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 13.1

Reflection and Absorption at Material Boundaries

Model what happens when light reaches a material boundary: some energy may return by reflection, some may continue by transmission, and some may become internal energy through absorption. Use ray geometry, energy fractions, wavelength-dependent response, and experimental evidence without confusing brightness with ray direction.

Learning Goals

  • Draw incident, reflected, transmitted, and normal lines at a boundary.
  • Apply the law of reflection with angles measured from the normal.
  • Distinguish specular reflection from diffuse reflection using local surface normals.
  • Use reflectance, transmittance, and absorptance to conserve optical energy.
  • Connect absorption to internal-energy change and predict heating.
  • Explain why material response depends on wavelength, surface condition, angle, and polarization.
  • Analyze mirror rotation, multiple boundaries, and measurement uncertainty.
\(R\)reflected fraction of incident power
\(T\)transmitted fraction of incident power
\(A\)absorbed fraction of incident power
\(\hat n\)surface normal used for ray angles

1. One Incident Beam, Several Energy Pathways

In the ray model, light travels along straight paths within a uniform medium and changes behavior at a boundary. At a passive material boundary, incident optical energy can be divided among:

  • Reflection: energy returns to the incident medium.
  • Transmission: energy crosses into the second material; its path may also refract.
  • Absorption: optical energy is transferred to the material's microscopic degrees of freedom.
  • Scattering: energy is redirected into many paths. Depending on the measurement boundary, scattered energy may be counted within reflection or transmission.

The ray widths in a sketch do not automatically represent power. Use labels, measured intensity, or energy fractions to communicate how much energy follows each path.

Trace geometry and energy separatelyThe reflected ray stays in the first medium, the transmitted ray enters the second, and absorbed energy remains in the material. Every ray angle is referenced to the local normal.

2. The Law of Reflection

The incident ray, reflected ray, and surface normal lie in one plane. The angle of reflection equals the angle of incidence:

\(\theta_r=\theta_i\)

Both angles are measured from the normal, not from the surface. If a problem gives an angle \(\alpha\) from the surface, then \(\theta_i=90^\circ-\alpha\). The law determines direction; it does not say that all incident energy is reflected.

Worked Example 1 · Reflection Geometry

Convert a Surface Angle Before Applying the Law

A ray makes a \(32^\circ\) angle with a flat mirror surface.

\(\theta_i=90^\circ-32^\circ=58^\circ\)
\(\theta_r=\theta_i=58^\circ\)

The reflected ray also makes \(32^\circ\) with the surface. Reporting \(32^\circ\) as the incidence angle would use the wrong reference line.

3. Specular Reflection

A surface smooth compared with the wavelength has nearly parallel local normals. Parallel incident rays therefore leave in an organized direction and can preserve spatial information needed for an image.

A narrow reflected beam can look very bright to an observer positioned in its path and nearly invisible to an observer elsewhere. That observation alone does not mean the surface reflects more total energy than every rough surface.

4. Diffuse Reflection

A rough surface has many differently oriented microscopic patches. Each patch still obeys \(\theta_r=\theta_i\) relative to its local normal, but the collection of reflected rays spreads over many directions.

Diffuse reflection lets a page or wall be seen from many viewing positions. Diffuse does not mean “the reflection law fails,” and it does not mean the light is necessarily absorbed.

5. Energy Fractions at a Passive Boundary

For incident power \(P_i\), define the dimensionless fractions

\(R=\frac{P_r}{P_i},\qquad T=\frac{P_t}{P_i},\qquad A=\frac{P_a}{P_i}\)

Energy conservation gives

\(P_i=P_r+P_t+P_a\qquad\Longrightarrow\qquad R+T+A=1\)

These quantities can also be defined using energy over a stated time interval. If intensity is used, compare powers crossing appropriate areas; at oblique incidence, using unmatched detector areas can produce a misleading ratio.

Energy accounting closes the boundary modelFor a passive sample, missing reflected or transmitted power is not destroyed; it is absorbed or scattered outside the detector's collection range.
Worked Example 2 · Three-Way Energy Split

Find Every Output Power

A beam delivers \(240\,\mathrm W\) to a sample. Measurements give \(R=0.35\) and \(A=0.20\).

\(T=1-R-A=1-0.35-0.20=0.45\)
\(P_r=(0.35)(240)=84\,\mathrm W\)
\(P_t=(0.45)(240)=108\,\mathrm W,\qquad P_a=(0.20)(240)=48\,\mathrm W\)

The three powers add to \(240\,\mathrm W\), providing a direct conservation check.

Worked Example 3 · Opaque Surface

Do Not Set Absorption to One Automatically

An opaque coating reflects \(62\%\) of incident light. Because it transmits essentially none, \(T\approx0\).

\(A=1-R-T=1-0.62-0=0.38\)

If \(500\,\mathrm J\) arrives, the coating absorbs \(E_a=(0.38)(500)=190\,\mathrm J\). Opaque means negligible transmission, not necessarily complete absorption.

6. What Absorption Does to the Material

Absorption transfers electromagnetic energy to matter. Depending on frequency and material structure, it can excite electrons, vibrations, rotations, or other microscopic motion. Much of that energy may ultimately become internal energy and later leave by thermal radiation, conduction, or convection.

\(P_a=A\,P_i,\qquad Q_{\mathrm{absorbed}}=P_a\Delta t\)

If heat losses and phase changes are negligible, a sample's temperature change can be estimated from

\(Q_{\mathrm{absorbed}}=mc\Delta T\)

Absorptance is a property of the complete situation: wavelength, incidence angle, polarization, material composition, thickness, surface coating, and temperature can all matter.

Worked Example 4 · Absorption and Heating

Connect Optical Power to Temperature Change

A \(0.25\,\mathrm{kg}\) plate with specific heat \(900\,\mathrm{J/(kg\cdot K)}\) receives \(40\,\mathrm W\) of light for \(90\,\mathrm s\). Its absorptance is \(0.70\). Neglect energy loss during the interval.

\(Q=(0.70)(40)(90)=2.52\times10^3\,\mathrm J\)
\(\Delta T=\frac{Q}{mc}=\frac{2.52\times10^3}{(0.25)(900)}=11.2\,\mathrm K\)

This is an ideal upper estimate. In a real experiment, thermal losses usually reduce the measured temperature rise.

7. Wavelength Selectivity Produces Color and Spectral Signatures

A material does not have one universal value of \(R\), \(T\), or \(A\). Write \(R(\lambda)\), \(T(\lambda)\), and \(A(\lambda)\) when the wavelength dependence matters. A red object under white light appears red because more red light reaches the observer than other visible wavelengths; the missing wavelengths may be absorbed or transmitted.

Illumination matters too. A red surface under blue-only light cannot reflect red photons that never arrive, so it may appear dark. A material that is opaque in visible light can be transparent at infrared or radio wavelengths.

Color is a wavelength-dependent boundary responseThe plotted material returns much more red light than blue or green. Its appearance still depends on the spectrum of the incident source and the observer or detector response.
Worked Example 5 · Spectral Energy Accounting

Predict Reflected Color and Absorbed Power

Equal \(12\,\mathrm W\) red, green, and blue components illuminate an opaque surface. Its reflectances are \(R_r=0.80\), \(R_g=0.10\), and \(R_b=0.10\).

\(P_{r,\mathrm{total}}=(0.80+0.10+0.10)(12)=12\,\mathrm W\)
\(P_{a,\mathrm{total}}=36\,\mathrm W-12\,\mathrm W=24\,\mathrm W\)

The reflected beam is dominated by \(9.6\,\mathrm W\) of red light, so the surface appears red under this source. Most green and blue power becomes internal energy.

8. Transparent, Translucent, and Opaque

  • Transparent: substantial transmission with limited scattering, so spatial information can be preserved.
  • Translucent: substantial transmitted light is scattered, so objects behind are not sharply resolved.
  • Opaque: negligible transmission at the wavelength and thickness being tested.

These labels describe transmission behavior, not a fixed molecular identity. Thickness and wavelength can change the classification.

9. Reflection Is Not the Same as Scattering

Specular and diffuse reflection both return energy to the incident side, but a detector aimed only at the mirror angle may collect the specular part and miss wide-angle scattering.

An apparent energy deficit may therefore reflect detector geometry rather than true absorption. A complete energy measurement uses detectors that collect all reflected and transmitted directions or explicitly states what was omitted.

10. Optional Boundary Model at Normal Incidence

For normal incidence between two nonabsorbing, nonmagnetic transparent media, a simplified Fresnel result estimates the reflected power fraction:

\(R=\left(\frac{n_1-n_2}{n_1+n_2}\right)^2\)

With no absorption, \(T=1-R\). This expression is not a universal reflection rule: oblique incidence requires polarization-dependent equations, and absorbing or coated materials require a more complete model.

Worked Example 6 · Air-to-Glass Surface

Estimate Normal-Incidence Reflectance

Light reaches glass with \(n_2=1.50\) from air with \(n_1=1.00\), normally incident. Treat both materials as nonabsorbing.

\(R=\left(\frac{1.00-1.50}{1.00+1.50}\right)^2=(-0.20)^2=0.040\)
\(T=1-R=0.960\)

About \(4.0\%\) of incident power reflects at that single idealized boundary. The negative field-amplitude sign hidden by the square relates to phase, not negative energy.

11. Multiple Boundaries Compound Losses

A window or slab has at least an entrance and an exit boundary. In a simple approximation that ignores absorption and repeated internal reflections, multiply the transmitted fraction at each boundary:

\(T_{\mathrm{two\ surfaces}}\approx(1-R_1)(1-R_2)\)

Thin-film interference, coatings, surface roughness, and repeated reflections can make the exact result different. State the approximation rather than silently treating two surfaces as one.

Worked Example 7 · Uncoated Slab Approximation

Apply Two Successive Surface Transmissions

Each surface of a nonabsorbing slab reflects \(4.0\%\). Ignore repeated internal reflections.

\(T_{\mathrm{slab}}\approx(0.960)(0.960)=0.922\)

About \(92.2\%\) reaches the far side in this approximation. Subtracting \(4\%+4\%=8\%\) gives \(92\%\), close but not exactly the same because the second surface receives only the power transmitted through the first.

12. Rotating a Mirror Rotates the Reflected Ray Twice as Much

If the incident ray direction stays fixed and a plane mirror rotates by a small angle \(\delta\), its normal also rotates by \(\delta\). The incidence angle changes by \(\delta\), and the equal reflection angle shifts on the other side of the new normal. The reflected ray therefore rotates by

\(\Delta\theta_{\mathrm{ray}}=2\delta\)

This is a geometric consequence of the reflection law and is widely used to amplify small rotations in optical measurements.

Worked Example 8 · Mirror Rotation

Track the Reflected Beam

A fixed laser strikes a plane mirror. The mirror rotates \(7.0^\circ\) while the laser direction remains unchanged.

\(\Delta\theta_{\mathrm{ray}}=2(7.0^\circ)=14.0^\circ\)

The reflected beam rotates \(14.0^\circ\). This result concerns the change in ray direction, not the new incidence angle by itself.

13. Ray, Wave, and Photon Models

The ray model predicts paths when boundaries and objects are large compared with wavelength. The wave model explains interference, polarization, and the detailed reflected amplitude. The photon model helps describe discrete energy transfer \(E_\gamma=hf\) during absorption.

Increasing intensity at fixed frequency increases the delivered energy per time and typically the photon arrival rate; it does not increase the energy of each photon.

14. Common Traps

  • Measuring reflection angles from the surface.
  • Assuming reflection means \(R=1\).
  • Setting \(A=1\) merely because a material is opaque.
  • Treating diffuse reflection as a violation of the reflection law.
  • Calling uncollected scattered light absorbed.
  • Using one \(R\) value for every wavelength and angle.
  • Adding percentage losses at successive surfaces without identifying an approximation.
  • Confusing a bright directional glint with large total reflected power.

15. Experimental Measurement and Uncertainty

Measure incident, reflected, and transmitted optical power with the same detector settings and controlled geometry. Dark-subtract ambient light, hold source distance fixed, and avoid saturating the sensor. For diffuse samples, a narrow detector misses most scattered light; an integrating collection geometry is more appropriate.

Repeat measurements at several angles and wavelengths. Report \(R=P_r/P_i\), \(T=P_t/P_i\), and the inferred \(A=1-R-T\) with uncertainty. A negative inferred \(A\) or a sum above one signals calibration, alignment, background, or collection errors rather than energy creation.

Mastery Check

1. A ray makes \(25^\circ\) with a mirror surface. What are \(\theta_i\) and \(\theta_r\)?

Show reasoning and answer

Angles are measured from the normal, so \(\theta_i=90^\circ-25^\circ=65^\circ\), and \(\theta_r=65^\circ\).

2. A rough surface sends light in many directions. Does each microscopic reflection violate \(\theta_r=\theta_i\)?

Show reasoning and answer

No. Each small patch obeys the law relative to its own local normal. Different normals create different reflected directions.

3. A sample has \(R=0.28\) and \(T=0.57\). Find \(A\).

Show reasoning and answer

\(A=1-R-T=1-0.28-0.57=0.15\). It absorbs \(15\%\) of incident power.

4. An opaque object looks white. Must it absorb nearly all visible light?

Show reasoning and answer

No. Opaque means \(T\approx0\). A white appearance is consistent with strong diffuse reflection across the visible spectrum and relatively low absorption.

5. A mirror turns clockwise by \(3^\circ\) while the incoming ray stays fixed. By how much does the reflected ray turn?

Show reasoning and answer

It turns \(2(3^\circ)=6^\circ\) in the corresponding direction.

6. A detector collects only the narrow specular beam from a rough surface. Can \(1-R_{\mathrm{measured}}-T_{\mathrm{measured}}\) safely be called absorptance?

Show reasoning and answer

Not without collecting or estimating diffuse scattering. Uncollected reflected or transmitted light would be falsely counted as absorption.

Investigation: Build an Optical Energy Ledger

Use the open educational PhET Bending Light simulation to trace incident, reflected, and transmitted rays. Predict the reflection direction before changing the laser angle, and measure all angles from the normal.

  1. Choose two media and verify \(\theta_r=\theta_i\) at several incident angles.
  2. Compare the displayed reflected and transmitted brightness qualitatively while keeping source intensity fixed.
  3. Change the material pair and record which observation changes: reflected amount, transmitted direction, or both.
  4. With real low-power light and a light sensor, compare a mirror, white card, dark card, clear sheet, and frosted sheet. Never aim a laser toward eyes or reflective hazards.
  5. Explain discrepancies using absorption, diffuse scattering, detector acceptance angle, ambient light, and alignment.

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