AP Physics 2 • Waves, Sound, and Physical Optics

Diffraction

How waves bend around obstacles and spread through openings, revealing the wave nature of light.

An atom with wave and electric energy symbols.
Inside this lesson 9 sections 5 questions 7 figures

SECTION 1

Historical Context & Motivation

For centuries, the nature of light was fiercely debated. Isaac Newton championed a corpuscular (particle) model, which neatly explained reflection and straight-line propagation but struggled to account for phenomena where light appeared to bend around corners or spread after passing through narrow openings. Diffraction — the bending and spreading of waves when they encounter obstacles or apertures — became a decisive piece of evidence for the wave theory of light, fundamentally reshaping our understanding of optics.

1665 • Grimaldi Observes Diffraction

Francesco Maria Grimaldi noticed that light passing through a small aperture produced a wider illumination pattern than geometric shadow predicted, coining the term diffractio ("breaking apart").

1678 • Huygens' Wave Principle

Christiaan Huygens proposed that every point on a wavefront acts as a source of secondary wavelets, providing a geometric framework that naturally predicted diffraction.

1801 • Young's Double-Slit Experiment

Thomas Young demonstrated interference fringes from two closely spaced slits, powerfully confirming the wave nature of light and showing that diffraction from each slit was essential to the result.

1818 • Fresnel's Wave Theory

Augustin-Jean Fresnel combined Huygens' wavelets with the principle of interference to produce a rigorous mathematical theory of diffraction that accurately predicted observed patterns.

The central question that diffraction addresses is deceptively simple: why doesn't light always travel in perfectly straight lines? The answer reveals that light is a wave, and all waves share the property of spreading when they interact with objects whose dimensions are comparable to their wavelength. Understanding diffraction is essential for AP Physics 2, as it connects the broader principles of wave behavior — superposition, interference, and Huygens' principle — to experimentally observable patterns on screens and detectors.

SECTION 2

Core Principles & Definitions

Diffraction occurs whenever a wave encounters an obstacle or opening whose size is on the order of the wave's wavelength. The effect is universal — it applies to sound, water waves, and all forms of electromagnetic radiation. Several foundational ideas govern how and when diffraction is observed.

1

Huygens' Principle

Every point on a wavefront serves as a source of spherical secondary wavelets. The new wavefront at a later time is the envelope (tangent surface) of all these wavelets. This geometric construction naturally produces bending at edges.
2

Wavelength-to-Aperture Ratio

Diffraction is most pronounced when the slit width or obstacle size is comparable to the wavelength (a ≈ λ). When a ≫ λ, waves travel nearly in straight lines; when a ≈ λ, waves spread broadly.
3

Single-Slit Diffraction

A single narrow slit produces a central bright maximum flanked by alternating dark and bright fringes. The central maximum is twice as wide as the secondary maxima, and the minima locations are governed by a sin θ = mλ.
4

Superposition & Interference

Diffraction patterns arise from the interference of many Huygens wavelets. Constructive interference creates bright regions; destructive interference creates dark regions (minima). Path-length differences determine which occurs.

SECTION 3

Visualizing Single-Slit Diffraction

Single-Slit Diffraction GeometryIncident plane wavesaScreenCentral maxθλ = wavelengtha = slit widthθ = angle to minimum
Plane waves (violet) arrive at a narrow slit of width a. After passing through, Huygens wavelets spread out (gold dashed lines). On the distant screen, an intensity pattern appears with a broad central maximum flanked by weaker secondary maxima. The angle θ is measured from the central axis to a dark minimum.

In the diagram above, notice how the central bright fringe is significantly wider than the secondary maxima on either side. This asymmetry is a hallmark of single-slit diffraction: the central maximum spans an angular width of 2λ/a (measured between the first minima on each side), while each secondary maximum is roughly half as wide. The intensity of the secondary maxima also drops off rapidly — the first secondary maximum has only about 4.5% of the central maximum's intensity. As the slit narrows (smaller a), the pattern spreads further, consistent with the idea that diffraction is most significant when the aperture is comparable to the wavelength.

SECTION 4

Mathematical Framework

The AP Physics 2 exam focuses primarily on locating the dark fringes (minima) in a single-slit diffraction pattern. The condition for destructive interference arises when pairs of Huygens wavelets from different parts of the slit cancel each other. Imagine dividing the slit into two equal halves: if the path-length difference between a wavelet from the top of the upper half and the corresponding wavelet from the top of the lower half equals λ/2, destructive interference occurs. This geometrical argument generalizes to the following equation.

SINGLE-SLIT MINIMA CONDITION

asinθ=mλ(m=±1,±2,±3,…)a sin θ = mλ (m = ±1, ±2, ±3, …)

a = slit width (m), θ = angle from the central axis to the m-th minimum, m = order number (nonzero integer), λ = wavelength (m). Note: m = 0 corresponds to the central maximum, not a minimum.

For small angles (θ < ~10°), the small-angle approximation sin θ ≈ tan θ ≈ y/L applies, where y is the distance from the center of the pattern to the minimum on the screen and L is the distance from the slit to the screen. This yields a convenient linear form for locating minima.

LINEAR POSITION OF MINIMA (SMALL ANGLE)

ym=mλL/ay_m = mλL / a

ym = distance from center of pattern to the m-th dark fringe on the screen, L = slit-to-screen distance (m). Valid when y ≪ L.

WIDTH OF CENTRAL MAXIMUM

w=2λL/aw = 2λL / a

w = full width of the central bright fringe measured between the first minima on either side. Since the first minima occur at m = ±1, the total spread is 2y₁ = 2λL/a.

ℹ
AP Exam Tip

SECTION 5

Diffraction Patterns & Classifications

Diffraction phenomena are broadly classified into two regimes depending on the geometry of the source, obstacle, and observation point. Fraunhofer (far-field) diffraction occurs when both the incoming waves and the observation screen are effectively at infinite distance from the aperture — achieved in practice by using plane waves and placing the screen far away (or using lenses). Fresnel (near-field) diffraction occurs when the source or screen is close to the aperture so that wavefront curvature matters. AP Physics 2 focuses almost exclusively on the Fraunhofer regime, which yields cleaner, more symmetric patterns amenable to the equations in Section 4.

Intensity Patterns: Narrow vs. Wide SlitNarrow Slit (a ≈ λ)Position on screenIWide Slit (a ≫ λ)Position on screenIComparison SummaryNarrow Slit (a ≈ λ)Wide Slit (a ≫ λ)Broad central maxNarrow central maxWide angular spreadMinimal angular spreadSecondary maxima visibleNearly geometric shadowθ₁ = arcsin(λ/a) is largeθ₁ = arcsin(λ/a) ≈ 0
Comparison of intensity distributions for a narrow slit (left, a ≈ λ) versus a wide slit (right, a ≫ λ). The narrow slit produces a broad diffraction pattern with clearly separated secondary maxima; the wide slit yields a sharp, nearly geometric image.

The key relationship to internalize is inverse: as slit width a decreases, the angular spread of the diffraction pattern increases. This is because sin θ₁ = λ/a, so a smaller a yields a larger θ₁. Conversely, increasing the wavelength λ while keeping a constant also broadens the pattern. This is why red light (λ ≈ 700 nm) produces wider diffraction patterns than blue light (λ ≈ 450 nm) through the same slit.

SECTION 6

Worked Example

Worked example

  1. 1

    Step 1 — Identify Given Values

    Monochromatic light of wavelength λ = 550 nm passes through a single slit of width a = 0.10 mm. A screen is placed L = 2.0 m from the slit. Find the width of the central bright maximum on the screen.
  2. 2

    Step 2 — Convert Units

    λ = 550 nm = 550 × 10⁻⁹ m = 5.50 × 10⁻⁷ m. a = 0.10 mm = 1.0 × 10⁻⁴ m. L = 2.0 m.
  3. 3

    Step 3 — Select the Appropriate Equation

    The width of the central maximum is the distance between the first minima on either side. Using the small-angle formula, the full width is w = 2λL / a.
  4. 4

    Step 4 — Substitute and Calculate

    w = 2 × (5.50 × 10⁻⁷ m) × (2.0 m) / (1.0 × 10⁻⁴ m) = 2 × (1.10 × 10⁻⁶) / (1.0 × 10⁻⁴) = 2.2 × 10⁻² m.
    w = 2.2 × 10⁻² m = 2.2 cm
  5. 5

    Step 5 — Interpret the Result

    The central bright fringe is about 2.2 cm wide, easily measurable with a ruler. Notice the small-angle approximation is valid: sin θ₁ = λ/a = 5.50 × 10⁻³, which is much less than 1. If the slit were narrower, the central maximum would spread further; if it were wider, it would shrink.

SECTION 7

Single-Slit vs. Double-Slit Patterns

Students often conflate single-slit diffraction with double-slit interference. While both produce fringe patterns, they arise from different physical setups and obey different equations. A double-slit experiment actually combines both effects: each slit individually diffracts light, and the overlapping diffracted waves from the two slits then interfere, creating a fine interference pattern modulated by the broader single-slit diffraction envelope.

FeatureSingle-Slit DiffractionDouble-Slit Interference
Key equationa sin θ = mλ (minima)d sin θ = mλ (maxima)
Parametera = slit widthd = slit separation
Equation gives...Dark fringesBright fringes
Central maximumTwice as wide as secondary maximaSame width as all other maxima
Intensity distributionRapidly decreasing secondary maximaRoughly equal bright fringes (modulated by single-slit envelope)
Effect of narrowing slit/decreasing dPattern broadensFringes spread apart

Comparison of single-slit diffraction and double-slit interference

SECTION 8

Connections to Diffraction Gratings & Resolution

Single-slit diffraction is the foundational case, but extending the concept to multiple slits leads to the diffraction grating — a device with hundreds or thousands of equally spaced slits. A grating produces extremely sharp, bright maxima at the same angles given by d sin θ = mλ, but with much higher angular resolution than a double slit because constructive interference from many slits is far more selective. Diffraction gratings are used in spectrometers to separate light into its component wavelengths with high precision.

ConceptAP Physics 2 LevelAdvanced / College Physics
Single slitLocate minima using a sin θ = mλFull intensity function I(θ) using sinc² function
Double slitLocate maxima using d sin θ = mλCombined intensity: interference × diffraction envelope
Diffraction gratingSame maxima equation; sharper peaksResolving power R = mN; spectral analysis
Circular apertureQualitative: Rayleigh criterion conceptθ_min = 1.22λ/D; telescope resolution limits

Diffraction topics at AP level vs. advanced physics

Another important extension is diffraction through a circular aperture, which produces a circular pattern of rings called an Airy disk. The Rayleigh criterion states that two point sources are just resolved when the central maximum of one Airy pattern falls on the first minimum of the other: θ_min = 1.22λ/D, where D is the aperture diameter. This principle sets the fundamental resolution limit for telescopes, microscopes, and cameras — a direct, practical consequence of diffraction.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A beam of monochromatic light passes through a single slit. If the slit width is decreased while the wavelength remains constant, what happens to the central bright maximum on the screen? A) It becomes narrower and brighter. B) It becomes wider and dimmer. C) It remains the same width but becomes dimmer. D) It becomes wider and brighter.
  • A. It becomes narrower and brighter.
  • B. It becomes wider and dimmer.
  • C. It remains the same width but becomes dimmer.
  • D. It becomes wider and brighter.
Show answer (B)
B is correct. The width of the central maximum is w = 2λL/a. Decreasing a increases w, so the pattern spreads out. Additionally, a narrower slit admits less total light energy, so the peak intensity decreases. A is wrong because a smaller slit produces a wider (not narrower) pattern. C is wrong because the width does change — it increases. D is wrong because while the pattern widens, less light passes through so it cannot become brighter.

Question 2: BASIC CALCULATION

Light of wavelength 600 nm passes through a single slit of width 0.20 mm and hits a screen 1.5 m away. What is the distance from the center of the pattern to the first dark fringe? A) 2.25 mm B) 4.50 mm C) 9.00 mm D) 0.45 mm
  • A. 2.25 mm
  • B. 4.50 mm
  • C. 9.00 mm
  • D. 0.45 mm
Show answer (B)
B is correct. Using y₁ = mλL/a with m = 1: y₁ = (1)(600 × 10⁻⁹ m)(1.5 m) / (2.0 × 10⁻⁴ m) = 9.0 × 10⁻⁴ / 2.0 × 10⁻⁴ = 4.5 × 10⁻³ m = 4.50 mm. A (2.25 mm) results from using half the correct formula (missing the L or using w/2 incorrectly). C (9.00 mm) is the full width of the central maximum (2y₁), not the distance to the first minimum. D (0.45 mm) results from a unit conversion error (dividing by 2.0 × 10⁻³ instead of 2.0 × 10⁻⁴).

Question 3: INTERMEDIATE

In a single-slit experiment, the second-order minimum for light of wavelength 500 nm occurs at angle θ = 1.43°. What is the slit width? A) 2.0 × 10⁻⁵ m B) 4.0 × 10⁻⁵ m C) 1.0 × 10⁻⁴ m D) 5.0 × 10⁻⁵ m
  • A. 2.0 × 10⁻⁵ m
  • B. 4.0 × 10⁻⁵ m
  • C. 1.0 × 10⁻⁴ m
  • D. 5.0 × 10⁻⁵ m
Show answer (B)
B is correct. From a sin θ = mλ, we solve for a: a = mλ / sin θ = (2)(500 × 10⁻⁹ m) / sin(1.43°). sin(1.43°) ≈ 0.02496. So a = 1.0 × 10⁻⁶ / 0.02496 ≈ 4.0 × 10⁻⁵ m. A results from using m = 1 instead of m = 2. C results from using m = 5 or an arithmetic error doubling the answer. D would correspond to m = 2.5, a non-integer order, which is physically meaningless for minima.

Question 4: APPLIED

A student performs a single-slit diffraction experiment to determine the wavelength of a laser. The slit width is a = 0.080 mm and the screen is L = 3.0 m from the slit. The student measures the distance between the first dark fringes on either side of the central maximum to be 4.2 cm. (a) Calculate the wavelength of the laser light. (b) The student replaces the slit with one that is half as wide. Predict how the distance between the first-order minima changes. Justify your answer. (c) If the student then uses the original slit but moves the screen to L = 1.5 m, what will be the new distance between the first dark fringes?
Show answer
(a) The distance between the first minima on either side is 2y₁ = 4.2 cm, so y₁ = 2.1 cm = 0.021 m. Using y₁ = λL/a → λ = y₁a/L = (0.021)(8.0 × 10⁻⁵) / 3.0 = 5.6 × 10⁻⁷ m = 560 nm. (b) Since w = 2λL/a, halving a doubles w. The new distance between first-order minima is 8.4 cm. This is because a narrower slit causes more diffraction, spreading the pattern further. (c) With L = 1.5 m and original a: 2y₁ = 2(560 × 10⁻⁹)(1.5)/(8.0 × 10⁻⁵) = 0.021 m = 2.1 cm. Halving L halves the pattern width.

Question 5: CRITICAL THINKING

Design an experiment using a laser pointer, an adjustable single slit, a ruler, and a screen to determine how the width of the central diffraction maximum depends on slit width. Your response should include: (a) A description of the experimental procedure, including what quantities are measured and how. (b) Identification of the independent, dependent, and at least two controlled variables. (c) A description of how to analyze the data graphically to confirm the expected relationship. (d) A discussion of one significant source of uncertainty and how it could be reduced.
Show answer
(a) Set up the laser so it passes through the adjustable slit and forms a diffraction pattern on a distant screen. Measure and record the slit-to-screen distance L. For at least five different slit widths, measure the distance between the two first-order dark fringes on either side of the central maximum; this distance equals the central maximum width w. (b) Independent variable: slit width a. Dependent variable: central maximum width w. Controlled variables: laser wavelength λ, slit-to-screen distance L, room darkness, and slit orientation. (c) Plot w on the vertical axis versus 1/a on the horizontal axis. If diffraction theory holds (w = 2λL/a), the data should form a straight line through the origin with slope equal to 2λL. A linear fit confirms the inverse relationship between w and a. (d) A significant source of uncertainty is determining the exact position of the dark fringes, since the intensity decreases gradually rather than dropping to zero at a sharp point. This can be reduced by using a light sensor (photodiode) mounted on a translation stage to locate the intensity minima quantitatively rather than relying on visual judgment.

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