ARCHETYPE: Instrument Magnification Numericals

By the end of the optical-instruments chapter you own half a dozen magnifying-power formulas — simple microscope, compound microscope, astronomical telescope, each in two adjustments, plus the two eye-correction powers — and an exam will hand you one number and ask for another. The trap is not the physics; it is picking the wrong row of the table under time pressure, or letting the microscope's short objective bleed into a telescope problem, or leaving a focal length in centimetres when a power in dioptres needs metres. This article is the drill that makes the choice automatic. There is really one master table and one reflex. Name the instrument and the adjustment, pick the matching row, substitute, and — for the "design" questions that separate top scorers — invert the formula to find the focal length that delivers a required magnification. A compound microscope reaches $M = 100$ from a $1\ \mathrm{cm}$ objective and a $5\ \mathrm{cm}$ eyepiece a tube apart; an astronomical telescope reaches $M = 20$ from a $100\ \mathrm{cm}$ objective and the same $5\ \mathrm{cm}$ eyepiece; a magnifying glass reaches $M = 6$ from a $5\ \mathrm{cm}$ lens; and a short-sighted eye is fixed by a $-0.5\ \mathrm{D}$ lens. Those four anchors — $100$ , $20$ , $6$ , $-0.5$ — are the spine of the whole problem set. The single idea underneath every formula is the one the eye taught us: instruments magnify the angle a thing subtends, $M = \beta/\alpha$ , not its true size. Keep that in view and the two opposite instincts fall into place — the microscope wants a short objective because it magnifies a near object, the telescope wants a long one because it magnifies a far angle — and the numericals become mechanical. We fix the master recipe, tabulate every row, then run the four canonical problems: a compound microscope, a telescope, a magnifier design, and an eye correction.

Key takeaways

One master table of magnifying powers: simple microscope $M = 1 + D/f$ (near) or $D/f$ (relaxed); compound microscope $M = (L/f_o)(D/f_e)$ (relaxed) or $(L/f_o)(1 + D/f_e)$ (near); astronomical telescope $M = f_o/f_e$ (normal) or $(f_o/f_e)(1 + f_e/D)$ (near), with tube length $L = f_o + f_e$ .
Plus eye-correction power: myopia $P = -1/(\text{far point})$ , hypermetropia $P = 1/0.25 - 1/(\text{near point})$ , in metres.
Compound microscope ( $f_o = 1$ , $f_e = 5$ , $L = 20\ \mathrm{cm}$ ): $M = 100$ (relaxed), $120$ (near point).
Astronomical telescope ( $f_o = 100$ , $f_e = 5\ \mathrm{cm}$ ): $M = 20$ (normal), $24$ (near point), $L = 105\ \mathrm{cm}$ .
Design inversions: a magnifier of $M = 6$ needs $f = 5\ \mathrm{cm}$ ( $+20\ \mathrm{D}$ ); a myope with far point $2\ \mathrm{m}$ needs $-0.5\ \mathrm{D}$ ; a hypermetrope with near point $0.5\ \mathrm{m}$ needs $+2.0\ \mathrm{D}$ .
Key idea: every instrument numerical is name-the-instrument, name-the-adjustment, pick-the-row, substitute — or invert the row for a design question — with focal lengths in the same units and lens powers computed with metres.

The core insight

Every optical-instrument numerical reduces to one master table of angular magnifications — simple and compound microscopes, the astronomical telescope, each in relaxed and near-point form, plus the two eye-correction powers — solved by naming the instrument and adjustment, reading the row, and substituting; design questions just invert the row to find the focal length that gives a target magnification, always keeping powers in dioptres with distances in metres.

Where this lives today

The master table is quietly present in every optical instrument you meet. A pair of $8\times42$ binoculars is a telescope with $M = f_o/f_e = 8$ and a $42\ \mathrm{mm}$ aperture, folded compact by prisms; a jeweller's loupe stamped " $10\times$ " is a simple microscope with $f = D/M \approx 2.5\ \mathrm{cm}$ ; a laboratory microscope's objective and eyepiece each carry their magnification etched on the barrel, and you multiply them exactly as equation (2) says. An optometrist's prescription is equation (4): $-0.5$ on the sphere line is the myopia lens of the worked example. Camera and telescope makers still design to these rows, then hand the last step to software — a phone's "digital zoom" is empty magnification past the lens's real limit, the same warning the microscope and telescope articles gave about resolving power. The formulas are two centuries old and utterly current: every time you focus binoculars, read a microscope's total magnification, or collect a pair of glasses, you are reading a row off this table.

The setup — one table, one reflex

An optical instrument does not make things bigger; it makes them subtend a bigger angle at your eye, and every magnifying power is the ratio $M = \beta/\alpha$ of the angle with the instrument to the angle without. That single idea, plus the near point $D = 25\ \mathrm{cm}$ and the thin-lens formula from earlier in this chapter, generates the entire table below. The reflex for any problem is four beats: name the instrument, name the adjustment (relaxed or near point for the eye's comfort; normal or near point for a telescope), pick the matching row, and substitute in consistent units. When a problem gives the magnification and asks for a focal length instead — a design question — you run the same row backwards.

Figure

The master table of magnifying powers: simple and compound microscopes and the astronomical telescope, each in its relaxed and near-point form, with the anchor numbers 6, 100 and 20

Two contrasts in the table earn their own mention because they are the traps. The microscope wants a short objective (its magnification rises as $1/f_o$ ); the telescope wants a long one (its magnification rises as $f_o$ ). And the "tube length" means different things: for the microscope $L$ is the objective-to-eyepiece image span that appears in $L/f_o$ , while for the telescope $L = f_o + f_e$ , the sum of focal lengths. Mix these up and a right method gives a wrong number.

The rows, in full

Here is the master table as formulas, with the near point $D = 25\ \mathrm{cm}$ throughout. The simple microscope (magnifying glass, one converging lens with the object inside its focus):

M = 1 + \frac{D}{f}\ \ (\text{near point}), \qquad M = \frac{D}{f}\ \ (\text{relaxed}) \qquad (1)

The compound microscope (short objective $f_o$ , eyepiece $f_e$ , tube length $L$ ):

M = \frac{L}{f_o}\cdot\frac{D}{f_e}\ \ (\text{relaxed}), \qquad M = \frac{L}{f_o}\left(1 + \frac{D}{f_e}\right)\ \ (\text{near point}) \qquad (2)

The astronomical telescope (long objective $f_o$ , eyepiece $f_e$ , tube length $L = f_o + f_e$ ):

M = \frac{f_o}{f_e}\ \ (\text{normal}), \qquad M = \frac{f_o}{f_e}\left(1 + \frac{f_e}{D}\right)\ \ (\text{near point}) \qquad (3)

And the eye corrections, powers in dioptres with all distances in metres — a diverging lens for myopia, a converging lens for hypermetropia:

P_{\text{myopia}} = -\frac{1}{x_f}, \qquad P_{\text{hypermetropia}} = \frac{1}{0.25} - \frac{1}{x_n} \qquad (4)

where: $x_f$ is the far point and $x_n$ the near point of the defective eye. That is the whole toolkit. Every numerical below is one of these rows read forward (given the focal lengths, find $M$ ) or backward (given $M$ , find a focal length). The near-point form of each instrument carries the small premium — the " $+1$ " or the $(1 + f_e/D)$ factor — that the relaxed/normal form drops.

Figure

The opposite priorities of the two instruments: a microscope uses a very short objective to magnify a near object, a telescope uses a long objective to magnify the angle of a far one

Derivation AThe master table, forward — compound microscope and telescope

The goal is to run the two hardest rows forward and land their anchors: the compound microscope at $M = 100$ (relaxed) and $120$ (near point), and the astronomical telescope at $M = 20$ (normal) and $24$ (near point). The strategy is the same $M = \beta/\alpha$ for both, treating each instrument as a two-stage angle magnifier, with the formulas quoted from the microscope and telescope articles and the arithmetic done cleanly.

Step 1. Compound microscope, relaxed. The objective's linear magnification is $m_o = L/f_o$ and the eyepiece's angular magnification is $M_e = D/f_e$ ; the total is their product. With $f_o = 1$ , $f_e = 5$ , $L = 20$ , $D = 25\ \mathrm{cm}$ : $M = \dfrac{L}{f_o}\cdot\dfrac{D}{f_e} = \dfrac{20}{1}\cdot\dfrac{25}{5} = 20 \times 5 = 100$ .

Step 2. Compound microscope, near point. The eyepiece gains its " $+1$ ": $M = \dfrac{L}{f_o}\left(1 + \dfrac{D}{f_e}\right) = \dfrac{20}{1}\left(1 + \dfrac{25}{5}\right) = 20 \times 6 = 120$ — a fifth more, at the cost of accommodating the eye.

Step 3. Astronomical telescope, normal. Magnification is the ratio of focal lengths, $M = \dfrac{f_o}{f_e} = \dfrac{100}{5} = 20$ , and the tube length is $L = f_o + f_e = 105\ \mathrm{cm}$ .

Step 4. Astronomical telescope, near point. The eyepiece acts as a near-point magnifier, adding the factor $(1 + f_e/D)$ : $M = \dfrac{f_o}{f_e}\left(1 + \dfrac{f_e}{D}\right) = \dfrac{100}{5}\left(1 + \dfrac{5}{25}\right) = 20 \times 1.2 = 24$ . Notice the two instruments' opposite objectives — the microscope's $f_o = 1\ \mathrm{cm}$ , the telescope's $f_o = 100\ \mathrm{cm}$ — with the same $5\ \mathrm{cm}$ eyepiece. Verified: compound M = 100 (relaxed) and 120 (near point); telescope M = 20 (normal), 24 (near point), L = 105 cm, are recomputed in verify_hs2-instrument-magnification.py.

Go deeper · Derivation B · 1 minThe master table, backward — magnifier design and eye correction

Derivation BThe master table, backward — magnifier design and eye correction

The goal is the inverse problems that separate top scorers: designing a magnifier to a required magnification ( $M = 6 \Rightarrow f = 5\ \mathrm{cm}$ , $P = +20\ \mathrm{D}$ ) and prescribing a corrective lens ( $-0.5\ \mathrm{D}$ for a myope, $+2.0\ \mathrm{D}$ for a hypermetrope). The strategy is identical for all inversions: write the forward row, substitute the known magnification or distances, and solve for the missing focal length or power.

Step 1. Magnifier design. A simple microscope must give $M = 6$ at the near point, so use $M = 1 + D/f$ and solve for $f$ : $6 = 1 + \dfrac{25}{f} \Rightarrow \dfrac{25}{f} = 5 \Rightarrow f = \dfrac{25}{5} = 5\ \mathrm{cm}$ .

Step 2. Convert to power: $P = 1/f = 1/0.05\ \mathrm{m} = +20\ \mathrm{D}$ — a strong converging lens, the jeweller's loupe. That is the whole design: the required magnification fixes the focal length, which fixes the power.

Step 3. Myopia correction. A short-sighted eye with far point $x_f = 2\ \mathrm{m}$ needs a diverging lens that images infinity at the far point: $P = -1/x_f = -1/2 = -0.5\ \mathrm{D}$ (focal length $-200\ \mathrm{cm}$ ). The minus sign marks the diverging lens.

Step 4. Hypermetropia correction. A long-sighted eye with near point $x_n = 0.5\ \mathrm{m}$ , wanting to read at $25\ \mathrm{cm}$ , needs a converging lens: $P = \dfrac{1}{0.25} - \dfrac{1}{0.5} = 4 - 2 = +2.0\ \mathrm{D}$ (focal length $+50\ \mathrm{cm}$ ). The plus sign marks the converging lens. Every design is one row solved for the unknown — never re-derived from scratch. Verified: magnifier M = 6 ⇒ f = 5 cm, P = +20 D; myopia P = -0.5 D; hypermetropia P = +2.0 D, are recomputed in verify_hs2-instrument-magnification.py.

Limiting cases and the row-picking map

Situation	Row to use	Watch for
magnifying glass, image at 25 cm	$M = 1 + D/f$	near point adds the $+1$
magnifying glass, relaxed eye	$M = D/f$	one less than the near-point value
compound microscope	$M = (L/f_o)(D/f_e)$	multiply the stages; short $f_o$
astronomical telescope	$M = f_o/f_e$	long $f_o$ ; tube length $f_o + f_e$
near-point telescope	$M = (f_o/f_e)(1 + f_e/D)$	the $\times 1.2$ premium
short sight (far point given)	$P = -1/x_f$	metres; negative (diverging)
long sight (near point given)	$P = 1/0.25 - 1/x_n$	metres; positive (converging)
design question ("find $f$ for $M$ ")	invert the matching row	solve for the missing symbol

Common trapPick the row before you compute — and never cross the microscope and the telescope

Almost every lost mark here is a row-selection error, not an arithmetic one. Three disciplines prevent it. First, the microscope's magnification uses the objective as $L/f_o$ and wants it short; the telescope's uses $f_o/f_e$ and wants it long — never carry one instrument's instinct into the other. Second, "tube length" is not one thing: for the telescope it is $f_o + f_e$ , for the compound microscope it is the image-throw $L$ that appears inside $L/f_o$ — do not compute $f_o + f_e$ for a microscope. Third, the eye-correction powers demand metres: a far point of $2\ \mathrm{m}$ gives $-0.5\ \mathrm{D}$ , but written as $200\ \mathrm{cm}$ it silently becomes $-0.005$ . And always state the adjustment beside the answer — relaxed or near point for a microscope, normal or near point for a telescope — because the " $+1$ " or the $(1 + f_e/D)$ factor changes the number ( $5$ vs $6$ , $20$ vs $24$ , $100$ vs $120$ ). Name the instrument, name the adjustment, pick the row, then substitute.

Worked examples

1 · Board level — a compound microscope in both settings (CBSE and NEET pattern)

A compound microscope has an objective of focal length $1\ \mathrm{cm}$ , an eyepiece of focal length $5\ \mathrm{cm}$ , and a tube length of $20\ \mathrm{cm}$ . Find its magnifying power (a) for a relaxed eye and (b) with the final image at the near point of $25\ \mathrm{cm}$ .

Step 1. Name the row: compound microscope, so $M = (L/f_o)(D/f_e)$ relaxed, or $(L/f_o)(1 + D/f_e)$ at the near point. All lengths are already in centimetres.

Step 2. Relaxed: $M = \dfrac{20}{1}\cdot\dfrac{25}{5} = 20 \times 5 = \boxed{100}$ .

Step 3. Near point: $M = \dfrac{20}{1}\left(1 + \dfrac{25}{5}\right) = 20 \times 6 = \boxed{120}$ .

Step 4. Sense-check: the objective contributes $L/f_o = 20\times$ , the eyepiece $5\times$ (relaxed) or $6\times$ (near), and their product is the instrument's power. The near-point setting is $20\%$ larger, the standard premium — and a single lens could never reach $100$ , which is the point of the two-stage design.

2 · Board level — an astronomical telescope (CBSE and ICSE pattern)

An astronomical telescope has an objective of focal length $100\ \mathrm{cm}$ and an eyepiece of focal length $5\ \mathrm{cm}$ . Find its angular magnification and tube length in normal adjustment, and its magnification at the near point.

Step 1. Name the row: astronomical telescope, so $M = f_o/f_e$ (normal) with $L = f_o + f_e$ , and $M = (f_o/f_e)(1 + f_e/D)$ at the near point.

Step 2. Normal adjustment: $M = \dfrac{100}{5} = \boxed{20}$ , tube length $L = 100 + 5 = \boxed{105\ \mathrm{cm}}$ .

Step 3. Near point: $M = \dfrac{100}{5}\left(1 + \dfrac{5}{25}\right) = 20 \times 1.2 = \boxed{24}$ .

Step 4. Contrast with Example 1: same $5\ \mathrm{cm}$ eyepiece, but the objective here is $100\ \mathrm{cm}$ (long) instead of $1\ \mathrm{cm}$ (short), and the magnification is a ratio $f_o/f_e$ rather than a product. That single swap — long objective, ratio formula — is the whole difference between the two instruments.

3 · Competitive level — design a magnifier to a required power (JEE-Main and NEET pattern)

A simple microscope (magnifying glass) is required to give a magnifying power of $6$ with the final image at the near point of $25\ \mathrm{cm}$ . Find the focal length of the lens and its power in dioptres.

Step 1. Name and invert the row: near-point simple microscope, $M = 1 + D/f$ , solved for $f$ given $M = 6$ .

Step 2. $6 = 1 + \dfrac{25}{f} \Rightarrow \dfrac{25}{f} = 5 \Rightarrow f = \dfrac{25}{5} = \boxed{5\ \mathrm{cm}}$ .

Step 3. Convert to power (focal length in metres): $P = \dfrac{1}{f} = \dfrac{1}{0.05} = \boxed{+20\ \mathrm{D}}$ — a strong converging lens, exactly a jeweller's loupe.

Step 4. Note the reflex: a design question fixes $M$ and asks for $f$ (or $P$ ). Write the forward row, substitute the magnification, solve the one unknown, and convert to metres before quoting a power. Do not re-derive from ray diagrams. Timing note: under a minute. The only slip is using the relaxed row $M = D/f$ (which would give $f = 25/6 = 4.17\ \mathrm{cm}$ ) when the question said near point; read the adjustment, then invert the right row.

4 · Competitive level — two eye corrections (JEE-Main and NEET pattern)

(a) A short-sighted person cannot see beyond $2\ \mathrm{m}$ . (b) A long-sighted person has a near point of $0.5\ \mathrm{m}$ and wants to read at $25\ \mathrm{cm}$ . Find the spectacle power each needs.

Step 1. Diagnose from the symptom: "cannot see beyond $2\ \mathrm{m}$ " is a finite far point — myopia, diverging lens; "near point at $0.5\ \mathrm{m}$ " is a receded near point — hypermetropia, converging lens.

Step 2. Myopia, part (a): $P = -\dfrac{1}{x_f} = -\dfrac{1}{2} = \boxed{-0.5\ \mathrm{D}}$ (focal length $-200\ \mathrm{cm}$ ). The minus confirms a diverging lens.

Step 3. Hypermetropia, part (b): $P = \dfrac{1}{0.25} - \dfrac{1}{x_n} = \dfrac{1}{0.25} - \dfrac{1}{0.5} = 4 - 2 = \boxed{+2.0\ \mathrm{D}}$ (focal length $+50\ \mathrm{cm}$ ). The plus confirms a converging lens.

Step 4. Read the signs back: negative for short sight, positive for long sight. These are the same two eye-defect anchors from the human-eye article, now sitting in the same master table as the microscope and telescope — a reminder that a spectacle lens is just another instrument row solved for its power. Timing note: about forty-five seconds for both. The two dangers are the concave/convex swap (decide the defect from the symptom first) and leaving a distance in centimetres before taking the reciprocal. Metres, then reciprocal, then read the sign.

Exam traps

Picking the wrong row. Most errors here are selection, not arithmetic. Name the instrument and the adjustment before you write anything, and pull the matching formula from the table. A telescope answered with a microscope's product formula is instantly wrong.

Crossing microscope and telescope. The microscope wants a short objective (magnification $\propto 1/f_o$ ); the telescope wants a long one (magnification $= f_o/f_e$ ). And "tube length" is $f_o + f_e$ for the telescope but the image-throw $L$ for the microscope. Keep the two instruments' priorities separate.

Forgetting the adjustment premium. Near point adds the " $+1$ " (simple and compound microscope) or the factor $(1 + f_e/D)$ (telescope). Relaxed/normal drops it. The numbers differ — $6$ vs $5$ , $120$ vs $100$ , $24$ vs $20$ — so state which adjustment your answer assumes.

Power in centimetres. The eye-correction powers need distances in metres: far point $2\ \mathrm{m} \to -0.5\ \mathrm{D}$ , near point $0.5\ \mathrm{m} \to$ the $1/0.5 = 2$ term. A stray centimetre gives a hundred-fold error.

The concave/convex swap. Myopia (short sight, finite far point) takes a diverging lens, negative power; hypermetropia (long sight, receded near point) takes a converging lens, positive power. Diagnose the defect first, and the sign follows.

Board scope. These instrument and eye-correction numericals are examined by CBSE, ICSE, JEE-Main, JEE-Advanced and NEET. They are not in AP Physics 2 (whose optics stops at single lenses and mirrors) or IGCSE 0625, and IB and A-Level carry no geometric optics of instruments. If your board is one of those, this drill is background — the Indian boards are its target.

Origins · 1 minRobert Hooke and how this was discovered

Robert Hooke

1635–1703 · London

Hooke is the man who put the compound microscope at the centre of science. His Micrographia (1665) — a sensation in its day, the first scientific bestseller — was built on his own compound microscope, and its huge fold-out engravings of a flea, a fly's eye and the "cells" of cork (a word he coined, from the tiny chambers his lenses revealed) showed the world what magnification could do. Hooke did not have the modern magnifying-power formulas — the tidy $(L/f_o)(D/f_e)$ and $f_o/f_e$ are later ray optics — but he was the pioneer who turned the compound instrument from a curiosity into a tool of science and quantified what it could resolve. He was also Newton's great rival, an inexhaustible experimenter who worked on elasticity (Hooke's law), gravitation, combustion and the wave nature of light. For this drill he stands for the instruments themselves: the microscope and, through his contemporaries, the telescope that these numericals put a magnification on.

History

1611

— Johannes Kepler, in Dioptrice, derives the magnifying power of a two-lens telescope.

1665

— Robert Hooke, in Micrographia, reveals what a compound microscope can magnify.

1758

— John Dollond's achromatic objective sharpens high-magnification instruments.

1782

— Jesse Ramsden builds his precision eyepiece for measuring instruments.

1873

— Ernst Abbe proves resolution, not magnification, sets a microscope's real limit.

Going further

Texts. NCERT Physics Class 12, Ch. 9 ("Ray Optics and Optical Instruments"), the microscope, telescope and eye sections and their exercises · H. C. Verma, Concepts of Physics Vol. 1, "Optical Instruments" · D. C. Pandey, Optics and Modern Physics, the combined instrument-magnification problem set for JEE and NEET.

Provenance. Standard optics. Recompute: simple $M = 1 + 25/5 = 6$ (near), $25/5 = 5$ (relaxed), power $1/0.05 = 20\ \mathrm{D}$ ; compound $M = (20/1)(25/5) = 100$ (relaxed), $(20/1)(1 + 25/5) = 120$ (near); telescope $M = 100/5 = 20$ (normal), $(100/5)(1 + 5/25) = 24$ (near), $L = 105\ \mathrm{cm}$ ; magnifier design $6 = 1 + 25/f \Rightarrow f = 5\ \mathrm{cm}$ ; myopia $P = -1/2 = -0.5\ \mathrm{D}$ (f = -200 cm); hypermetropia $P = 1/0.25 - 1/0.5 = +2.0\ \mathrm{D}$ (f = +50 cm).

Prerequisites. Simple and Compound Microscopes and Refracting and Reflecting Telescopes — the concept articles that derive the magnifying powers this gym drills; and The Human Eye and Its Defects, whose two correction powers sit in the same table. The thin-lens formula and dioptre power were established earlier in this chapter. Leads to. The university Optical Instruments seam, where these magnifications meet their diffraction limits.

Go deeper (UG). The full instrument optics — exit pupils, eye relief, field of view, aberration balancing, and the diffraction limits that cap useful magnification — is developed in Optical Instruments in the university optics track. Nothing here needs it; every row of the table is quoted ray optics, and the numericals are algebra.

How do I decide quickly which formula to use?

Run the four-beat reflex: name the instrument, name the adjustment, pick the row, substitute. "Instrument" is one of magnifying glass, compound microscope, astronomical telescope, or corrective lens. "Adjustment" is relaxed or near point for a microscope, normal or near point for a telescope. Once you have both, the row is fixed and the rest is arithmetic. If the question gives you a magnification and asks for a focal length, it is a design question — use the same row backwards. The whole skill is in the first two beats: pick the right row before touching the calculator, and you will not confuse a microscope's product formula with a telescope's ratio.

Why does the microscope want a short objective but the telescope a long one?

Because they magnify different things. The compound microscope magnifies a near object, and its objective contributes $L/f_o$ — so a shorter $f_o$ magnifies more. The telescope magnifies the angle of a far object, and its magnification is $f_o/f_e$ — so a longer $f_o$ magnifies more. It is the single most common cross-over error: carrying the microscope's "short objective" instinct into a telescope problem, or vice versa. Anchor on the formulas: microscope $(L/f_o)(D/f_e)$ has $f_o$ in the denominator of a product (short wins), telescope $f_o/f_e$ has $f_o$ on top of a ratio (long wins).

What is the difference between "tube length" in a microscope and in a telescope?

They are genuinely different quantities with the same name. In a compound microscope, the tube length $L$ is essentially the distance between the objective's back focus and the eyepiece's front focus — the span over which the objective throws its real image — and it appears in the magnification as $L/f_o$ . In an astronomical telescope in normal adjustment, the length is $L = f_o + f_e$ , the sum of the two focal lengths, because the objective's focal plane coincides with the eyepiece's. So for a microscope you are given $L$ directly and use it in $L/f_o$ ; for a telescope you compute $L = f_o + f_e$ . Using the telescope rule on a microscope (or the reverse) is a classic slip.

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

Because the near-point setting brings the final image as close as the eye can focus — $25\ \mathrm{cm}$ — where it subtends the largest possible angle, and magnifying power is all about angle. Mathematically it adds a premium: the simple and compound microscopes gain the " $+1$ " ( $1 + D/f$ instead of $D/f$ ; $1 + D/f_e$ instead of $D/f_e$ ), and the telescope gains the factor $(1 + f_e/D)$ . The cost is that the eye must accommodate to focus a near image, which tires it over long use, so the relaxed/normal setting (image at infinity) is preferred for comfort even though it magnifies slightly less. The exam point is always to state which adjustment your number assumes, because they differ.

When I design a lens, when do I get a focal length in centimetres versus a power in dioptres?

Keep them separate and convert deliberately. Magnification formulas ( $M = 1 + D/f$ , $M = f_o/f_e$ , etc.) work in whatever length unit you keep consistent — usually centimetres, since $D = 25\ \mathrm{cm}$ — and give you a focal length in centimetres. Power, however, is always $P = 1/f$ with $f$ in metres, so once you have $f$ in centimetres you divide by $100$ before taking the reciprocal: $f = 5\ \mathrm{cm} = 0.05\ \mathrm{m}$ gives $P = 20\ \mathrm{D}$ . The eye-correction formulas go straight to power and so demand metres from the start: a far point of $2\ \mathrm{m}$ gives $-0.5\ \mathrm{D}$ . The rule: magnifications in centimetres are fine, but any dioptre needs metres.

Are these the same formulas as in the microscope, telescope and eye articles?

Yes — this article does not introduce new physics; it collects the magnifying-power and eye-correction results derived in those concept articles into one table and drills the numericals and design inversions. The simple and compound microscope rows come from the microscope article, the telescope rows from the telescope article, and the two power formulas from the human-eye article. What is new here is the skill: choosing the right row fast, keeping the microscope and telescope from bleeding into each other, and running any row backward to design a lens to a required magnification. If a formula here feels unfamiliar, revisit the concept article that derived it — this gym assumes you have seen all of them.