How can I calculate an image distance for a thin converging lens?
Quick answer
For an ideal thin converging lens, use 1/dᵢ + 1/dₒ = 1/f, rearranged as dᵢ = 1/(1/f − 1/dₒ). Put the focal length f and object distance dₒ in the same units, then calculate dᵢ in those units. For example, if f = +12 cm and dₒ = +36 cm, then dᵢ = +18 cm. Under this sign convention, a positive image distance indicates a real image opposite the object. [S1]
The question, the details, the next step.
I have a converging lens with a stated focal length and an object distance. What equation should I use to find the image distance?
Use the ideal thin-lens equation:
1/dᵢ + 1/dₒ = 1/f
Here, dᵢ is image distance, dₒ is object distance, and f is focal length. Solving for the unknown image distance gives:
dᵢ = 1/(1/f − 1/dₒ)
For a converging lens in the convention used here, enter its focal length as positive; the example in the source also uses a positive object distance. Keep the distances in matching units. The equation describes an ideal thin-lens model, so treat the answer as a model calculation rather than a measurement of a particular real lens. [S1]
Can you show the arithmetic with actual values? I want to see how the reciprocals work.
Suppose, as a worked example, the stated values are f = +12 cm and dₒ = +36 cm. Substitution gives:
1/dᵢ = 1/12 − 1/36
Use a common denominator:
1/dᵢ = 3/36 − 1/36 = 2/36 = 1/18
Therefore dᵢ = +18 cm. The units remain centimeters because both input distances were in centimeters. This result follows by rearranging the thin-lens equation. [S1]
What does the plus sign on 18 cm tell me, and how do I avoid a sign mistake?
For the sign convention described in the source, a positive image distance means the image is on the side of the lens opposite the object—a real image. [S1] Before calculating, write down the values with their signs and use the convention your class or problem specifies. Then check that you subtracted 1/dₒ from 1/f, not the reverse. Follow the stated sign convention when applying the equation. [S1]
Could you give me a reusable setup I can copy for my own numbers, plus a quick check?
Copy this and replace the bracketed values:
f = [focal length with sign] [unit] dₒ = [object distance with sign] [same unit]
1/dᵢ = 1/f − 1/dₒ 1/dᵢ = 1/[f] − 1/[dₒ] = [result in 1/unit] dᵢ = [reciprocal result] [unit]
For a quick check, put your answer back into 1/dᵢ + 1/dₒ and see whether it equals 1/f, allowing for rounding. If the given distances use different units, convert them to one unit before taking reciprocals. Under the stated convention, a positive dᵢ means a real image opposite the object. [S1] This calculation does not by itself establish how accurately a real lens follows the ideal thin-lens model.
A practical checklist
Record f and dₒ with the signs required by the problem's convention.
Convert both distances to the same unit before calculating.
Substitute into 1/dᵢ = 1/f − 1/dₒ.
Take the reciprocal to obtain dᵢ, retaining the distance unit.
Substitute the result into the original equation as an arithmetic check.
Interpret the sign using the stated convention, not an assumed one.
A few more useful answers
Should I change centimeters to meters first?
Not if both input distances are already in centimeters: using centimeters consistently gives the image distance in centimeters. If the inputs use different units, convert them to one unit first. [S1]
What if my problem does not explain its sign convention?
Check the definitions provided with the assignment or ask which convention to use. The equation should be applied using the stated sign convention. [S1]
Sources
References used for this guide.
- [S1] OpenStax University Physics Volume 3, Section 2.4: Thin Lenses openstax.org · Checked