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HLT-OPTI68-02 · Lesson 4 · 15 clock hours · theory

Spherical Lenses: Power, Focal Length, and Image Formation

Objective. Students will calculate the power and focal length of a thin spherical lens, construct ray diagrams for convex and concave lenses, and predict image location, size, and orientation.

⚠ Pending professional review

This lesson, its worked examples and its practice questions has not yet been verified by a licensed clinician in the field or by a certification or approval body. It is a study draft and may contain errors. Do not use it to make decisions about any patient or medication. Always follow your instructor, your supervising professional, and current official references and regulations.

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What Is a Spherical Lens?

A spherical lens has surfaces that are portions of a sphere. The two most important types for opticianry are convex (converging) lenses, which are thicker at the center than at the edge, and concave (diverging) lenses, which are thinner at the center. Understanding how light behaves when it passes through these lenses is the foundation of every prescription you will interpret and every pair of eyeglasses you will fabricate.

Lens Power and Focal Length

Lens power (P) describes how strongly a lens bends light. It is measured in diopters (D). A higher power means stronger bending. Power and focal length (f) are directly related by the thin-lens power equation:

P = 1 / f

where f is measured in meters. Rearranging:

f = 1 / P

A +2.00 D convex lens has a focal length of 0.50 m (50 cm). A −4.00 D concave lens has a focal length of −0.25 m (−25 cm). The negative sign for a concave lens is not an error — it signals that the lens diverges light and produces a virtual focal point on the same side as the incoming light.

Convex lenses have positive power; concave lenses have negative power. This sign convention runs throughout all optical prescriptions.

The Thin-Lens Equation

Once you know focal length, you can predict exactly where an image will form using the thin-lens equation:

1/f = 1/do + 1/di

Solving for image distance:

1/di = 1/f − 1/do

Image Characteristics

Images are described along four dimensions:

1. Real or virtual — A real image forms where rays actually converge; it can be projected on a screen. A virtual image appears where rays seem to diverge from; it cannot be projected.

2. Erect or inverted — Is the image right-side-up or flipped?

3. Magnified, same size, or diminished — Is the image larger, equal to, or smaller than the object?

4. Location — Described by di.

Magnification (m) links image size to object size:

m = −di / do

A negative m means the image is inverted; magnitude greater than 1 means magnified.

Ray Diagrams for Convex Lenses

Three principal rays are drawn from the tip of an object arrow:

1. Ray 1 travels parallel to the principal axis, then refracts through the far focal point (F₂).

2. Ray 2 passes through the optical center of the lens undeviated.

3. Ray 3 travels through the near focal point (F₁), then exits parallel to the principal axis.

Where any two of these rays cross (or appear to diverge from) is where the image tip forms.

Three key object positions for a convex lens:

Ray Diagrams for Concave Lenses

For a concave lens, Ray 1 (parallel to axis) refracts as if it came from the near focal point. Ray 2 passes through the center undeviated. Ray 3 aims toward the far focal point but exits parallel. The rays diverge; extending them back gives a virtual, erect, diminished image on the same side as the object — regardless of where the object is placed. This is the basis for minus-power lenses used to correct myopia.

Safety and Scope-of-Practice Notes

Connecting This to the ABO Exam

The American Board of Opticianry exam tests your ability to apply the thin-lens equation, interpret plus and minus powers, and understand how image formation relates to refractive error correction. Mastering these concepts now builds the quantitative fluency you will use every day on the dispensing floor.

Key terms9

Diopter (D)The unit of lens power, equal to the reciprocal of focal length in meters (P = 1/f). A +2.00 D lens has a focal length of 0.50 m.
Focal length (f)The distance from the optical center of a thin lens to the point where parallel rays converge (convex) or appear to diverge from (concave), measured in meters.
Principal axisThe imaginary straight line passing through the center of a lens perpendicular to its surface, serving as the reference line for ray diagrams and sign conventions.
Thin-lens equation1/f = 1/do + 1/di; the equation relating focal length, object distance, and image distance for a lens thin enough that its thickness is ignored.
Real imageAn image formed where refracted rays actually converge; it appears on the opposite side of the lens from the object and can be projected onto a screen.
Virtual imageAn image formed where refracted rays appear to diverge from when traced backward; it appears on the same side as the object and cannot be projected on a screen.
Magnification (m)The ratio of image size to object size, calculated as m = −di / do. A negative value indicates an inverted image; magnitude > 1 indicates enlargement.
Converging lensA convex lens with positive power that bends parallel rays toward a real focal point on the far side; used to correct hyperopia and presbyopia.
Diverging lensA concave lens with negative power that spreads parallel rays apart as if they originated from a virtual focal point on the near side; used to correct myopia.

Worked examples4

1. A lens has a power of +4.00 D. What is its focal length in centimeters?
  1. Write the formula: f = 1 / P
  2. Substitute: f = 1 / 4.00 = 0.25 m
  3. Convert to centimeters: 0.25 m × 100 = 25 cm
Answer: f = 25 cm (positive, confirming a converging lens)
2. An object is placed 40 cm (0.40 m) from a +5.00 D convex lens. Find the image distance and describe the image.
  1. Find focal length: f = 1 / 5.00 = 0.20 m
  2. Write the thin-lens equation: 1/di = 1/f − 1/do
  3. Substitute: 1/di = 1/0.20 − 1/0.40 = 5.00 − 2.50 = 2.50
  4. Solve: di = 1 / 2.50 = 0.40 m = 40 cm
  5. Calculate magnification: m = −di / do = −0.40 / 0.40 = −1.0
Answer: di = 40 cm on the far side of the lens. The image is real, inverted, and the same size as the object (m = −1.0). The object was at 2F, confirming the expected result.
3. An object is placed 10 cm (0.10 m) from a +5.00 D convex lens — between F and the lens. Find the image distance and describe the image.
  1. Find focal length: f = 1 / 5.00 = 0.20 m
  2. Write the thin-lens equation: 1/di = 1/f − 1/do
  3. Substitute: 1/di = 1/0.20 − 1/0.10 = 5.00 − 10.00 = −5.00
  4. Solve: di = 1 / (−5.00) = −0.20 m = −20 cm
  5. The negative sign means the image is on the same side as the object (virtual).
  6. Calculate magnification: m = −(−0.20) / 0.10 = +2.0
Answer: di = −20 cm (virtual image, same side as object). The image is virtual, erect, and magnified by a factor of 2. This is the magnifying-glass condition.
4. A concave lens has a power of −2.50 D. An object is placed 1.00 m away. Find the image distance.
  1. Find focal length: f = 1 / (−2.50) = −0.40 m
  2. Write the thin-lens equation: 1/di = 1/f − 1/do
  3. Substitute: 1/di = 1/(−0.40) − 1/1.00 = −2.50 − 1.00 = −3.50
  4. Solve: di = 1 / (−3.50) = −0.286 m ≈ −28.6 cm
Answer: di ≈ −28.6 cm (virtual image on the same side as the object, erect and diminished — consistent with all concave-lens images of real objects).

The math in this lesson2

Learners convert focal length 0.25 m to 25 cm and compute magnification −di/do.
N-Q.1A-CED.4
Brush up in the math halls at newSchoolWorks.com:
In class: Instructor-led ray-diagram construction on grid paper for three object distances relative to a convex lens (beyond 2F, at 2F, between F and lens); students replicate each diagram and label image characteristics. · Problem set (20 problems): calculate focal length from power and vice versa; determine image distance using the thin-lens equation; identify real/virtual, erect/inverted, magnified/diminished images. · Lab preview activity: each student uses a demonstration convex lens (+5.00 D) and a ruler to locate the focal point of a projected light beam, confirming the calculated focal length within 5 mm.
Materials: Grid paper, rulers, pencils, demonstration +5.00 D convex lens, penlight or desk lamp, meter stick, scientific calculators, thin-lens equation reference card
Check: Scored problem set: 17/20 correct (85%); ray diagrams checked for correct construction of at least 2 of 3 cases; lab preview measurement within 5 mm accepted.

Practice check8

Answer each question, then check it. Nothing is recorded; this is your own practice.

ABO — Optics and Lens Theory
1. A lens has a focal length of 0.50 m. What is its power in diopters?
ABO — Optics and Lens Theory
2. Using the thin-lens equation, an object is placed 25 cm from a +4.00 D lens. What is the image distance?
ABO — Optics and Lens Theory
3. An object is placed between the focal point and a convex lens. What kind of image is formed?
ABO — Optics and Lens Theory
4. A concave lens always produces what type of image for a real object?
ABO — Optics and Lens Theory
5. A +5.00 D demonstration lens is used in a lab to project a light beam. Approximately where should the focal point be found?
ABO — Optics and Lens Theory
6. Which principal ray passes through the optical center of a lens without changing direction?
ABO — Optics and Lens Theory
7. An object is placed at exactly 2F from a convex lens. Which statement correctly describes the image?
ABO — Optics and Lens Theory
8. In the thin-lens equation 1/f = 1/do + 1/di, a negative value for di indicates what?
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