Thin Lens Equation Calculator
Solve classic optics problems using the thin lens equation: 1/f = 1/dₒ + 1/dᵢ. Pick what you want to find (f, dₒ, dᵢ, or m), use clean sign conventions, and instantly get image type: real vs. virtual, upright vs. inverted, and magnified vs. reduced. Includes steps, quick picks, and a mini ray / axis visual.
Background
The thin lens model treats the lens as “thin,” so refraction happens at a single plane. Using object distance dₒ, image distance dᵢ, and focal length f, we can predict where the image forms and whether it’s real or virtual. Magnification is m = -dᵢ/dₒ (and also m = hᵢ/hₒ if you include heights).
How to use this calculator
- Choose what you want to solve for (dᵢ, dₒ, f, or m).
- Enter the known values in the same distance units.
- Click Calculate to get the answer, image type callouts, and step-by-step.
How this calculator works
- Thin lens equation: 1/f = 1/dₒ + 1/dᵢ.
- Solve algebraically for the missing variable (e.g., dᵢ = 1 / (1/f - 1/dₒ)).
- Magnification: m = -dᵢ/dₒ (or m = hᵢ/hₒ if heights provided).
- Image interpretation: dᵢ > 0 real, dᵢ < 0 virtual; sign of m tells upright vs inverted.
Formula & Equation Used
Thin lens: 1/f = 1/dₒ + 1/dᵢ
Image distance: dᵢ = 1 / (1/f − 1/dₒ)
Object distance: dₒ = 1 / (1/f − 1/dᵢ)
Focal length: f = 1 / (1/dₒ + 1/dᵢ)
Magnification: m = −dᵢ/dₒ = hᵢ/hₒ
Example Problem & Step-by-Step Solution
Example 1 — Find image distance
A converging lens has f = 10 cm. An object is placed at dₒ = 25 cm. Find dᵢ and the magnification.
- Thin lens: 1/dᵢ = 1/f − 1/dₒ
- 1/dᵢ = 1/10 − 1/25 = 0.1 − 0.04 = 0.06
- dᵢ = 1/0.06 ≈ 16.7 cm (real image)
- Magnification: m = −dᵢ/dₒ = −16.7/25 ≈ −0.667 (inverted, reduced)
Example 2 — Magnifying glass (virtual image)
A converging lens has f = 10 cm. An object is placed at dₒ = 6 cm (inside the focal length). Find dᵢ and m.
- Use 1/dᵢ = 1/f − 1/dₒ.
- 1/dᵢ = 1/10 − 1/6 = 0.1 − 0.1667 = −0.0667.
- dᵢ = 1/(−0.0667) ≈ −15 cm (virtual image).
- m = −dᵢ/dₒ = −(−15)/6 = +2.5 (upright, magnified).
Example 3 — Diverging lens (always virtual)
A diverging lens has f = −12 cm. An object is at dₒ = 30 cm. Find dᵢ and describe the image.
- 1/dᵢ = 1/f − 1/dₒ = 1/(−12) − 1/30 = −0.0833 − 0.0333 = −0.1167
- dᵢ = 1/(−0.1167) ≈ −8.57 cm (virtual).
- m = −dᵢ/dₒ = −(−8.57)/30 ≈ +0.286 (upright, reduced).
Frequently Asked Questions
Q: When is the thin lens equation valid?
When the lens is thin compared to the object and image distances, and rays make small angles (paraxial approximation).
Q: What does a negative image distance mean?
It means the image is virtual and appears on the same side of the lens as the object.
Q: What does the sign of magnification mean?
m < 0 → image is inverted. m > 0 → image is upright.