The thin lens equation, explained.
The thin-lens equation relates focal length to object and image distance. An object outside the focal length forms a real inverted image. Inside the focal length, the outgoing rays diverge and their backward extensions meet at an upright virtual image.
The relationship to remember
Key formulas
- Thin lens equation
- 1/f = 1/dₒ + 1/dᵢ
- Real-is-positive sign convention
- Image distance
- dᵢ = f·dₒ / (dₒ − f)
- Negative dᵢ means a virtual image
- Magnification
- M = −dᵢ/dₒ = hᵢ/hₒ
- Negative M means inverted
- Lens power
- P = 1/f
- In dioptres when f is in metres
Worked example
Problem. An object 4 cm tall stands 15 cm from a converging lens with a focal length of 10 cm. Where is the image, how tall is it, and what type is it?
- 1/dᵢ = 1/f − 1/dₒ = 1/10 − 1/15 = 1/30, so dᵢ = 30 cm.
- Magnification: M = −dᵢ/dₒ = −30 / 15 = −2.
- Image height: hᵢ = M × hₒ = −2 × 4 = −8 cm.
- A positive dᵢ means a real image on the far side of the lens; the negative M means it is inverted.
Answer. A real, inverted image forms 30 cm behind the lens and is 8 cm tall, twice the object’s height.
A quick check
With f = 10 cm and object distance 25 cm, the image distance is about 16.67 cm and magnification is −0.67. The minus sign means inverted.
Inside the default experiment
- 1/dᵢ = 1/10 − 1/25 = 0.06 cm⁻¹.
- Image distance = 16.67 cm. Magnification = −dᵢ/dₒ = -0.67.
- Image height = M × hₒ = -2 cm.
When this model applies
Ideal thin converging lens and paraxial rays. A positive object distance represents a real object; negative image distance represents a virtual image.
Make a prediction. Then test it.
Move the object inside the focal length. How does the image change?
Try it: Converging lens simulator ↗Frequently asked questions
Wondering about the why? Start here.
What is a virtual image?
It appears where backward extensions of light rays meet. The rays do not actually converge there, so that image cannot be projected directly onto a screen.
What does negative magnification mean?
The image is inverted relative to the object. Its absolute value gives the ratio of image height to object height.