Lens Equation Calculator (Thin Lens and Mirror)
Choose a convex or concave lens or mirror, enter the focal length and how far the object is, and read off where the image forms and how it looks.
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Formula
- Thin lens and mirror equation:
1/f = 1/do + 1/di, sodi = f × do ÷ (do − f). - Magnification:
m = −di ÷ do. Image height:hi = m × ho.
Sign convention (real is positive): f > 0 for convex lenses and concave mirrors, f < 0 for concave lenses and convex mirrors. di > 0 is a real, inverted image; di < 0 is a virtual, upright image. The magnitude of m tells you whether the image is enlarged (above 1) or reduced (below 1).
The lens and mirror formulas are the same; only where the image appears differs. A real image forms behind a lens and in front of a mirror.
Worked example
Convex lens, f = 10 cm, object 30 cm away and 5 cm tall.
di = 10 × 30 ÷ (30 − 10) = 15 cm (real), m = −15 ÷ 30 = −0.5, hi = −0.5 × 5 = −2.5 cm: a real, inverted, reduced image.
Move the object to 5 cm: di = 10 × 5 ÷ (5 − 10) = −10 cm, m = 2; the image is virtual, upright and twice as large.
Frequently asked questions
What is the thin lens equation?
1/f = 1/do + 1/di, where f is the focal length, do the object distance and di the image distance. Rearranged, di = f × do ÷ (do − f). For f = 10 cm and do = 30 cm, di = 10 × 30 ÷ 20 = 15 cm.
What sign convention does this calculator use?
Real is positive. Real objects have do > 0. A converging device (convex lens, concave mirror) has f > 0 and a diverging one (concave lens, convex mirror) has f < 0; you enter the size of f and the device type supplies the sign. A positive di means a real image (light really meets there: behind a lens, in front of a mirror) and a negative di means a virtual image (on the object's side of a lens, behind a mirror).
How is magnification found?
m = −di ÷ do. With di = 15 cm and do = 30 cm, m = −0.5: the image is half the size and inverted. The image height is m × the object height, so a 5 cm object gives −2.5 cm.
What happens when the object is inside the focal length?
A converging device then makes a virtual, upright, enlarged image, which is how a magnifying glass works. For f = 10 cm and do = 5 cm, di = −10 cm and m = 2.
Why is there an error when the object is exactly at the focal point?
The outgoing rays are parallel, so the image forms at infinity: do − f = 0 and the equation divides by zero. Move the object slightly.