Q.What are the sign conventions followed for lenses?

The Thin Lens Formula: Why a Lens Behaves the Way It Does

When you hold a magnifying glass and move it toward a piece of paper, something dramatic happens. At first the image is blurry, then suddenly a sharp, bright spot appears — and if you hold it still, the paper can catch fire. That spot is the image of the sun, and the distance from the lens to the paper is the focal length. The thin lens formula is the mathematical rule that governs exactly where that image will form for any object you place in front of the lens.

The Intuition: Bending Light Systematically

A thin lens works by bending every ray of light that passes through it. The key idea is that the bending is predictable — it depends only on where the ray hits the lens and the lens's own power (its focal length). For a thin lens, we assume the lens is so thin that we can ignore its thickness and treat all bending as happening at a single plane through its centre.

Imagine an object placed at some distance $u$ from the lens. Light from each point on the object spreads out in all directions. The lens intercepts this light and redirects it so that all rays from a single point on the object meet again at a single point on the other side — that meeting point is the image. The distance from the lens to that image is $v$.

The relationship between $u$, $v$, and the focal length $f$ is not arbitrary. It comes from geometry: similar triangles formed by the rays and the lens surface give a clean, simple equation.

The Precise Statement

For a thin lens, the object distance $u$, image distance $v$, and focal length $f$ are related by:

This is the thin lens formula in its Cartesian sign convention form.

If you use the older "real is positive" convention, the formula looks like $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$. The physics is identical — only the signs change. Stick to one convention and use it consistently.

What the Formula Tells You

The formula says that the curvature of the wavefront (the reciprocal of distance) changes by a fixed amount as light passes through the lens. That fixed amount is $1/f$, the lens's power measured in dioptres.

• If the object is very far away ($u \to -\infty$), then $1/u \to 0$, so $1/v = 1/f$, meaning $v = f$. The image forms at the focal point — this is why you can burn paper with sunlight.
• If the object is at twice the focal length ($u = -2f$), then $1/v = 1/f + 1/(-2f) = 1/(2f)$, so $v = 2f$. The image is the same size as the object and inverted. …