Library / Foundations: Physics / FPH-47
Foundations: Physics
Magnification of a thin lens
Ratio of image size to object size for a thin lens, with a sign for orientation.
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- FPH-47
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- v1
- Field
- Foundations: Physics
- Result
- m
Magnification is the ratio of image size to object size, m = −v/u. The minus sign records that a real image is inverted: m is negative for inverted and positive for upright. The size ratio is |m|. Using the lens equation to eliminate v gives m = −f/(u − f). An object at the focal point gives infinite magnification at infinite distance, so u ≠ f.
Ratio of image size to object size for a thin lens, with a sign for orientation. A negative value means the image is inverted; the size ratio is the magnitude of m.
Calculator
- Image distance v15 cm
- Size ratio |m| (image height / object height)0.5
- Image height for a 10 mm object5 mm
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- An object 30 cm from a 10 cm lens has magnification −0.5: the image is 0.5 times the object’s size and inverted. A 10 mm object gives a 5 mm image.
- For m = −1 the object must be at 2f (here 20 cm): the image is then the same size as the object, inverted, at 2f on the other side.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing () | Local step | Change in result |
|---|---|---|---|---|---|---|
| f Focal length (converging lens, positive) | 1 to 50 cm | −23.32 | 76 | 99.32 | +10% | +15.8 % |
| u Object distance (real object, positive) | 1 to 200 cm | −2.344 | 4.225 | 6.569 | +10% | −13 % |
Sweep of Object distance (real object, positive)
| Object distance (real object, positive) (cm) | Magnification |
|---|---|
| 1 | 1.111 |
| 25.88 | −0.6299 |
| 50.75 | −0.2454 |
| 75.63 | −0.1524 |
| 100.5 | −0.1105 |
| 125.4 | −0.08667 |
| 150.3 | −0.0713 |
| 175.1 | −0.06056 |
| 200 | −0.05263 |
Sweep one variable
Variables and how each one contributes
Magnification
What the equation gives you. Everything else feeds it.
Focal length (converging lens, positive)
The focal length of the converging lens, in cm. It is positive. Makes the magnification larger in size (more negative for an object beyond the focal point), because a stronger lens forms a bigger image. Raising it by 1 cm (to 11 cm) moves the result by +15.8 %, from −0.5 to −0.5789.
Object distance (real object, positive)
The distance from the lens to the object, in cm, for a real object (positive). Shrinks the size of the magnification as the object moves away: a distant object forms a small image. Raising it by 5 cm (to 35 cm) moves the result by −20 %, from −0.5 to −0.4.
Worked example with the values above
- Image distance: v = fu/(u − f) = 15 cm.
- Ratio: −v/u = −15 / 30.
- m = −0.5, so the image is inverted and smaller than the object.
Assumptions and limits
- A thin lens in the paraxial approximation: rays close to the axis and small angles.
- A real object (u > 0) and a converging lens (f > 0); a virtual image gives a negative v. The object must not be at the focal point, where the image is at infinity.
Common mistakes
- Forgetting the minus sign, so inverted images appear upright.
- Using m = v/u without the sign.
Statistics and uncertainty
The magnification changes quickly near the focal point, so it is very sensitive to the object distance there. At the defaults (where u/(u − f) = 1.5) a 1 % error in u changes m by about 1.5 %, and near u = f the sensitivity grows without limit.
Where it is used
- Choosing the object distance for a required image size in a camera or projector.
- Judging whether an image will be inverted or upright.
Maths you need
- Focal length: distance at which parallel rays meet
- Real and virtual images: on opposite sides of the lens