Simple & Compound Microscopes

A single lens held over a stamp is the humblest instrument in physics, and one of the most revealing. Slide it until the print leaps up, erect and enlarged, and you have built a simple microscope — a magnifying glass — whose entire theory is one line: $M = 1 + D/f$ . Put a $5\ \mathrm{cm}$ lens over the page and it enlarges six times. That is the ceiling for one lens; to go further you must let one lens magnify the image of another, and the compound microscope is born — two lenses in a tube, the second enlarging what the first has already enlarged, so their powers multiply. A $1\ \mathrm{cm}$ objective and a $5\ \mathrm{cm}$ eyepiece a tube-length apart reach $M = 100$ from parts you could hold in one hand. The physics of both is the same idea the eye taught us: what matters to vision is not an object's true size but the angle it subtends at your eye. A magnifier wins by letting you bring the object far inside the $25\ \mathrm{cm}$ near point — impossibly close for the naked eye — while still seeing it sharply, so it fills a far larger angle. Every magnifying-power formula in this article is a ratio of two such angles, $M = \beta/\alpha$ , and the small-angle approximation turns each ratio into a tidy formula you can derive from the thin-lens relation you already own. But there is a hard limit the formulas hide, and it is the deepest idea here: more magnification is not more detail. Past a point, a compound microscope enlarges the blur along with the image, and what caps the useful magnification is not the lenses' focal lengths but resolving power — the finest separation the objective's aperture can distinguish, set by the wavelength of light itself. That limit, named for Ernst Abbe, is why an optical microscope stops at about a five-hundredth of a millimetre no matter how you stack the lenses, and why the electron microscope had to be invented. We will build the simple and compound instruments, derive their magnifications, and then meet the wall that magnification cannot climb.

Key takeaways

A simple microscope (magnifying glass) is one converging lens with the object inside its focus, giving an erect, virtual, magnified image; magnifying power $M = 1 + D/f$ (final image at the near point $D = 25\ \mathrm{cm}$ ) or $M = D/f$ (relaxed eye, image at infinity).
A $f = 5\ \mathrm{cm}$ magnifier gives $M = 1 + 25/5 = 6$ (near point) or $M = 25/5 = 5$ (relaxed).
A compound microscope stacks a short-focus objective and an eyepiece: the objective forms a real magnified image, the eyepiece magnifies that. Total $M = (L/f_o)(D/f_e)$ (relaxed) or $(L/f_o)(1 + D/f_e)$ (near point), with $L$ the tube length.
With $f_o = 1\ \mathrm{cm}$ , $f_e = 5\ \mathrm{cm}$ , $L = 20\ \mathrm{cm}$ , $D = 25\ \mathrm{cm}$ : $M = (20/1)(25/5) = 100$ relaxed, or $120$ at the near point.
High magnification wants short focal lengths; the two-stage magnification is a product, not a sum.
Key idea: a microscope multiplies angular size, not real size — the simple lens gives 1 + D/f, the compound gives the product of objective and eyepiece stages — but the true limit on what you can see is resolving power set by the wavelength of light, not magnification.

The core insight

Both microscopes work by enlarging the angle an object subtends at the eye: the simple microscope gives $M = 1 + D/f$ (or $D/f$ relaxed), and the compound microscope multiplies an objective magnification $L/f_o$ by an eyepiece magnification $D/f_e$ to reach $M = (L/f_o)(D/f_e)$ — but past a point extra magnification only enlarges the blur, because resolving power, fixed by the wavelength of light, sets the real limit on detail.

Where this lives today

The compound microscope on a school bench and the one in a pathology lab share the exact optics of equation (3): a short objective, a magnifying eyepiece, their powers multiplying. But the frontier moved to Abbe's limit and then past it. Oil-immersion objectives push the numerical aperture toward $1.4$ to squeeze the resolving detail down to about $0.2\ \mathrm{micrometres}$ , the best visible light can do. The electron microscope, using electron waves thousands of times shorter than light, resolves individual atoms — the direct descendant of Abbe's insight that wavelength, not magnification, sets the limit. And in a twist Abbe would have relished, twenty-first-century super-resolution fluorescence microscopy (STED, PALM, STORM — Nobel Prize, 2014) beats the diffraction limit with visible light by clever tricks of switching molecules on and off, seeing detail finer than $\lambda/2\,\text{NA}$ once thought absolute. Every one of these still begins with the humble bargain of this article: bring the object impossibly close, and let a lens keep it in focus.

The physical picture

Vision is about angle. A coin at arm's length and the Moon can subtend the same angle at your eye, and to the eye they then look the same size. So to make something look bigger, you do not have to make it bigger — you have to make it subtend a bigger angle. The naked eye's best trick is to bring the object to its near point, $25\ \mathrm{cm}$ , the closest it can focus; nearer than that and the image blurs. A grain of pollen at $25\ \mathrm{cm}$ subtends a hopelessly tiny angle, and that angle is the ceiling on the detail the unaided eye can extract. A magnifying glass breaks the ceiling by a simple bargain. Place the object inside the focal length of a converging lens and the lens forms an erect, virtual, magnified image far away, which the relaxed eye can view comfortably — while the object itself sits just a few centimetres from your eye, far inside the near point. You are, in effect, allowed to hold the object at $5\ \mathrm{cm}$ instead of $25\ \mathrm{cm}$ , so it subtends a much larger angle, and the lens keeps it in focus. The magnifying power is precisely the factor by which that angle grows.

Figure

A simple microscope: the object sits just inside the focal point of one converging lens, which forms an erect, enlarged, virtual image far away that the eye views at ease

To go beyond a single lens you cannot simply pick a shorter $f$ forever — a very short-focus lens becomes a tiny bead impossible to look through, and aberrations swamp it. The trick is to magnify twice. A first lens, the objective, of very short focal length, forms a real, enlarged, inverted image of the object a good way down a tube. A second lens, the eyepiece, is then used as a magnifying glass on that real image. Because the second stage enlarges the already-enlarged image, the two magnifications multiply, and the compound microscope reaches magnifications no single lens could.

The simple microscope

Set the object just inside the focal point of a converging lens and adjust so the final virtual image sits at the near point $D$ — the most enlarged view a relaxed-but-focused eye can accept. The magnifying power, the ratio of the angle the image subtends to the angle the object would subtend at the unaided eye at $D$ , is

M = 1 + \frac{D}{f} \qquad (1)

where: $D = 25\ \mathrm{cm}$ is the near point and $f$ the lens's focal length (same units). Read it as "one, plus how many focal lengths fit in the near point." A $f = 5\ \mathrm{cm}$ magnifier gives $M = 1 + 25/5 = 6$ : the letters look six times taller. Shorter $f$ means more magnification, which is why a jeweller's loupe is a fat little lens of $f$ around $2.5\ \mathrm{cm}$ . If instead you relax fully and let the final image go to infinity — easier on the eye for long use — the object sits exactly at the focus and the magnifying power drops slightly to

M = \frac{D}{f} \qquad (2)

the same $f = 5\ \mathrm{cm}$ lens now giving $M = 25/5 = 5$ . The difference between (1) and (2) is the single " $+1$ ", and it is worth exactly one accommodation: the near-point setting squeezes a little more angle out of the lens at the cost of tiring the eye. Boards expect you to state which adjustment a number belongs to. In dioptres, that $f = 5\ \mathrm{cm}$ lens has power $P = 1/0.05 = +20\ \mathrm{D}$ — a strong converging lens.

The compound microscope

Now the two-stage instrument. The objective (focal length $f_o$ , short) forms a real, inverted, magnified image inside the tube; its linear magnification is $m_o$ . The eyepiece (focal length $f_e$ ) then views that real image as a simple microscope, adding an angular magnification $M_e$ . The instrument's total magnifying power is the product:

M = m_o \times M_e = \frac{L}{f_o}\cdot\frac{D}{f_e} \qquad (3)

where: $L$ is the tube length — the distance between the objective's back focus and the eyepiece's front focus, essentially the separation over which the objective throws its image — and the final image is at infinity (relaxed eye). If instead the final image is brought to the near point, the eyepiece stage gains its " $+1$ " and

M = \frac{L}{f_o}\left(1 + \frac{D}{f_e}\right) \qquad (4)

Put in the standard teaching set $f_o = 1\ \mathrm{cm}$ , $f_e = 5\ \mathrm{cm}$ , $L = 20\ \mathrm{cm}$ , $D = 25\ \mathrm{cm}$ . Relaxed, equation (3) gives $M = (20/1)(25/5) = 20 \times 5 = 100$ . At the near point, equation (4) gives $M = (20/1)(1 + 25/5) = 20 \times 6 = 120$ . One hundred times, from a $1\ \mathrm{cm}$ lens doing $20\times$ and a $5\ \mathrm{cm}$ lens doing $5\times$ — the multiplication is the whole gain, and it is why the objective is made as short-focus as the optics allow.

Figure

A compound microscope in relaxed adjustment: the short-focus objective forms a real inverted image inside the tube, and the eyepiece magnifies that image into a large virtual image at infinity — the two magnifications multiply

Resolving power — the limit magnification cannot beat

Here is the idea that separates someone who has memorised the formulas from someone who understands the instrument. You can always raise the magnification of a compound microscope — swap in a shorter eyepiece, lengthen the tube — but past a point the extra magnification shows you nothing new. It enlarges the image and the blur together, giving "empty magnification": a bigger, fuzzier picture with no more detail. What sets the real limit is resolving power: the smallest separation between two points that the instrument can still show as two, rather than one smudge. Because light is a wave, an objective of finite aperture cannot focus a point to a point — it focuses it to a small disc — and two points closer than about half a wavelength's worth of that disc merge. Ernst Abbe put a number on it in 1873:

d_{\min} \approx \frac{\lambda}{2\,\text{NA}} \qquad (5)

where: $\lambda$ is the wavelength of light, $\text{NA}$ is the objective's numerical aperture (a measure of the cone of light it collects), and $d_{\min}$ is the finest resolvable detail. The formula is quoted here — its proof belongs to wave optics, the next chapter — but its message is decisive. For green light, $\lambda \approx 550\ \mathrm{nm}$ , and the best oil-immersion objectives reach $\text{NA} \approx 1.4$ , so $d_{\min} \approx 550/(2 \times 1.4) \approx 196\ \mathrm{nm}$ , about $0.2\ \mathrm{micrometres}$ — a five-thousandth of a millimetre, and no arrangement of glass lenses does better. To see finer, you must use a shorter wavelength: the electron microscope, using electron waves thousands of times shorter than light, is the answer, and it is why magnification alone is never the figure of merit. This limit reappears, in angular form, as the resolving power of a telescope in the next article.

Derivation AThe simple microscope magnifying power M = 1 + D/f

The goal is to derive the magnifying power of a single converging lens used as a magnifier, $M = 1 + D/f$ for the final image at the near point (and $M = D/f$ relaxed), landing the anchor $M = 6$ for $f = 5\ \mathrm{cm}$ . The strategy is to write $M$ as the ratio of two visual angles — the angle the image subtends through the lens, over the angle the bare object would subtend at the near point — and use the small-angle approximation and the thin-lens formula to reduce it. Magnifying power is always an angular comparison; that is the physical heart of every instrument formula.

Step 1. Fix the reference angle. Without the lens, the closest you can hold an object of height $h$ and still see it sharply is the near point $D$ , where it subtends $\alpha \approx h/D$ (small angle, so $\tan\alpha \approx \alpha$ ). This is the best the naked eye can do, and it is the denominator of $M$ .

Step 2. Insert the lens with the object at distance $|u|$ inside the focus, arranged so the virtual image lands at the near point. The eye, close to the lens, sees the object subtend $\beta \approx h/|u|$ — the object is now only $|u|$ away instead of $D$ . So $M = \beta/\alpha = (h/|u|)/(h/D) = D/|u|$ .

Step 3. Eliminate $|u|$ with the thin-lens formula. The virtual image at the near point means $v = -D$ ; the converging lens has $f > 0$ . From $\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$ with $v = -D$ and $u = -|u|$ : $-\dfrac{1}{D} + \dfrac{1}{|u|} = \dfrac{1}{f}$ , so $\dfrac{1}{|u|} = \dfrac{1}{f} + \dfrac{1}{D}$ .

Step 4. Substitute back: $M = \dfrac{D}{|u|} = D\left(\dfrac{1}{f} + \dfrac{1}{D}\right) = \dfrac{D}{f} + 1 = 1 + \dfrac{D}{f}$ . With $D = 25\ \mathrm{cm}$ , $f = 5\ \mathrm{cm}$ : $M = 1 + 5 = 6$ . For the relaxed eye the image goes to infinity, the object sits at the focus ( $|u| = f$ ), and the " $+1$ " disappears: $M = D/f = 25/5 = 5$ . The whole difference between the two settings is that single accommodation. Verified: M = 1 + 25/5 = 6 (near point) and M = 25/5 = 5 (relaxed), and P = 1/0.05 = 20 D, are recomputed in verify_hs2-microscopes.py.

Go deeper · Derivation B · 1 minThe compound microscope magnifying power M = (L/f_o)(D/f_e)

Derivation BThe compound microscope magnifying power M = (L/f_o)(D/f_e)

The goal is to show that the compound microscope's total magnifying power is the product of an objective linear magnification $L/f_o$ and an eyepiece angular magnification $D/f_e$ , giving $M = 100$ for the standard set. The strategy is to treat the two lenses in turn: the objective makes a real image whose linear magnification we read off, and the eyepiece then works exactly as the simple microscope of the previous derivation on that real image. The magnifications multiply because the second stage acts on the output of the first.

Step 1. The objective's job. The object sits just beyond the objective's focus, so the objective forms a real, inverted image at a distance $v_o$ down the tube, with linear magnification $m_o = v_o/|u_o|$ . Because the object is placed just outside the focus, $|u_o| \approx f_o$ , and the image forms near the far end of the tube, $v_o \approx L$ (the tube length). Hence $m_o \approx L/f_o$ — for the standard set, $20/1 = 20$ .

Step 2. Hand the real image to the eyepiece. That real, magnified image now plays the role of "object" for the eyepiece, which is used exactly as a simple microscope. From the previous derivation, the eyepiece contributes an angular magnification $M_e = D/f_e$ with the final image at infinity (relaxed), or $M_e = 1 + D/f_e$ at the near point. For the standard set, relaxed $M_e = 25/5 = 5$ .

Step 3. Multiply the stages. The overall magnifying power is the objective's linear magnification times the eyepiece's angular magnification, because the eyepiece magnifies the image the objective already enlarged: $M = m_o \times M_e = \dfrac{L}{f_o}\cdot\dfrac{D}{f_e}$ . This product is the defining feature — two lenses of modest power reach a magnification neither could alone.

Step 4. Evaluate. Relaxed: $M = (20/1)(25/5) = 20 \times 5 = 100$ . At the near point the eyepiece gains its " $+1$ ": $M = (20/1)(1 + 25/5) = 20 \times 6 = 120$ . The near-point setting buys a fifth more magnification at the cost of accommodating the eye — the same trade as the simple microscope, now applied to the eyepiece stage only. Verified: m_o = L/f_o = 20, M = (20/1)(25/5) = 100 relaxed and (20/1)(1 + 25/5) = 120 near point are recomputed in verify_hs2-microscopes.py.

Limiting cases

Case	Result	Meaning
simple, near point vs relaxed	$M = 1 + D/f$ vs $D/f$	the near-point setting adds exactly $+1$ to the magnification
$f \to$ small	$M$ large	shorter focal length, more magnification — the jeweller's fat loupe
$f \geq D$	$M \leq 2$	a weak lens barely magnifies; a magnifier needs $f \ll D$
compound, $f_o \to$ small	$M$ large	the objective is made as short-focus as optics allow
magnification raised past the Abbe limit	no new detail	"empty magnification" — the blur enlarges with the image
$\lambda \to$ shorter (electrons)	$d_{\min}$ smaller	finer resolution — the electron microscope's whole reason to exist

Common trapMore magnification is not more detail — and the compound magnification multiplies, it does not add

Two misconceptions cost marks and understanding here. The first is believing that magnification is the microscope's figure of merit — that a "1000×" instrument beats a "400×" one at showing detail. It does not, past the point where the aperture's resolving power runs out: beyond about $0.2\ \mathrm{micrometres}$ for visible light, extra magnification only enlarges the same blur, which is why microscope makers quote numerical aperture, not just magnification, and why the electron microscope exists at all. The second is arithmetic: the compound magnification is a product of the two stages, $M = (L/f_o)(D/f_e)$ , never a sum. A student who writes $L/f_o + D/f_e$ turns a $100\times$ instrument into a $25\times$ one and loses the whole question. Keep the two ideas straight — resolving power caps usefulness, and the two lens stages multiply — and both the conceptual mark and the numerical mark are secure. State, too, which adjustment (near point or relaxed) each number belongs to, since the " $+1$ " changes the answer.

Worked examples

Board level — a magnifier and a compound microscope (CBSE and NEET pattern)

(a) Find the magnifying power of a converging lens of focal length $5\ \mathrm{cm}$ used as a simple microscope, for a final image (i) at the near point and (ii) at infinity. (b) A compound microscope has objective focal length $1\ \mathrm{cm}$ , eyepiece focal length $5\ \mathrm{cm}$ and tube length $20\ \mathrm{cm}$ . Find its magnifying power for a relaxed eye and for the near-point setting. Take $D = 25\ \mathrm{cm}$ .

Step 1. Simple microscope, near point: $M = 1 + D/f = 1 + 25/5 = \boxed{6}$ . Relaxed: $M = D/f = 25/5 = \boxed{5}$ . The difference is the single " $+1$ ", the accommodation the near-point view demands.

Step 2. Compound microscope, relaxed: apply $M = (L/f_o)(D/f_e) = (20/1)(25/5) = 20 \times 5 = \boxed{100}$ .

Step 3. Compound microscope, near point: apply $M = (L/f_o)(1 + D/f_e) = (20/1)(1 + 25/5) = 20 \times 6 = \boxed{120}$ .

Step 4. Sense-check the split. The objective alone does $L/f_o = 20\times$ ; the eyepiece does $5\times$ (relaxed) or $6\times$ (near); their product is the instrument's power. Notice that a single lens could never reach $100\times$ — that is the entire point of stacking two.

Competitive level — design the objective from the magnification (JEE-Main and NEET pattern)

A compound microscope is required to give a magnifying power of $100$ for a relaxed eye, using an eyepiece of focal length $5\ \mathrm{cm}$ and a tube length of $20\ \mathrm{cm}$ . Find the focal length of the objective and the linear magnification it must provide. Take $D = 25\ \mathrm{cm}$ .

Step 1. Write the relaxed formula and put in what you know: $M = \dfrac{L}{f_o}\cdot\dfrac{D}{f_e}$ , so $100 = \dfrac{20}{f_o}\cdot\dfrac{25}{5} = \dfrac{20}{f_o}\times 5 = \dfrac{100}{f_o}$ .

Step 2. Solve for the objective: $f_o = \dfrac{100}{100} = \boxed{1\ \mathrm{cm}}$ — a very short-focus lens, exactly as the design demands.

Step 3. Read off the objective's own linear magnification: $m_o = L/f_o = 20/1 = \boxed{20}$ . So the objective supplies $20\times$ and the eyepiece the remaining $5\times$ , and $20 \times 5 = 100$ closes the loop.

Step 4. Note the recurring move: a "design" question fixes $M$ and one or two focal lengths and asks for the missing one. Write the magnifying-power formula, substitute, and solve the single unknown — never re-derive from ray diagrams under time pressure. Timing note: about a minute. The trap that eats time is forgetting which adjustment the target $M$ refers to; "relaxed" uses $D/f_e$ , "near point" uses $1 + D/f_e$ , and using the wrong one throws $f_o$ off. Read the adjustment first, then solve.

Exam traps

Adding the stages instead of multiplying. The compound magnification is $M = (L/f_o)(D/f_e)$ , a product. Writing $L/f_o + D/f_e$ is the single most common error and collapses $100$ to $25$ . Two lenses in series multiply.

Confusing the two adjustments. Near point uses the " $+1$ ": simple $M = 1 + D/f$ , compound $M = (L/f_o)(1 + D/f_e)$ . Relaxed drops it: $D/f$ and $(L/f_o)(D/f_e)$ . State which you are using, because the numbers differ ( $6$ vs $5$ , $120$ vs $100$ ).

Treating tube length as the sum of focal lengths. In a microscope $L$ is the objective-to-eyepiece image separation (the tube length), not $f_o + f_e$ . That " $f_o + f_e$ " rule belongs to the telescope, next article — do not carry it here.

Believing bigger magnification means finer detail. Past the resolving-power limit, extra magnification is empty. Detail is capped by the wavelength of light and the objective's aperture, not by the focal lengths — quote numerical aperture, not just " $\times$ ".

Board scope. The compound-microscope magnification is examined by CBSE, ICSE, JEE-Main, JEE-Advanced and NEET. IGCSE 0625 covers the magnifying glass qualitatively (a single converging lens, $m =$ image height over object height) but not the compound formula. AP Physics 2 carries no optical-instrument magnification (its geometric optics stops at lenses and mirrors), and IB and A-Level have no geometric optics of instruments. If your board is one of those, the compound-microscope formula is background, not syllabus.

Origins · 1 minErnst Abbe and how this was discovered

Ernst Abbe

1840–1905 · Jena

Abbe is the physicist who turned microscope-making from a craft into a science. Hired in the 1860s by the instrument-maker Carl Zeiss, whose lenses were superb but designed by trial and error, Abbe worked out the theory of image formation in the microscope — and in 1873 published the result that reframed the whole field: the finest detail an objective can resolve is set not by its magnification but by diffraction, $d \approx \lambda/2\,\text{NA}$ , where NA is the numerical aperture he named. He derived the Abbe sine condition for aberration-free imaging and defined the numerical aperture that microscope objectives are still labelled with today. His work explains exactly why this article's magnifying-power formulas, true as they are, cannot be the last word: past the diffraction limit, magnification is empty. The elementary $M = (L/f_o)(D/f_e)$ is older ray optics; Abbe is why resolution, not magnification, is the real limit — and, with Zeiss and the glass chemist Otto Schott, he built the company that made that theory into the world's microscopes.

History

1595

— Zacharias Janssen is credited with an early compound microscope in the Netherlands.

1665

— Robert Hooke's Micrographia popularises the compound microscope and coins the word cell.

1676

— Antonie van Leeuwenhoek observes micro-organisms through a single powerful lens.

1873

— Ernst Abbe formulates the diffraction limit that caps a microscope's useful resolution.

1931

— Ernst Ruska builds the first electron microscope, passing the light-diffraction limit.

Going further

Texts. NCERT Physics Class 12, Ch. 9 ("Ray Optics and Optical Instruments"), the simple and compound microscopes and their magnifying powers · H. C. Verma, Concepts of Physics Vol. 1, "Optical Instruments" · D. C. Pandey, Optics and Modern Physics, the microscope problem set for JEE and NEET.

Provenance. Standard optics. Recompute: simple $M = 1 + 25/5 = 6$ (near point) and $25/5 = 5$ (relaxed), magnifier power $1/0.05 = 20\ \mathrm{D}$ ; compound $m_o = 20/1 = 20$ , $M = (20/1)(25/5) = 100$ (relaxed) and $(20/1)(1 + 25/5) = 120$ (near point); the design inversion $100 = (20/f_o)(25/5) \Rightarrow f_o = 1\ \mathrm{cm}$ , $m_o = 20$ ; and the Abbe resolving limit $d = 550\ \mathrm{nm}/(2 \times 1.4) \approx 196\ \mathrm{nm} \approx 0.2\ \mathrm{micrometres}$ .

Prerequisites. The Human Eye and Its Defects — the near point $D = 25\ \mathrm{cm}$ and the idea of visual angle that every magnifying power is built on. The thin-lens formula $1/v - 1/u = 1/f$ and lens power in dioptres were established earlier in this chapter and are quoted here. Leads to. Refracting and Reflecting Telescopes — the same two-lens idea turned outward to magnify the angle of far objects, where the objective is made long, not short · Instrument Magnification Numericals — the drill that runs every microscope and telescope case.

Go deeper (UG). The wave-optical theory of resolution — Abbe's diffraction limit derived, the Rayleigh criterion, numerical aperture and the point-spread function — is developed in Optical Instruments in the university optics track, and the wave physics behind it opens in the next chapter of this subject. Nothing on this page needs it; the magnifying-power formulas are pure ray optics.

What is the difference between magnification and resolving power?

Magnification is how much bigger the image looks; resolving power is how much finer detail you can actually distinguish. They are independent, and confusing them is the deepest misconception in the topic. You can magnify an image as much as you like by choosing shorter focal lengths, but if two features are closer than the resolving limit — about $0.2\ \mathrm{micrometres}$ for visible light — no magnification will ever separate them; you just get a bigger blur. Resolving power is set by the wavelength of light and the objective's aperture (Abbe's $d \approx \lambda/2\,\text{NA}$ ), not by the lenses' focal lengths. A good microscope is specified by both: enough magnification to present the detail comfortably, and enough numerical aperture to resolve it in the first place.

Why is the objective's focal length made so short?

Because the objective's contribution to the magnification is $L/f_o$ , so the shorter $f_o$ is, the more it magnifies. A $1\ \mathrm{cm}$ objective over a $20\ \mathrm{cm}$ tube gives $20\times$ on its own; halving $f_o$ would double that. The eyepiece cannot be made arbitrarily short (a very tiny lens is impossible to look through and swamped by aberrations), so the heavy lifting is done by the objective. This is the exact opposite of the telescope in the next article, where the objective is made as long-focus as possible — because a telescope's magnification is $f_o/f_e$ , and there a long objective wins. Same two lenses, opposite priorities, because one instrument magnifies near objects and the other magnifies angles of far ones.

Why is the image in a compound microscope inverted, and does it matter?

The objective forms a real, inverted image (as any converging lens does for an object beyond its focus), and the eyepiece then presents that inverted image erect-relative-to-itself but still inverted relative to the original object — so what you see is upside down and left-right reversed. For looking at pollen or a cell it does not matter, which is why biological microscopes leave it inverted and save the cost and light-loss of an extra erecting lens. It matters when you need a correctly-oriented view — dissecting instruments and telescopes for terrestrial use add an erecting system. For exams, just be ready to state that the compound microscope's final image is virtual, inverted and magnified.

What is tube length, exactly, and is it f_o + f_e?

No — and mixing this up with the telescope is a classic slip. In a compound microscope the tube length $L$ is essentially the distance between the back focus of the objective and the front focus of the eyepiece, which is where the objective throws its real image. It is the span over which the first-stage magnification builds up, and it appears in the formula as $L/f_o$ . It is not $f_o + f_e$ : that sum is the tube length of a telescope in normal adjustment, a different instrument with a different geometry. When you see a microscope problem, use $L$ as given; when you see a telescope problem, the length is $f_o + f_e$ .

Why does the near-point setting give more magnification than the relaxed one?

Because the near-point setting squeezes the image to the closest distance the eye can still focus, $25\ \mathrm{cm}$ , which subtends the largest possible angle — and magnifying power is all about angle. Mathematically it adds the " $+1$ ": the simple microscope goes from $D/f$ to $1 + D/f$ , and the compound's eyepiece from $D/f_e$ to $1 + D/f_e$ . The cost is that your eye must accommodate (squeeze its own lens) to focus a near image, which tires it over long use. The relaxed setting sends the final image to infinity, slightly less magnification but comfortable for hours at the bench. Neither is "right" — state which one a given number assumes, because boards mark the distinction.

Is a magnifying glass really a microscope?

Yes — a simple microscope is exactly a magnifying glass: one converging lens with the object held inside its focal length. The grand name just distinguishes it from the compound microscope, which uses two lenses in series. Physically it does the humblest version of the same job: it lets you hold an object far closer than the $25\ \mathrm{cm}$ near point while keeping it in focus, so it subtends a bigger angle and looks larger. Its magnifying power, $1 + D/f$ (or $D/f$ relaxed), is modest — a $5\ \mathrm{cm}$ lens gives only $6\times$ — which is why serious magnification needs the two-stage compound instrument. But the underlying idea, magnifying an angle rather than a size, is identical.