both real images on the available screen and rail. There is no universal best $L$: focusing uncertainty itself depends on aperture, transverse magnification, optical aberrations and the target. Close to $d=0$, the nonnegative measured separation and the quadratic term in $d$ also make first-order propagation of uncertainty unreliable if its error is comparable to its value. Repeatability, controls and the principal plane correction determine whether the elegant two-distance formula is an adequate measurement model.

KIT's teaching apparatus uses a glass spoke-wheel or test negative as the object, a halogen lamp with an illuminating optical lens, an imaging optical lens, a screen and a triangular rail with a position scale. The illuminating optical lens concentrates the lamp filament's image on the imaging optical lens's centre; translucent paper checks that illumination. This reduces the contribution of peripheral light rays and their spherical aberration. The instructions require fixed object-screen separation exceeding $4f$, at least ten displacement measurements, and separate red- and blue-filter measurements to test chromatic aberration.^[ref]

The following measurement protocol elaborates that apparatus rather than reporting additional KIT results. Align the target centre, optical lens and screen normal along one optical axis; measure $L$ between the target pattern plane and receiving screen face. Lock the target and screen. Slide the imaging optical lens to obtain the enlarged real image, then the reduced real image, reading the same carrier reference mark at each setting. Record $r_{-},r_{+}$ and $d=r_{+}-r_{-}$ for each pair. Record corresponding target and image intervals to test the reciprocal transverse magnifications independently; merely observing different image sizes is a weaker test.

Use the same central target detail to judge best optical focus. Bracket each setting by moving through both visibly blurred sides, repeat from both travel directions, and record the accepted position range as well as repeat scatter. Rail divisions set a readout limit but do not alone specify focusing uncertainty. Check scale calibration, carrier play and drift; a common additive carrier offset cancels in $d$, whereas scale-factor error remains. Keep illumination, aperture, target, screen and optical lens orientation fixed within a series. Change the aperture and colour filter in separate control series: spherical aberration and chromatic aberration can shift the apparent best optical focus, while a very small aperture introduces stronger diffraction and makes focus discrimination harder. Record these settings alongside the inferred focal length, rather than interpreting all disagreement as random measurement error.

For a rigid thick lens in air, let $h=x_{H^{\prime }}-x_H$ be the signed separation of its rear and front principal planes along the direction of propagation. Then the positive object and image distances measured from the principal planes obey $u+v=L-h$. Translating the whole assembly without reversing it moves both principal planes equally, so the measured carrier displacement still equals the difference of the two solutions for $u$. The same algebra therefore givesHere $f$ is the effective focal length measured from the principal planes. This deduction uses the principal-plane imaging equation, not the uncorrected thin lens formula.^[ref] Mechanical centre thickness is not generally $h$. Record thickness and optical lens orientation, but use a justified bound or model for $h$ if it matters at the claimed precision. The two readings $L,d$ alone cannot determine both $f$ and unknown $h$; cancelling a carrier offset does not cancel this limitation.

The Bessel lens displacement method measures a converging thin lens's focal length by moving only the optical lens between a fixed object and screen. If their separation is $L>4f$, two settings give sharp real images: one enlarged, the other reduced. Read their separation $d$ on an optical bench, then calculateThe two settings exchange object and image distances, so their signed transverse magnifications are reciprocals. This makes the experiment a visible test of conjugate optical planes as well as a focal length measurement.^[ref]

For an original stipulated example, put the object at $x=0$, the screen at $x=L=45\mathrm{cm}$, and take an ideal thin lens with $f=10\mathrm{cm}$. Its two settings are $x_{-}=15\mathrm{cm}$ and $x_{+}=30\mathrm{cm}$, with $d=15\mathrm{cm}$. The transverse magnifications are $m_{-}=-2$ and $m_{+}=-1/2$: both real images are inverted. An object $0.50\mathrm{cm}$ tall would give signed image heights $-1.00\mathrm{cm}$ and $-0.25\mathrm{cm}$. Conversely, stipulated readings $L=45\mathrm{cm}$ and $d=15\mathrm{cm}$ give $f=(45^2-15^2)/(4\times 45)=10\mathrm{cm}$. These are model calculations, not measured data or a manufacturer's calibration.

A fixed offset between a carrier's reference mark and an ideal thin lens plane cancels when subtracting the two carrier readings. That useful cancellation does not make a thick lens equivalent to one unknown optical centre: its two principal planes can be separated. The Bessel lens displacement experiment below makes this limitation explicit.

Place a converging thin lens at coordinate $x$ between the object at $0$ and the screen at $L$. In air and the paraxial approximation, the required object and image distances are $u=x$ and $v=L-x$. The thin lens equation gives a quadratic equation:Its discriminant is $L^2-4Lf$. For $L>4f>0$, the two roots and their separation areBoth roots lie inside $(0,L)$, and both object and image distances exceed $f$. The two roots therefore describe finite real images on the screen. Exchanging the roots exchanges $u,v$; from transverse magnification $m=-v/u$,The image-height ratio between the enlarged and reduced settings is $|m_{-}|/|m_{+}|=|m_{-}{|}^2$, rather than $|m_{-}|$.

At $L=4f$, the roots coincide: $d=0$, $x=L/2=2f$, $u=v=2f$ and $m=-1$. There is one setting, not two separately measurable settings. For $0<L<4f$ the discriminant is negative, so no position between the fixed object and screen gives a sharp real image there. This does not forbid a real image elsewhere if the screen is moved. Another short proof observes that $x(L-x)$ has maximum $L^2/4$ at $x=L/2$; the focusing requirement $x(L-x)=Lf$ is possible exactly when $L\ge 4f$.

Two conjugate optical planes are object and image planes related by an imaging system: points in one are imaged onto corresponding points in the other in the paraxial approximation. For a converging thin lens forming a real image in air, their distances $u,v>0$ obey $1/u+1/v=1/f$. Exchanging $u$ and $v$ preserves this relation; this symmetry gives the two settings of the Bessel lens displacement method.

Ernst Ruska and Max Knoll developed electron optics while improving the writing spot of a cathode-ray oscilloscope.^[ref] A structured object tests image formation more strongly than a single bright spot: each feature must retain its own image position.

On 7 April 1931, a cold cathode source and two short magnetic coils imaged aperture grids at $50\mathrm{kV}$. A platinum grid preceded the first coil; a bronze grid preceded the second. The fluorescent screen showed one-stage grid images and the first grid reimaged through both stages. Comparing these patterns tested successive image formation, not merely beam concentration. Heating could destroy even metal grids. This prototype did not establish resolution beyond an optical microscope.^[ref]

In 1933, Ruska measured $3\mathrm{mm}$ focal lengths at $75\mathrm{kV}$ with a pole-piece lens, then built an instrument giving $12000$-fold magnification.^[ref] TU Berlin dates images surpassing conventional optical microscope resolution to 1933.^[ref]

A fluorescent screen converts the spatial electron distribution into visible light.^[ref] A collector reporting only total electric current would discard that spatial pattern. In a two-stage image relay, the first lens makes an intermediate real image; the second treats that image as its object. The diagram below illustrates this geometry with invented distances and an ideal beam.

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.

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].

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.

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].

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]

For a converging thin lens forming a real image of a real object, let $u,v$ be positive distances from the lens and $f$ the focal length. Set $u^{\prime }=u-f$ and $v^{\prime }=v-f$, measuring from the neighboring focal points. Substituting into $1/u+1/v=1/f$ gives $(f+u^{\prime })(f+v^{\prime })=f(2f+u^{\prime }+v^{\prime })$, so $u^{\prime }{v}^{\prime }=f^2$. Other object-image geometries require an explicitly consistent signed-distance convention.

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.

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.

Substituting the stated refractive index and thickness coefficient, $\alpha =2{\mathrm{m}}^{-1}$, givesThe thin lens equation then givesThe positive image distance places a real image on the outgoing side of the optical lens, at $x=+1.75\mathrm{m}$. The object is beyond the focal point, so outgoing light rays actually converge there rather than only their backward extensions meeting.

The real image is upside down. A converging optical lens produces negative transverse magnification on the screen: a point above the optical axis sends a central light ray through the optical lens centre to a point below the optical axis. The same happens in the horizontal direction. Viewed directly from behind, the screen therefore shows a picture rotated through $180^{\circ }$ relative to the scene.

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 two equal lenses with a virtual intermediate object, use positive distances for real objects on a lens's incident side and real images on its outgoing side. The first lens gives $1/v_1=1/f-1/u_1=1/30-1/60$, hence $v_1=60\mathrm{cm}$. At the second lens these rays are already converging toward a point $50\mathrm{cm}$ to its right: its object distance is therefore $u_2=-50\mathrm{cm}$.Thus the final image is real, $28.75\mathrm{cm}$ to the right of lens 1. Its transverse magnification is $(-60/60)(-18.75/[-50])=-3/8$: it is inverted. The intermediate focus is not a real object placed before lens 2.