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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).

Enter values

Tip: Use consistent sign convention (chosen below). The calculator will also tell you the image type.

Any unit works — just keep f, dₒ, and dᵢ in the same unit.

We’ll still accept any signed inputs you enter. This setting mainly affects the helper text + interpretation callouts.

Converging lens: f > 0. Diverging lens: f < 0.

Usually dₒ > 0 when the object is on the incoming-light side.

Real image: dᵢ > 0. Virtual image: dᵢ < 0.

If you provide heights, we’ll compute m = hᵢ/hₒ. Otherwise we use m = -dᵢ/dₒ.

Options

Toggle a principal ray. For virtual images, we also show a dashed back-trace.

Chips prefill values and calculate immediately.

Result

No results yet. Enter values and click Calculate.

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.

  1. Thin lens: 1/dᵢ = 1/f − 1/dₒ
  2. 1/dᵢ = 1/10 − 1/25 = 0.1 − 0.04 = 0.06
  3. dᵢ = 1/0.06 ≈ 16.7 cm (real image)
  4. 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.

  1. Use 1/dᵢ = 1/f − 1/dₒ.
  2. 1/dᵢ = 1/10 − 1/6 = 0.1 − 0.1667 = −0.0667.
  3. dᵢ = 1/(−0.0667) ≈ −15 cm (virtual image).
  4. 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. 1/dᵢ = 1/f − 1/dₒ = 1/(−12) − 1/30 = −0.0833 − 0.0333 = −0.1167
  2. dᵢ = 1/(−0.1167) ≈ −8.57 cm (virtual).
  3. 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.