Playground
Interactive thin lens ray diagram: move the object distance slider to see the image form, flip from real to virtual, and display magnification.
Variables
| Symbol | Name | SI | Dimension | Range |
|---|---|---|---|---|
| Focal length Distance from lens center to focal point (positive for converging, negative for diverging) | m | L | -0.5 – 0.5 | |
| Object distance Distance from object to the lens center | m | L | 0.01 – 2 | |
| Image distanceoutput Distance from lens center to the image (positive for real, negative for virtual) | m | L | -2 – 2 |
Deep dive
Derivation
Using similar triangles from ray diagrams: a ray through the center is undeviated, and a ray parallel to the axis refracts through the focal point. For a thin lens, equating the two triangles gives h_i/h_o = d_i/d_o (magnification) and h_i/h_o = (d_i - f)/f. Setting these equal: d_i/d_o = (d_i - f)/f → f·d_i = d_o·d_i - d_o·f → d_o·f + d_i·f = d_o·d_i. Dividing by d_o·d_i·f: 1/f = 1/d_o + 1/d_i.
Experimental verification
Verified by measuring object and image distances with calibrated optical benches. Modern automated lens testing systems confirm focal lengths to < 0.1% accuracy using collimated laser sources.
Common misconceptions
- Covering half the lens removes half the image — it dims the whole image, doesn't cut it in half.
- A negative image distance means no image — it means a virtual image on the same side as the object.
- The formula works for thick lenses — it only works when lens thickness << focal length (thin lens approximation).
Real-world applications
- Camera autofocus: adjusts d_o by moving the lens to satisfy 1/f = 1/d_o + 1/d_i.
- Eyeglasses: diverging lenses (f < 0) correct myopia by shifting the image back to the retina.
- Projectors: place the slide at d_o slightly > f to get a large real image on a distant screen.
- Magnifying glasses: object inside f produces a virtual, magnified, upright image.
Worked examples
Converging lens image
Given:
- f:
- 0.1
- d_o:
- 0.15
Find: d_i
Solution
d_i = f*d_o/(d_o - f) = 0.10 × 0.15 / (0.15 - 0.10) = 0.015/0.05 = 0.30 m
Diverging lens virtual image
Given:
- f:
- -0.2
- d_o:
- 0.3
Find: d_i
Solution
d_i = f*d_o/(d_o - f) = (-0.20)(0.30)/(0.30 - (-0.20)) = -0.06/0.50 = -0.12 m
Scenarios
What if…
- scenario:
- What if d_o = f exactly?
- answer:
- 1/d_i = 0, so d_i → ∞. Rays emerge parallel — the image forms at infinity. This is how collimated beams are made.
- scenario:
- What if d_o < f for a converging lens?
- answer:
- d_i becomes negative — a virtual, upright, magnified image forms. This is the magnifying glass regime.
- scenario:
- What if the lens is in water instead of air?
- answer:
- The focal length changes because refraction depends on the ratio of lens-to-medium refractive indices. The lensmaker's equation must be used with the medium's n.
Limiting cases
- condition:
- d_o → ∞
- result:
- d_i → f
- explanation:
- Parallel rays (distant object) converge at the focal point.
- condition:
- d_o = f
- result:
- d_i → ∞
- explanation:
- Object at focal point produces parallel rays — image at infinity.
- condition:
- d_o = 2f
- result:
- d_i = 2f
- explanation:
- Object at 2f produces a same-size inverted image at 2f.
Context
Multiple contributors · 1693
Developed through contributions by Kepler, Barrow, and Halley. The modern thin lens formula was formalized in the late 17th century.
Hook
How far must you hold a magnifying glass from a bug to project its image on a wall?
A thin converging lens (f = 10 cm) views an object at 15 cm. Find the image distance using 1/f = 1/do + 1/di.
Dimensions:
- lhs:
- 1/f → [L⁻¹]
- rhs:
- 1/d_o + 1/d_i → [L⁻¹] + [L⁻¹] = [L⁻¹]
- check:
- Both sides are [L⁻¹] = diopters (m⁻¹). ✓
Validity: Valid for thin lenses (thickness << focal length) in the paraxial approximation (small angles). Breaks down for thick lenses and wide-angle rays (aberrations).