Formula of Lens: 1/v − 1/u = 1/f, Derivation and Examples

The formula of lens, also called the thin lens equation, is $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ . In it, u stands for how far the object sits from the optical centre, v for how far away the image forms and f for the focal length. Know any two and you can find the third, for convex and concave lenses alike.

Signs come from the New Cartesian convention. Light travels left to right, distances to the right of the lens are positive and distances to the left are negative, so u for any real object comes out below zero. A convex lens has a positive f, a concave lens a negative f. Many US books flip the sign of the object term and add the two reciprocals, entering the object distance as a positive number. Pick one system for a whole problem and never switch halfway.

Quick facts

Thin lens equation
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
Formula of focal length
$f = \frac{uv}{u - v}$
Image distance
$v = \frac{uf}{u + f}$
Convex lens
f positive; real images have v positive
Concave lens
f negative; image always virtual, v negative
Magnification
$m = \frac{v}{u}$
Unit
metre in SI; cm is fine if all three distances use it

Terms in the formula of lens

Lengths taken from the optical centre O
Symbol	Measured to	Usual sign
$u$	The object	Negative (object on the incoming side)
$v$	The image	Positive if real, negative if virtual
$f$	The principal focus	Positive for convex, negative for concave

The focal length is a property of the lens alone. It depends on the glass and the shape of the faces, as the lens maker formula shows, and not on where the object is placed.

Derivation of lens formula

Place an object AB upright on the axis of a convex lens, beyond the focus F. A is on the axis and B is its top. A real, inverted image A′B′ forms on the other side. Use two rays from B: one through the optical centre O, which goes straight on, and one parallel to the axis, which meets the lens at C and then passes through F to B′. Because the lens is thin, OC equals AB.

Triangles ABO and A′B′O are similar, so

$\frac{A'B'}{AB} = \frac{OA'}{OA}$

Triangles OCF and A′B′F are similar, and OC = AB, so

$\frac{A'B'}{AB} = \frac{FA'}{OF} = \frac{OA' - OF}{OF}$

Equate and insert signs. OA′ = +v, OA = −u, OF = +f:

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

Cross multiplying gives $vf = -uv + uf$ . Divide every term by $uvf$ :

$\frac{1}{u} = -\frac{1}{f} + \frac{1}{v} \quad\Rightarrow\quad \frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

Write the sign of every distance on your diagram before you substitute. That single habit prevents most lost marks in this derivation.

Worked examples

Example 1: a convex lens forming a real image

A toy is placed 36 cm from a convex lens with f = 24 cm. Where is the image?

Signs first: u = −36 cm and f = +24 cm.

$\frac{1}{v} = \frac{1}{24} + \frac{1}{-36} = \frac{3 - 2}{72} = \frac{1}{72} \Rightarrow v = +72\ \text{cm}$

Real, 72 cm on the far side; m = 72/(−36) = −2, so inverted and doubled.

Example 2: a concave lens

A coin lies 48 cm from a diverging lens whose focal length is 16 cm. Locate its image.

With signs: object at −48 cm, focal length −16 cm.

$\frac{1}{v} = -\frac{1}{16} - \frac{1}{48} = -\frac{4}{48} \Rightarrow v = -12\ \text{cm}$

You get a virtual image 12 cm from the lens on the coin's side.

Example 3: a camera, with millimetres and metres

A camera lens has f = 50 mm and photographs a person 2.0 m away. How far behind the lens must the sensor be?

Convert: f = 0.050 m, u = −2.0 m.

$v = \frac{uf}{u + f} = \frac{(-2.0)(0.050)}{-2.0 + 0.050} = \frac{-0.100}{-1.95} = 0.0513\ \text{m}$

About 51.3 mm, just over the focal length. Focusing a camera means moving the lens by these small amounts.

Example 4: formula of focal length from an experiment

On an optical bench, a sharp image appears 40 cm behind a lens when the object is 40 cm in front. Find f.

$f = \frac{uv}{u - v} = \frac{(-40)(40)}{-40 - 40} = \frac{-1600}{-80} = +20\ \text{cm}$

Equal object and image distances mean the object is at 2F, a quick way to measure the focal length in the lab.

Rearranged forms and image rules

Which rearrangement to use
To find	Use
Image distance v	$v = \frac{uf}{u + f}$
Object distance u	$u = \frac{vf}{f - v}$
Focal length f	$f = \frac{uv}{u - v}$
Power P	$P = \frac{1}{f}$ with f in metres

The calculator on this page uses exactly these three forms, in metres. The power of lens formula page explains dioptres, and the magnification of lens formula turns v into an image size. For two thin lenses in contact, add the reciprocals: $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$ .

Lens versus mirror and the usual slips

The lens relation subtracts 1/u. The mirror formula adds it. Swapping them is the most common error in this chapter.
A real object has a negative u. Entering +30 cm for an object 30 cm away flips the answer.
A concave lens never forms a real image of a real object. A positive v from a concave lens means a sign is wrong.
Using different units for u, v and f. Convert first.

For the bigger picture of how lenses, mirrors and prisms fit with wave ideas, start at the waves and optics hub, which gathers the whole topic in one place.

Frequently asked questions

What is the formula of lens?

Write 1/v − 1/u = 1/f. Every distance starts at the optical centre: u runs to the object, v to the image, f to the focus.

What is the formula of focal length?

From the lens relation, f = uv/(u − v). From the shape of the lens, the lens maker's equation ties f to the glass and to how sharply each face curves.

Why is the lens formula different from the mirror formula?

A lens sends light through to the other side while a mirror sends it back. With the same sign rules, that changes the plus sign between 1/v and 1/u into a minus.

Can the lens formula be used for a concave lens?

Yes. The same equation works for both lenses. Just enter f as a negative number for a concave lens.

What is the SI unit of focal length?

The metre. In practice centimetres or millimetres are common, which is fine as long as u, v and f all use the same unit.