Power of a lens

Hand an optician your eye-test numbers and you will not hear "your lens needs a focal length of 25 centimetres" — you will hear "you need a plus-four lens." That single number, in dioptres, is the power of a lens: how strongly it bends a beam of parallel light toward a point (convex) or spreads it apart (diverging, concave). Power is nothing more than the reciprocal of focal length, but writing it that way turns an awkward length into a clean, additive "strength" rating that doctors, opticians and lens-makers actually use. This article takes the lens formula and magnification you already know from earlier in this chapter and adds the one new idea — power — that the next chapter, on the human eye and its defects, leans on directly. By the end you should be able to convert a focal length into a power (and back), read the sign correctly, and use power in a numerical without hesitation.

The core insight

The power of a lens, $P = 1/f$ with $f$ in metres and $P$ measured in dioptres (D), tells you how strongly the lens converges (positive, convex) or diverges (negative, concave) light, with shorter focal lengths giving larger powers.

The core idea

Definition. Power is defined as the reciprocal of the focal length, provided the focal length is expressed in metres:

P = \frac{1}{f}

where: $P$ is the power of the lens and $f$ is its focal length measured in metres, following the usual sign convention (distances measured from the optical centre, in the direction of incident light taken as positive).

Unit. The SI unit of power is the dioptre, symbol D. A lens has a power of exactly one dioptre when its focal length is one metre:

1\ \text{D} = 1\ \text{m}^{-1}

So a spectacle lens rated "+2 D" simply means its focal length is $1/2 = 0.5\ \text{m} = 50\ \text{cm}$ .

Sign follows the lens type. A convex (converging) lens has positive focal length, so its power is positive; a concave (diverging) lens has negative focal length, so its power is negative. The sign is not decoration — it is the only way to tell, from the power alone, whether a lens converges or diverges light.

Units matter before you divide. Because the formula demands $f$ in metres, a focal length quoted in centimetres must first be converted:

f(\text{in m}) = \frac{f(\text{in cm})}{100}

Skipping this step is the single most common slip students make with this formula — it produces an answer that is a hundred times too large.

Why opticians prefer dioptres. Power grows as focal length shrinks: a strongly curved (short-focal-length) lens bends light sharply and has a large power, while a gently curved (long-focal-length) lens bends light weakly and has a small power. Dioptres are also additive — two thin lenses held close together combine to a power that is (approximately) the sum of their individual powers — which is exactly the practical, combinable "strength" rating an optician needs when stacking or comparing corrective lenses.

Figure

A number line of focal length in centimetres below a matching number line of power in dioptres, showing that shorter focal length corresponds to larger power for a convex lens

Derivation APower of a lens from its focal length

The goal is to derive the power–focal-length relation $P = 1/f$ in dioptres from the plain statement "power is the reciprocal of the focal length in metres," and to show precisely why the centimetre-to-metre conversion is not optional but forced by the definition of the dioptre itself, using nothing beyond unit arithmetic that a Class 10 student already has.

Step 1. Start from the definition. Power is defined so that a lens which brings parallel rays to a sharp focus at a distance of exactly one metre is assigned a power of exactly one dioptre. Writing this as a proportionality, doubling the focal length must halve the power (a weaker lens focuses farther away), which is precisely the reciprocal relationship $P \propto 1/f$ .

Step 2. Fix the constant of proportionality using the defining case. At $f = 1\ \text{m}$ , $P = 1\ \text{D}$ by definition, so $P = k/f$ requires $k = 1$ when $f$ is in metres. Hence $P = 1/f$ with no leftover constant — but only if $f$ is measured in metres, since that is the unit in which the defining case ( $f=1\ \text{m} \Rightarrow P=1\ \text{D}$ ) was fixed.

Step 3. Show why centimetres break the formula. Suppose $f$ were left in centimetres, say $f = 100\ \text{cm}$ (the same lens as above). Naively computing $1/100$ gives $0.01$ , not $1$ — because the constant $k=1$ was calibrated for metres, not centimetres. This is exactly why every power calculation opens with $f(\text{m}) = f(\text{cm})/100$ : it restores the units the formula was defined in.

Step 4. State the final relation with its sign rule attached. $P = 1/f$ , $f$ in metres, $P$ in dioptres; $f$ positive for a convex lens gives $P$ positive, $f$ negative for a concave lens gives $P$ negative. This single formula, correctly unit-converted and correctly signed, is what the worked example below applies.

Worked example

Two spectacle lenses are prescribed together: a convex lens of focal length 25 cm for reading, and a concave lens of power $-2.5$ D for distance vision. Find the power of the convex lens and the focal length of the concave lens, and state which one is more powerful.

Step 1. Convert the convex lens's focal length to metres and apply the sign convention. It is convex, so $f = +25\ \text{cm} = +0.25\ \text{m}$ .

Step 2. Apply $P = 1/f$ to the convex lens. $P = \dfrac{1}{0.25} = +4\ \text{D}$ .

Step 3. Find the concave lens's focal length by rearranging the formula, $f = 1/P$ , and substituting the given power. $f = \dfrac{1}{-2.5} = -0.4\ \text{m} = -40\ \text{cm}$ . The negative sign confirms it is concave, as stated.

Step 4. Compare magnitudes to decide which lens is "more powerful." $|{+4\ \text{D}}| = 4\ \text{D}$ for the convex lens versus $|{-2.5\ \text{D}}| = 2.5\ \text{D}$ for the concave lens, so the convex reading lens (+4 D, focal length 25 cm) is more powerful than the concave distance lens (−2.5 D, focal length 40 cm), even though the concave lens has the larger-magnitude focal length in centimetres.

Common traps

Forgetting to convert focal length from centimetres to metres before applying $P = 1/f$ — the resulting power comes out a hundred times too large or too small.
Dropping the sign and reporting only the magnitude of power — the sign is essential, since it is the only way to tell convex from concave from the power value alone.
Confusing "more powerful" with "larger focal-length magnitude" — power is inversely related to focal length, so a smaller focal length always means a larger power, not the reverse.
Mixing up power with magnification — power is a fixed property of the lens itself (how strongly it bends light), while magnification also depends on where the object is placed.

Ask the tutor: Want to see how this same $P = 1/f$ formula is used to work out the exact spectacle-lens power needed to correct a real case of near-sightedness or far-sightedness?

Why do opticians use dioptres instead of just quoting focal length in centimetres?

Dioptres are a more practical "strength" scale. A focal length in centimetres does not combine sensibly — you cannot simply add two focal lengths to find the effect of stacking two lenses. Powers in dioptres, however, add up (approximately) when thin lenses are placed close together, which is exactly what happens when a spectacle lens sits close to your eye's own lens. That additive property, plus a single signed number that also tells you convex from concave, is why the entire optical industry prescribes lenses in dioptres.

Does a "stronger" glasses prescription always mean a physically bigger or thicker lens?

Not necessarily, and this is a common misconception. Power depends only on focal length, which is a geometric and optical property tied to how curved the lens surfaces are and what material it is made of — not on the lens's overall size or thickness in a simple, direct way. A high-power lens (short focal length) is usually more strongly curved, but modern high-index lens materials can achieve a strong power in a thinner, flatter lens than an older material would need. Never judge power just by looking at a lens.

I calculated a huge power like 400 D — what went wrong?

This is almost always the forgotten unit conversion. If you plug a focal length in centimetres directly into $P = 1/f$ without dividing by 100 first, every answer comes out exactly a hundred times too large (or, if you accidentally multiply instead of divide, a hundred times too small). Always write $f$ in metres as an explicit first step — for example, $f = 25\ \text{cm} \Rightarrow f = 0.25\ \text{m}$ — before touching the reciprocal. Then a power of a few dioptres, not a few hundred, is the sanity check that you got it right.

Power of a lens

The core idea

Worked example

Common traps

Before this

Where this leads