Lens formula and magnification

A convex lens can throw a large, upside-down image of a candle flame onto a screen a metre away, while a concave lens tucked into a pair of spectacles always shows a small, upright image no matter where you hold it. Ray diagrams explain why this happens, but they are slow to draw and hard to read off precisely. The lens formula and magnification formula replace the drawing with two short equations: feed in the object distance and focal length, and out comes the exact image position, size, and nature — real or virtual, erect or inverted, magnified or diminished. Every numerical on spherical lenses in this chapter reduces to applying these two relations correctly, together with the Cartesian sign convention, so mastering the bookkeeping of signs matters as much as knowing the formulas themselves.

The core insight

The lens formula $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ locates the image and the magnification formula $m=\frac{h'}{h}=\frac{v}{u}$ sizes and characterises it, with every quantity's sign fixed by the Cartesian convention.

The core idea

For a thin spherical lens, all distances are measured from the optical centre along the principal axis, following the Cartesian sign convention: distances measured against the incident light (to the left, for the usual left-to-right ray direction) are negative, and distances measured with it (to the right) are positive. Heights above the axis are positive, heights below are negative. With the object distance $u$ , image distance $v$ , and focal length $f$ all measured this way, the lens formula connects them:

\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 in the same length unit and all carrying sign. The magnification compares image height to object height, and turns out to equal the ratio of image distance to object distance as well:

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

where: $h$ is the object height (positive, since real objects sit above the axis) and $h'$ is the image height, positive if the image is erect and negative if it is inverted. Four facts about signs make the two formulas usable:

$f$ is positive for a convex (converging) lens, negative for a concave (diverging) lens.
$u$ is negative whenever the object is real, i.e. placed on the incoming-light side of the lens — which is every case in this chapter.
A positive $m$ signals a virtual, erect image; a negative $m$ signals a real, inverted image.
$|m|>1$ means magnified (larger than the object), $|m|<1$ means diminished, and $|m|=1$ means same size.

The CBSE syllabus asks you to state and apply both formulas, not derive them — the derivation from lens geometry is reserved for later study. What you must be fluent in is reading the sign of $v$ and $m$ off a calculation and translating that directly into "real or virtual," "erect or inverted," and "magnified or diminished," exactly as a ray diagram would show.

Figure

A convex lens with an object to its left beyond the focal point and the real inverted image formed to its right, with u v and f marked along the axis

Derivation ARearranging the lens formula to solve for any missing quantity

The lens formula $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ has exactly three quantities in it, and most numericals give you two of the three and ask for the third, so the real skill being tested is fast, sign-safe algebraic rearrangement rather than any new physics, and this derivation walks through obtaining all three working forms from the one memorised equation.

Step 1. Start from the memorised form and isolate whichever term is unknown by moving the other two to the right-hand side. If $v$ is unknown, write $\frac{1}{v} = \frac{1}{f} + \frac{1}{u}$ . If $u$ is unknown, write $\frac{1}{u} = \frac{1}{v} - \frac{1}{f}$ . If $f$ is unknown, write $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$ . Each is just the original equation with terms moved across the equals sign — no new physics enters at this step.

Step 2. Combine the two fractions on the right into a single fraction using a common denominator, then take the reciprocal to get an explicit formula. For $f$ unknown, $\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{u - v}{uv}$ , so

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

This single-shot formula is handy when a question directly gives both $u$ and $v$ and asks only for $f$ , skipping the two-line LCM computation every time.

Step 3. Always substitute signed values, not magnitudes. A common error is plugging in $u=30$ and $v=20$ (dropping the negative sign on $u$ ) into the Step 2 formula — this silently changes which lens (convex or concave) the answer describes. Keep every distance exactly as the sign convention gives it, right through to the final reciprocal.

Step 4. Sanity-check the result against the sign rules: a positive $f$ from this calculation must correspond to a convex lens, a negative $f$ to a concave lens; if the physical setup described in the question (e.g. "a lens that produces a virtual image of every object") contradicts the sign you obtained, re-check the substitution before trusting the arithmetic.

Worked example

A convex lens of focal length 12 cm forms an image of an object placed 30 cm in front of it. Find the image position, the magnification, and the nature of the image; then find what focal length a different lens would need to place that same object's image at $v=+15\ \text{cm}$ instead.

Step 1. Write the knowns with sign convention: $u=-30\ \text{cm}$ , $f=+12\ \text{cm}$ .

Step 2. Apply the lens formula: $\frac{1}{v} = \frac{1}{f}+\frac{1}{u} = \frac{1}{12}+\frac{1}{-30}$ .

Step 3. Use the LCM of 12 and 30, which is 60: $\frac{1}{v} = \frac{5}{60}-\frac{2}{60} = \frac{3}{60} = \frac{1}{20}$ , so $v=+20\ \text{cm}$ .

Step 4. Find the magnification: $m = \frac{v}{u} = \frac{20}{-30} = -0.67$ . Since $v$ is positive the image is real, and since $m$ is negative and $|m|<1$ it is inverted and diminished — 20 cm on the far side of the lens, matching an object placed beyond $2F$ of a convex lens.

Step 5. Now use the Step 2 rearrangement $f=\dfrac{uv}{u-v}$ to answer the second part, with the same object at $u=-30\ \text{cm}$ but the new target $v=+15\ \text{cm}$ : $f = \dfrac{(-30)(15)}{(-30)-(15)} = \dfrac{-450}{-45} = +10\ \text{cm}$ .

Step 6. Interpret: a convex lens (positive $f$ ) of focal length 10 cm is needed to move this same object's real image from 20 cm to 15 cm behind the lens — a shorter focal length pulls the image closer, which is worth remembering as a quick physical check on any such answer.

Common traps

Substituting $u$ and $v$ without their signs, which gives a numerically plausible but conceptually wrong answer about whether the image is real or virtual.
Confusing the lens formula $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ with the mirror formula $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ — the minus sign is what makes it a lens equation.
Reporting only the magnitude of $m$ and skipping the sign, which is the one thing that tells you erect vs. inverted.
Expecting a concave lens to ever form a real or magnified image of a real object — with a single diverging lens the image is always virtual, erect, and diminished, whatever $u$ is.

Ask the tutor: Want a few more mixed convex/concave numericals with signs deliberately tricky, worked step by step?

Why does u always come out negative in these problems?

Because every object in this chapter is a real object placed on the side the light arrives from, and the Cartesian sign convention fixes the optical centre as the origin with distances measured against the incoming light as negative. Since light is drawn travelling left to right and the object sits to the left, $u$ is always negative. You would only ever see a positive $u$ in more advanced setups with a virtual object, which is outside this syllabus.

How do I decide the sign of f before I even start the calculation?

Sign of $f$ depends only on the type of lens, not on where the object is: a convex (converging) lens always has $f$ positive, and a concave (diverging) lens always has $f$ negative. Fix this sign first, from the question's wording, before touching $u$ or $v$ — getting $f$ 's sign wrong flips the entire answer, including whether the final image reads as real or virtual.

If v comes out negative, does that always mean the image is virtual?

Yes, for lenses under this sign convention, a negative $v$ means the image forms on the same side as the object, which is the definition of a virtual image here. This happens for every object placed within the focal length of a convex lens, and for every object placed anywhere in front of a concave lens. A positive $v$ always means a real image forms on the far side, where a screen could actually catch it.

Lens formula and magnification

The core idea

Worked example

Common traps

Before this

Where this leads