Optics Calculator
The thin lens equation predicts exactly where an image forms and how magnified it is, from just the lens's focal length and how far the object sits from it.
Inputs
Result
Image distance 15 cm
Magnification -0.5
How it works
1/f = 1/u + 1/v
How the optics calculator works
1/f = 1/u + 1/v, rearranged here to solve for image distance: v = 1 ÷ (1/f − 1/u).
Magnification = −v ÷ u, where a negative sign conventionally indicates an inverted image.
Worked example: 10 cm focal length, object at 30 cm
- 1/v = 1/10 − 1/30 = 3/30 − 1/30 = 2/30, so v = 15 cm.
- Magnification = −15 ÷ 30 = −0.5, meaning the image is inverted and half the size of the object.
Common mistakes to avoid
Mixing up sign conventions between different optics textbooks
This calculator uses a standard convention where a negative magnification means an inverted image — some textbooks and regions use different sign conventions for object/image distance, so double-check against your specific course's convention before comparing results directly.
Entering an object distance smaller than the focal length without expecting a different image type
When the object is closer to the lens than its focal length, the lens produces a virtual, upright, magnified image rather than a real inverted one — the formula's output (a negative or unusual-looking v) reflects this different physical situation.
Frequently asked questions
What does a negative magnification value mean?
The image is inverted relative to the object — this is the normal case for a real image formed by a converging lens when the object is beyond the focal length.
What's the difference between a real and virtual image in this context?
A real image can be projected onto a screen (light actually converges there); a virtual image cannot (light only appears to diverge from that point) — which type forms depends on where the object sits relative to the focal length.
How would this change for a diverging (concave) lens instead of a converging one?
A diverging lens has a negative focal length by convention, which changes the sign pattern of the result — always producing a virtual, upright, reduced image regardless of object distance.
Can this be used for mirrors as well as lenses?
The mirror equation has an identical mathematical form (1/f = 1/u + 1/v), so the same calculation applies, though sign conventions for mirrors can differ slightly by convention from those used for lenses.
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