Thin Lens Equation

Equivalent forms

d_i = \frac{f \cdot d_o}{d_o - f}

M = -\frac{d_i}{d_o}

The reciprocal form unifies converging and diverging lenses with a single sign convention.

Symbol	Name	SI unit	Dimension	Range
$f$	Focal length Distance from lens center to focal point (positive for converging, negative for diverging)	m	L	-0.5 – 0.5
$d_o$	Object distance Distance from object to the lens center	m	L	0.01 – 2
$d_i$	Image distanceoutput Distance from lens center to the image (positive for real, negative for virtual)	m	L	-2 – 2

Unit systems

Symbol	SI	CGS	Imperial
$f$	m	cm	in
$d_o$	m	cm	in
$d_i$	m	cm	in

Where it holds

Valid for thin lenses (thickness << focal length) in the paraxial approximation (small angles). Breaks down for thick lenses and wide-angle rays (aberrations).

Dimensional analysis

1/f \to [L^{-1}]

1/d_o + 1/d_i \to [L^{-1}] + [L^{-1}] = [L^{-1}]

Both sides are

[L^{-1}]

= diopters

(m^{-1})

.

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Discovery

Multiple contributors · 1693

Developed through contributions by Kepler, Barrow, and Halley. The modern thin lens formula was formalized in the late 17th century.

Try this

How far must you hold a magnifying glass from a bug to project its image on a wall?

A thin converging lens (f = 10 cm) views an object at 15 cm. Find the image distance using 1/f = 1/do + 1/di.

Research status: stable

Real-world applications

Camera autofocus: adjusts d_o by moving the lens to satisfy $1/f = 1/d_o + 1/d_i$ .
Eyeglasses: diverging lenses (f < 0) correct myopia by shifting the image back to the retina.
Projectors: place the slide at d_o slightly > f to get a large real image on a distant screen.
Magnifying glasses: object inside f produces a virtual, magnified, upright image.

Common misconceptions

Covering half the lens removes half the image — it dims the whole image, doesn't cut it in half.
A negative image distance means no image — it means a virtual image on the same side as the object.
The formula works for thick lenses — it only works when lens thickness << focal length (thin lens approximation).

Experimental verification

Verified by measuring object and image distances with calibrated optical benches. Modern automated lens testing systems confirm focal lengths to < 0.1% accuracy using collimated laser sources.

Derivation

Using similar triangles from ray diagrams: a ray through the center is undeviated, and a ray parallel to the axis refracts through the focal point.

For a thin lens, equating the two triangles gives

h_i/h_o = d_i/d_o

(magnification) and

h_i/h_o = (d_i - f)/f

.

Setting these equal:

d_i/d_o = (d_i - f)/f \to f\cdot d_i = d_o\cdot d_i - d_o\cdot f \to d_o\cdot f + d_i\cdot f = d_o\cdot d_i

.

Dividing by

d_o\cdot d_i\cdot f

:

1/f = 1/d_o + 1/d_i

.

Limiting cases

d_o \to \infty

⟶

d_i \to f

Parallel rays (distant object) converge at the focal point.

d_o = f

⟶

d_i \to \infty

Object at focal point produces parallel rays — image at infinity.

d_o = 2f

⟶

d_i = 2f

Object at 2f produces a same-size inverted image at 2f.

What if…

What if

d_o = f

exactly?

$1/d_i = 0$ , so $d_i \to \infty$ . Rays emerge parallel — the image forms at infinity. This is how collimated beams are made.

What if d_o < f for a converging lens?

d_i becomes negative — a virtual, upright, magnified image forms. This is the magnifying glass regime.

What if the lens is in water instead of air?

The focal length changes because refraction depends on the ratio of lens-to-medium refractive indices. The lensmaker's equation must be used with the medium's n.

1

Converging lens image

Given ·

f:: 0.1
d o:: 0.15

Find · d_i

Steps

Rearrange: $1/d_i = 1/f - 1/d_o$
$1/d_i = 1/0.10 - 1/0.15 = 10 - 6.667 = 3.333$
$d_i = 1/3.333 = 0.30\,\mathrm{m}$
Positive $d_i \to$ real, inverted image at 30 cm

Result ·

d_i = f*d_o/(d_o - f) = 0.10 \times 0.15 / (0.15 - 0.10) = 0.015/0.05 = 0.30\,\mathrm{m}

2

Diverging lens virtual image

Given ·

f:: -0.2
d o:: 0.3

Find · d_i

Steps

$1/d_i = 1/f - 1/d_o = 1/(-0.20) - 1/0.30$
$1/d_i = -5.0 - 3.333 = -8.333$
$d_i = -0.12\,\mathrm{m}$
Negative $d_i \to$ virtual image 12 cm on the same side as the object

Result ·

d_i = f*d_o/(d_o - f) = (-0.20)(0.30)/(0.30 - (-0.20)) = -0.06/0.50 = -0.12\,\mathrm{m}

Snell's Law→Law of Reflection→