Image formation by lenses

Hold a convex lens close to a page of text and the letters appear large, upright, and readable — a magnifying glass. Pull the same lens farther away and, past a certain distance, the image flips upside down and shrinks. A concave lens, meanwhile, never does anything so dramatic: whatever you look through it, the image stays small, upright, and stuck close to the lens. The difference is not the lens being fussy — it is a strict, predictable rule connecting where the object sits, where the two focal points $F_1$ and $F_2$ lie, and what kind of image forms. This article covers exactly what CBSE Class 10 expects: the six standard object-position cases for a convex lens, the single unvarying case for a concave lens, and the lens formula, magnification formula, and power of a lens that let you calculate — not just describe — every such image. The maths mirrors the mirror formula from the previous topic, but with two sign flips that trip up almost everyone at first.

The core insight

Image formation by a lens is governed entirely by object position relative to the focal points $F_1$ and $F_2$ , and is calculated using the lens formula $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ , the magnification $m=\frac{v}{u}$ , and the power $P=\frac{1}{f}$ (in dioptres, with $f$ in metres).

The core idea

Convex (converging) lens. As the object moves from infinity toward the lens, the image behaviour runs through six stages, the pattern closely resembling a concave mirror but built around two focal points instead of one:

Object position	Image position	Size	Nature
At infinity	At focus $F_2$	Point-sized	Real, inverted
Beyond $2F_1$	Between $F_2$ and $2F_2$	Diminished	Real, inverted
At $2F_1$	At $2F_2$	Same size	Real, inverted
Between $F_1$ and $2F_1$	Beyond $2F_2$	Enlarged	Real, inverted
At focus $F_1$	At infinity	Highly enlarged	Real, inverted
Between optical centre and $F_1$	Same side as object	Enlarged	Virtual, erect

Concave (diverging) lens. There is no table to memorise here: wherever the object is placed, the image is always virtual, erect, and diminished, formed on the same side of the lens as the object, sitting between the optical centre and the focus. This is exactly analogous to a convex mirror's single, unvarying rule.

The lens formula connects object distance $u$ , image distance $v$ , and focal length $f$ , all measured from the optical centre using the New Cartesian Sign Convention (distances measured against the direction of incident light are negative, and distances measured with it are positive):

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

where: $u$ is object distance, $v$ is image distance, $f$ is focal length — all in the same length unit, with $f$ positive for a convex lens and negative for a concave lens. Note the minus sign, unlike the mirror formula's plus sign; this single difference is the most common source of numerical slips when moving between the two chapters.

Magnification for a lens is

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

where: $h$ is object height, $h'$ is image height. Unlike the mirror's magnification, there is no minus sign here. A positive $m$ means the image is erect and virtual; a negative $m$ means it is inverted and real; $|m|>1$ means enlarged and $|m|<1$ means diminished.

Power of a lens measures how strongly it bends (converges or diverges) light, defined as the reciprocal of focal length with $f$ in metres:

P = \frac{1}{f}

where: $P$ is power in dioptres (D), the SI unit, and $f$ is in metres. A convex lens has positive power (it converges light), a concave lens has negative power (it diverges light) — consistent with the sign convention already used for $f$ .

Figure

Ray diagram of a convex lens with an object between F1 and 2F1 producing an enlarged inverted real image beyond 2F2 on the far side

Derivation AUsing the lens formula to find image position and applying it to magnification

Given the focal length of a convex lens and the distance of an object placed in front of it, use the lens formula together with the sign convention to find the image distance $v$ , then combine that with the object distance to compute magnification and describe the image completely, exactly as required for CBSE numericals on this topic.

Step 1. Assign signs to all known quantities using the New Cartesian Sign Convention. The object distance $u$ is always negative (object is in front of the lens, against the incident light direction). The focal length $f$ is positive for a convex lens and negative for a concave lens.

Step 2. Substitute the known values of $u$ and $f$ into the lens formula $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ and rearrange to isolate $\frac{1}{v}$ :

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

Step 3. Combine the two fractions on the right using their LCM, simplify to a single fraction equal to $\frac{1}{v}$ , and invert it to get $v$ in the same length unit as $u$ and $f$ .

Step 4. Read the nature of the image from the sign of $v$ : a positive $v$ (same side as the outgoing light, opposite the object) means a real image; a negative $v$ (same side as the object) means a virtual image.

Step 5. Compute magnification with $m=\frac{v}{u}$ . A negative $m$ confirms a real, inverted image; a positive $m$ confirms a virtual, erect image. Compare $|m|$ to $1$ to state whether the image is enlarged, diminished, or the same size, and multiply $|m|$ by the object height $h$ to get the image height $h'$ .

Worked example

An object 2 cm tall is placed 30 cm in front of a convex lens of focal length 20 cm. Find the position, nature, and size of the image, and the power of the lens.

Step 1. By sign convention: $u = -30\ \text{cm}$ , $f = +20\ \text{cm}$ (convex lens), $h = 2\ \text{cm}$ .

Step 2. Apply the lens formula:

\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{20} + \frac{1}{-30} = \frac{1}{20} - \frac{1}{30}

Step 3. LCM of 20 and 30 is 60, so

\frac{1}{v} = \frac{3}{60} - \frac{2}{60} = \frac{1}{60} \implies v = 60\ \text{cm}

Since $v$ is positive, the image forms 60 cm on the far side of the lens — a real image.

Step 4. Magnification:

m = \frac{v}{u} = \frac{60}{-30} = -2

Negative $m$ means the image is inverted; $|m| = 2 > 1$ means it is enlarged, so $h' = m \times h = -2 \times 2 = -4\ \text{cm}$ (twice the object's height, inverted).

Step 5. Power of the lens, converting $f = 20\ \text{cm} = 0.2\ \text{m}$ :

P = \frac{1}{f} = \frac{1}{0.2} = 5\ \text{D}

The image is real, inverted, enlarged, formed 60 cm from the lens on the far side, with height 4 cm, and the lens has a power of +5 D.

Competitive stretch

A concave lens used for a myopic (short-sighted) correction has focal length 40 cm. By the sign convention, $f = -40\ \text{cm} = -0.4\ \text{m}$ , so its power is $P = \frac{1}{f} = \frac{1}{-0.4} = -2.5\ \text{D}$ . The negative sign confirms it is diverging; whatever the object distance, the image it forms stays virtual, erect, and diminished, between the optical centre and the focus.

Common traps

Confusing the mirror formula's plus sign with the lens formula's minus sign. Mirror: $1/v + 1/u = 1/f$ ; lens: $1/v - 1/u = 1/f$ — this is the single most common numerical error in this chapter.
Adding a minus sign to lens magnification. Mirror magnification is $m=-v/u$ ; lens magnification is $m=+v/u$ , no minus sign.
Forgetting to convert focal length to metres before computing power. Power in dioptres needs $f$ in metres; using centimetres gives an answer 100 times too large.
Assuming a concave lens can ever give a real or magnified image. Like a convex mirror, a concave lens always gives a virtual, erect, diminished image, for every object position, with no exceptions.

Ask the tutor: If the six convex-lens cases or the sign conventions still feel shaky, ask the AI tutor to drill you with random object distances and have you name the image type each time.

Why does the lens formula have a minus sign while the mirror formula has a plus sign?

The two formulas come from the same underlying sign convention (New Cartesian Sign Convention) but are written for geometrically different setups — light passing through a lens versus reflecting off a mirror. At Class 10 level you are only expected to state and correctly apply both formulas, not derive why the signs differ; just memorise mirror as plus and lens as minus, and double-check every numerical answer against the sign rule for u and f.

Can a convex lens ever produce a virtual image, or a concave lens ever produce a real one?

Yes for the convex lens: when the object sits between the optical centre and focus $F_1$ , the convex lens produces an enlarged, virtual, erect image — this is exactly how a magnifying glass works. A concave lens, however, can never produce a real image; it is always virtual, erect, and diminished for every object position, with no exception, because a diverging lens can never actually bring rays to a real crossing point.

How is power of a lens different from magnification, since both compare something about the image?

They measure completely different things. Magnification compares the size of the image to the size of the object, and depends on where the object is placed. Power measures how strongly the lens itself bends light, and depends only on the lens's focal length, not on any particular object position. A given lens has one fixed power in dioptres, but its magnification changes every time you move the object.

Image formation by lenses

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads