Optical microscope

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 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 $(D-e)/u$.

A microscope that uses controlled incident and analysed polarization of an electromagnetic wave to study specimens. Crossed polarizers reveal orientation-dependent phase retardation in a birefringent specimen.

Rotating a uniform crystallite around a fixed propagation direction changes the crossed-polar intensity factor ${\sin }^2(2\alpha )$. Extinctions repeat every90degrees and maxima occur45degrees away. Similar interference colour can coexist with different brightness if retardation is similar but azimuth differs.

An extinction pattern with four dark arms produced, for example, by a radial arrangement of local optical principal directions between crossed polarizers. The local intensity factor ${\sin }^2(2\phi )$ vanishes along the polarizer axes. Such a radial-material pattern need not be a conoscopic interference pattern.

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

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

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

The 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.

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.

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

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

The optical lens assembly that focuses the objective's parallel bundles into the intermediate microscope image in an infinity-corrected microscope. Its focal length participates in the objective-system transverse magnification $-f_t/f_o$.

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]

A ruled scale viewed together with the intermediate microscope image. Its intervals acquire specimen length units through microscope scale calibration.

A specimen-stage reference bearing a scale of known length, imaged through the microscope objective during microscope scale calibration.

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

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

Phase-contrast microscopy converts specimen-dependent phase differences into visible light intensity differences. It can reveal structures in living, unstained biological cells, allowing their changes to be followed without first making an absorbing preparation.^[ref] The key question is how an optical microscope can read a phase when an ordinary detector records light intensity.

Consider a transparent specimen with complex amplitude transmission $t(x,y)=e^{i\varphi (x,y)}$. In ideal in-focus coherent imaging with all relevant spatial frequencies transmitted, its image has $I=|t{|}^2=1$: varying $\varphi$ alone produces no brightness detail. This is a particular model, not a claim that all transparent specimens are invisible. Diffraction, defocus, absorption and a finite optical pupil can already change a real image.

A phase-contrast microscope images an illuminating annulus onto a phase plate ring near the objective's rear focal plane. The aligned ring attenuates and shifts direct light relative to most diffracted light; their interference changes image brightness.^[ref]

The method is beautiful because the bright direct beam becomes its own reference: a small specimen-dependent complex amplitude can interfere with it and change the image at first order. Merely magnifying an invisible phase pattern would leave $|t{|}^2$ unchanged.

An optical microscope configured for phase-contrast microscopy.

A specimen whose leading optical effect is a spatially varying phase, rather than absorption. A thin specimen with refractive index $n_s$ in a medium of refractive index $n_m$ has, in a straight-path approximation,Here $\lambda$ is the vacuum wavelength; the integral is an optical path length difference. For the chosen values $\lambda =550\mathrm{nm}$, thickness $1\mu \mathrm{m}$ and $n_s-n_m=0.005$, the phase is $\pi /55\simeq 0.0571$ radians. This calculation does not identify thickness separately from refractive index contrast.^[ref]

A phase object for which the relevant relative phase variations are much smaller than one radian. The Taylor expansion $e^{i\varphi }=1+i\varphi +O({\varphi }^2)$ separates a direct reference term from a small specimen term. A constant overall phase can be removed before making this approximation: phase-contrast microscopy measures spatial variations, not an absolute common phase.

An optical filter that changes selected light's phase relative to other transmitted light. In phase-contrast microscopy, its ring can also attenuate the direct reference; the amplitude transmission and light intensity transmission must be distinguished. An amplitude transmission $a=1/2$ gives an intensity transmission $a^2=1/4$.

The following original model isolates the conversion mechanism. Use a periodic image window, unit illumination, ideal unit magnification, and a phase plate that changes only the zero spatial frequency. Write $c_0=⟨e^{i\varphi }⟩$ for the spatial mean. Multiplying that single Fourier coefficient by $a{e}^{i\theta }$, with $0<a\le 1$, gives the exact image field in this model:This formula follows by splitting $e^{i\varphi }=c_0+(e^{i\varphi }-c_0)$: the second term has zero mean and is left unchanged. A real annular, partly coherent instrument alters a finite band of directions, so this calculation is a mechanism model rather than its full image-formation law.

For a weak phase object with $⟨\varphi ⟩=0$, one has $c_0=1+O({\varphi }^2)$ and henceThe cross term is $2\mathrm{Re}(a{e}^{-i\theta }i\varphi )=2a\varphi \sin \theta$. Thus $\theta =\pi /2$ makes positive $\varphi$ brighter and $\theta =-\pi /2$ makes it darker under this stated phase convention for complex amplitudes. The sign comes from this convention and chosen plate; a manufacturer’s positive/negative contrast labels should not be inferred from the sign of $\theta$ alone.

For example, choose $a=0.5$ and a zero-mean periodic pattern whose peak phase is $0.084$. The first-order peak intensities are $0.25\pm 0.084$, compared with the uniform ideal bright-field microscopy value $1$. The relative change about the reduced background is approximately $2\varphi /a$ for a quarter-cycle shift. Attenuating the reference increases relative contrast while reducing the background photon count; it is not light amplification or a universal improvement of signal-to-noise ratio.

Two checks expose what the model can and cannot measure. A spatially constant phase gives $E_{\mathrm{out}}=a{e}^{i\theta }{e}^{i\varphi }$ and $I_{\mathrm{out}}=a^2$, so no absolute phase is recovered. Orthogonality of the constant and zero-mean parts also givesFor $a\le 1$ this cannot exceed the incident mean light intensity. The image redistributes brightness and loses some reference light rather than creating energy.

A finite phase plate also intercepts diffracted light, altering low spatial frequencies. Halos and shade-off can therefore mimic borders or obscure interiors.^[ref] In the coherent zero-mode model of phase contrast, the separation is perfect by assumption, so that model cannot predict those artifacts.

Frits Zernike traced the idea to unwanted companion lines of a diffraction grating. Around 1930 he found that visible striping disappeared when a telescope focused exactly on the grating. A phase strip placed over the principal spectral line changed its relation to the companion light and restored contrast. He then applied the separation of direct and diffracted light to a microscope and demonstrated it at Zeiss in 1932.^[ref]

His later absorbing and annular strips improved contrast and the appearance of halos; commercial Zeiss accessories followed in 1941.^[ref] The route from a grating defect to biological observation illustrates why apparently unwanted instrumental effects can be productive experimental clues.

Transmitted-light imaging with an optical microscope in which the directly transmitted illumination contributes the bright background. An absorbing specimen can reduce that background locally. The ideal lossless phase object calculation in coherent zero-mode model of phase contrast shows why ordinary brightness detection need not reveal every optical variation.