With a ruler, a pencil and the three special rays you can already draw the image a converging lens makes — where it lands, which way up it is, roughly how big. But "roughly" is not enough to build a camera, grind a pair of spectacles, or design a telescope. An optician does not sketch a ray diagram for your eyes; they write down a number. This page turns the picture into arithmetic: one small equation that pins down the image exactly.

You already know the geometry from ray diagrams and image formation: a converging lens of focal length f, an object a distance u in front of it, and an image a distance v on the far side. Those three lengths are not free — they are locked together by the thin-lens equation.

The thin-lens equation

Trace the two easy rays for an object at a general distance and work out where they cross (it is a short exercise with similar triangles). The crossing point always obeys one relation between the object distance u, the image distance v and the focal length f:

An equation with signed quantities needs a rule for the signs, and any consistent rule will do so long as you never mix two. Throughout this page we use the "real-is-positive" convention, the one used most in A-level work:

• A real object and a real image have positive distances (u > 0, v > 0).
• A virtual image — one on the same side as the object — has a negative image distance (v < 0).
• A converging lens has positive f; a diverging lens has negative f.

So the whole story of real-versus-virtual is told by the sign of v that drops out of the equation: work out v, and its sign tells you which kind of image you have — no separate ray diagram needed.

Worked example 1 — an object beyond the focus

A converging lens has focal length f = 10\ \text{cm}. A candle stands u = 30\ \text{cm} in front of it. Where is the image?

Step 1 — write the equation and put in the numbers.

Step 2 — combine the fractions. Over a common denominator of 30,

Step 3 — invert.

v is positive, so the image is real — it forms 15\ \text{cm} beyond the lens on the far side, where you could catch it on a screen. Because v < u it is smaller than the candle, and (as always for a real image from a single lens) inverted. This is exactly the camera photographing a distant scene.

Worked example 2 — inside the focus: a virtual image

Same lens, f = 10\ \text{cm}, but now push the object inside the focal length: u = 6\ \text{cm}. This is how you hold a magnifying glass.

so

The equation has handed us a negative v. In the real-is-positive convention that means the image is virtual: it sits 15\ \text{cm} out on the same side as the object, it is upright, and — as we are about to compute — it is magnified. No screen can catch it; you must look through the lens to see it. The maths gave us all of that from a single minus sign, without drawing a thing.

Magnification

The equation locates the image; the linear magnification m tells us its size. It compares image height with object height, and — from the similar triangles of the ray through the lens centre — that ratio also equals the ratio of the two distances:

Back to Example 1 (u = 30, v = 15):

so a 4\ \text{cm} candle makes a 4 \times 0.5 = 2\ \text{cm} image — half height, inverted, real.

And Example 2 (u = 6, |v| = 15):

the magnifying glass makes everything 2.5\times taller — enlarged, upright, virtual. Same lens, opposite behaviour, decided entirely by whether the object sat outside or inside f.

The power of a lens

Opticians and lens-makers rarely quote a focal length. They quote a lens's power, P — literally how hard the lens bends light. A short focal length means strong bending, so power is simply the reciprocal of the focal length:

Getting the metres right. A lens of focal length f = 20\ \text{cm} = 0.20\ \text{m} has power

Run it backwards: a 2\ \text{D} lens has f = 1/2 = 0.5\ \text{m} = 50\ \text{cm} — a gentle, weakly curved lens. A +10\ \text{D} lens has f = 10\ \text{cm}: stronger, and visibly fatter.

Worked example 3 — combining lenses (spectacles). A reading lens of +1.5\ \text{D} is held against a magnifier of +2.0\ \text{D}. Their combined power is

a single effective lens of focal length f = 1/3.5 \approx 0.29\ \text{m} \approx 29\ \text{cm}. Stacking a diverging lens instead subtracts: +3\ \text{D} next to -1\ \text{D} gives +2\ \text{D}. This adding of powers is exactly how an optician fine-tunes a prescription.

A diverging lens

For a diverging lens the only new thing is the sign: its focal length is negative. Take f = -15\ \text{cm} and a real object at u = 10\ \text{cm}.

v is negative, so the image is virtual, upright, and with m = 0.6 it is diminished — closer to the lens than the object. This is always what a diverging lens does to a real object: a small, upright, virtual image. It is the lens in glasses for short sight, shrinking the world just enough to bring it into focus.

See the equation move

Here is a converging lens of focal length f = 10\ \text{cm} (the focal points F and the 2F points are marked). Drag the object distance u. For every position the image is placed at v = \dfrac{uf}{u - f} straight from the thin-lens equation, and the readout shows the live values of u, v and m = v/u.

Watch the sign of v. Beyond the focus the image is real (solid, inverted, on the far side) and v is positive; slide the object inside f and v flips negative — the image jumps to the object's own side, turns upright, and swells. The picture and the arithmetic tell the exact same story.

Reading real or virtual straight off v

This is the payoff of the sign convention. Once you have solved the equation you do not need a ray diagram to classify the image — the number v already carries the answer:

• v > 0 → a real image, on the far side of the lens, inverted, catchable on a screen.
• v < 0 → a virtual image, on the same side as the object, upright, seen only by looking through the lens.

A converging lens gives whichever the object asks for — real for u > f, virtual for u < f. A diverging lens, with its negative f, always returns v < 0 for a real object: it can only make a virtual, upright, diminished image. The equation knows all of this before you have drawn a single ray.