Images Formed by Lenses | Varsity Tutors

SECTION 1

Historical Context & Motivation

The ability to bend light using transparent materials has fascinated natural philosophers for millennia. Ancient Romans observed that glass spheres filled with water could magnify text, and medieval scholars in the Islamic world systematically studied refraction — the change in direction of light as it passes between media with different optical densities. The development of lenses transformed human civilization, enabling eyeglasses, microscopes, telescopes, and ultimately the cameras and projectors that define modern imaging technology. Understanding how lenses form images remains central to optics and is a core topic in AP Physics 2.

~1000

Ibn al-Haytham's Kitāb al-Manāẓir

Often called the "father of optics," Ibn al-Haytham published the Book of Optics, establishing that light travels in straight lines and explaining refraction through curved surfaces.

1286

Invention of Spectacles

Italian craftsmen in Florence produced the first wearable eyeglasses using convex lenses, correcting presbyopia and demonstrating the practical power of lens optics.

1608

The Refracting Telescope

Hans Lippershey filed the first patent for a telescope combining a converging and diverging lens. Galileo refined the design the following year to observe Jupiter's moons.

1621

Snell's Law of Refraction

Willebrord Snell discovered the mathematical relationship n₁ sin θ₁ = n₂ sin θ₂, providing the quantitative foundation for predicting how lenses redirect light rays.

1733

Achromatic Lens

Chester Moor Hall combined crown and flint glass to create the first achromatic doublet, correcting chromatic aberration and advancing practical lens design.

The fundamental question this lesson addresses is: given an object placed at a known distance from a lens of known focal length, where does the image form, how large is it, and is it real or virtual? Answering this requires mastery of both ray-tracing techniques and the thin-lens equation, tools that unify the behavior of converging and diverging lenses under a single mathematical framework.

SECTION 2

Core Principles & Definitions

A lens is a piece of transparent material — typically glass or plastic — bounded by two refracting surfaces, at least one of which is curved. In the thin-lens approximation, the thickness of the lens is negligible compared to the object and image distances, allowing us to treat all refraction as occurring at a single plane. This simplification underlies the AP Physics 2 treatment of lens optics and yields remarkably accurate predictions for most everyday optical systems.

Converging (Convex) Lens

Thicker at the center than at the edges. Parallel rays converge to a real focal point on the far side. Focal length f is positive.

Diverging (Concave) Lens

Thinner at the center than at the edges. Parallel rays diverge as though emanating from a virtual focal point on the near side. Focal length f is negative.

Real vs. Virtual Images

A real image forms where refracted rays actually converge and can be projected onto a screen. A virtual image forms where rays appear to diverge from; it cannot be captured on a screen.

Principal Axis & Focal Points

The principal axis is the horizontal line through the center of the lens. A thin lens has two symmetric focal points, each a distance |f| from the center.

Sign Convention

Object distance d_o is positive when the object is on the incoming-light side. Image distance d_i is positive for real images (opposite side) and negative for virtual images (same side as object).

✦ KEY TAKEAWAY

Think of a converging lens as a funnel that channels parallel light rays to a single point, much like a satellite dish focuses radio waves. A diverging lens does the opposite — it spreads rays apart as though they originated from a point behind the lens, similar to how a sprinkler head disperses water outward from a single pipe.

SECTION 3

Ray Diagrams for Converging Lenses

Ray diagrams provide a powerful graphical method for locating images. For a converging lens, three principal rays drawn from the tip of the object are sufficient to determine the image location and characteristics. The diagram below shows the case where the object is placed beyond 2f, producing a real, inverted, and reduced image between f and 2f on the opposite side.

Three principal rays for a converging lens with the object beyond 2F. Ray 1 travels parallel to the axis and refracts through F'. Ray 2 passes through the lens center undeviated. Ray 3 passes through F and exits parallel to the axis. The three rays converge at the image location — a real, inverted, reduced image between F' and 2F'.

The behavior of the image depends critically on where the object is placed relative to the focal point. When the object is between f and 2f, the image is real, inverted, and enlarged beyond 2f. When the object is inside f, the rays diverge after passing through the lens and the observer traces them back to find a virtual, upright, and magnified image on the same side as the object — this is exactly how a magnifying glass works.

💡 AP Exam Tip

On the AP Physics 2 exam, you may be asked to draw or interpret ray diagrams. Always draw at least two of the three principal rays. The intersection point (or apparent intersection for virtual images) determines the image position. Remember that a virtual image requires extending the refracted rays backward using dashed lines.

SECTION 4

Mathematical Framework

The quantitative analysis of thin lenses rests on two equations that connect focal length, object distance, image distance, and magnification. These equations use the standard sign convention: positive image distances correspond to real images on the far side of the lens, and negative image distances correspond to virtual images on the same side as the object.

THIN-LENS EQUATION

1/f = 1/dₒ + 1/dᵢ

f = focal length (positive for converging, negative for diverging); dₒ = object distance (positive when on the incoming-light side); dᵢ = image distance (positive for real images, negative for virtual images).

MAGNIFICATION

M = hᵢ/hₒ = −dᵢ/dₒ

M = lateral magnification; hᵢ = image height; hₒ = object height. When |M| > 1 the image is enlarged; when |M| < 1 it is reduced. A negative M indicates an inverted image; a positive M indicates an upright image.

LENSMAKER'S EQUATION (REFERENCE)

1/f = (n − 1)(1/R₁ − 1/R₂)

n = index of refraction of the lens material; R₁ and R₂ = radii of curvature of the two surfaces. This equation explains why focal length depends on both the lens shape and the material, though on the AP exam you will typically be given f directly.

These equations are algebraically equivalent to the ray-diagram approach but offer exact numerical answers. Notice that the thin-lens equation has the same form as the mirror equation — this is not a coincidence, as both describe image formation by focusing devices. The sign convention, however, differs: for mirrors, positive dᵢ is on the same side as the object (reflected light), whereas for lenses positive dᵢ is on the opposite side (transmitted light).

📝 Sign Convention Summary

Converging lens: f > 0. Diverging lens: f < 0. Real image: dᵢ > 0 (opposite side from object). Virtual image: dᵢ < 0 (same side as object). Object distance: dₒ > 0 (virtually always positive for single-lens problems).

SECTION 5

Image Characteristics by Object Position

The nature of the image formed by a lens changes dramatically as the object moves relative to the focal point. The table below summarizes all important cases for both converging and diverging lenses. Mastering this table is essential for the AP Physics 2 exam, where qualitative reasoning about image characteristics appears frequently.

Summary of image characteristics for converging and diverging lenses

Lens Type	Object Position	Image Position	Image Type	Orientation & Size
Converging	dₒ > 2f	f < dᵢ < 2f	Real	Inverted, reduced
Converging	dₒ = 2f	dᵢ = 2f	Real	Inverted, same size
Converging	f < dₒ < 2f	dᵢ > 2f	Real	Inverted, enlarged
Converging	dₒ = f	dᵢ → ∞	No image	Rays emerge parallel
Converging	dₒ < f	\|dᵢ\| > dₒ (same side)	Virtual	Upright, enlarged
Diverging	Any dₒ > 0	\|dᵢ\| < \|f\| (same side)	Virtual	Upright, reduced

A diverging lens always produces a virtual, upright, and reduced image regardless of object position. The dashed lines show that refracted rays diverge but appear to originate from a point on the same side as the object.

✦ KEY TAKEAWAY

A diverging lens behaves like a "one-trick pony" — it always produces a virtual, upright, and reduced image no matter where you place the object. A converging lens, by contrast, is versatile: it can produce real or virtual images, enlarged or reduced, depending on object placement relative to the focal point.

SECTION 6

Worked Example

Let us work through a complete problem that requires both the thin-lens equation and the magnification equation, illustrating the full analytic workflow expected on the AP exam.

Converging Lens: Object Between f and 2f

Step 1 — Identify Given ValuesA 4.0 cm tall object is placed 18.0 cm from a converging lens of focal length 12.0 cm. We want to find the image distance dᵢ, magnification M, image height hᵢ, and whether the image is real or virtual.

dₒ = 18.0 cm, f = +12.0 cm, hₒ = 4.0 cm

Step 2 — Apply the Thin-Lens EquationSubstitute into 1/f = 1/dₒ + 1/dᵢ: 1/12.0 = 1/18.0 + 1/dᵢ. Solving for 1/dᵢ: 1/dᵢ = 1/12.0 − 1/18.0 = 3/36 − 2/36 = 1/36.

dᵢ = +36.0 cm

Step 3 — Interpret the Sign of dᵢBecause dᵢ is positive, the image forms on the opposite side of the lens from the object. This means the image is real and can be projected onto a screen.

Step 4 — Calculate MagnificationM = −dᵢ/dₒ = −36.0/18.0 = −2.0. The negative sign indicates the image is inverted, and |M| = 2.0 means the image is twice the height of the object.

M = −2.0 (inverted, enlarged)

Step 5 — Find Image Heighthᵢ = M × hₒ = (−2.0)(4.0 cm) = −8.0 cm. The negative value confirms inversion; the absolute image height is 8.0 cm.

hᵢ = −8.0 cm (8.0 cm tall, inverted)

Step 6 — Verify with Ray Diagram LogicThe object is at 18.0 cm, which is between f = 12.0 cm and 2f = 24.0 cm. According to our classification table, this should produce a real, inverted, enlarged image beyond 2f. Indeed, dᵢ = 36.0 cm > 2f = 24.0 cm, confirming consistency.

SECTION 7

Converging vs. Diverging Lenses

Converging and diverging lenses often appear side by side on the AP exam, and students must quickly distinguish their properties. The following table highlights the key contrasts and helps build the intuition needed for rapid qualitative analysis during the multiple-choice section.

Key differences between converging and diverging lenses

Property	Converging (Convex)	Diverging (Concave)
Shape	Thicker at center	Thinner at center
Focal length sign	f > 0	f < 0
Parallel rays	Converge to real focus	Diverge from virtual focus
Can produce real images?	Yes, when dₒ > f	No (always virtual)
Can produce virtual images?	Yes, when dₒ < f	Always
Image orientation (virtual)	Upright	Upright
Typical application	Magnifying glass, camera, projector	Peephole, correcting nearsightedness

✦ KEY TAKEAWAY

The thin-lens equation applies identically to both lens types — the sign of f encodes all the optical personality of the lens. A positive f guarantees that converging behavior is possible, while a negative f ensures that real focal convergence never occurs. When in doubt, plug in the signs and let the algebra tell you whether the image is real or virtual.

SECTION 8

Connection to Advanced Optics

The thin-lens model is a powerful first approximation, but real optical systems involve complications such as aberrations, multi-element lens assemblies, and thick-lens corrections. Understanding where the thin-lens model breaks down prepares you for college-level optics courses and clarifies the limits of the AP Physics 2 framework.

Thin-lens model vs. advanced treatments

Feature	Thin-Lens Model (AP Physics 2)	Advanced Optics
Lens thickness	Neglected; refraction at single plane	Accounted for using principal planes
Aberrations	Ignored	Spherical, chromatic, coma, astigmatism analyzed
Multiple lenses	Image of first lens becomes object for second	Matrix (ray-transfer) methods for complex systems
Wave effects	Not considered (geometric optics)	Diffraction limits resolution; physical optics required
Equation	1/f = 1/dₒ + 1/dᵢ	Same form but f depends on position for thick lenses

For multi-lens systems — which do appear on the AP exam — the key technique is sequential application of the thin-lens equation. The image formed by the first lens serves as the object for the second lens, with the new object distance calculated from the second lens. If that object falls on the opposite side of the second lens from the incoming light, the object distance is negative (a virtual object). This cascading approach extends naturally to systems of three or more lenses and forms the conceptual basis for compound microscopes and telescopes.

SECTION 9

Practice Problems

An object is placed at a distance of 1.5f from a converging lens of focal length f. Which of the following correctly describes the image?

A diverging lens has a focal length of −15.0 cm. An object is placed 30.0 cm from the lens. What is the image distance?

A 3.0 cm tall object is placed 10.0 cm from a converging lens. The image formed is virtual, upright, and 6.0 cm tall. What is the focal length of the lens?

PROBLEM 4 — APPLIED

A physics student wants to determine the focal length of a converging lens experimentally. Describe a procedure using a light source, a screen, and a ruler. Explain what data should be collected, how it should be analyzed, and identify a major source of uncertainty.

PROBLEM 5 — CRITICAL THINKING

A converging lens of focal length 15.0 cm is placed 40.0 cm to the left of a diverging lens of focal length −10.0 cm. An object 5.0 cm tall is placed 25.0 cm to the left of the converging lens. (a) Determine the position of the final image relative to the diverging lens. (b) Determine the overall magnification and describe the final image. (c) Is the final image real or virtual? Justify your answer.

SUMMARY

Lesson Summary

Lenses form images by refracting light at curved surfaces. A converging (convex) lens has a positive focal length and can produce real or virtual images depending on whether the object is beyond or within the focal point. A diverging (concave) lens has a negative focal length and always produces a virtual, upright, reduced image. The thin-lens equation 1/f = 1/dₒ + 1/dᵢ and the magnification equation M = −dᵢ/dₒ provide quantitative predictions, while ray diagrams using three principal rays offer geometric verification.

For AP Physics 2, remember the sign conventions: positive dᵢ means real image (opposite side), negative dᵢ means virtual image (same side). Positive M means upright, negative M means inverted. In multi-lens systems, the image of the first lens becomes the object for the second, and the total magnification is the product of individual magnifications.