Lens Power & Magnification

When an optician writes your prescription, they do not write a focal length. They write a number like " $+2.5$ " and a unit you may never have questioned: the dioptre. That number is the power of the lens you need, and it is nothing but the reciprocal of the focal length in metres:

P = \frac{1}{f}

A short-focus lens bends light hard, so it has a large power; a long-focus lens bends gently and has a small one. The dioptre exists because power is the natural currency of a lens — as the next article shows, powers simply add when you stack lenses, which focal lengths do not. Positive power converges, negative power diverges, and the size of the number is how strong the lens is. Power answers "how strong?". Its partner, magnification, answers "how big?":

m = \frac{v}{u} = \frac{h'}{h}

the ratio of image height to object height, carrying a sign that tells you the orientation — negative for an inverted, real image, positive for an erect, virtual one — and a magnitude that tells you the scale. A single New Cartesian identity, $m = f/(f + u)$ , lets you find it straight from the object distance and the focal length, without first computing $v$ . This page pins down both numbers exactly. It fixes the one unit slip that costs the most marks — power needs $f$ in metres, so a $20$ cm lens is $+5$ dioptres, not $+0.05$ — derives the magnification identity, and works the classic bench method, due to Bessel, that measures a lens's focal length (and hence its power) from two sharp-image positions. Power and magnification are the two "how much" numbers of a lens, and every optical prescription and every camera spec is written in them.

Key takeaways

Power $P = \dfrac{1}{f}$ , with $f$ in metres, measured in dioptres (D): positive for a converging lens, negative for a diverging one; a strong (short- $f$ ) lens has a large $|P|$ .
The commonest slip: $f$ must be in metres. A $+20$ cm lens is $+5$ D (not $+0.05$ D); a $-20$ cm lens is $-5$ D.
Magnification $m = \dfrac{v}{u} = \dfrac{h'}{h}$ : sign gives orientation ( $m < 0$ inverted/real, $m > 0$ erect/virtual), magnitude gives scale.
A handy identity, $m = \dfrac{f}{f + u}$ , finds the magnification from $f$ and $u$ alone — for $f = +20$ , $u = -30$ it gives $m = -2$ , matching $v/u$ .
From a prescription: $P = +2.5$ D means $f = \dfrac{1}{2.5}\ \mathrm{m} = 40$ cm; the anchors $+5$ , $+4$ , $+2$ D correspond to $f = 20$ , $25$ , $50$ cm.
Key idea: power $P = 1/f$ (in metres, dioptres) says how strongly a lens bends light, and magnification $m = v/u = h'/h$ says how large and which way up the image is — the two numbers that specify a lens in use.

The core insight

The power of a lens is $P = 1/f$ with $f$ in metres, measured in dioptres and signed by the lens type, while the linear magnification is $m = v/u = h'/h = f/(f+u)$ , whose sign gives the image's orientation and whose magnitude gives its scale.

Where this lives today

Power and magnification are the working language of every place light is focused for a purpose. An optician's prescription is a power in dioptres — a myope might get $-2.5$ D (diverging, $f = -40$ cm), a hyperope $+2.5$ D (converging) — chosen so that, added to the eye's own power, the combined focus lands on the retina; this "powers add" logic is why the number is a dioptre and not a focal length. A camera is specified the other way, by focal length in millimetres, because photographers care about field of view and magnification rather than adding lenses to an eye; a " $50$ mm" lens and a " $200$ mm" telephoto are a $+20$ D and a $+5$ D lens respectively, and the telephoto's larger $|v/u|$ is exactly why it magnifies distant subjects. Microscopes and telescopes are quoted in magnification — " $\times 400$ ", " $\times 10$ " — the linear or angular size ratio their lens combinations achieve. And the phoropter, the mask of clicking lenses an optometrist swings in front of you, is a rack of calibrated dioptres, adding power a quarter-dioptre at a time until the chart snaps into focus. Every one of these is $P = 1/f$ or $m = v/u$ , read off an instrument.

The physical picture

A lens is described, in use, by two numbers pulling in different directions. One is intrinsic to the lens — how hard it bends light, set once and for all by its focal length. The other depends on where you put the object — how large the image comes out. Power captures the first, magnification the second, and keeping them apart is half the battle.

Power turns the focal length inside out. A lens with a short focal length gathers parallel light to a near point, bending each ray through a large angle; a lens with a long focal length bends gently. So "strength" runs opposite to focal length, and the tidy way to say it is $P = 1/f$ : halve the focal length and you double the power. Measuring $f$ in metres makes the unit the dioptre, and the choice is not arbitrary — it is exactly the unit in which the powers of stacked lenses add, which is why every optician, and the next article, works in dioptres rather than centimetres.

Magnification is the ratio of the sizes, $m = h'/h$ , and the thin-lens geometry makes it equal to the ratio of the distances, $m = v/u$ — the ray through the optical centre proves it, as the derivation shows. Its sign is doing real work: because a real image has $u$ and $v$ of opposite sign, $m = v/u$ comes out negative, and negative means inverted; a virtual image has them of the same sign, so $m > 0$ , erect. The number thus reports orientation and scale in one stroke, which is why boards ask for it alongside the image distance.

Figure

The dioptre number line ties power to focal length: diverging lenses have negative power on the left, converging lenses positive power on the right, and each dioptre value corresponds to a focal length by P equals one over f in metres. Minus 5 D is a diverging lens of 20 cm, plus 2 D a converging lens of 50 cm, plus 5 D a converging lens of 20 cm.

Power, in dioptres

The definition is one line, but the unit hides a trap, so state it carefully:

P = \frac{1}{f} \qquad (1)

where: $f$ is the focal length in metres, and $P$ is the power in dioptres (D), one dioptre being one reciprocal metre ( $1\ \mathrm{D} = 1\ \mathrm{m^{-1}}$ ). The sign of $P$ is the sign of $f$ : a converging (convex) lens has $f > 0$ , hence $P > 0$ ; a diverging (concave) lens has $f < 0$ , hence $P < 0$ . So a lens's power carries both its strength (the magnitude) and its type (the sign) in a single signed number — which is exactly what a prescription needs. The metres are not optional. A focal length of $20$ cm is $0.20$ m, so its power is $P = 1/0.20 = +5\ \mathrm{D}$ — a clean, human-sized number. Leave the focal length in centimetres and you get $1/20 = 0.05$ , off by a factor of a hundred, and the optician's chart makes no sense. Run it the other way, from a prescription: a power of $+2.5\ \mathrm{D}$ means $f = 1/2.5 = 0.40$ m $= 40$ cm. Always convert to metres before you divide. The reason the dioptre is worth a special unit — rather than just quoting $f$ — is that power adds. Put two thin lenses together and their powers sum, $P = P_1 + P_2$ , which is the whole of the next article and the reason opticians, who stack corrective lenses onto the eye's own, think entirely in dioptres. Focal lengths do not add like that; powers do.

Magnification, and a shortcut

The magnification is the ratio of the heights, which the thin-lens geometry makes equal to the ratio of the distances:

m = \frac{h'}{h} = \frac{v}{u} \qquad (2)

where: $h$ is the object height, $h'$ the image height, $u$ and $v$ the object and image distances in New Cartesian signs. Note again the absence of a minus — this is a lens, so $m = +v/u$ , unlike the mirror's $m = -v/u$ from The Mirror Formula. The sign still classifies the image for free: $m < 0$ inverted (and, for a real image, this is automatic since $u < 0$ , $v > 0$ ); $m > 0$ erect; $|m| > 1$ magnified, $|m| < 1$ diminished. Often you know $u$ and $f$ but not yet $v$ , and there is a shortcut that skips the extra step. Starting from the thin-lens formula and eliminating $v$ (the derivation does it), the magnification is

m = \frac{f}{f + u} \qquad (3)

where: $u$ is signed. For a converging lens $f = +20$ cm with an object at $u = -30$ cm, this gives $m = \dfrac{20}{20 + (-30)} = \dfrac{20}{-10} = -2$ — a real, inverted, doubled image — agreeing exactly with $m = v/u = 60/(-30) = -2$ from the full method. Equation (3) is the fast route in a competitive exam; equation (2) is the one to fall back on when you already have $v$ .

Derivation APower in dioptres, and its sign

The goal is to establish $P = 1/f$ with $f$ in metres, to fix why the unit is the dioptre and why its sign matches the lens type, and to run the definition both ways — focal length to power and prescription to focal length — on the anchor numbers. This is short, but the unit discipline it installs prevents the single most common power mistake.

Step 1. Define power as the reciprocal focal length, $P = 1/f$ . The physical motive: a short focal length means strong bending, so "strength" should be large when $f$ is small, and the reciprocal delivers exactly that. Choosing $f$ in metres sets the unit of $P$ as the reciprocal metre, named the dioptre: $1\ \mathrm{D} = 1\ \mathrm{m^{-1}}$ .

Step 2. Carry the sign through. Since $P$ and $f$ are reciprocals, they share a sign: a converging lens has $f > 0$ so $P > 0$ ; a diverging lens has $f < 0$ so $P < 0$ . Thus a single signed number, like $+2.5$ or $-1.5$ D, states both the strength and the type of the lens — the compactness that makes it a prescription unit.

Step 3. Compute forward. $f = +20\ \mathrm{cm} = 0.20\ \mathrm{m}$ gives $P = 1/0.20 = +5\ \mathrm{D}$ ; $f = -20\ \mathrm{cm}$ gives $-5\ \mathrm{D}$ ; $f = +25\ \mathrm{cm}$ gives $+4\ \mathrm{D}$ ; $f = +50\ \mathrm{cm}$ gives $+2\ \mathrm{D}$ . The metres matter: in centimetres you would wrongly get $0.05$ instead of $5$ .

Step 4. Compute backward. A prescription is a power, so invert: $P = +2.5\ \mathrm{D}$ gives $f = 1/2.5 = 0.40\ \mathrm{m} = 40\ \mathrm{cm}$ , converging. This is exactly the calculation behind every pair of spectacles — and because the eye's own lens contributes its own power, the two add, which is why the next article's "powers add" rule is what makes prescriptions work. Verified: the powers $+5$ , $-5$ , $+4$ , $+2$ D and the prescription $+2.5\ \mathrm{D} \to 40$ cm are recomputed in verify_hs2-lens-power-magnification.py.

Go deeper · Derivation B · 1 minMagnification from the thin-lens formula

Derivation BMagnification from the thin-lens formula

The goal is to derive the shortcut $m = f/(f + u)$ from the thin-lens formula and $m = v/u$ , and to verify it gives $m = -2$ two independent ways for the anchor case. The identity is worth having because it delivers the magnification directly from $f$ and $u$ , which are usually what a problem gives you, without the detour through $v$ .

Step 1. Start from the two lens equations: $\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$ and $m = \dfrac{v}{u}$ . The plan is to solve the first for $v$ and substitute into the second.

Step 2. Solve for $v$ . Rearrange $\dfrac{1}{v} = \dfrac{1}{f} + \dfrac{1}{u} = \dfrac{u + f}{f u}$ , so $v = \dfrac{f u}{u + f}$ . This is the image distance in terms of $f$ and $u$ alone.

Step 3. Substitute into $m = v/u$ : $m = \dfrac{1}{u}\cdot\dfrac{f u}{u + f} = \dfrac{f}{u + f} = \dfrac{f}{f + u}$ . The object distance $u$ has cancelled once, leaving the magnification as a function of $f$ and $u$ only — equation (3).

Step 4. Verify on the anchor. With $f = +20$ , $u = -30$ : the shortcut gives $m = \dfrac{20}{20 - 30} = -2$ . The long way gives $v = \dfrac{20 \times (-30)}{-30 + 20} = \dfrac{-600}{-10} = +60$ cm, then $m = v/u = 60/(-30) = -2$ . Identical — a real, inverted, doubled image, and the height follows as $h' = -2 \times 5 = -10$ cm. Verified: the identity $m = f/(f+u)$ and the two-road agreement $m = -2$ are recomputed in verify_hs2-lens-power-magnification.py.

Figure

The magnification comes from the ray through the optical centre, which travels straight and makes two similar triangles: object height h at object distance u on the left, image height h prime at image distance v on the right. Their similarity gives the ratio of heights equal to the ratio of distances, so m equals h prime over h equals v over u.

Limiting cases

Case	Result	Meaning
$f$ small (short focus)	$\|P\|$ large	a strong lens; a thick spectacle or a microscope objective
$f \to \infty$ (flat glass)	$P \to 0$	zero power — no bending, a plain window
$u \to -\infty$ (distant object)	$m = f/(f + u) \to 0$	a tiny image at the focal plane — the camera
$u = -2f$ (object at $2F$ )	$m = -1$	real, inverted, same size
$u \to 0$ (object at the lens)	$m \to 1$	image same size, at the lens — no magnification
$u = -f$ (object at focus)	$m \to \infty$	image at infinity; magnification undefined

Common trapMetres for power; a positive m is not a magnified image

Two habits protect almost every mark on this page. The first: power needs $f$ in metres. A $20$ cm lens is $+5$ dioptres, not $+0.05$ — forget the conversion and every power is a hundredfold wrong. Write " $f = 0.20\ \mathrm{m}$ " as your first line, then divide. The second is subtler: a positive magnification does not mean the image is magnified. The sign of $m$ reports orientation, not size — $m > 0$ means erect (and hence virtual), while $m < 0$ means inverted (and, for a lens, real). It is the magnitude $|m|$ that reports scale: $|m| > 1$ enlarged, $|m| < 1$ shrunk. So a diverging lens with $m = +0.4$ gives an erect image that is smaller than the object, and a converging lens with $m = -2$ gives an inverted image larger than it. Read the sign for which-way-up and the magnitude for how-big, separately, every time — conflating them is a classic distractor on NEET and JEE.

Worked examples

Board level — powers, prescriptions, and an image height (CBSE and NEET pattern)

(a) Find the power of lenses of focal length $+20$ cm, $-20$ cm, $+25$ cm and $+50$ cm. (b) An optician prescribes $+2.5$ D; what focal length is that? (c) A $5$ cm object is placed $30$ cm in front of the $+20$ cm lens; find the magnification and image height.

Step 1. Powers, converting each $f$ to metres first. $+20\ \mathrm{cm} = 0.20\ \mathrm{m} \to P = 1/0.20 = \boxed{+5\ \mathrm{D}}$ . $-20\ \mathrm{cm} \to P = 1/(-0.20) = \boxed{-5\ \mathrm{D}}$ (diverging). $+25\ \mathrm{cm} = 0.25\ \mathrm{m} \to P = \boxed{+4\ \mathrm{D}}$ . $+50\ \mathrm{cm} = 0.50\ \mathrm{m} \to P = \boxed{+2\ \mathrm{D}}$ . Notice the pattern: the shorter the focal length, the larger the power.

Step 2. The prescription runs the definition backwards: $f = 1/P = 1/2.5 = 0.40\ \mathrm{m} = \boxed{40\ \mathrm{cm}}$ , a converging lens (positive power).

Step 3. For part (c) use the shortcut, equation (3): $f = +20$ cm, $u = -30$ cm, so $m = \dfrac{f}{f + u} = \dfrac{20}{20 - 30} = \dfrac{20}{-10} = \boxed{-2}$ . The image is real (the minus) and twice as tall.

Step 4. Image height: $h' = m \times h = -2 \times 5 = \boxed{-10\ \mathrm{cm}}$ . The minus records that it is inverted; the magnitude, $10$ cm, that it is twice the object's size. A quick cross-check with the full method: the lens formula gives $v = +60$ cm, so $m = v/u = 60/(-30) = -2$ , the same. The two roads agree, as they must.

Competitive level — Bessel's method measures the power (JEE pattern)

An illuminated object and a screen are fixed $90$ cm apart. A converging lens placed between them forms a sharp image on the screen at two distinct positions, $30$ cm apart. Find the focal length of the lens and its power.

Step 1. Recognise the setup — this is Bessel's method. When the object-to-screen distance $D$ exceeds $4f$ , there are exactly two lens positions that focus a sharp image, and the object and image distances simply swap between them. Their separation $d$ satisfies $d = \sqrt{D(D - 4f)}$ , which rearranges to a formula for $f$ that needs only the two measured lengths.

Step 2. Solve for $f$ : $f = \dfrac{D^2 - d^2}{4D}$ . Substitute $D = 90$ cm, $d = 30$ cm: $f = \dfrac{90^2 - 30^2}{4 \times 90} = \dfrac{8100 - 900}{360} = \dfrac{7200}{360} = \boxed{20\ \mathrm{cm}}$ .

Step 3. Convert to power: $f = 0.20\ \mathrm{m}$ , so $P = 1/0.20 = \boxed{+5\ \mathrm{D}}$ . The bench has measured the lens's power without ever knowing where its optical centre is — which is exactly Bessel's point, since a real lens is thick and its centre is hard to locate.

Step 4. Sanity check the two positions. With $D = 90$ and $d = 30$ , the lens sits at $30$ cm from the object in one position (giving $v = 60$ , a magnified image) and at $60$ cm in the other (giving $v = 30$ , a diminished image) — object and image distances swapped, both consistent with $f = 20$ cm. The symmetry is the signature of the method. Timing note: about 90 seconds. The competitive trick is to recognise "two sharp positions a distance $d$ apart" as Bessel's method and reach straight for $f = (D^2 - d^2)/(4D)$ — deriving it from scratch under time pressure is the slow way to lose the question. Learn the one formula and the setup solves itself.

Exam traps

Focal length in centimetres for power. The single biggest slip. $P = 1/f$ needs $f$ in metres: $20$ cm $\to 0.20$ m $\to +5$ D. A power quoted as $0.05$ D for a normal lens is this error every time.

Sign of the power. $P$ inherits the sign of $f$ : converging $+$ , diverging $-$ . A prescription of $-1.5$ D is a diverging lens for short sight; do not drop the minus.

Positive $m$ means erect, not magnified. The sign is orientation, the magnitude is scale. A diverging lens has $m > 0$ yet always shrinks the image; a real image has $m < 0$ yet may be enlarged.

Confusing $|m|$ and signed $m$ . When a question asks "how many times magnified", it wants $|m|$ ; when it asks "describe the image", it wants the sign too. Give both.

Board scope. Linear magnification $m = v/u = h'/h$ is examined by CBSE, ICSE, JEE Main, JEE Advanced, NEET and IGCSE 0625 — IGCSE does linear magnification for a converging lens, but power in dioptres is beyond the IGCSE core, so an IGCSE student learns $m$ here and should treat the dioptre as enrichment. Power in dioptres is examined by the Indian boards and, among A-Levels, by OxfordAQA International A-Level only; CIE 9702 does no optics and current Edexcel IAL does no lens power. AP Physics 2 does magnification but not the dioptre framing. Know which of the two numbers your board wants.

Origins · 1 minFriedrich Wilhelm Bessel and how this was discovered

Friedrich Wilhelm Bessel

1784–1846 · Königsberg

Bessel was the astronomer who made measurement itself an art, and the bench method for a lens's focal length is a small piece of that obsession with precision. Self-taught in astronomy, he computed a comet's orbit that won Olbers's notice and became director of the Königsberg Observatory. In 1838 he made the first measurement of a star's distance, the parallax of 61 Cygni. The same demand for certainty gave optics Bessel's method: to find a lens's focal length without locating its hard-to-find optical centre, fix an object and screen a set distance apart and record the two lens positions that both throw a sharp image; the focal length, and hence the power, follow from those positions alone. The dioptre as a name came later, from Ferdinand Monoyer in 1872; Bessel gave the measurement.

History

1801

— Thomas Young estimates the eye's focusing power, an early quantitative handling of optical strength.

1840

— Friedrich Wilhelm Bessel develops the two-position bench method for a lens's focal length, and hence its power.

1841

— Carl Friedrich Gauss casts paraxial imaging in the conjugate form that yields the magnification identity.

1872

— Ferdinand Monoyer introduces the dioptre as the unit of lens power, one reciprocal metre.

1873

— Ernst Abbe founds the modern theory of the microscope objective and its magnifying power at Zeiss.

Going further

Texts. NCERT Physics Class 12, Ch. 9 ("Ray Optics and Optical Instruments"), §9.5, power of a lens and magnification · H. C. Verma, Concepts of Physics Vol. 1, "Geometrical Optics" · Cambridge IGCSE 0625, linear magnification · OxfordAQA International A-Level, power of a lens.

Provenance. Standard paraxial optics. Recompute: $P = 1/f$ with $f$ in metres — $+20\ \mathrm{cm} \to +5$ D, $-20\ \mathrm{cm} \to -5$ D, $+25\ \mathrm{cm} \to +4$ D, $+50\ \mathrm{cm} \to +2$ D; prescription $+2.5\ \mathrm{D} \to f = 40$ cm. Magnification (converging $f = +20$ , $u = -30$ ): $m = f/(f+u) = 20/(-10) = -2$ , matching $v/u = 60/(-30) = -2$ , and $h' = -2 \times 5 = -10$ cm. Bessel's method: $D = 90$ cm, $d = 30$ cm $\to f = (D^2 - d^2)/(4D) = 7200/360 = 20$ cm $\to P = +5$ D, with the two lens positions at $30$ and $60$ cm.

Prerequisites. The Thin Lens Formula — the image equation this page draws $m$ and the shortcut from. Leads to. Lenses in Contact & Combination, where "powers add" is derived and used.

Go deeper (UG). Angular magnification, magnifying power, and the eye-plus-instrument analysis are developed in Image Formation and Magnification in the university optics track. It is enrichment; every result here follows from the thin-lens formula and the definition $P = 1/f$ .

Why does power use metres and not centimetres?

Because the dioptre is defined as one reciprocal metre, so the arithmetic only comes out in dioptres if the focal length is in metres. A $20$ cm lens has $f = 0.20$ m and power $1/0.20 = 5$ D — a comfortable, human-sized number, which is part of why the metre was chosen. If you leave $f$ in centimetres you compute $1/20 = 0.05$ , which is a power in "reciprocal centimetres", a unit nobody uses, and it is a hundred times too small. The fix is a habit: whenever you see the word "power" or "dioptre", convert the focal length to metres on the very first line, before dividing.

Why do we bother with power when we already have focal length?

Because power adds and focal length does not. When two thin lenses sit together, their combined power is simply $P = P_1 + P_2$ — a sum you can do in your head — whereas the combined focal length obeys the clumsier $1/F = 1/f_1 + 1/f_2$ . Opticians live in this world: your spectacle lens adds its power to your eye's own lens, so a diopter is the natural bookkeeping unit for the eye-plus-glasses system. The next article is built entirely on "powers add", and once you have seen it you understand why the dioptre earns its own name instead of everyone just quoting focal lengths.

Does a positive magnification mean the image is bigger?

No, and this catches people constantly. The sign of $m$ tells you the orientation, not the size: $m > 0$ means the image is erect (and therefore virtual, for a single lens), while $m < 0$ means inverted (and, for a lens, real). The magnitude $|m|$ tells you the size: $|m| > 1$ is enlarged, $|m| < 1$ is reduced. So a diverging lens can have $m = +0.4$ — positive, yet the image is smaller than the object. Always read the two things separately: sign for which-way-up, magnitude for how-big. Only when $|m| > 1$ is the image genuinely magnified, regardless of the sign.

What is Bessel's method actually measuring, and why is it clever?

It measures a lens's focal length without needing to know where the lens's optical centre is. For a thick, real lens the centre is buried inside the glass and hard to pin down, so measuring $u$ and $v$ from it is error-prone. Bessel's trick: fix the object and the screen a distance $D$ apart (with $D > 4f$ ) and slide the lens; you find two positions that both give a sharp image, separated by $d$ . Because those two positions are related by simply swapping the object and image distances, the focal length comes out as $f = (D^2 - d^2)/(4D)$ — built only from the two lengths $D$ and $d$ , both measured on the bench, with the troublesome centre never appearing. It is a beautiful example of designing a measurement so that the quantity you cannot measure cancels out.

How is the power of the eye related to spectacles?

The relaxed human eye has a total power of about $+60$ dioptres, most of it from the cornea and the rest from the internal lens, focusing distant light onto the retina. If the eye is too powerful or too weak for its length, the image lands in front of or behind the retina, and a spectacle lens corrects it by adding its power to the eye's. A short-sighted (myopic) eye is too powerful, so it needs a diverging lens of negative power to subtract a little; a long-sighted (hyperopic) eye needs a converging lens of positive power to add some. Because powers add, the optician just prescribes the dioptres needed to bring the total to the right value — which is exactly the lens-combination rule of the next article applied to your own eye.

Can magnification be negative and less than one at the same time?

Yes — the two properties are independent. A magnification of $m = -0.5$ describes an image that is inverted (the minus) and half the size (the magnitude). This is a very common case: a converging lens with the object well beyond $2F$ — the camera regime — produces exactly this, a small inverted real image, which is why a photograph of a distant scene is upside-down on the sensor before the electronics flip it. So do not assume "negative means big" or "less than one means erect"; the sign and the magnitude each carry their own separate piece of information.