What is AP Physics 2 unit 14?
Unit 14 is the wave and optics unit of AP Physics 2. It builds from the basic idea that waves transfer energy without transferring matter, then develops the tools needed to analyze how waves behave at boundaries, how they interfere constructively and destructively, and how those interference effects produce the bright and dark patterns seen in diffraction and thin-film problems.
Unit 14 covers wave properties, periodic waves, boundary behavior, polarization, electromagnetic waves, the Doppler effect, interference, standing waves, single-slit diffraction, double-slit interference, diffraction gratings, and thin-film interference. It accounts for 12-15% of the AP Physics 2 exam.
Wave fundamentals
Waves transfer energy without transferring matter. Mechanical waves need a medium; electromagnetic waves do not. Key quantities are amplitude, wavelength, frequency, period, and wave speed, connected by v = f lambda and T = 1/f. Transverse and longitudinal waves differ in the direction of the disturbance relative to propagation.
Interference and diffraction
When waves overlap, superposition determines the net displacement. Constructive interference occurs when path length difference equals m lambda; destructive when it equals (m + 1/2) lambda. Single-slit dark fringes follow a sin theta = m lambda. Double-slit bright fringes follow d sin theta = m lambda. Standing waves form when waves reflect and overlap in a confined region.
Thin-film interference
Light reflecting off the two surfaces of a thin film interferes based on film thickness t, index of refraction n, and phase shifts at each boundary. A 180-degree phase shift occurs when reflecting from a medium with a higher index. The optical path difference is 2nt, and whether that produces constructive or destructive interference depends on how many phase inversions occur.
Light as a wave explains observable patternsThe central thread of Unit 14 is that treating light and sound as waves with well-defined wavelength and frequency explains a wide range of phenomena: why soap bubbles show color, why a siren changes pitch as it passes, why a narrow slit produces a spread-out pattern of bright and dark bands, and why antireflection coatings work. Every major topic in the unit is an application of wave superposition and the conditions for constructive or destructive interference.
Unit 14 review notes
14.1
Wave Properties and Periodic Waves
A wave pulse is a single disturbance; a periodic wave is a continuous, repeating disturbance described by wavelength lambda, frequency f, period T, and amplitude A. Wave speed depends on the medium: for a string, v = sqrt(F_T / (m/L)); for all EM waves in vacuum, c = 3.00 x 10^8 m/s. The fundamental relationships are T = 1/f and v = f lambda. Amplitude is independent of frequency; wave energy increases with frequency.
- Transverse vs. longitudinal: In transverse waves the disturbance is perpendicular to propagation (strings, EM waves); in longitudinal waves it is parallel (sound, compressions and rarefactions).
- v = f lambda: Wave speed equals frequency times wavelength. Speed is set by the medium; changing medium changes wavelength but not frequency.
- T = 1/f: Period and frequency are reciprocals. A 440 Hz sound wave has a period of about 2.27 ms.
- Amplitude and energy: Amplitude measures maximum displacement from equilibrium. Louder sound has greater pressure amplitude; higher-frequency waves carry more energy.
- Mechanical vs. EM waves: Mechanical waves (sound, string waves) require a medium. Electromagnetic waves propagate through vacuum at c.
Given a wave with f = 500 Hz traveling at 340 m/s in air, what is its wavelength? (lambda = v/f = 0.68 m)
| Property | Mechanical wave | Electromagnetic wave |
|---|
| Requires medium | Yes | No |
| Example | Sound in air | Visible light |
| Speed in vacuum | N/A | c = 3.00 x 10^8 m/s |
| Wave type | Transverse or longitudinal | Transverse only |
14.3
Boundary Behavior and Polarization
When a wave reaches a boundary between two media, it partially reflects and partially transmits. The key rule for reflection inversion: if the wave slows down in the new medium, the reflected wave is inverted (phase shift of 180 degrees); if it speeds up, the reflected wave is not inverted. Frequency never changes at a boundary; wavelength changes because speed changes. Only transverse waves can be polarized; longitudinal waves cannot.
- Inversion rule: Reflected wave inverts when moving from a faster medium to a slower one (analogous to a fixed-end reflection on a string).
- Frequency conservation: Frequency is set by the source and does not change when a wave crosses a boundary. Wavelength adjusts via lambda = v/f.
- Polarization: A polarized transverse wave oscillates in a single plane. Passing unpolarized light through a polarizer reduces intensity by half.
- Polarization methods: Transverse waves can be polarized by reflection, refraction, or passing through a polarizing filter. Longitudinal waves (sound) cannot be polarized.
A wave on a light string hits a junction with a heavier string. Is the reflected pulse inverted or not? (Inverted, because the wave slows down in the heavier string.)
| Scenario | Reflected wave inverted? |
|---|
| Wave moves from fast to slow medium | Yes |
| Wave moves from slow to fast medium | No |
| Fixed end of string | Yes |
| Free end of string | No |
14.4
Electromagnetic Waves
Electromagnetic waves consist of oscillating electric and magnetic fields that are perpendicular to each other and to the direction of propagation, making them transverse waves. They require no medium and all travel at c = 3.00 x 10^8 m/s in vacuum. The EM spectrum is ordered by wavelength from longest to shortest: radio, microwave, infrared, visible (red to violet), ultraviolet, X-ray, gamma ray. Higher frequency means shorter wavelength and more energy per photon.
- E and B orientation: The electric field E, magnetic field B, and propagation direction k are mutually perpendicular. E and B oscillate in phase.
- EM spectrum order: Radio (longest lambda) > microwave > infrared > visible > ultraviolet > X-ray > gamma ray (shortest lambda, highest f).
- Visible light: Within visible light, red has the longest wavelength (~700 nm) and violet the shortest (~400 nm). ROYGBV in decreasing wavelength order.
- c = f lambda: All EM waves in vacuum satisfy c = f lambda. A 600 nm orange photon has f = c/lambda = 5.0 x 10^14 Hz.
Which EM wave carries more energy per photon: a microwave or an X-ray? (X-ray, because it has higher frequency.)
| Region | Approximate wavelength range |
|---|
| Radio | > 1 mm |
| Visible | 400-700 nm |
| X-ray | 0.01-10 nm |
| Gamma ray | < 0.01 nm |
14.5
The Doppler Effect
The Doppler effect is the change in observed frequency when a source and observer move relative to each other. AP Physics 2 requires only qualitative reasoning: no Doppler formula calculations. When source and observer approach each other, observed frequency is higher than the rest frequency. When they move apart, observed frequency is lower. If they move together at the same velocity, observed frequency equals rest frequency.
- Approaching: f_obs > f_rest: Wavefronts bunch together in the direction of approach, so the observer detects more cycles per second. Example: ambulance siren pitch rises as it approaches.
- Receding: f_obs < f_rest: Wavefronts spread out behind the source, so the observer detects fewer cycles per second. Example: siren pitch drops as the ambulance passes.
- No relative motion: f_obs = f_rest: If source and observer move at the same velocity in the same direction, no Doppler shift occurs.
- Applications: Doppler radar measures vehicle speed; medical ultrasound uses Doppler shifts to measure blood flow; astronomical redshift indicates galaxies receding from Earth.
A fire truck moves away from a stationary listener. Is the observed pitch higher, lower, or equal to the emitted frequency? (Lower, because the source is receding.)
| Relative motion | Observed frequency vs. rest frequency |
|---|
| Source approaches observer | Higher |
| Source recedes from observer | Lower |
| Source and observer move together | Equal |
| Observer approaches stationary source | Higher |
14.6
Wave Interference and Standing Waves
Superposition states that overlapping waves add their displacements. Constructive interference occurs when path difference = m lambda (waves in phase); destructive when path difference = (m + 1/2) lambda (waves out of phase). Beats arise when two slightly different frequencies interfere: f_beat = |f1 - f2|. Standing waves form when two identical waves travel in opposite directions in a confined region, producing fixed nodes (zero amplitude) and antinodes (maximum amplitude). Only specific wavelengths fit the boundary conditions, giving a harmonic series.
- Superposition: Net displacement at any point equals the sum of individual wave displacements at that point.
- Nodes and antinodes: Nodes are points of permanent zero displacement; antinodes are points of maximum displacement. Adjacent nodes are separated by lambda/2.
- Fixed-fixed string harmonics: f_n = nv / 2L for n = 1, 2, 3, ... The fundamental (n=1) has one antinode; each higher harmonic adds one more.
- Open-closed pipe harmonics: f_n = nv / 4L for odd n only (1, 3, 5, ...). A closed end is a node; an open end is an antinode.
- Beats: Two sources at 440 Hz and 444 Hz produce beats at f_beat = 4 Hz, heard as a pulsing amplitude variation.
A string of length 0.80 m has wave speed 320 m/s. What is the fundamental frequency? (f1 = v/2L = 320/(1.6) = 200 Hz)
| Boundary type | Harmonic series | Formula |
|---|
| Fixed-fixed string or open-open pipe | All harmonics | f_n = nv/2L |
| Open-closed pipe | Odd harmonics only | f_n = nv/4L (odd n) |
14.7
Single-Slit Diffraction
Diffraction is the spreading of a wave around edges or through an opening. It is most pronounced when the opening width a is comparable to the wavelength lambda. For single-slit diffraction, dark fringes (minima) occur where a sin theta = m lambda (m = +/-1, +/-2, ...). The central bright fringe is the widest and brightest band; secondary maxima are much dimmer. Using the small-angle approximation, the position of the mth dark fringe on a screen at distance L is y_min = m lambda L / a.
- Diffraction condition: Spreading is most significant when a is close to lambda. Very large openings produce little diffraction; very small openings spread the wave widely.
- Single-slit dark fringe: a sin theta = m lambda gives the angles of destructive interference for m = +/-1, +/-2, ...
- Central maximum width: The central bright fringe spans from the m = -1 to m = +1 dark fringes, making it twice as wide as secondary maxima.
- Effect of changing a or lambda: Narrower slit (smaller a) or longer wavelength produces a wider diffraction pattern. Wider slit or shorter wavelength narrows the pattern.
Light of wavelength 500 nm passes through a 0.10 mm slit onto a screen 2.0 m away. Where is the first dark fringe? (y = lambda L / a = (500e-9)(2.0)/(0.10e-3) = 0.010 m = 1.0 cm)
14.8
Double-Slit Interference and Diffraction Gratings
In Young's double-slit experiment, two slits separated by distance d produce an interference pattern. Bright fringes occur where d sin theta = m lambda; dark fringes where d sin theta = (m + 1/2) lambda. For small angles, fringe spacing is y = lambda L / d. A diffraction grating has many slits and produces sharper, more widely spaced bright maxima using the same equation d sin theta = m lambda, where d is the grating spacing. Gratings separate wavelengths more effectively than double slits.
- Path length difference: Delta D = d sin theta. When Delta D = m lambda, waves arrive in phase (bright fringe). When Delta D = (m+1/2) lambda, waves cancel (dark fringe).
- Fringe spacing: y = lambda L / d for small angles. Larger slit separation d produces more closely spaced fringes; longer lambda spreads them out.
- Diffraction grating: Many parallel slits with spacing d. Same condition d sin theta = m lambda applies, but maxima are much sharper and brighter than in double-slit patterns.
- White light through a grating: Different wavelengths diffract at different angles, spreading white light into a spectrum. Longer wavelengths (red) diffract more than shorter ones (violet).
Double slits with d = 0.20 mm are illuminated by 600 nm light. The screen is 1.5 m away. What is the fringe spacing? (y = lambda L / d = (600e-9)(1.5)/(0.20e-3) = 4.5 mm)
| Feature | Double slit | Diffraction grating |
|---|
| Number of slits | 2 | Hundreds to thousands |
| Bright fringe sharpness | Broad | Very sharp |
| Condition for maxima | d sin theta = m lambda | d sin theta = m lambda |
| Best use | Demonstrating interference | Separating wavelengths precisely |
14.9
Thin-Film Interference
Thin-film interference occurs when light reflects off the top and bottom surfaces of a film whose thickness t is comparable to the wavelength of light. The two reflected rays interfere based on: (1) the optical path difference 2nt, where n is the film's index of refraction, and (2) any 180-degree phase shifts at each reflection. A phase shift occurs when reflecting from a medium with a higher index of refraction; no phase shift occurs when reflecting from a lower index. Counting the number of phase inversions determines whether 2nt = m lambda gives constructive or destructive interference.
- Phase shift rule: Reflection from a higher-n medium causes a 180-degree phase shift. Reflection from a lower-n medium causes no phase shift.
- Optical path difference: The extra distance traveled by the ray reflecting off the bottom surface is 2nt, where n is the film index and t is the film thickness.
- One phase inversion (e.g., air-soap-air): Only the top surface reflection inverts. Constructive: 2nt = (m + 1/2) lambda; destructive: 2nt = m lambda.
- Two phase inversions (e.g., air-film-glass with n_film < n_glass): Both surfaces invert, so inversions cancel. Constructive: 2nt = m lambda; destructive: 2nt = (m + 1/2) lambda.
- Antireflection coating: A coating with index between air and glass causes one phase inversion. Thickness t = lambda / (4n) produces destructive interference in reflection, minimizing glare.
A soap film (n = 1.33) in air has one phase inversion at the top surface. What minimum thickness gives constructive reflection for 532 nm light? (2nt = lambda/2, so t = lambda/(4n) = 532/(4 x 1.33) = 100 nm)
| Phase inversions | Constructive condition | Destructive condition | Example |
|---|
| One (air-film-air) | 2nt = (m+1/2) lambda | 2nt = m lambda | Soap bubble |
| Two (air-film-glass, n_film < n_glass) | 2nt = m lambda | 2nt = (m+1/2) lambda | Antireflection coating |
| Zero (film on lower-n substrate) | 2nt = m lambda | 2nt = (m+1/2) lambda | Oil on water (rare geometry) |
Practice AP Physics 2 unit 14 questions
Try stimulus-based AP practice questions and written prompts after you review the notes.
A graph shows the observed frequency of a sound source as a function of time as recorded by a stationary observer. The source moves along a straight road that passes directly by the observer. The source emits sound at a constant rest frequency f₀, indicated by a dashed line on the graph.
QuestionWhich of the following correctly compares the observed frequency at time t₁ (before the source passes) to the observed frequency at time t₃ (after the source passes), and explains what occurs at time t₂?
f(t₁) > f₀ > f(t₃); at t₂ the source is at the point of closest approach and the observed frequency rapidly shifts from above f₀ to below f₀.
f(t₁) < f₀ < f(t₃); at t₂ the source is at the point of closest approach and the observed frequency rapidly shifts from below f₀ to above f₀.
f(t₁) > f₀ > f(t₃); at t₂ the observed frequency equals zero because the source is moving perpendicular to the observer.
f(t₁) = f(t₃) = f₀; at t₂ the observed frequency is at its maximum because the source is closest to the observer.
A thin film of oil (n = 1.45) floats on water (n = 1.33). The figure shows the cross-section of the film with incident light rays and the two reflected rays labeled R1 and R2. The film thickness is t.
QuestionWhich of the following correctly describes the phase shifts experienced by rays R1 and R2 at their respective interfaces, and what does this imply about the condition for constructive interference?
R1 undergoes a 180° phase shift; R2 undergoes no phase shift. Constructive interference requires 2t = (m + ½)λ_film.
R1 undergoes no phase shift; R2 undergoes a 180° phase shift. Constructive interference requires 2t = mλ_film.
Both R1 and R2 undergo 180° phase shifts. Constructive interference requires 2t = mλ_film.
Neither R1 nor R2 undergoes a phase shift. Constructive interference requires 2t = (m + ½)λ_film.
4. A uniform, transparent soap film of thickness t=450 nm and index of refraction nf=1.33 is suspended in air, as shown in Figure 1. Monochromatic light in air is incident normally on the film. The reflected light from the top surface of the film and the reflected light from the bottom surface of the film interfere and determine whether the film appears bright or dark in reflection for a given wavelength. Assume the film is nonabsorbing and that the index of refraction of air is na=1.00.
Figure 1. Normal-incidence reflection from a uniform soap film in air showing the two reflected rays (top-surface reflection and bottom-surface reflection after a round trip inside the film).
Indicate whether the student's claim is correct or incorrect. Without manipulating equations, justify your answer by referencing (i) the relative phase changes that occur upon reflection at each boundary and (ii) the relative number of wavelengths that fit into the round-trip path inside the film for each wavelength.
3. In Experiment 1, a student uses a laser pointer to produce an interference pattern on a screen with a double-slit. The student is asked to design an experiment to determine the wavelength of the laser light and then use that wavelength to predict the appearance of reflected light from a thin soap film.
Figure 1. Double-slit interference setup for measuring laser wavelength
Figure 2. Graph grid for linear plot to determine laser wavelength
Fringe order, m | Distance from central bright fringe to mth bright fringe, y (m) |
|---|
1 | 0.0060 |
2 | 0.0121 |
3 | 0.0181 |
4 | 0.0242 |
5 | 0.0302 |
i. Indicate two quantities, either measured quantities from Table 1 or additional calculated quantities, that could be graphed to produce a straight line that could be used to determine λ.
Vertical axis: Horizontal axis:
ii. On Figure 2, create a graph of the quantities indicated in part C(i) that can be used to determine λ.
• Use Table 2 to record the data points or calculated quantities that you will plot.
• Clearly label the axes, including units as appropriate.
• Plot the points you recorded in Table 2.
iii. Draw a best-fit line for the data graphed in part C(ii).
2. A student investigates wave behavior using a ripple tank and a monochromatic laser. A mechanical wave in the ripple tank travels across the surface of water toward boundaries and openings. The student can also shine a laser through an adjustable slit and onto a screen. In a later trial, the student places a thin, transparent soap film directly over the slit opening so the laser light passes through the film immediately after passing through the slit. The laser has vacuum wavelength λ0=532 nm.
Figure 1. Ripple-tank pulse reflection at two boundary types and a laser single-slit setup (with optional soap film over the slit).
(i) On Figure 1, consider the rigid wall (fixed end). Draw the reflected pulse on the left side of the wall at an instant just after reflection.
(ii) On Figure 1, consider the flexible absorbing barrier that behaves like a free end for the surface disturbance. Draw the reflected pulse on the left side of the barrier at an instant just after reflection.
In each case, your drawing must indicate the direction of motion of the reflected pulse and the pulse orientation (inverted or not inverted).
Figure 2. Snapshot of a periodic transverse surface wave: displacement versus position, with wavelength and amplitude labeled.
Derive an expression for the period T of the wave in terms of v and the wavelength λ, and then determine the numerical value of T using the data in Figure 2. Begin your derivation by writing a fundamental relationship for periodic waves.
Figure 3. Axes for sketching single-slit diffraction intensity I(y) on a screen; y = 0 at the central maximum.
Use the small-angle approximation and the single-slit minimum condition. Clearly indicate y=0 and the locations of the first minima on your sketch.
Constructive in reflection
Destructive in reflection
Briefly justify your answer by using the phase changes upon reflection and the path difference through the film. If you use a condition for interference, state it clearly.