Lens formula and power

A ray diagram can show you roughly where a lens forms an image, but it cannot hand you an exact number — for that you need the lens formula, the algebraic twin of the ray diagram. Once you fix a consistent sign convention, one equation connects object distance, image distance and focal length for any lens, convex or concave, and a second equation reads off how big and which way up the image is. A third quantity, power, turns focal length into a single number opticians and lens-makers actually use, in dioptres. This article builds the New Cartesian sign convention, the lens formula $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ , magnification, and power — the full quantitative toolkit for the ICSE Class 10 "Refraction Through a Lens" numericals, with every step checked against the ray-diagram rules you already know.

The core insight

Under the New Cartesian sign convention, the single equation $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ links object distance, image distance and focal length for any lens, while $m=\frac{v}{u}$ gives the image's size and orientation, and $P=\frac{1}{f}$ (in metres) gives the lens's power in dioptres.

The core idea

Every lens numerical starts with the New Cartesian sign convention. All distances are measured from the optical centre $O$ . Distances measured in the direction of the incident light (taken left to right) are positive; distances measured against it are negative. Heights measured upward from the principal axis are positive, downward are negative. Because the object is almost always placed to the left of the lens, the object distance $u$ is almost always negative. Focal length follows a fixed rule too: a convex (converging) lens has $f$ positive; a concave (diverging) lens has $f$ negative. Get these two signs wrong and every later number in the calculation is wrong, even if the arithmetic is perfect. With the convention fixed, the lens formula connects the three key distances:

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

where: $u$ is the object distance, $v$ is the image distance, and $f$ is the focal length, all measured from the optical centre with the sign convention above. This single equation holds for convex and concave lenses alike, and for real or virtual images — you never need a different formula for a different case, only careful signs.

Linear magnification tells you how the image compares to the object, both in size and orientation:

m = \frac{v}{u} = \frac{h_i}{h_o}

where: $h_i$ is image height and $h_o$ is object height. The sign of $m$ tells you orientation — positive means erect (same way up as the object), negative means inverted. The magnitude tells you size — $|m|>1$ means magnified, $|m|<1$ means diminished, $|m|=1$ means same size. Reading sign and magnitude separately, rather than as one blended idea, avoids most magnification mistakes. Finally, power measures how strongly a lens bends light toward or away from the axis — a short, strongly curved focal length bends light sharply and has high power, a long focal length bends it gently and has low power:

P = \frac{1}{f} \quad (f \text{ in metres}, P \text{ in dioptres, D})

A convex lens converges light and has positive power; a concave lens diverges light and has negative power. A lens with $f=0.5\ \text{m}$ has $P=+2\ \text{D}$ ; halve the focal length and the power doubles.

Figure

Ray diagram of a convex lens forming a real inverted image beyond the focal point with object and image distances marked

Derivation ADeriving the lens formula from similar triangles

The goal is to show, using only the geometry of two pairs of similar triangles in the standard convex-lens ray diagram, that the object distance $u$ , image distance $v$ and focal length $f$ must satisfy $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ , which is why the formula is trustworthy rather than a rule to memorise blindly.

Step 1. Draw an object $AB$ on the axis with a ray $AP$ parallel to the axis striking the lens at $P$ and refracting through the second focal point $F$ to meet the axis at the image $A'B'$ . Triangles $OPF$ and $A'B'F$ are similar, because $OP$ is parallel to $A'B'$ (both perpendicular to the axis) and they share the vertical angle at $F$ . So $\frac{A'B'}{OP} = \frac{A'F}{OF}$ . Since $OP$ equals the object height $AB$ (the ray leaves the lens at the same height it entered), this gives $\frac{A'B'}{AB} = \frac{A'F}{OF}$ .

Step 2. Now use a second ray, $AO$ , passing straight through the optical centre without bending. Triangles $ABO$ and $A'B'O$ are similar, because $AB$ is parallel to $A'B'$ and they share the vertical angle at $O$ . This gives $\frac{A'B'}{AB} = \frac{OB'}{OB}$ , which in magnitude is $\frac{v}{u}$ .

Step 3. Equate the two expressions for $\frac{A'B'}{AB}$ from Steps 1 and 2: $\frac{OB'}{OB} = \frac{A'F}{OF}$ . Write $A'F = OB' - OF$ (the image distance minus the focal length), so $\frac{v}{u} = \frac{v-f}{f}$ .

Step 4. Cross-multiply and simplify: $vf = u(v-f) = uv - uf$ . Divide every term by $uvf$ : $\frac{1}{u} = \frac{1}{f} - \frac{1}{v}$ , which rearranges to $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ — exactly the lens formula, once the New Cartesian signs of $u$ , $v$ , $f$ are substituted into this magnitude relation.

Worked example

An object is placed $30\ \text{cm}$ in front of a convex lens of focal length $15\ \text{cm}$ . Find the image distance and magnification, and describe the image.

Step 1. Apply the sign convention: $u = -30\ \text{cm}$ , and since the lens is convex, $f = +15\ \text{cm}$ .

Step 2. Substitute into the lens formula and solve for $v$ : $\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{15} + \frac{1}{-30} = \frac{2}{30} - \frac{1}{30} = \frac{1}{30}$ , so $v = 30\ \text{cm}$ .

Step 3. Find the magnification: $m = \frac{v}{u} = \frac{30}{-30} = -1$ .

Step 4. Interpret: $v$ is positive, so the image is real; $m$ is negative, so it is inverted; $|m|=1$ , so it is the same size as the object. The image forms $30\ \text{cm}$ on the far side of the lens — this matches the ray-diagram rule for an object placed at $2F$ ( $u=-2f$ ), confirming the calculation.

Competitive stretch

The same object is now placed in front of a concave lens of focal length $20\ \text{cm}$ at $u=-40\ \text{cm}$ . Then $\frac{1}{v} = \frac{1}{-20} + \frac{1}{-40} = -\frac{3}{40}$ , giving $v \approx -13.3\ \text{cm}$ and $m = \frac{v}{u} \approx 0.33$ — a virtual, erect, diminished image, with power $P = \frac{1}{-0.20} = -5\ \text{D}$ . Every concave-lens numerical for a real object lands in this same virtual-erect-diminished pattern, so it is worth recognising on sight.

Common traps

Forgetting to convert focal length to metres before finding power — an easy slip that throws the answer off by a factor of $100$ .
Mixing up the sign of $f$ for convex (positive) versus concave (negative) lenses, which corrupts every later step.
Misreading the sign of $m$ as "size" instead of orientation — a negative $m$ means inverted, not smaller; always check sign and magnitude separately.
Reaching for the lens-maker's formula (in terms of radii of curvature) — it is beyond the ICSE syllabus; only $u$ , $v$ , $f$ are used here.

Ask the tutor: Want to try a few more lens-formula numericals with mixed convex and concave lenses to lock in the sign conventions?

Why is the object distance u almost always negative in these problems?

Because the New Cartesian convention measures every distance from the optical centre, taking the direction of incident light as positive. An object is placed on the side the light is coming from, which is against the positive direction, so $u$ comes out negative almost every time you set up a numerical. The only distances that end up positive on the object side are unusual cases like a virtual object, which do not appear at ICSE level, so in practice you can expect $u$ to be negative and should double-check your answer if it is not.

Does the lens formula change for a concave lens, or do I use the same equation?

You use exactly the same equation, $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ , for both lens types — nothing about the formula itself changes. What changes is the sign you substitute for $f$ : positive for convex, negative for concave. Get that one sign right at the start and the rest of the algebra automatically produces a virtual, erect, diminished image for a concave lens with a real object, matching what the ray diagram already tells you.

How is power different from focal length if P is just 1/f?

Power and focal length carry the same information but answer different questions. Focal length tells you a distance — where parallel rays converge or appear to diverge from. Power tells you strength on a fixed scale, in dioptres, so lenses of very different focal lengths can be compared directly: a $10\ \text{cm}$ lens ( $P=+10\ \text{D}$ ) is far more powerful than a $100\ \text{cm}$ lens ( $P=+1\ \text{D}$ ), even though both are simply "convex." Opticians prescribe lenses in dioptres precisely because powers of different lenses in contact simply add, which focal lengths do not.

Lens formula and power

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads