Image formation by lenses

Hold a magnifying glass over a page and the letters swell into giant, upright shapes. Hold it further back, past its focal length, and the same lens throws an upside-down image onto a wall. Same glass, same curved surfaces — completely different pictures, depending on nothing but how far the object sits from the lens. That single idea, object distance versus focal length, is the whole of "image formation by lenses." A lens is a piece of transparent material with at least one curved surface, and it refracts light twice — entering and leaving — to bend rays toward or away from the principal axis. A convex lens converges parallel rays to a real focus; a concave lens diverges them from a virtual one. This article turns the ray-diagram picture you already know into an algebraic tool, the thin-lens formula, so you can find an image's exact position and size by calculation, not just by sketching.

The core insight

For a thin lens, the image distance $v$ , object distance $u$ and focal length $f$ are linked by $\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$ , and the same sign convention that fixes this formula also fixes the magnification $m = v/u$ .

The core idea

A convex (converging) lens is thicker at the centre than at the edges; parallel rays bend inward and meet at a real principal focus $F$ on the far side. A concave (diverging) lens is thinner at the centre; parallel rays bend outward and only appear to come from a virtual focus on the near side. The optical centre $O$ is the point through which any ray passes undeviated, and the focal length $f$ is the distance from $O$ to $F$ . To calculate rather than just sketch, ICSE uses the New Cartesian sign convention: all distances are measured from the optical centre $O$ ; distances measured in the direction of the incident light (usually taken as left to right) are positive, distances measured against it are negative; heights above the axis are positive, below are negative. Under this convention a convex lens has a positive $f$ and a concave lens has a negative $f$ ; the object, being on the incoming side, always has $u$ negative for a real object. With that convention, the position of any image follows from the thin-lens formula:

\frac{1}{v} - \frac{1}{u} = \frac{1}{f}

where: $v$ is the image distance from the optical centre, $u$ is the object distance from the optical centre, and $f$ is the focal length — all measured with signs, in the same units (commonly centimetres). The companion quantity is linear magnification,

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

where: $h$ is the object height, $h'$ is the image height, and $m$ is a pure ratio. A negative $m$ means the image is inverted relative to the object; a positive $m$ means it is erect. $|m| > 1$ means magnified, $|m| < 1$ means diminished, and $|m| = 1$ means the same size — exactly the same three outcomes the six-position ray-diagram table already gives you, now reached by arithmetic instead of a drawing.

Figure

Ray diagram of a convex lens forming a real inverted image of an object placed beyond twice the focal length

Derivation AThe thin-lens formula from similar triangles

The goal is to derive $\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$ from similar triangles, using the two rays whose paths are already fixed by definition — a ray through the optical centre travelling straight, and a ray parallel to the axis that bends through the far focus — so that the algebra matches exactly the ray diagram a student would draw by hand.

Step 1. Set up the diagram. Place an object of height $h$ on the axis at distance $u$ (negative, to the left of $O$ ) from a convex lens of focal length $f$ (positive). Draw the ray from the top of the object through the optical centre $O$ , continuing straight and undeviated, and the ray from the top of the object travelling parallel to the axis until it strikes the lens, then bending to pass through the far focus $F'$ . Where these two rays meet is the top of the image, height $h'$ , at distance $v$ from $O$ .

Step 2. Use the undeviated ray through $O$ . The object triangle (height $h$ , base $|u|$ ) and the image triangle (height $h'$ , base $v$ ) are similar, because the ray through $O$ is a straight line crossing the axis. So $\dfrac{h'}{h} = \dfrac{v}{|u|}$ , which with the sign convention becomes $\dfrac{h'}{h} = \dfrac{v}{u}$ — this is the magnification relation, obtained as a by-product before the main formula is even reached.

Step 3. Use the ray that bends through $F'$ . This ray crosses the axis at $F'$ , a distance $f$ from $O$ , and travels on to meet the first ray at the image point, a distance $v - f$ further along, at height $h'$ below the axis by the time it crosses. Comparing the small triangle at the lens (height $h$ , the incoming ray's height above the axis as it crosses the lens) with the triangle it forms with the axis out to the image gives $\dfrac{h'}{h} = \dfrac{v - f}{f}$ , since both triangles share the same apex angle at $F'$ and are similar by AA.

Step 4. Equate the two expressions for $h'/h$ from Steps 2 and 3, since both describe the same ratio: $\dfrac{v}{u} = \dfrac{v - f}{f}$ . Cross-multiplying, $vf = u(v - f) = uv - uf$ . Divide every term by $uvf$ : $\dfrac{1}{u} = \dfrac{1}{f} - \dfrac{1}{v}$ , which rearranges to the standard form $\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$ . The same steps, carried through with the sign convention, work unchanged for a concave lens (negative $f$ ) or a virtual image (negative $v$ ).

Worked example

A convex lens of focal length $f = 15\ \text{cm}$ has an object placed at $u = -30\ \text{cm}$ , i.e. exactly at $2F$ on the incoming side. Find the image position and describe its nature.

Step 1. Write the lens formula with the given signs: $\dfrac{1}{v} - \dfrac{1}{-30} = \dfrac{1}{15}$ , so $\dfrac{1}{v} = \dfrac{1}{15} - \dfrac{1}{30}$ .

Step 2. Combine the fractions over a common denominator of $30$ : $\dfrac{1}{v} = \dfrac{2}{30} - \dfrac{1}{30} = \dfrac{1}{30}$ , giving $v = 30\ \text{cm}$ . The positive sign means the image forms on the outgoing side of the lens, at $2F'$ — a real image.

Step 3. Find the magnification: $m = v/u = 30 / (-30) = -1$ . The negative sign means the image is inverted; the magnitude $1$ means it is the same size as the object. This matches the standard ray-diagram result for an object at $2F$ : real, inverted, same size, formed at $2F$ on the other side.

Step 4 (competitive stretch). Now move the same object to $u = -20\ \text{cm}$ , between $F$ (at $15\ \text{cm}$ ) and $2F$ (at $30\ \text{cm}$ ). Then $\dfrac{1}{v} = \dfrac{1}{15} - \dfrac{1}{20} = \dfrac{4 - 3}{60} = \dfrac{1}{60}$ , so $v = 60\ \text{cm}$ , and $m = 60/(-20) = -3$ . The image is real, inverted, three times magnified, and forms $60\ \text{cm}$ away, beyond $2F'$ — consistent with the ray-diagram rule that an object between $F$ and $2F$ gives a magnified real image beyond $2F$ .

Common traps

Plugging in $u$ or $f$ as positive numbers out of habit — the New Cartesian convention makes the object distance $u$ negative for every real object, and a concave lens's $f$ negative; skipping the signs gives a wrong image position or the wrong real/virtual verdict.
Writing the formula as $1/u - 1/v = 1/f$ (swapped) instead of $1/v - 1/u = 1/f$ — the order matters because of the sign convention, not just as a memory trick.
Forgetting that $m = v/u$ can come out negative even when the arithmetic for $v$ looks "normal" — always read the sign of $m$ before describing the image as erect or inverted.
Applying the six-position convex-lens table to a concave lens: a concave lens with a real object always gives a virtual, erect, diminished image, and the lens formula will return a negative $v$ smaller in magnitude than $u$ confirming exactly that, every time.

Ask the tutor: Want to work through the thin-lens formula for a concave lens numerically, to see why it always lands on virtual, erect, diminished no matter what value of u you pick?

Why is u always negative even though the object is a real, physical thing right in front of the lens

The New Cartesian sign convention measures every distance from the optical centre, treating the direction of the incident light as positive. A real object sits on the side the light comes from, which is defined as the negative direction. It is not saying the object is somehow "negative" in reality — it is a bookkeeping rule that makes one formula work for every case: real or virtual image, convex or concave lens, without needing six separate formulas.

How does the lens formula connect to the ray-diagram table I already memorised

They describe the same physics two ways. Every row of the six-position table for a convex lens — object at infinity, beyond 2F, at 2F, between F and 2F, at F, and between F and O — is just the lens formula evaluated at that particular $u$ . Plugging $u = -2f$ always gives $v = 2f$ and $m = -1$ ; plugging $u$ between $-f$ and $-2f$ always gives $v$ beyond $2f$ and $|m| > 1$ . The formula is the general rule; the table is its six most useful special cases.

My calculated v came out negative for a convex lens — did I make a mistake

Not necessarily. A negative $v$ from a convex lens is correct exactly when the object is placed between the optical centre and the focus (u between 0 and -f), which is the case that produces a virtual, magnified, erect image — the magnifying-glass case. Check your object distance first: if $|u| < f$ , a negative $v$ is the expected, correct answer, not an error.

Image formation by lenses

The core idea

Worked example

Common traps

Before this

Where this leads