Thin lens formula 1/v - 1/u = 1/f — derivation with sign convention

Question

Derive the thin lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ using the refraction at two surfaces of a thin convex lens. State the sign convention used.

(NCERT Class 12, Chapter 9 — essential for boards and NEET)

Solution — Step by Step

Sign convention (New Cartesian)

All distances measured from the optical centre of the lens.
Distances in the direction of incident light are positive.
Object distance $u$ is negative (object on the left), image distance $v$ is positive for real image.

Refraction at the first surface

Light goes from medium of refractive index 1 (air) to medium of refractive index $n$ (glass). Using the single-surface refraction formula:

$\frac{n}{v_1} - \frac{1}{u} = \frac{n - 1}{R_1}$

Here $v_1$ is the image distance for the first refraction (this image acts as the object for the second surface).

Refraction at the second surface

Light goes from glass ( $n$ ) to air (1). The object for this surface is at $v_1$ :

$\frac{1}{v} - \frac{n}{v_1} = \frac{1 - n}{R_2}$

Add the two equations

$\frac{1}{v} - \frac{1}{u} = (n-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$

The right side is $\frac{1}{f}$ by the lens maker's equation. Therefore:

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

The $v_1$ terms cancel when we add the two equations — this is why the "thin lens" approximation works (we assume the two surfaces are at essentially the same point).

Why This Works

A thin lens has negligible thickness compared to the radii of curvature. This means the image formed by the first surface serves as the object for the second surface at the same location. When we add the two refraction equations, the intermediate image distance $v_1$ cancels out cleanly.

The beauty of the thin lens formula is its simplicity: once you know $f$ , you only need $u$ to find $v$ . The entire complexity of two refracting surfaces and the glass medium is absorbed into a single number — the focal length.

Alternative Method

Using similar triangles: draw two rays from the object — one through the centre (undeviated) and one parallel to the axis (passes through focus after refraction). The two triangles formed give the magnification $m = v/u$ . Combined with the geometry at the focal point, you get $1/v - 1/u = 1/f$ .

💡 Expert Tip

For NEET, always write the sign convention before substituting values. Common assignments: for a convex lens with a real object on the left, $u$ is negative, $f$ is positive, and $v$ is positive for a real image. Forgetting the negative sign on $u$ is the number one source of wrong answers.

Common Mistake

⚠️ Common Mistake

Many students write $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ (with a plus sign). This is the mirror formula, not the lens formula. For a lens, it is $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ . The difference arises because for mirrors, both object and image are on the same side, while for lenses, they are on opposite sides. Mixing up the mirror and lens formulas is extremely common — and extremely costly.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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