Power of a lens

Hand an optician two lenses and ask "which one bends light more sharply?" and she will not measure focal length with a ruler — she will read a single signed number off the lens, in dioptres, and know instantly. A lens of $+6$ D bends parallel light into focus in under 17 cm; a lens of $+0.5$ D barely bends it at all, needing two full metres to converge. That number is the power of a lens, and it is the quantity written on every spectacle prescription on the planet — not focal length, not curvature, not glass type, just a plus or minus sign and a number of dioptres. Power exists because focal length, while physically fundamental, is clumsy to work with. It is measured in centimetres or metres, it does not add simply when lenses are stacked, and it hides the sign story (converging versus diverging) inside an algebraic convention that is easy to fumble. Power fixes all three problems in one stroke: it is defined as the reciprocal of focal length in metres, it carries the sign automatically, and — most usefully — when thin lenses are placed in contact, their powers simply add. That additive property is the whole reason optometrists, camera-lens designers and physics examiners reach for power rather than focal length whenever more than one lens is involved. This section builds power of a lens from the definition you already have — the thin lens formula and the lens maker's formula — connects it to the sign convention you learned for mirrors and lenses, and works two questions in full: one a direct dioptre calculation, the other a reverse-engineering problem that asks what curvature of glass would grind a given prescription. By the end you should be able to convert fluently between focal length and power in either direction, read the sign of a prescription as physics rather than jargon, and see exactly why the next topic — combining lenses in contact — becomes almost trivial once you think in dioptres instead of centimetres.

Key takeaways

Power is the reciprocal of focal length in metres: $P = 1/f$ , with the SI unit the dioptre (D), where $1\ \text{D} = 1\ \text{m}^{-1}$ .
Sign follows the Cartesian convention on $f$ . A convex (converging) lens has positive power; a concave (diverging) lens has negative power — the sign is not optional decoration, it is the physics of which way the lens bends light.
Larger $|P|$ means a shorter focal length, not a longer one — power and focal length move in opposite directions because they are reciprocals of each other.
Power connects directly to lens geometry through the lens maker's formula: $P=(n-1)\left(\dfrac1{R_1}-\dfrac1{R_2}\right)$ , so more curved surfaces or higher refractive index glass both raise $|P|$ .
Powers of thin lenses in contact simply add, $P = P_1+P_2+\cdots$ — this is the single biggest reason power, not focal length, is the preferred unit whenever more than one lens is involved (explored fully in the next topic).
Spectacle and contact-lens prescriptions are written directly in dioptres, with the sign telling you at a glance whether the correction is for myopia (short sight, negative power) or hypermetropia (long sight, positive power).
Key idea: power P = 1/f (in dioptres, f in metres) turns a lens's bending strength into one signed number — positive for converging convex lenses and negative for diverging concave lenses — that is directly readable from lens geometry via the lens maker's formula and adds simply across lenses in contact.

The core insight

The power of a lens, $P=1/f$ with $f$ in metres and $P$ in dioptres, is a signed measure of how strongly it bends parallel light — positive and short-focal-length for a converging convex lens, negative for a diverging concave lens — and because power (unlike focal length) adds simply for lenses in contact, it is the quantity actually used on every real spectacle prescription.

Where it fits in the syllabus

Within Unit IX, "Ray Optics and Optical Instruments," power of a lens sits right after the thin lens formula $\frac1v-\frac1u=\frac1f$ and the lens maker's formula $\frac1f=(n-1)\left(\frac1{R_1}-\frac1{R_2}\right)$ , and it is the direct stepping stone into the next prescribed topic, combination of thin lenses in contact, $\frac1f=\frac1{f_1}+\frac1{f_2}$ , which is far easier to state and use in terms of power. You are expected to know and apply the definition $P=1/f$ (with $f$ in metres, $P$ in dioptres), to reason about the sign of power for converging versus diverging lenses using the Cartesian sign convention, and to compute the power of a single lens — or, once you reach the next topic, a stack of lenses in contact — from given focal lengths or radii of curvature. This is squarely inside the 2025-26 CBSE prescription; nothing about power of a lens has been rationalised out, and no aberration or lens-cutting content is needed to use it correctly.

The physical picture

Start from a fact you already know: a converging (convex) lens brings a parallel beam of light to a real focus at distance $f$ in front of it (measuring from the optical centre), while a diverging (concave) lens spreads a parallel beam apart so that the rays only appear to come from a point at distance $f$ behind it. The shorter that distance $f$ , the more sharply the lens has bent the rays inward or outward. Power is simply this bending strength turned into a number:

P=\frac{1}{f}

where: $f$ is the focal length of the lens measured in metres, using the same Cartesian sign convention you already use for mirrors and lenses (distances measured from the optical centre, with the direction of incident light taken as positive), and $P$ is the power in dioptres. The physical reading is direct: a lens that focuses parallel light at $f=0.1$ m (10 cm) is bending light far more sharply than one that focuses it at $f=1$ m, and dividing 1 by a small number gives a large result — so short focal length automatically becomes large power, and long focal length automatically becomes small power. Power is therefore not a new physical effect; it is a smarter way of stating a fact you already had. The SI unit of power is the dioptre, symbol D, defined so that $1\ \text{D} = 1\ \text{m}^{-1}$ — a lens of power exactly $1$ D has a focal length of exactly $1$ m. Dioptres are used precisely because focal lengths of everyday lenses (spectacles, camera lenses, magnifiers) span from a few centimetres to a few metres, and stating " $0.04$ m" or " $2.5$ m" is far less convenient on a prescription slip than stating " $+25$ D" or " $+0.4$ D." Because $f$ carries a sign under the convention — positive for a convex/converging lens, negative for a concave/diverging lens — power inherits exactly the same sign. A convex lens therefore has positive power and a concave lens has negative power, with no separate rule to memorise: the sign of $P$ is nothing but the sign of $f$ , carried through the reciprocal. This is also exactly the convention printed on real prescriptions: "+2.50 D" unambiguously means a converging correction lens, and " $-3.75$ D" unambiguously means a diverging one, without the optician needing to add the word "convex" or "concave" at all.

Figure

Two lenses side by side bending parallel incoming rays, a short thick lens with a short focal length labeled high power, and a long thin lens with a long focal length labeled low power, showing that shorter focal length means larger power

The main reason power, rather than focal length, is the preferred quantity when more than one lens is involved is that powers of thin lenses placed in contact simply add:

P = P_1+P_2+P_3+\cdots

where: each $P_i$ is the power (in dioptres, with its own sign) of one thin lens in the stack, and $P$ is the net power of the combination. Focal lengths do not add this simply — you cannot write $f=f_1+f_2$ — which is precisely why the next syllabus topic, combination of thin lenses in contact, is stated and used in terms of power rather than focal length. Power's definition here is what makes that later addition rule work; the two topics are really one idea told in two steps.

Derivation APower of a lens from the lens maker's formula

The goal of this derivation is to show that the power of a thin lens is not just an abstract reciprocal of focal length but is fixed directly by the lens's own physical geometry — the curvature of its two surfaces and the refractive index of its glass — by combining the definition of power with the lens maker's formula, so that you can compute or estimate a lens's power straight from how it is ground, without ever separately calculating focal length as an intermediate step.

Step 1. Start from the lens maker's formula for a thin lens of refractive index $n$ (relative to the surrounding medium, usually air) with two spherical surfaces of radii $R_1$ and $R_2$ , both measured with the Cartesian sign convention from the optical centre: $\dfrac1f=(n-1)\left(\dfrac1{R_1}-\dfrac1{R_2}\right)$ . This formula is quoted directly from the prescribed syllabus content; it packages Snell's law applied at each of the two curved surfaces in turn, and here it is simply the starting relation we already have available.

Step 2. Recall the definition of power, $P=1/f$ , and note that the left-hand side of the lens maker's formula above, $1/f$ , is exactly this same quantity. There is nothing to "solve" here in the usual algebraic sense — the two expressions are already staring at each other on either side of the same equals sign, because power was defined as nothing more than the reciprocal of $f$ .

Step 3. Substitute directly: replacing $1/f$ by $P$ turns the lens maker's formula into an expression for power itself, $P=(n-1)\left(\dfrac1{R_1}-\dfrac1{R_2}\right)$ . This single line is now a working formula that predicts a lens's power from its manufacturing geometry alone — refractive index and the two radii of curvature — with no separate focal-length calculation needed in between.

Step 4. Read what the formula says physically. Power grows in magnitude when $(n-1)$ is larger (higher-index glass bends light more sharply at each surface) and when the surfaces are more strongly curved, i.e. when $R_1$ and $R_2$ are small in magnitude (tight curves) rather than large (nearly flat surfaces). This is exactly the intuition of a lens grinder: a thicker, more strongly curved blank of dense glass produces a higher-power lens, while a nearly flat, low-index blank produces a low-power one — and this derivation is the algebra behind that shop-floor intuition.

Worked examples

Board level - power of a convex lens and a concave lens from given focal lengths

A convex lens has a focal length of 25 cm. Calculate its power. A concave lens has a focal length of magnitude 50 cm. Calculate its power.

Step 1. Convert the convex lens's focal length to metres before touching the power formula: $f = 25\ \text{cm} = 0.25\ \text{m}$ , and because the lens is convex (converging), this value is taken as positive, $f=+0.25$ m.

Step 2. Apply $P=1/f$ for the convex lens: $P=\dfrac1{0.25}=4\ \text{D}$ . The positive sign confirms it is a converging lens, consistent with it being convex.

Step 3. Repeat for the concave lens, again converting to metres first and keeping the sign that the Cartesian convention assigns to a diverging lens: $f = -50\ \text{cm} = -0.5\ \text{m}$ (negative, since the lens diverges light). Then $P=\dfrac1{-0.5}=-2\ \text{D}$ .

Step 4. Compare the two results. The convex lens has power $+4$ D, a fairly strong converging lens; the concave lens has power $-2$ D, a moderate diverging lens. Notice the sign alone tells you converging versus diverging — no need to separately state "convex" or "concave" once the power is known, and the lens with the larger magnitude of power (the $4$ D convex lens) is also the one with the shorter focal length, exactly as the reciprocal relationship demands.

Competitive level - designing the curvature that would grind a given prescription

An optometrist needs to prescribe a corrective lens of power $-2.5$ D for a patient with myopia. A lens-grinding workshop plans to shape it as an equiconcave lens (both surfaces curved outward with equal magnitude of radius, $R_1=-R$ and $R_2=+R$ ) from glass of refractive index $n=1.5$ . What focal length does the prescription require, and what radius of curvature $R$ should the workshop grind on each surface?

Step 1. Read off the focal length directly from the prescribed power, since $P=1/f$ inverts cleanly: $f=\dfrac1P=\dfrac1{-2.5}=-0.4\ \text{m}=-40\ \text{cm}$ . The negative sign confirms this must be a diverging (concave) lens, which matches the physics of correcting myopia — a diverging lens moves the eye's over-converged focal point backward onto the retina.

Step 2. Set up the lens maker's formula for the equiconcave shape described, using $R_1=-R$ and $R_2=+R$ for a lens whose two surfaces curve away from each other with equal magnitude: $P=(n-1)\left(\dfrac1{R_1}-\dfrac1{R_2}\right)=(n-1)\left(\dfrac1{-R}-\dfrac1{R}\right)=(n-1)\left(-\dfrac2R\right)$ .

Step 3. Substitute the known power and refractive index and solve for $R$ : $-2.5=(1.5-1)\left(-\dfrac2R\right)=-\dfrac1R$ , so $\dfrac1R=2.5$ , giving $R=0.4\ \text{m}=40\ \text{cm}$ . Answer: the prescription requires a focal length of $-40$ cm, ground as an equiconcave lens with each surface having a radius of curvature of 40 cm from glass of refractive index 1.5. Timing note: well under two minutes once you trust the shortcut. Rather than working focal length and then separately invoking the lens maker's formula as two disconnected steps, chain them as one substitution, $P=(n-1)(1/R_1-1/R_2)$ , and solve directly for whichever geometric quantity is unknown. The one trap here is the sign convention on $R_1$ and $R_2$ for an equiconcave lens — get $R_1$ and $R_2$ 's signs backwards and you will compute a positive power for what must physically be a diverging lens, a mistake that a quick sign check against the given $-2.5$ D catches immediately.

Common traps

Forgetting to convert focal length to metres before computing power — using centimetres directly gives an answer off by a factor of 100 (e.g., $f=20$ cm should give $P=5$ D, not $P=0.05$ D, and $f=20$ cm should never be substituted as " $20$ " straight into $P=1/f$ ).
Assuming power is always positive — power carries the sign of focal length, so diverging (concave) lenses have negative power, and dropping that sign on a prescription is not a rounding error, it is prescribing the wrong lens entirely.
Thinking a "stronger" lens (larger $|P|$ ) always has a longer focal length — it is the reverse: larger power means shorter focal length, since $P$ and $f$ are reciprocals of each other, not proportional to each other.
Mixing up the signs of $R_1$ and $R_2$ when going from the lens maker's formula to power, especially for concave or equiconcave shapes, which flips the computed sign of $P$ and can make a diverging lens look converging on paper.
Confusing power of a lens (optical bending strength, in dioptres) with power in the everyday sense of energy delivered per unit time (watts, as in electrical power) — the same English word names two entirely different physical quantities, and an exam answer that reaches for $P=VI$ or $P=W/t$ here is answering the wrong chapter.

Common trapPower is not "how strong the lens looks" — it is a signed reciprocal, and the sign is half the physics

The single most expensive mistake with power of a lens is treating it as a magnitude-only "strength rating," the way you might rate a magnifying glass as "quite strong" without further thought. Power is a signed quantity, and the sign is not a footnote — it is the entire statement of whether the lens converges or diverges light. A student who correctly computes $|P|=2.5$ D but drops the minus sign for a concave lens has not made a small numerical slip; they have described a completely different optical device, one that would focus rather than spread the very beam the concave lens is meant to spread. On an exam, a bare number without the correct sign for power (or for focal length) is routinely marked as an incomplete or wrong answer, precisely because the sign carries as much physical information as the magnitude does. Always ask, before writing a final numeric answer: is this lens convex or concave, and does my sign match that?

You should be able to…

Define power of a lens as $P=1/f$ with $f$ in metres, and state its SI unit, the dioptre (D), where $1\ \text{D}=1\ \text{m}^{-1}$ .
Convert fluently between focal length in any everyday length unit and power in dioptres, always working in metres for the calculation itself.
Explain, from the Cartesian sign convention, why convex (converging) lenses have positive power and concave (diverging) lenses have negative power.
Derive or quote the relation $P=(n-1)\left(\dfrac1{R_1}-\dfrac1{R_2}\right)$ linking power directly to a lens's refractive index and radii of curvature, and use it in either direction (find $P$ from geometry, or find geometry from a required $P$ ).
Interpret a real spectacle or contact-lens prescription — its sign and magnitude — in terms of lens type (converging or diverging) and the corresponding focal length.
Explain, in one sentence, why power rather than focal length is the quantity used when combining more than one thin lens, setting up the next topic.

Ask the tutor: want to practise converting a stack of spectacle-shop focal lengths into dioptres, or see how today's power formula turns into the combination-of-lenses addition rule in the next topic — ask the AI tutor to walk through a few together.

Why do we even need dioptres — why not just always say the focal length in centimetres?

Because focal length does not add simply when you put lenses together, and dioptres were invented precisely to fix that. If you stack a $+25$ cm lens and a $-50$ cm lens in contact, you cannot add $25$ and $-50$ to get anything meaningful — but you can add their powers, $+4$ D and $-2$ D, to instantly get a combined power of $+2$ D. That additive shortcut, plus the fact that a plain signed number like " $-2.50$ D" fits neatly on a small prescription slip and encodes both magnitude and lens type at once, is why opticians, lens designers and this syllabus all prefer power over focal length whenever more than a single lens is being described.

Why is a concave lens's power negative — power sounds like it should always be a positive amount of something?

Power inherits its sign entirely from focal length, and focal length's sign is a convention chosen to encode direction, not magnitude. Under the Cartesian sign convention, a real focus formed on the side the light is travelling toward (the case for a convex, converging lens) is taken as positive distance, while the virtual focus of a diverging concave lens, which forms on the same side as the incoming light, is taken as negative. Power, being $1/f$ , simply carries that same directional bookkeeping. So "negative power" does not mean "less optical effect" — a $-5$ D lens diverges light just as forcefully as a $+5$ D lens converges it; the sign only tells you which of the two behaviours you are looking at.

Does a lens with bigger power always look thicker or magnify more?

Not necessarily, and this is a common exam misconception. Power tells you how strongly a lens bends parallel rays, which is a statement about focal length, not directly about lens thickness or the magnification it produces for a particular object. A thin lens ground from very high refractive index glass can have a short focal length (large power) while remaining physically thin, and magnification for a given object also depends on the object's distance from the lens, not on power alone. In everyday spectacle lenses, however, higher-power corrections (large $|P|$ , especially strong myopia or hypermetropia prescriptions) usually do need more curved, and therefore often visibly thicker or more strongly curved, lens blanks — but that is a practical consequence of the lens maker's formula, not a separate rule about power itself.

How does power of a lens connect to the next topic, combining lenses?

Directly, and that is exactly why this topic is placed right before it in the syllabus. When two or more thin lenses are held in contact, their individual powers add algebraically, $P=P_1+P_2+\cdots$ , to give the power of the whole stack, and you can then invert that combined power to get the effective focal length of the system if you need it, $f=1/P$ . This works cleanly only because power was defined as a reciprocal quantity in the first place; if the next topic instead tried to add focal lengths directly, the arithmetic would not work out, which is exactly why the syllabus introduces power first as the tool that makes lens combinations tractable.