Images Formed by Lenses
Use near and far focal points to construct lens images and interpret their distances and magnification.
AP Physics 2 Topic 13.4 · College Board CED printed p. 100, 101 / PDF p. 107, 108 · See this topic in the course outline
A lens changes ray direction twice, at its two surfaces. The thin-lens model combines those interactions into a midline rule. Let actual ray intersections determine whether the image is real before interpreting the algebra.
Before you start
- Snell’s law qualitatively
- Signed reciprocals
- Height ratios
Two focal points belong to a thin lens
For an ordinary transparent lens in a lower-index surrounding medium such as air, a thin convex lens is converging: rays parallel to its principal axis transmit and meet at the focal point on the far side. A thin concave lens is diverging: transmitted rays spread as though they came from the focal point on the incident side. Lens behavior depends on shape and the indices of lens and surrounding medium; the convex/converging and concave/diverging names here assume the usual AP setup.
A lens has a focal point on each side. The curved surfaces and their material determine its focusing behavior. In the usual thin-lens model with the same medium on both sides, the two focal points are equally far from the midline. Reversing the direction of incident parallel light swaps which side plays the incoming and outgoing role. Do not use one focal point for every principal ray regardless of its direction.
Thin-lens distances and real or virtual images
A real lens image occurs when actual transmitted rays from one object point intersect. It can be formed on a screen on the outgoing side. A virtual image occurs when diverging transmitted rays have backward extensions meeting at an apparent origin, typically on the object side for a single lens with a real object. These extensions are geometry; light does not travel backward along them.
Use the real-is-positive sign convention: a real incident-side object has d_o > 0; an outgoing-side real image has d_i > 0; an incident-side virtual image has d_i < 0. A converging lens has f > 0; a diverging lens has f < 0. Measure distances from the lens midline and apply . This thin-lens approximation neglects thickness and is most reliable for rays close to the principal axis. Notice that a positive mirror image is on the reflected side, whereas a positive lens image is on the transmitted side; both are positive because they are real.
The signed transverse magnification is . The CED’s size ratio uses . With upright positive h_o, positive m is upright and negative m is inverted.
Build a lens image from principal rays
Draw an axis, the lens midline, both focal points, and an object arrow. From its tip draw: (1) a parallel ray, which refracts through the far focus for a converging lens or diverges as if from the near focus for a diverging lens; (2) a ray through the center of the thin lens, which proceeds approximately straight; (3) a ray through the near focus of a converging lens, which exits parallel. For a diverging lens, aim the third incoming ray toward the far focus; it exits parallel. Two rays locate the image tip, and the third checks it.
Use solid lines for actual paths, with arrows from the object through the lens. Use dashed backward extensions only when locating a virtual image. For a converging lens with the object at 2f, the transmitted rays meet at 2f on the other side, below the axis. The image is real, inverted, and the same size.
Track the image as an object moves
For a real object outside a converging lens’s focal length, the image is real and inverted; moving the object from far away toward f sends the image farther away and increases its size. At d_o = f the outgoing rays are parallel and no finite image plane exists. Inside f the image becomes virtual, upright, and enlarged. A diverging lens gives a real object an upright reduced virtual image on the incident side, between the lens and its near focus.
| f (cm) | d_o (cm) | d_i (cm) | m | Image |
|---|---|---|---|---|
| +10 | 30 | +15 | −0.50 | Real, inverted, reduced |
| +10 | 20 | +20 | −1.0 | Real, inverted, same size |
| +10 | 15 | +30 | −2.0 | Real, inverted, enlarged |
| +10 | 5 | −10 | +2.0 | Virtual, upright, enlarged |
| −10 | 20 | −6.67 | +0.333 | Virtual, upright, reduced |
A screen’s failure to show a sharp image can indicate a virtual image, an object at the focus, misalignment, or a screen not at the correct plane. Diagnose it with the ray construction and distances; do not conclude that the lens forms no image at all.
Locate a projected lens image
A 2.00 cm tall object stands 30.0 cm to the left of a thin converging lens with f = +12.0 cm. Light travels left to right. Find the screen position and image height.
Use d_o = +30.0 cm and 1/d_i = 1/12.0 − 1/30.0 = 1/20.0 cm⁻¹.
Why this step? The real-object thin-lens model uses the signed reciprocal equation.
d_i = +20.0 cm, so the screen goes 20.0 cm to the right, on the transmitted side.
Why this step? A positive lens image distance locates an actual ray intersection.
m = −20.0/30.0 = −0.667; h_i = −1.33 cm.
Why this step? The minus sign describes inversion, and the magnitude describes height.
A parallel ray bends through the far focus and a center ray continues straight; their intersection is below the axis at the screen.
Why this step? The construction checks the sign and reduced size.
Answer and check
Screen 20.0 cm right of the lens; image real, inverted, 1.33 cm tall.
Interpret a diverging-lens image
An upright object is 24.0 cm from a thin diverging lens with f = −8.00 cm. Find and describe its image.
1/d_i = −1/8.00 − 1/24.0 = −1/6.00 cm⁻¹, so d_i = −6.00 cm.
Why this step? The diverging focal length is negative, and the object is real.
m = −(−6.00)/24.0 = +0.250.
Why this step? The sign indicates upright orientation; magnitude gives one-quarter size.
The parallel ray diverges as if from the near focus, and the center ray remains straight. Their backward extensions meet 6.00 cm on the object side.
Why this step? The virtual ray construction agrees with the negative distance.
Answer and check
Virtual image 6.00 cm on the incident side, upright, one-quarter as tall.
Use a converging lens inside its focal length
A reading lens has f = +15.0 cm and the page is 10.0 cm from it. Locate and classify the image.
1/d_i = 1/15.0 − 1/10.0 = −1/30.0 cm⁻¹, so d_i = −30.0 cm.
Why this step? Inside the focal length the transmitted rays diverge, so a negative image distance is expected.
m = −(−30.0)/10.0 = +3.00. A ray diagram’s backward extensions meet on the page side.
Why this step? Positive magnification gives upright orientation and the magnitude gives enlargement.
Answer and check
Virtual upright image 30.0 cm from the lens on the page side, three times as tall.
Follow actual rays before extending them backward
A converging thin lens has a focus on both sides when the surrounding medium is the same. With light traveling left to right, a parallel ray exits through the far focus, a center ray proceeds approximately straight, and a ray through the near focus exits parallel. The three rays come from the same object point; their intersection locates its image.
Move a real object inside the positive focal length and those transmitted rays diverge. Extend their outgoing lines backward using dashes. The extensions meet on the incident side, locating an upright enlarged virtual image. The solid arrows still show light traveling through the lens toward the observer; they never carry light backward to that virtual point. Use the signed equation and magnification to check this geometrical construction, while retaining the thin, near-axis approximation.
Inspect diagram labels
The diagram keeps its full size here. Focus the view and use the arrow keys, or scroll across it, to inspect every label.
Inspect diagram labels
The diagram keeps its full size here. Focus the view and use the arrow keys, or scroll across it, to inspect every label.
Make all three transmitted rays meet
A thin converging lens in air has f = +10.0 cm. Light goes left to right from a real object 25.0 cm away and 1.50 cm tall.
Inspect diagram labels
The diagram keeps its full size here. Focus the view and use the arrow keys, or scroll across it, to inspect every label.
- 1/dᵢ = 1/10.0 − 1/25.0 = 0.0600 cm⁻¹, so dᵢ = +16.7 cm on the outgoing side.
- m = −16.7/25.0 = −0.667, so hᵢ = −1.00 cm.
- The parallel ray bends through the far focus, the center ray continues straight, and the near-focus ray leaves parallel. The plotted actual rays meet at the same inverted image tip.
Answer and check
A screen at 16.7 cm on the transmitted side receives a real, inverted, reduced image 1.00 cm tall.
Make the model respond
Focused Physics 2 model
Refraction and thin-lens ray models
Predict first. Choose a model. Predict the direction of refraction or whether the lens image is real or virtual.
n₁ sin θ₁ = n₂ sin θ₂; θreflection = θ₁; sin θc = n₂/n₁ when n₁ > n₂
On a narrow screen, scroll within a figure to read its labels. The table and explanation provide the same physics in text.
θ₂ = 25.37°. The transmitted ray bends toward the normal.
Read the model data and comparisons
| Quantity | Value | Unit | Meaning or limit |
|---|---|---|---|
| Incident-medium index n₁ | 1 | ||
| Transmitted-medium index n₂ | 1.5 | ||
| Incident angle | 40 | ° | All angles are from the normal |
| Reflected angle | 40 | ° | |
| Transmitted angle | 25.37 | ° | |
| Critical angle | not applicable (n₁ ≤ n₂) | ° |
Indices set wave speed v = c/n and wavelength λ = v/f in each medium; frequency stays constant across the boundary. For oblique transmission, a larger index gives a smaller angle to the normal. At normal incidence, direction does not change even when the index changes; speed and wavelength still change with index. Total internal reflection requires incidence from higher to lower index and an angle above the critical angle. At exactly the critical angle, the refracted ray is tangent to the boundary. This diagram includes reflection direction but does not calculate reflected or transmitted intensities.
Assumptions and limits. Geometric optics, flat interface, homogeneous transparent media, positive real refractive indices, incidence from medium 1 to medium 2. Angles are from the normal, not the surface. No absorption, Fresnel intensity coefficients, polarization, or curved surfaces.
Try it from memory
Write your answer and reason before opening the worked answer. Your self-assessment is a study decision; it is not an automatically graded score.
Question 1
A parallel ray travels left to right through a converging lens. Which focus does it pass through afterward? What about a ray initially passing through the near focus?
Worked answer and reason
The parallel ray bends through the far/right focus. The near-focus incoming ray exits parallel to the axis.
The two sides’ focal points play distinct roles in the principal-ray construction.
Question 2
A lens gives d_i = −8 cm for a real object with d_o = 16 cm. Find m and describe the image.
Worked answer and reason
m = +0.50: upright, virtual, reduced to half the height.
Use the signed distance ratio, then interpret the sign and magnitude separately.
Question 3
An object is exactly at a converging lens’s near focal point. Where should a screen be put for a finite sharp image in the ideal model?
Worked answer and reason
No finite screen position works because outgoing rays are parallel.
The thin-lens equation gives 1/d_i = 0; this is a limiting case, not d_i = 0.
Question 4
Why does the center principal ray go straight in this course, and when can that construction become inaccurate?
Worked answer and reason
It uses an ideal thin-lens approximation near the principal axis. Thick lenses or strongly off-axis rays need a more detailed optical model.
The ray rule belongs to stated approximation conditions.
JavaScript enables saved drafts and self-assessment. You can always work on paper.
Catch the tempting mistake
- A convex lens always forms a real image.
- A real object inside its focal length gives an upright enlarged virtual image.
- Both lens focal points are used the same way for every ray.
- For a converging lens, parallel incident rays head through the far focus; near-focus incident rays exit parallel.
- The image is negative because it is left on the drawing.
- Optical signs distinguish real and virtual; the drawing can be reversed without changing those signs.
Put it to work
Give the physical reason as well as the calculation. The worked solutions are on a separate page.
- Problem 7 · calculation and three-ray construction
- Problem 8 · comparison and virtual-ray construction
- Problem 10 · comparison table and ray translation
- Problem 11 · data graph, experimental design, and prediction
Explain and apply the model
Try these problems before opening the solutions. Show the physical model, your calculation or sketch, and the reason your conclusion follows.
A screen image and an apparent image from the same lens
A thin converging lens in air has f = +12.0 cm with the same medium on both sides. Light travels left to right from an upright 2.00 cm object. First dₒ = +30.0 cm; then the object is moved to dₒ = +8.00 cm. Use real-is-positive optical distances.
(a) For the 30.0 cm object position, calculate dᵢ, m, and hᵢ and construct an arrowed three-principal-ray diagram marking both foci. Determine image side, type, orientation and size, and say where a screen should go. Explain how the actual rays justify the type.
13.4.A.313.4.A.513.4.A.5.i13.4.A.5.ii13.4.A.613.4.A.713.4.A.7.i
(b) For the 8.00 cm object position, calculate dᵢ, m, and hᵢ and construct all three principal rays again. Distinguish solid actual transmitted paths from dashed backward extensions. Use the diagram to classify the image and explain how an observer can see it despite the failure of a screen on the image side.
Written working workspace
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Scientific calculator workspace
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Scientific calculator
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For the teacher
- Begin with a labeled object, lens midline, both foci, and travel direction.
- Pair a screen projection outside f with direct viewing through the lens inside f.
- The rays model supports thin-lens and Snell/TIR cases; use static sketches for mirrors and diverging-lens details.
Source objectives and scope
- 13.4.A — Describe the image formed by a lens.
Suggested skills are teaching guidance. The course applies all science practices across its objectives.
Science references for this teaching
- College Board AP Physics 2 CED, Unit 13 (supplied pinned source) — Official Unit 13 requirements, plane/spherical mirror scope, and magnitude form of magnification
- OpenStax College Physics 2e: The Ray Aspect of Light — Ray and wavefront geometry; geometric-optics approximation
- OpenStax College Physics 2e: The Law of Reflection — Normal-angle law and local diffuse/specular reflection
- OpenStax College Physics 2e: Image Formation by Mirrors — Paraxial spherical mirrors, mirror distances, magnification
- OpenStax College Physics 2e: The Law of Refraction — Refractive index, speed, and Snell angles
- OpenStax College Physics 2e: Total Internal Reflection — Index-order condition and critical-angle boundary
- OpenStax College Physics 2e: Image Formation by Lenses — Thin-lens images, sign convention, and principal rays
Glossary and deeper reading · 6 refreshers
These editorial refreshers supplement the reading. Definitions and examples are added guidance; original source wording and recorded answers remain in their original context.
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Snell’s law
Snell’s law relates the incidence and refraction angles to the refractive indices of two media. Both angles are measured from the normal to the boundary, rather than from the surface. For example, entering a higher-index material at oblique incidence bends a light ray toward the normal.
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Focal point
A focal point is where rays initially parallel to the principal axis converge or appear to diverge after reflection or refraction. The principal-axis description uses the paraxial model, in which rays remain close to the axis. For example, a converging thin lens brings parallel incoming rays to its far focal point.
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Virtual image
A virtual image is located where backward extensions of outgoing rays appear to meet. The rays do not physically converge at that location, so placing a screen there does not capture that image. For example, a plane mirror produces an image that appears behind the mirror.
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Real image
A real image forms where light rays physically converge after leaving an optical element. It can be projected onto a screen placed at the image location. For example, a projector’s lens forms a real image of its source on a screen.
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Optical magnification
Optical magnification is the ratio of image height to object height in the chosen signed convention. Its magnitude gives the size ratio, while its sign indicates upright or inverted orientation. For example, an inverted image half the object’s height has negative magnification with magnitude one half.
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Focal length
Focal length is the signed distance from the relevant lens or mirror reference point to its principal focal point in the chosen convention. It determines how strongly the optical element converges or diverges paraxial rays. For example, a converging lens with a shorter positive focal length bends initially parallel rays to a nearer focus.