Magnification Calculator
Enter the object and image heights, or their distances from a lens or mirror, to get the linear magnification, the image size, and whether the image is upright or inverted.
Find the Magnification of a Lens or Mirror
Magnification Formula
Where:
- \(m\) — linear (transverse) magnification, a dimensionless number
- \(h'\) — height of the image; a negative value means the image is inverted
- \(h\) — height of the object (always positive for a real object)
- \(v\) — image distance from the lens or mirror; positive for a real image, negative for a virtual image
- \(u\) — object distance from the lens or mirror; positive for a real object
The formula expresses the fact that similar triangles formed by the object, the image and the rays through the optical centre (or the mirror vertex) make the ratio of heights equal to the ratio of distances. The minus sign in \(m = -v/u\) follows the real-is-positive sign convention and automatically makes magnification negative for an inverted, real image.
Magnification – Calculation Example
Example: an object 5 cm tall stands in front of a converging lens and forms an inverted image 10 cm tall (entered as a negative image height, since it is upside down).
\[ m = \frac{h'}{h} = \frac{-10 \text{ cm}}{5 \text{ cm}} = -2 \]
Result: \(m = -2\). The image is inverted, twice as tall as the object (200 % of its size), i.e. 10 cm tall. Because the magnification is negative and this is a single thin lens, the image is also real.
Check with distances: the same setup gives an object distance of 20 cm and an image distance of 40 cm (real image, so \(v > 0\)):
\[ m = -\frac{v}{u} = -\frac{40 \text{ cm}}{20 \text{ cm}} = -2 \]
Result: both methods agree: \(m = -2\), image height \(h' = m \cdot h = -2 \times 5\text{ cm} = -10\text{ cm}\).
How to Use the Magnification Calculator
- Open the Calculation mode dropdown and choose Object and image height or Object and image distance, depending on which pair of values you know.
- In height mode, enter the Object height (h) and the Image height (h′) with their units. Type the image height as a negative number if the image is inverted.
- In distance mode, enter the Object distance (u) and the Image distance (v). Keep \(v\) positive for a real image and negative for a virtual one. Optionally fill in the object height so the calculator also reports the image size.
- Read the Magnification Results: the signed magnification \(m\), its absolute value \(|m|\), the image size as a percentage of the object, and — when an object height is given — the image height in several units.
What Is Magnification?
Magnification tells you how much larger or smaller an image is compared with the object that produced it, and in which orientation it appears. The linear (transverse) magnification \(m\) is a ratio of heights, so it has no units: \(m = 3\) means the image is three times as tall as the object, and \(m = 0.5\) means it is half as tall.
The sign carries the orientation. A negative \(m\) means the image is inverted (upside down); a positive \(m\) means it is upright. For a single thin lens or mirror this sign also tells you the nature of the image: an inverted image is real — it can be projected onto a screen — while an upright one is virtual and can only be seen by looking through the lens or into the mirror.
The two equivalent forms of the formula come from geometry. Rays drawn from the top of the object through the optical centre of the lens (or reflecting off a mirror) create two similar triangles, so the height ratio equals the distance ratio: \(h'/h = -v/u\). The minus sign exists because a real image, which lies on the opposite side of a lens (or in front of a mirror), is inverted relative to the object.
In practice, magnification matters everywhere optics appears: choosing the objective of a microscope or telescope, designing a projector, sizing a magnifying glass, or checking the image formed by a camera lens. Combined with the image distance, it also lets you predict the size of the image on a sensor or screen from the size of the object alone.
Interpreting the Result
| Magnification | Image size | Orientation | Nature (single lens/mirror) |
|---|---|---|---|
| \(|m| < 1\) | Reduced (smaller than the object) | Depends on sign | — |
| \(|m| = 1\) | Same size as the object | Depends on sign | — |
| \(|m| > 1\) | Enlarged (larger than the object) | Depends on sign | — |
| \(m < 0\) | — | Inverted | Real |
| \(m > 0\) | — | Upright | Virtual |
Assumptions, Limitations and Tips
- The formulas apply to a single thin lens or a spherical mirror under the paraxial (small-angle) approximation. Thick lenses, compound systems and wide-angle rays introduce aberrations the simple ratio cannot capture.
- This calculator uses the real-is-positive sign convention: real objects and real images get positive distances. Some textbooks use the Cartesian convention, where the formula reads \(m = v/u\) instead of \(m = -v/u\) — check which convention your course uses before comparing answers.
- In height mode, the real-versus-virtual nature of the image cannot be decided from the heights alone; the calculator states the usual rule (inverted → real, upright → virtual for a single element), but multi-lens systems can break it.
- The linear magnification computed here is different from the angular magnification quoted for microscopes and telescopes, which compares viewing angles rather than heights.
- Keep the units consistent: the calculator converts each input from its chosen unit, so mixed units (e.g. object in cm, image in inches) are handled correctly.
For more physics tools, try our Kinematic Equations Solver or the Conservation of Energy Calculator.
Frequently Asked Questions
How do I calculate magnification by hand without a calculator?
Divide the image height by the object height: \(m = h'/h\). For example, a 2 cm image of a 4 cm object gives \(m = 0.5\) — the image is half the object's size. If you only know distances, use \(m = -v/u\) (real-is-positive convention). Keep the sign of the image height: enter it as negative when the image is upside down, otherwise you will lose the orientation information.
What does a negative magnification mean?
A negative magnification means the image is inverted (upside down) relative to the object. The absolute value still tells you the size ratio: \(m = -2\) means the image is twice as tall and flipped. For a single thin lens or spherical mirror, a negative \(m\) also indicates a real image that can be projected on a screen. A positive \(m\) means an upright image, which for a single element is virtual.
Does magnification have units? Can I mix cm and inches?
No — magnification is a ratio of two lengths, so it is dimensionless; \(m = 3\) simply means three times as large. However, when computing an image height from \(h' = m \cdot h\), the units of the result match the units of the object height. This calculator converts each input (mm, cm, m, in, ft) internally, so you can enter the object in inches and the image in centimetres without converting by hand.
Why does my textbook's formula have no minus sign?
There are two common sign conventions. The real-is-positive convention used here gives \(m = -v/u\). The Cartesian convention, common in many physics textbooks, measures distances from the lens with signs based on direction and writes \(m = v/u\) directly. Both give the same physical answer when applied consistently. Before comparing results, check which convention your course uses — a sign mismatch is the most frequent source of confusion.
What is the difference between linear and angular magnification?
Linear (transverse) magnification compares actual heights, \(m = h'/h\), and applies to images of extended objects on screens or sensors. Angular magnification compares the angle an object subtends at the eye with and without an instrument — that is what the "10×" on binoculars or a microscope eyepiece means. The two numbers are different, so don't compare a telescope's angular rating with a linear \(m\) from this calculator.
What happens if the magnification equals 1 or 0?
\(m = 1\) means the image is exactly the same size as the object; for a single element this happens when the object sits at twice the focal length. \(m = -1\) gives a same-size but inverted image, used in some optical setups. A magnification near zero means the image is tiny — as when the object is very far away — and \(m = 0\) exactly corresponds to an object at infinity, imaged as a point at the focal plane.
Can I use this calculator for two lenses or a compound system?
Only with care. The formula \(m = -v/u\) is derived for one thin lens or one spherical mirror. For multiple elements, compute each stage separately: the image of the first lens becomes the object of the second, and the total magnification is the product \(m_{total} = m_1 \cdot m_2 \cdot ...\). Also note that in compound systems an inverted image need not be real, so the simple sign rules from a single element no longer hold.