AP Course

AP Physics 2

Study AP Physics 2 Units 9–15, with the Fall 2026 radioactive-decay clarification applied and official exam updates linked separately.

Study Units 9–15 through original models, derivations, experiments, quizzes, and practice sets.

Lessons
Particle Motion Behind Temperature and PressureConnecting Gas Variables with the Ideal-Gas ModelHeat Flow and the Approach to Thermal BalanceEnergy Accounting for Thermodynamic ProcessesMaterial Response to Heating and Heat ConductionEntropy, Probability, and the Direction of Thermal ChangeCharge Interactions and Coulomb-Force ModelsTracking Charge During Contact and InductionMapping Electric Fields from Source ChargesEnergy of Configurations of ChargesElectric Potential and Equipotential ReasoningCapacitance and Energy Stored in Electric FieldsConserving Energy in Charged-Particle MotionCharge Flow and Conventional CurrentModeling Sources, Wires, and Loads in Simple NetworksHow Geometry and Material Set ResistanceElectrical Energy Transfer and PowerReducing Series-Parallel DC NetworksLoop Equations from Energy ConservationJunction Equations from Charge ConservationTransient Charging and Discharging in RC NetworksSources, Direction, and Strength of Magnetic FieldsMagnetic Forces and Curved Paths of ChargesForces Between Fields and Current-Carrying ConductorsChanging Magnetic Flux and Induced EMFReflection and Absorption at Material BoundariesLocating Images with Mirror Ray ModelsRefraction, Index, and Total Internal ReflectionLocating Images with Thin-Lens ModelsPulses, Wave Types, and Propagation SpeedPeriodic-Wave Measures and GraphsBoundary Changes and PolarizationElectromagnetic-Wave BehaviorFrequency Shifts from Relative MotionSuperposition, Interference, and Standing WavesDiffraction from Openings and EdgesTwo-Slit and Grating Pattern GeometryPhase Change and Thin-Film ColorQuantum Models and Dual Wave-Particle EvidenceEnergy Levels in a Hydrogen-Like AtomConnecting Spectral Lines to Energy TransitionsThermal Spectra and Blackbody CurvesPhoton Thresholds in the Photoelectric EffectPhoton-Electron Scattering and MomentumMass-Energy Accounting in Fission and FusionRandom Nuclear Decay, Activity, and Half-Life
Quizzes
Practice Problems Formula notes, diagrams, and practice sets will be added here.

AP Physics 2 · Unit 13 · Topic 13.4

Locating Images with Thin-Lens Models

Use refraction through shaped surfaces to explain why lenses converge or diverge light. Locate images with principal rays, verify them with the thin-lens equation and magnification, and track intermediate images through multi-lens systems.

Learning Goals

  • Distinguish converging and diverging lenses by their effect on parallel rays.
  • Use three principal-ray rules to locate real and virtual images.
  • Apply the thin-lens equation with a consistent sign convention.
  • Use signed magnification to determine orientation and relative size.
  • Predict how image properties change as a real object crosses \(2F\) and \(F\).
  • Relate focal length to optical power and identify model limitations.
  • Treat the image of one lens as the object for a second lens.
\(d_o\)signed object distance
\(d_i\)signed image distance
\(f\)signed focal length
\(m\)signed lateral magnification

1. A Lens Redirects Light at Two Boundaries

A lens is transparent material bounded by refracting surfaces. Each ray normally changes direction once while entering and again while leaving. The combined effect depends on surface curvature and on the index contrast between the lens and its surroundings.

  • Converging lens: rays initially parallel to the optical axis meet at a real focal point on the outgoing side. In air, it is commonly thicker at the center.
  • Diverging lens: parallel rays spread after the lens and appear to originate from a virtual focal point on the incoming side. In air, it is commonly thinner at the center.

Shape alone is not a universal rule: changing the surrounding medium can alter the index contrast and even the converging or diverging behavior.

2. The Thin-Lens Approximation

The thin-lens model replaces refraction at two separated surfaces with one ideal change at a central plane. It works best when lens thickness is small compared with object and image distances and when rays remain near the optical axis.

A lens has focal points on both sides. For the same surrounding medium, their distances from the lens have equal magnitude \(f\). Optical power is

\(P=\frac{1}{f}\)

Focal length must be in meters, giving power in diopters \((\mathrm D=\mathrm{m^{-1}})\). A converging lens has \(f>0\) and \(P>0\); a diverging lens has \(f<0\) and \(P<0\).

Worked Example 1 · Optical Power

Convert Focal Length Before Inverting

A converging lens has \(f=+20.0\,\mathrm{cm}=+0.200\,\mathrm m\).

\(P=\frac{1}{0.200\,\mathrm m}=+5.00\,\mathrm D\)

The positive sign identifies a converging lens. Using \(20.0\) without converting centimeters to meters would give the wrong value and unit.

3. Principal Rays for a Converging Lens

  1. A ray parallel to the axis exits through the far focal point.
  2. A ray through the near focal point exits parallel to the axis.
  3. A ray through the optical center continues approximately straight.

4. Principal Rays for a Diverging Lens

  1. A parallel ray exits as if it came from the near focal point.
  2. A ray aimed toward the far focal point exits parallel.
  3. A ray through the optical center continues approximately straight.

5. Reliable Ray-Diagram Procedure

  1. Draw the optical axis, thin-lens plane, and both focal points to scale.
  2. Place the object arrow on the incoming side and begin every construction ray at the same object point.
  3. Trace two principal rays through the lens; use a third as a check.
  4. If refracted rays physically cross, mark a real image with solid paths.
  5. If they diverge, extend them backward with dashed lines to locate a virtual image.
  6. Read location, orientation, and relative height from the diagram before calculating.
Converging lens with \(d_o>f\)The parallel ray exits through the far focus while the central ray continues straight. Their physical intersection locates a real inverted image.

6. Thin-Lens Equation and Magnification

For paraxial rays, the distances satisfy

\(\frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i}\)

Similar triangles formed by the central ray give the signed lateral magnification:

\(m=\frac{h_i}{h_o}=-\frac{d_i}{d_o}\)

The lens equation locates the image; magnification then gives its orientation and size. A ray diagram should predict the signs before numbers are substituted.

QuantityPositiveNegative
\(d_o\)real object on incoming sidevirtual object on outgoing side
\(d_i\)real image on outgoing sidevirtual image on incoming side
\(f\) and \(P\)converging lensdiverging lens
\(m\) and \(h_i\)upright imageinverted image
Worked Example 2 · Converging Lens Beyond \(2F\)

Find a Reduced Real Image

A \(6.0\,\mathrm{cm}\)-tall object is \(60\,\mathrm{cm}\) from a converging lens with \(f=+20\,\mathrm{cm}\).

\(\frac{1}{d_i}=\frac{1}{20}-\frac{1}{60}=\frac{1}{30}\quad\Rightarrow\quad d_i=+30\,\mathrm{cm}\)
\(m=-\frac{30}{60}=-0.50,\qquad h_i=(-0.50)(6.0)=-3.0\,\mathrm{cm}\)

The image lies between \(F\) and \(2F\) on the outgoing side, is real, inverted, and half-size.

Worked Example 3 · Object at \(2F\)

Use a Symmetric Calibration Case

A converging lens has \(f=+18\,\mathrm{cm}\), and an object is placed at \(d_o=36\,\mathrm{cm}=2f\).

\(\frac{1}{d_i}=\frac{1}{18}-\frac{1}{36}=\frac{1}{36}\quad\Rightarrow\quad d_i=36\,\mathrm{cm}\)
\(m=-\frac{36}{36}=-1\)

The image is at \(2F\) on the far side, real, inverted, and the same size. This is a useful check for a scale ray diagram.

7. Converging-Lens Image Regimes

Object positionImage positionType and orientationSize
\(d_o>2f\)between \(F\) and \(2F\)real, invertedreduced
\(d_o=2f\)at \(2F\)real, invertedsame size
\(fbeyond \(2F\)real, invertedenlarged
\(d_o=f\)at infinity in ideal modelemerging rays parallelno finite image
\(0object sidevirtual, uprightenlarged
The focal point separates real and virtual image branchesFrom \(d_i=fd_o/(d_o-f)\), the finite image moves very far away near \(d_o=f\), then changes sides as \(d_o\) crosses the focus.
Worked Example 4 · Magnifying Lens

Place the Object Inside the Focus

A \(2.0\,\mathrm{cm}\)-tall object is \(8.0\,\mathrm{cm}\) from a converging lens with \(f=+12\,\mathrm{cm}\).

\(\frac{1}{d_i}=\frac{1}{12}-\frac{1}{8}=-\frac{1}{24}\quad\Rightarrow\quad d_i=-24\,\mathrm{cm}\)
\(m=-\frac{-24}{8.0}=+3.0,\qquad h_i=(3.0)(2.0)=6.0\,\mathrm{cm}\)

The virtual image is on the object side, upright, and three times the object height. A screen cannot capture it at \(d_i=-24\,\mathrm{cm}\).

8. What Happens at the Focal Point?

At \(d_o=f\), the lens equation gives \(1/d_i=0\). Rays from each object point leave parallel in the paraxial model, so no finite screen position produces a focused image. As an object approaches \(F\) from outside, the real image moves far away and grows; crossing inside produces a distant upright virtual image.

The mathematical divergence does not create infinite brightness or size in a real apparatus. Finite aperture, diffraction, aberrations, and the limited field of view constrain what can be observed.

9. Diverging Lenses Have One Regime for Real Objects

A diverging lens has \(f<0\). For a real object \(d_o>0\), the lens equation always gives \(d_i<0\). The virtual image lies between the lens and the near focal point, and

\(0

Thus the image is upright and reduced. Refracted rays do not pass through the virtual image; only their backward extensions meet there.

A diverging lens forms a reduced virtual imageThe parallel ray exits as if it came from the near focus, while the central ray continues straight. Backward extensions meet between the lens and focus.
Worked Example 5 · Diverging Lens

Interpret the Negative Image Distance

A diverging lens has \(f=-15\,\mathrm{cm}\), and a real object is \(30\,\mathrm{cm}\) away.

\(\frac{1}{d_i}=\frac{1}{-15}-\frac{1}{30}=-\frac{1}{10}\quad\Rightarrow\quad d_i=-10\,\mathrm{cm}\)
\(m=-\frac{-10}{30}=+\frac13\)

The image is \(10\,\mathrm{cm}\) from the lens on the object side, upright, virtual, and one third the object's height.

10. Real Images, Screens, Cameras, and Eyes

A real image exists where rays converge whether or not a screen is present. A screen, film sensor, or retina simply intercepts the organized light there. If the screen is displaced from the image plane, each object point produces a blur spot rather than a point.

A camera focuses objects at different distances by changing the lens-to-sensor spacing or effective focal length. For a very distant object, \(d_o\to\infty\) and \(d_i\to f\).

Worked Example 6 · Camera Image

Find Sensor Distance and Image Height

A \(1.80\,\mathrm m\)-tall person stands \(2.00\,\mathrm m\) from a camera lens with \(f=+5.00\,\mathrm{cm}\). Use centimeters.

\(\frac{1}{d_i}=\frac{1}{5.00}-\frac{1}{200}=0.195\,\mathrm{cm^{-1}}\quad\Rightarrow\quad d_i=5.13\,\mathrm{cm}\)
\(m=-\frac{5.13}{200}=-0.0256\)
\(h_i=(-0.0256)(180\,\mathrm{cm})=-4.62\,\mathrm{cm}\)

The sensor should be about \(5.13\,\mathrm{cm}\) behind the lens. The real image is inverted and \(4.62\,\mathrm{cm}\) tall.

11. Solve Backward from Image Evidence

Experimental problems may provide object and screen positions rather than focal length. If a sharp real image forms, use positive \(d_o\) and \(d_i\), then solve

\(f=\frac{d_od_i}{d_o+d_i}\)

The same focal length should describe multiple trials within uncertainty. Plotting suitable reciprocal quantities or fitting the full thin-lens model avoids relying on a single measurement.

Worked Example 7 · Infer Focal Length

Use a Sharp Screen Image

An object is \(30\,\mathrm{cm}\) from a lens, and a sharp image appears on a screen \(60\,\mathrm{cm}\) on the other side.

\(f=\frac{(30)(60)}{30+60}=20\,\mathrm{cm}=0.200\,\mathrm m\)
\(P=\frac{1}{0.200}=+5.00\,\mathrm D\)

A projectable image makes \(d_i>0\), and the resulting positive focal length identifies a converging lens.

12. Aperture Controls Brightness and Detail, Not Ideal Image Location

Every unobstructed part of an ideal lens receives rays from every object point. Covering the top half therefore does not remove the top half of the image. The complete image remains but receives less light and can be dimmer.

A smaller aperture also changes diffraction and reduces rays far from the axis, often reducing aberration. Image location predicted by the ideal thin-lens equation remains approximately unchanged.

13. Two-Lens Systems Must Be Solved Sequentially

For separated lenses, the image made by lens 1 becomes the object for lens 2. First calculate \(d_{i1}\). Use the lens separation to locate that intermediate image relative to lens 2, assign the new signed \(d_{o2}\), and calculate \(d_{i2}\).

\(m_{\mathrm{total}}=m_1m_2\)

If the intermediate image lies on the incoming side of lens 2, it is a real object for lens 2. If it lies beyond lens 2 before rays arrive there, lens 2 has a virtual object and \(d_{o2}<0\).

Each lens gets its own object and image distancesDo not substitute one global distance into both lens equations. Locate the intermediate image, then reset the coordinate reasoning at lens 2.
Worked Example 8 · Two Converging Lenses

Track an Intermediate Image

Lens 1 has \(f_1=+10\,\mathrm{cm}\), with an object \(15\,\mathrm{cm}\) to its left. Lens 2 has \(f_2=+15\,\mathrm{cm}\) and is \(50\,\mathrm{cm}\) to the right of lens 1.

\(\frac{1}{d_{i1}}=\frac{1}{10}-\frac{1}{15}=\frac{1}{30}\quad\Rightarrow\quad d_{i1}=30\,\mathrm{cm}\)

The intermediate image is \(50-30=20\,\mathrm{cm}\) to the left of lens 2, so \(d_{o2}=+20\,\mathrm{cm}\).

\(\frac{1}{d_{i2}}=\frac{1}{15}-\frac{1}{20}=\frac{1}{60}\quad\Rightarrow\quad d_{i2}=+60\,\mathrm{cm}\)
\(m_1=-\frac{30}{15}=-2,\qquad m_2=-\frac{60}{20}=-3,\qquad m_{\mathrm{total}}=+6\)

The final real image is \(60\,\mathrm{cm}\) to the right of lens 2 and upright relative to the original because two inversions produce positive total magnification.

14. Lens Maker's Relation

For a thin lens in air, surface curvatures and refractive index determine focal length:

\(\frac{1}{f}=(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\)

The signs of \(R_1\) and \(R_2\) depend on a stated convention. The relation explains why stronger curvature or greater index contrast generally increases optical power.

15. Model Limits and Aberrations

The ideal model assumes thin elements, paraxial rays, and one effective focal length. Real lenses can show spherical aberration, coma, and chromatic aberration because ray bending depends on position and wavelength.

A numerical answer with many digits is not physically precise if the thin-lens assumptions or measured distances are only approximate.

16. Reliable Lens Workflow

  1. Classify the lens and predict the image regime.
  2. Sketch lens, axis, focal points, object, and two rays.
  3. Assign signs to \(f\), \(d_o\), and \(d_i\).
  4. Solve the lens equation symbolically.
  5. Use \(m=-d_i/d_o\) and \(h_i=mh_o\).
  6. Translate signs into location, type, orientation, and size.
  7. Check ray geometry, limits, units, and screen feasibility.

17. Common Traps

  • Using \(f>0\) for a diverging lens.
  • Drawing a central ray bending in the thin-lens model.
  • Extending virtual rays with solid lines.
  • Forgetting the minus sign in magnification.
  • Calling an upright virtual image projectable.
  • Thinking half a lens makes half an image.
  • Using centimeters in \(P=1/f\) and labeling the result diopters.
  • Using the original object distance again at lens 2.
  • Trusting algebra that contradicts the predicted ray regime.
Mastery Check

1. A real object lies beyond \(2F\) of a converging lens. Describe the image.

Show reasoning and answer

It lies between \(F\) and \(2F\) on the far side and is real, inverted, and reduced.

2. What happens when a real object is exactly at the focal point of a converging lens?

Show reasoning and answer

Emerging rays are parallel and the ideal image is at infinity, so no finite screen position gives a sharp image.

3. A result gives \(d_i=-18\,\mathrm{cm}\) and \(m=+2.0\). Interpret it.

Show reasoning and answer

The image is virtual on the object side, upright, and twice the object's height.

4. What image does a diverging lens form for a real object?

Show reasoning and answer

An upright, reduced, virtual image between the lens and the near focal point.

5. Half of a lens is covered. Is half the image lost?

Show reasoning and answer

No. Rays from every object point pass through the remaining aperture, so the whole image remains, typically dimmer and with altered diffraction or aberration.

6. A lens has power \(-4.0\,\mathrm D\). Find its focal length and type.

Show reasoning and answer

\(f=1/P=-0.250\,\mathrm m\). The negative focal length identifies a diverging lens.

Investigation: Test Ray Diagrams and the Thin-Lens Equation

Use the open educational PhET Geometric Optics lens simulation. Predict image location, orientation, size, and projectability before each trial.

  1. Place a converging-lens object beyond \(2F\), at \(2F\), between \(2F\) and \(F\), near \(F\), and inside \(F\).
  2. Record \(d_o\), \(d_i\), \(h_o\), and \(h_i\); compare \(1/f\) with \(1/d_o+1/d_i\).
  3. Switch to a diverging lens and verify its single real-object regime.
  4. Change aperture and decide which quantities change: brightness, image location, image size, or sharpness.
  5. Fit several trials rather than treating one screen position as exact; report measurement uncertainty.

For a classroom bench, use an illuminated object, lens holder, and white screen. Do not focus direct sunlight: a converging lens can create hazardous intensity and heat.

Official curriculum reference: College Board AP Physics 2 course page. The explanation and worked example are independently written for this study site.