Optical Instruments

Point a lens at the Moon and nothing about the Moon changes — it is 3500 km across and a quarter of a million miles away, exactly as before. What the instrument changes is the angle the Moon's light makes as it enters your eye, and angle is the only thing your retina can read as size. This is the quiet secret of every visual instrument: a magnifier, a microscope, a telescope does not enlarge the object, it enlarges the angle its rays subtend, and the eye reports the swollen angle as a swollen object. That reframing splits the whole subject in two. A camera or projector cares about the physical height of the image landing on a sensor or screen — its figure of merit is the transverse magnification $M_T=-s_i/s_o$ from image formation. A magnifier, microscope, or telescope cares instead about angular magnification $M=\theta'/\theta$ , the ratio of the angle its final image subtends to the angle the object subtends when viewed directly at the near point. The two numbers answer different questions, and confusing them is the classic error of the subject. From this single idea the working formulas of an entire optical bench fall out by composition. A simple magnifier lets you hold an object closer than the 25 cm near point and view it relaxed, buying $M=25\,\mathrm{cm}/f$ . A compound microscope stacks two stages — an objective that forms a real, enlarged intermediate image, then an eyepiece that magnifies that — and multiplies their powers. A telescope collapses a distant scene onto its focal plane and re-collimates it, magnifying angles by $-f_o/f_e$ . Each is the thin-lens equation and the definition of angular magnification, applied twice.

Key takeaways

An optical instrument is a purposeful arrangement of lenses, mirrors, and stops that controls where an image forms and, for visual use, how large it appears — not how large it physically is.
Two magnifications must be kept apart: transverse $M_T=-s_i/s_o$ (physical image height, for cameras and projectors) and angular $M=\theta'/\theta$ (apparent size, for magnifiers, microscopes, telescopes).
The relaxed eye focuses parallel light onto the retina; instruments in "normal adjustment" deliver their final image at infinity so the eye views without strain, referenced to the standard near point $D=25\ \mathrm{cm}$ .
A simple magnifier gives $M=D/f$ ; a compound microscope multiplies objective transverse magnification by eyepiece angular magnification, $M=-\dfrac{L}{f_o}\dfrac{D}{f_e}$ ; a telescope gives $M=-f_o/f_e$ .
The Galilean telescope (convex objective, concave eyepiece) yields an erect image; the Keplerian (two convex lenses) yields an inverted one but a wider, brighter field — the design that won.
Key idea: For a visual instrument the useful quantity is angular magnification $M=\theta'/\theta$ — the ratio of the angle the final image subtends to the angle the object subtends viewed directly at the 25 cm near point — obtained by composing thin-lens imaging with the definition of angular size, never the physical size of any intermediate image.

The core insight

For a visual instrument the useful quantity is the angular magnification $M=\theta'/\theta$ : the ratio of the angle subtended by the final (usually at-infinity) image to the angle the object subtends when viewed directly at the near point $D=25\ \mathrm{cm}$ . A magnifier gives $D/f$ , a microscope $-(L/f_o)(D/f_e)$ , a telescope $-f_o/f_e$ — each a composition of the thin-lens equation with the small-angle relation between size and angle.

Where this lives today

The two-stage logic of the microscope and the afocal logic of the telescope run unbroken through modern instruments. A research microscope still stacks an objective and an eyepiece (or, now, a tube lens and a camera), but its objective is corrected for a dozen aberrations and its resolution is pushed to and past the diffraction limit by confocal, structured-illumination, and single-molecule-localisation methods. Giant telescopes keep the Keplerian skeleton — a light-gathering objective (now a segmented mirror up to $\sim40\ \mathrm{m}$ across) feeding an eyepiece-turned-instrument — while adaptive optics reshapes the wavefront in real time to beat atmospheric blur. The phone in your pocket is a fixed-focus camera whose $f$ -number and sensor size set its light-gathering and depth of field exactly as equation (5) says, and every laser-spectroscopy or cold-atom experiment builds a bespoke telescope or relay to deliver and collect light. The formulas are three centuries old; the engineering around them is not.

The physical picture

Your eye is an angle meter. A coin held at arm's length and the full Moon can subtend the same half-degree, and so "look the same size," even though one is centimetres and the other thousands of kilometres across. Perceived size is set by the retinal image, and the retinal image is set by the angle the incoming cone makes with the axis. Every visual instrument is therefore an angle amplifier: it re-aims the object's rays so they enter the pupil more steeply, and the retina duly reports a larger object. The near point $D=25\ \mathrm{cm}$ is the reference. That is the closest a relaxed young eye can focus, so it is the largest angle you can honestly get from an object without help. A magnifier cheats the limit: it forms a virtual image far away while letting you place the object at its focal point, so you view a big angle with a relaxed eye. The gain, angle-with-instrument over best-angle-without, is $D/f$ .

Figure

Keplerian telescope in normal adjustment: the objective (focal length f_o) forms a small inverted image at the shared focal plane; the eyepiece (f_e) re-collimates it, so a ray bundle entering at angle θ leaves at the larger angle θ′, with M = −f_o/f_e

Think of the compound microscope as two acts. The objective, a short-focus lens close to the specimen, throws a real and already enlarged image up the tube — pure transverse magnification. The eyepiece then works as a plain magnifier on that real image, boosting its angular size for the eye. The total power is the product of the two, which is why a $40\times$ objective and a $10\times$ eyepiece make a $400\times$ instrument. The telescope runs the same machinery backwards. A distant object is effectively at infinity, so the objective forms a tiny real image at its focal plane; the eyepiece re-collimates it. The image is small but sits close to the eye, so it is seen under a far larger angle than the object itself — angular magnification $f_o/f_e$ . Bigger objective focal length or shorter eyepiece focal length both raise the power.

Figure

Punchline — telescope angular magnification |M| = f₀/f_e versus eyepiece focal length for a fixed objective f₀ = 1200 mm: a 25 mm eyepiece gives 48×, a 50 mm eyepiece 24×; shorter eyepieces buy power but shrink the exit pupil and field

The core equations

Everything below is the definition of angular magnification, $M=\theta'/\theta$ , applied to a particular stack of lenses. First the reference: an object of height $h$ viewed at the near point $D=25\ \mathrm{cm}$ subtends $\theta\approx h/D$ . That is the honest baseline every instrument is measured against. A simple magnifier is a single converging lens with the object at (or just inside) its front focal point, so the virtual image sits at infinity and the eye stays relaxed. The ray from the object top now enters the eye at $\theta'\approx h/f$ :

M=\frac{\theta'}{\theta}=\frac{h/f}{h/D}=\frac{D}{f}\qquad\big(\text{image at }\infty\big),\qquad M=1+\frac{D}{f}\quad\big(\text{image at }D\big) \qquad (1)

where: $D=25\ \mathrm{cm}$ — standard near point; $f$ — magnifier focal length (same units as $D$ ); the two forms differ by whether the virtual image is placed at infinity (relaxed eye) or pulled in to the near point (maximum magnification, tenser eye). A 5 cm magnifier gives $M=5\times$ relaxed, $6\times$ strained. A compound microscope cascades an objective and an eyepiece. The objective, of focal length $f_o$ , forms a real image a tube length $L$ beyond its back focal point, with transverse magnification $m_o=-L/f_o$ ; the eyepiece then magnifies that real image as a simple magnifier, $M_e=D/f_e$ . The powers multiply:

M=m_o\,M_e=-\frac{L}{f_o}\,\frac{D}{f_e} \qquad (2)

where: $L$ — optical tube length (objective back focus to eyepiece front focus, standardised at $160\ \mathrm{mm}$ ); $f_o,f_e$ — objective and eyepiece focal lengths; $D=25\ \mathrm{cm}=250\ \mathrm{mm}$ ; the minus sign records the inverted image. Objectives and eyepieces are sold by their separate powers ( $m_o=L/f_o$ , $M_e=D/f_e$ ), and the instrument's power is their product. An astronomical (Keplerian) telescope is afocal: parallel light in gives parallel light out. The objective forms the distant object's image at its back focal plane, which coincides with the eyepiece front focal plane, so the two focal lengths simply add to the tube length. Tracing the chief ray through both lenses gives:

M=-\frac{f_o}{f_e},\qquad \text{tube length}=f_o+f_e \qquad (3)

where: $f_o,f_e$ — objective and eyepiece focal lengths (both positive for a Keplerian instrument); the minus sign marks the inverted image. A Galilean telescope replaces the eyepiece with a diverging lens ( $f_e<0$ ), giving $M=-f_o/f_e>0$ (erect) and a shorter tube $f_o-|f_e|$ , at the cost of a narrow field. Because the objective rim is the aperture stop of a telescope, its image through the eyepiece is the exit pupil, and matching that pupil to the eye is what makes the collected light usable:

d_{\mathrm{XP}}=\frac{D_o}{|M|}=\frac{f_e}{f_o}\,D_o \qquad (4)

where: $D_o$ — objective (entrance-pupil) diameter; $|M|$ — angular magnification; $d_{\mathrm{XP}}$ — exit-pupil diameter. Larger magnification means a smaller exit pupil; a $7\times50$ binocular gives $50/7\approx7.1\ \mathrm{mm}$ , matched to the dark-adapted eye. A camera is the non-visual member of the family: it maps the world onto a fixed sensor, focusing by moving the lens and controlling exposure through the $f$ -number $N=f/D$ , so image irradiance scales as $1/N^2$ :

N=\frac{f}{D},\qquad E\propto\frac{1}{N^{2}} \qquad (5)

where: $N$ — focal ratio; $f$ — focal length; $D$ — entrance-pupil (aperture) diameter; $E$ — image-plane irradiance. Each "stop" on the aperture ring changes $N$ by $\sqrt2$ , halving or doubling the light.

Derivation AThe compound-microscope magnification as a product of two stages

We derive the total angular magnification of a compound microscope by treating it explicitly as two composed imaging stages — an objective that forms a real intermediate image and an eyepiece that views it as a magnifier — and multiplying the magnification of each. The strategy is to compute the objective's transverse magnification from the thin-lens geometry, recognise that the intermediate image is the object for the eyepiece, and apply the magnifier result $D/f_e$ to it; the product is the number engraved on the instrument, and it shows why microscope power is a simple product of objective and eyepiece markings.

Step 1. The specimen sits just outside the objective front focus, so the objective forms a real, inverted, magnified image a distance $\approx L$ (the tube length) beyond its back focal point. By the Newtonian form of image formation the transverse magnification is $m_o=-L/f_o$ . Why the tube length appears: with the object essentially at the front focus, the image distance measured from the back focal point is the tube length $L$ , and transverse magnification of a lens equals $-x_i/f=-L/f_o$ .

Step 2. That real intermediate image is now the object for the eyepiece, placed at the eyepiece front focal point so the final virtual image is at infinity and the eye is relaxed. The eyepiece therefore acts as a simple magnifier of angular magnification $M_e=D/f_e$ , with $D=25\ \mathrm{cm}$ . Why it is a magnifier, not an imager: the eyepiece converts the intermediate image's height into an angle at the eye, exactly the magnifier situation of equation (1).

Step 3. Angular magnification of the whole instrument is the angle the final image subtends divided by the angle the specimen would subtend at the near point. Because the objective first multiplies the height by $m_o$ and the eyepiece then multiplies the resulting angle by $M_e$ , the two compose by multiplication: $M=m_o\,M_e=-\dfrac{L}{f_o}\dfrac{D}{f_e}$ . Why they multiply and don't add: transverse magnification scales height, the eyepiece scales angle-per-height; chaining a height gain into an angle gain is a product.

Step 4. Put numbers to it: a $40\times$ objective ( $f_o=4\ \mathrm{mm}$ , $L=160\ \mathrm{mm}$ ) with a $10\times$ eyepiece ( $f_e=25\ \mathrm{mm}$ ) gives $M=-(160/4)(250/25)=-(40)(10)=-400\times$ . Why the markings already carry it: manufacturers stamp $m_o=L/f_o$ and $M_e=D/f_e$ on the barrels, so the user just multiplies. *Verified in verify_optical-instruments.py: $m_o=-40$ , $M_e=10$ , $M=-400\times$ for the standard $160\ \mathrm{mm}$ tube.*

Go deeper · Derivation B · 1 minThe telescope's angular magnification from the afocal geometry

Derivation BThe telescope's angular magnification from the afocal geometry

We prove that an astronomical telescope in normal adjustment magnifies angles by $-f_o/f_e$ , working directly from the afocal condition that parallel light enters and parallel light leaves. The strategy is to send in a bundle of parallel rays tilted at angle $\theta$ from the axis, follow the ray through the objective to the shared focal plane, and read off the angle $\theta'$ at which the eyepiece re-launches it; the ratio is the magnification, and its sign reveals the inverted image that a star-gazer accepts but a birdwatcher does not.

Step 1. A distant point off-axis by angle $\theta$ sends parallel rays into the objective, which focuses them to a point in its back focal plane at height $h=f_o\tan\theta\approx f_o\,\theta$ . Why the focal plane: parallel rays converge in the focal plane, and the off-axis tilt $\theta$ maps to a transverse height $f_o\theta$ there.

Step 2. The eyepiece front focal plane coincides with this plane (the afocal condition), so it collects the rays from the height- $h$ point and re-collimates them into a parallel bundle leaving at angle $\theta'=h/f_e=f_o\theta/f_e$ . Why re-collimation: a source at the front focal point of a lens emerges as parallel light, tilted by height-over-focal-length.

Step 3. The angular magnification is $M=\theta'/\theta=f_o/f_e$ in magnitude; a careful sign accounting (the image is on the opposite side of the axis) gives $M=-f_o/f_e$ , an inverted image. Why inverted: the objective forms a real inverted image which the eyepiece presents as-is, so astronomical telescopes show the sky upside-down.

Step 4. For $f_o=1200\ \mathrm{mm}$ and $f_e=25\ \mathrm{mm}$ , $M=-48\times$ ; a Galilean design ( $f_e<0$ ) restores an erect image and shortens the tube to $f_o-|f_e|$ , which is why opera glasses are Galilean. Why Keplerian won for astronomy: the real intermediate image admits a field stop and crosshairs, and gives a wider, brighter field despite the inversion. *Verified in verify_optical-instruments.py: $\theta'=f_o\theta/f_e$ , $M=-48\times$ for $f_o=1200,\ f_e=25\ \mathrm{mm}$ ; Galilean tube $=f_o-|f_e|$ .*

Limiting cases

Case	Result	Meaning
magnifier, image at $\infty$	$M=D/f$	relaxed-eye viewing; a 25 mm lens gives $10\times$
magnifier, image at near point	$M=1+D/f$	tenser eye, one extra unit of magnification
$f_e\to f_o$ (telescope)	$M\to-1$	no angular gain — a "unit-power" finder, image inverted
Galilean, $f_e<0$	$M=-f_o/f_e>0$	erect image, short tube, narrow field — opera glasses
camera stop $\times\sqrt2$	$E\to E/2$	each f-stop halves image irradiance ( $E\propto1/N^2$ )
exit pupil $>$ eye pupil	light wasted	outer rays strike the iris; magnify more to shrink $d_{\mathrm{XP}}$

Common trap“A telescope magnifies the object” — as if it made the star physically bigger

The object is untouched; what grows is the angle its light subtends at your eye, and hence the retinal image. This is why the only honest figure of merit for a magnifier, microscope, or telescope is angular magnification $M=\theta'/\theta$ , referenced to the $25\ \mathrm{cm}$ near point — not the transverse magnification $M_T=-s_i/s_o$ of any intermediate image. A magnifier can throw a virtual image metres tall and yet its transverse magnification is modest; its whole worth is the angle. And the " $150\times$ " on a telescope box is angular and eyepiece-dependent: swap the eyepiece and the number changes, while the objective sets the light-gathering and resolution that actually limit what you can see.

Worked examples

1 · A standard compound microscope (confidence builder)

A microscope has a $40\times$ objective of focal length $f_o=4\ \mathrm{mm}$ on a $160\ \mathrm{mm}$ tube and a $10\times$ eyepiece of focal length $f_e=25\ \mathrm{mm}$ . Find the total magnification and confirm the barrel markings.

Step 1. Objective transverse magnification: $m_o=-L/f_o=-160/4=\boxed{-40}$ — exactly the " $40\times$ " on the barrel, the minus recording inversion.

Step 2. Eyepiece angular magnification: $M_e=D/f_e=250/25=\boxed{10}$ — the " $10\times$ ", using $D=25\ \mathrm{cm}=250\ \mathrm{mm}$ .

Step 3. Total: $M=m_o\,M_e=(-40)(10)=\boxed{-400\times}$ . The instrument's power is the product of the two engraved numbers; the specimen appears $400$ times its near-point angular size, inverted.

Step 4. Sanity check on brightness and resolution: the power is $400\times$ but the resolution is set by the objective's numerical aperture, not by piling on eyepiece power — pushing $M$ past roughly $1000\times\mathrm{NA}$ gives "empty magnification," bigger but no sharper.

2 · A refractor and its exit pupil (exam level)

A refracting telescope has objective focal length $f_o=1200\ \mathrm{mm}$ , objective diameter $D_o=100\ \mathrm{mm}$ , and a $25\ \mathrm{mm}$ eyepiece. Find the angular magnification, the exit-pupil diameter, and whether it suits a dark-adapted eye.

Step 1. Angular magnification: $M=-f_o/f_e=-1200/25=\boxed{-48\times}$ — the sky appears 48 times larger in angle, and inverted.

Step 2. Exit-pupil diameter: $d_{\mathrm{XP}}=D_o/|M|=100/48=\boxed{2.08\ \mathrm{mm}}$ , the bright disc of light hovering behind the eyepiece where you place your pupil.

Step 3. A $2.08\ \mathrm{mm}$ exit pupil is smaller than the $\sim7\ \mathrm{mm}$ dark-adapted pupil, so the eye captures all the delivered light; the image is bright enough, and the eye relief places the pupil at the exit pupil.

Step 4. Drop to a $50\ \mathrm{mm}$ eyepiece and $M=-24\times$ , $d_{\mathrm{XP}}=4.17\ \mathrm{mm}$ — lower power, wider field, larger exit pupil. The eyepiece is the design knob that trades power against field and pupil, exactly the punchline curve above.

Physics of the portrait

Galileo did not invent the telescope, but he was the first to point a good one at the sky and reason quantitatively about what it showed — turning a Dutch novelty into the instrument that broke open the heavens.

Origins · 1 minGalileo Galilei and how this was discovered

Galileo Galilei

1564–1642 · Pisa & Padua

Hearing in 1609 of a Dutch device that made distant things look near, Galileo reconstructed it from optical principles and rapidly improved it from about $3\times$ to nearly $30\times$ — a convex objective with a concave eyepiece, the design still called Galilean. Turned skyward, it delivered a torrent of discoveries in months: four moons orbiting Jupiter, the phases of Venus, mountains and craters on the Moon, sunspots, and the resolution of the Milky Way into countless stars, published in Sidereus Nuncius (1610). The phases of Venus were decisive: they are impossible in a strictly Earth-centred cosmos and natural in a Sun-centred one, and Galileo pressed the point until it cost him his freedom. He later built a compound microscope — which he called the occhiolino, and for which Giovanni Faber coined the name microscope in 1625 — extending the same lens-craft to the very small. His instruments embody this article's lesson — that an arrangement of two lenses, understood through their focal lengths, turns a faint angle into a visible world.

History

1608

— Hans Lippershey applies in the Netherlands for a patent on a refracting telescope

1609

— Galileo builds his own telescopes, reaching nearly 30× magnification

1610

— Galileo publishes Sidereus Nuncius, reporting Jupiter's moons, lunar relief, and the stars of the Milky Way

1611

— Kepler describes the two-convex-lens (Keplerian) telescope in Dioptrice

1625

— Giovanni Faber coins the name microscope for Galileo's compound instrument (which Galileo called the occhiolino)

1665

— Robert Hooke publishes Micrographia, the first great work of microscopy

1676

— Antonie van Leeuwenhoek observes microorganisms with single-lens microscopes

Going further

Texts. Hecht, Optics, 5th ed., Ch. 5–6 (the eye, magnifier, microscope, telescope, stops) · Saleh & Teich, Fundamentals of Photonics, 3rd ed., Ch. 1 (ray optics, imaging systems) · Foot, Atomic Physics, appendices on optical systems for spectroscopy · Demtröder, Laser Spectroscopy, Vol. 1 (spectrographs, apertures, etendue).

Provenance. Definitions and magnifications follow Hecht Ch. 5–6 and Saleh & Teich Ch. 1: the simple-magnifier $M=D/f$ (relaxed) / $1+D/f$ (near point), the compound-microscope $M=-(L/f_o)(D/f_e)$ with the standard $160\ \mathrm{mm}$ tube, the telescope $M=-f_o/f_e$ , exit pupil $d_{\mathrm{XP}}=D_o/|M|$ , and camera $N=f/D$ with $E\propto1/N^2$ . verify_optical-instruments.py reproduces: $m_o=-40$ , $M_e=10$ , $M=-400\times$ ; $M=-48\times$ and $d_{\mathrm{XP}}=2.08\ \mathrm{mm}$ for $f_o=1200,\ D_o=100,\ f_e=25\ \mathrm{mm}$ ; the $7\times50\to7.1\ \mathrm{mm}$ exit pupil; magnifier $25/5=5\times$ ; the Galilean tube $f_o-|f_e|$ . The near point $D=25\ \mathrm{cm}$ is the conventional relaxed-eye value.

Builds on: Thin Lens Equation and Lensmaker's Equation · Image Formation and Magnification · Ray Model of Light · Thick Lenses and Principal Planes. Leads to: Dispersion and Prisms · Plane and Spherical Mirrors.

Why is the near point taken as 25 cm, and does the magnification change for someone with different eyes?

The $25\ \mathrm{cm}$ near point is a convention for a relaxed young adult eye — the closest most people can focus comfortably. It is the reference angle every visual magnification is quoted against, so the " $10\times$ " on an eyepiece assumes it. If your own near point differs (it lengthens with age), the apparent gain you personally experience shifts slightly, but the instrument's stated magnification does not, because it is defined against the standard $25\ \mathrm{cm}$ . This is also why microscope and telescope powers are angular, comparative numbers, not absolute physical sizes.

What is "empty magnification," and why can't I just use a stronger eyepiece to see more detail?

Magnification and resolution are different things. A stronger eyepiece enlarges the image but adds no new information: the finest detail is set by the objective's numerical aperture (microscope) or aperture diameter (telescope) through diffraction. Once the magnification exceeds roughly $500$ – $1000\times\mathrm{NA}$ for a microscope, you are enlarging a blur — "empty magnification," bigger but no sharper. To see more real detail you need a higher-NA objective or a larger telescope aperture, which is why light-gathering optics, not eyepieces, define an instrument's class.

What exactly is the exit pupil, and why do binoculars quote numbers like 7×50?

The exit pupil is the small bright disc of light floating behind the eyepiece — the image of the objective (the aperture stop) formed by the eyepiece. Its diameter is $d_{\mathrm{XP}}=D_o/|M|$ . In " $7\times50$ ," the $7$ is angular magnification and $50\ \mathrm{mm}$ is the objective diameter, so the exit pupil is $50/7\approx7.1\ \mathrm{mm}$ — matched to the fully dark-adapted human pupil, so no gathered light is wasted. Placing your eye's pupil at the exit pupil (the "eye relief" distance) is what lets you see the full, bright field.

Why did the Keplerian (two-convex-lens) design displace Galileo's own telescope for astronomy?

Galileo's telescope uses a concave eyepiece, which gives an erect image and a compact tube — great for opera glasses — but a cramped field of view and no accessible real image. Kepler's two-convex design forms a real, inverted intermediate image inside the tube. That real image is priceless: you can place a field stop and a crosshair there for precise measurement, and the design delivers a wider, brighter field. Astronomers happily accept an upside-down sky in exchange, which is why every serious refractor and the eyepiece side of every reflector is Keplerian.

How does the camera fit in if it isn't a "visual" instrument judged by angular magnification?

A camera images the world onto a fixed sensor, so its relevant magnification is transverse — how many millimetres of sensor a given scene fills — set by focal length and object distance. Its exposure is governed by the $f$ -number $N=f/D$ : image irradiance goes as $1/N^2$ , so each full stop ( $N\times\sqrt2$ ) halves the light. Focal length also sets the angular field the sensor captures (short $f$ = wide angle). The camera is the transverse-magnification member of the family, a useful contrast that sharpens why magnifiers, microscopes, and telescopes are instead judged by angle.

If the telescope's magnification is

-f_o/f_e

, why not just make the eyepiece focal length tiny for huge power?

Two limits bite. First, the exit pupil $d_{\mathrm{XP}}=D_o/|M|$ shrinks as you raise power; below about $0.5\ \mathrm{mm}$ the image dims and the eye's own aberrations and "floaters" intrude. Second, resolution is fixed by the objective diameter through diffraction ( $\theta_{\min}\approx1.22\lambda/D_o$ ), so beyond a magnification of roughly $D_o$ in millimetres you again reach empty magnification — larger but no more detailed. Practical maximum power is around $2\times$ the objective diameter in millimetres; past that you are trading field, brightness, and comfort for nothing.