Lateral Magnification

Equivalent forms

m = \frac{h_i}{h_o}

m = -\frac{v}{u}

A single signed ratio captures both how big the image is and whether it stands upright or inverted.

Symbol	Name	SI unit	Dimension	Range
$d_o$	Object distance Distance from the object to the lens (positive for a real object)	m	L	0.5 – 20
$d_i$	Image distance Distance from the lens to the image (positive for a real image)	m	L	0.5 – 20
$m$	Magnificationoutput Ratio of image height to object height (signed)	dimensionless	1	-10 – 10

Unit systems

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

Where it holds

Valid for thin lenses and mirrors under the paraxial (small-angle) approximation. Thick lenses, large apertures, and off-axis points introduce aberrations that distort the simple ratio.

Discovery

Johannes Kepler · 1611

In his treatise Dioptrice, Kepler laid out the geometric optics of real images formed by converging lenses, establishing the proportionality between image and object distances that underlies the magnification rule.

Try this

Why does a magnifying glass flip the world upside-down once you pull it far enough from the page?

An object sits 4 cm in front of a lens that forms an image 6 cm behind it. What is the magnification, and is the image upright or inverted?

Research status: stable

Real-world applications

Camera lenses set magnification to fit a scene onto a fixed sensor.
Microscopes and telescopes chain magnifications across multiple lenses.
Projectors invert and enlarge a slide onto a distant screen.
Magnifying glasses produce upright virtual images for reading fine print.

Common misconceptions

Magnification is always greater than 1 — many imaging setups (cameras) produce |m| < 1, a reduced image.
A negative magnification means the image is smaller — the sign indicates inversion, not size.
Virtual images cannot be magnified — a magnifying glass forms an upright, enlarged virtual image with m > 1.

Experimental verification

On an optical bench, measure object and image distances for a converging lens and compare the ratio d_i/d_o to the measured height ratio on a screen; the signed values agree, including the observed inversion of real images.

Derivation

Trace the ray through the lens center: it passes undeviated.

Similar triangles formed by the object height h_o at distance d_o and image height h_i at distance d_i give

h_i/h_o = -d_i/d_o

, where the minus sign encodes the inversion of a real image.

By definition

m = h_i/h_o

, hence

m = -d_i/d_o

.

Limiting cases

d_i = d_o

⟶

m = -1

Equal distances give a same-size, inverted real image (object at 2f for a thin lens).

d_i \ll d_o⟶ |m| \ll 1A distant object forms a tiny image near the focal plane, like a camera photographing a landscape.

What if…

What if the object is inside the focal length?

The image becomes virtual, d_i is negative, so $m = -d_i/d_o$ is positive and greater than 1 — an upright, enlarged image, exactly how a magnifying glass works.

What if you double the object distance?

For a fixed lens the image moves closer to the focal point and shrinks; the magnification magnitude drops toward zero.

1

Real image from a converging lens

Given ·

d o:: 4
d i:: 6

Find · m

Steps

Apply $m = -d_i/d_o$ .
Substitute $d_i = 6\,\mathrm{cm}$ , $d_o = 4\,\mathrm{cm}$ .
$m = -6/4 = -1.5$ (inverted, enlarged).

Result ·

m = -d_i/d_o = -6/4 = -1.5

. The image is inverted and 1.5x the object's height.

Thin Lens Equation→Mirror Equation→Lensmaker's Equation→