2.2 Vergence, Lens Power & Focal Length
Key Takeaways
- Vergence in diopters equals the reciprocal of the distance in meters; converging light is plus (+), diverging light is minus (-).
- Focal length and power are reciprocals: F = 1/f and f = 1/F, with f in meters; a +4.00 D lens has a 0.25 m (25 cm) focal length.
- The fundamental vergence equation is L' = L + F, where L is incident (object) vergence, F is lens power, and L' is exiting (image) vergence.
- Nominal (thin-lens) power equals the sum of the two surface powers: F = F1 + F2, where each surface power is F = (n-1)/r.
- A positive image vergence L' gives a real image on the far side of the lens; a negative L' gives a virtual image on the same side as the object.
Vergence and the Diopter
Vergence describes how strongly a bundle of light rays is converging or diverging as it travels. It is measured in diopters (D), defined as the reciprocal of the distance (in meters) to the point where the rays meet or from which they appear to diverge:
Vergence (D) = 1 / distance (m)
By convention, converging light has plus (+) vergence and diverging light has minus (-) vergence. Rays leaving a point source diverge, so they carry minus vergence; the farther the source, the flatter the wavefront and the smaller the vergence magnitude. At infinity, rays are parallel and vergence is zero (0.00 D) — the reason distance prescriptions are measured with effectively parallel incoming light.
Worked example: a point of light 25 cm (0.25 m) in front of a lens sends diverging light to it. Its vergence at the lens is 1 / (-0.25) = -4.00 D. A source 50 cm away arrives with 1 / (-0.50) = -2.00 D — a flatter, weaker divergence, confirming that nearer objects diverge more strongly.
Focal Length and Lens Power
A lens changes vergence by a fixed amount called its power (F), measured in diopters. Power is the reciprocal of the focal length (f) in meters:
F = 1 / f and f = 1 / F
The focal length is the distance from the lens to the point where parallel incoming light is brought to focus (plus lens) or appears to diverge from (minus lens). Worked examples: a +4.00 D lens has f = 1 / 4.00 = 0.25 m (25 cm); a +2.50 D lens has f = 1 / 2.50 = 0.40 m (40 cm); a -5.00 D lens has f = 1 / (-5.00) = -0.20 m, a virtual focal point 20 cm in front of the lens on the same side as the incoming light.
Converging vs. Diverging Lenses
Plus (convex) lenses converge light and are thickest in the center; they can form real images that can be projected on a screen. Minus (concave) lenses diverge light, are thinnest in the center, and form virtual images that appear on the same side as the object and cannot be projected. Plus lenses correct hyperopia and presbyopia (add power); minus lenses correct myopia.
Surface Power and Nominal Lens Power
Each refracting surface contributes power according to:
F(surface) = (n - 1) / r
where n is the lens index and r is the radius of curvature in meters. The nominal (thin-lens) power of a lens is simply the sum of its two surface powers:
F = F1 (front) + F2 (back)
Worked example (from lab practice): a finished lens must be -6.00 D and is made on a +4.00 D front base curve. Rearranging, the required back surface power = total - front = -6.00 - (+4.00) = -10.00 D. The lab grinds a -10.00 D concave back curve to net the ordered power. This nominal model ignores lens thickness; true (effective) power adds a small thickness correction covered later.
The Vergence Equation: L' = L + F
The single most useful relationship in geometric optics tracks vergence through a lens:
L' = L + F
where L is the incident (object) vergence entering the lens, F is the lens power, and L' is the exiting (image) vergence. The image distance is then 1 / L' (in meters).
Worked example 1 (real image): an object sits 50 cm in front of a +5.00 D lens. Incident vergence L = 1 / (-0.50) = -2.00 D. Then L' = L + F = -2.00 + 5.00 = +3.00 D. Image distance = 1 / 3.00 = +0.333 m (33.3 cm) behind the lens. Because L' is positive, the light exits converging and forms a real image on the far side.
Worked example 2 (virtual image): the same object 50 cm in front of a +1.00 D lens. L = -2.00 D, so L' = -2.00 + 1.00 = -1.00 D. Image distance = 1 / (-1.00) = -1.00 m — one meter in front of the lens, on the object's side. The negative L' signals a virtual image; the weak plus lens could not overcome the object's divergence.
Real vs. Virtual Images
| Feature | Real image | Virtual image |
|---|---|---|
| Exiting vergence L' | Positive (+) | Negative (-) |
| Location | Opposite side from object | Same side as object |
| Can project on screen? | Yes | No |
| Typical lens | Plus (strong enough) | Minus, or weak plus |
Worked example 3 (minus lens, always virtual): an object 33 cm (0.33 m) in front of a -4.00 D lens. L = 1 / (-0.33) = -3.00 D, so L' = -3.00 + (-4.00) = -7.00 D. Image distance = 1 / (-7.00) = -0.143 m (14.3 cm) in front of the lens — an upright, minified virtual image. A diverging lens with a real (diverging) object can never form a real image, a useful sanity check.
Sign Conventions and Common Mistakes
The vergence equation only works if signs are applied consistently. Three errors cost candidates points:
- Forgetting to convert to meters. Vergence is 1/distance in meters; an object at 40 cm is 0.40 m, giving 2.50 D, not 40.
- Dropping the object's minus sign. Real objects diverge light, so incident vergence L is negative. Omitting the sign flips the whole answer.
- Misreading the result. A positive L' means a real image on the far side; a negative L' means a virtual image on the near (object) side. The sign of L', not the sign of F, tells you the image type.
Mastering L' = L + F lets you predict image position and character for any object distance and lens power — the backbone of nearly every optics calculation on the Advanced exam.
A +2.50 D lens has a focal length of approximately:
An object is placed 50 cm in front of a +5.00 D lens. Using L' = L + F, where is the image and what type is it?
A finished lens must total -6.00 D and is fabricated on a +4.00 D front base curve. What back surface power does the lab grind?