Diffraction
Connect opening size and path cancellation with single-slit diffraction patterns.
AP Physics 2 Topic 14.7 · College Board CED printed p. 122, 123 / PDF p. 129, 130 · See this topic in the course outline
A slit acts as a spread-out set of wave sources. Its parts can reinforce or cancel on a distant screen.
Before you start
- Path difference
- Wavelength
- Trigonometry
Edges spread waves; geometry sets the pattern
Diffraction is spreading around an obstacle or through an opening. It is most prominent when the opening size is comparable to wavelength. A doorway is comparable to some sound wavelengths, so sound reaches around it readily; ordinary visible wavelengths are tiny compared with the doorway, so light mainly travels straight except near edges.
Wavefront contributions from different locations across a single opening overlap and interfere. A long narrow slit produces bands mainly across its narrow dimension. A circular opening makes a circular central bright region and rings. The shape of the aperture matters, so do not apply a rectangular-slit equation to every opening.
Read dark bands from path cancellation
For monochromatic light of wavelength λ through a slit of width a, the edge-to-edge path difference toward angle θ is . At θ = 0 all parts are in phase, giving a broad central bright band. At the first minimum, Δ = λ: pairing points separated by a/2 gives λ/2 path differences, so each pair cancels. More generally the dark minima satisfy . There is no dark minimum at m = 0.
With screen distance L and |θ| below about 10°, . The central bright band's width between first minima is about 2λL/a; the side bands are narrower and generally dimmer. Single-slit bright side maxima are not exactly at integer multiples of λ in the edge-to-edge path difference. Narrowing a or increasing λ widens the pattern.
| Order m | y (mm) | Meaning |
|---|---|---|
| −2 | −10 | second dark minimum |
| −1 | −5 | first dark minimum |
| 0 | 0 | central maximum, not a minimum |
| 1 | 5 | first dark minimum |
| 2 | 10 | second dark minimum |
Plot an intensity curve with a broad central peak from −5 to +5 mm and smaller side peaks between zeros. The positions reveal λ/a; an absolute intensity scale also depends on the incident light and aperture.
Infer a slit width from dark bands
A 600 nm laser crosses a single slit. At L = 1.5 m the first dark minima are at y = ±4.5 mm. Find a and the central bright width.
|y|/L = 0.0030, so the angle is well below 10°. Use m = 1 for the first dark minimum.
Why this step? Verify the approximation and count the correct feature.
.
Why this step? The first-minimum path difference across the slit is one wavelength.
The central bright band extends from −4.5 mm to +4.5 mm, a width of 9.0 mm.
Why this step? The width includes both sides of the central maximum.
Answer and check
a = 0.20 mm and central width = 9.0 mm.
A single slit has many interfering contributions
One opening can make interference because different points across its width emit contributions that reach the same distant point by different paths. Project the edge-to-edge separation onto the outgoing direction. The right triangle gives Δ = a sin θ; this is a path difference between contributions from one opening, not a separation between two slits.
On the central axis Δ = 0 and all points reinforce. At the first dark minimum Δ = λ. Pair a point in the top half with a corresponding point in the bottom half, a/2 away: the pair has Δ/2 = λ/2 path difference, so the equal-amplitude contributions oppose. All such pairs cancel in the ideal uniformly illuminated far-field model. The diagram exposes the geometry that an intensity curve alone does not show.
Inspect diagram labels
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Use the projected difference to predict a dark band
A uniformly illuminated single slit has a = 0.200 mm; monochromatic λ = 500 nm light travels to a screen L = 2.00 m away in air. Use the far-field model.
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.
- At the first minimum, a sin θ = λ, so sin θ = (5.00 × 10⁻⁷)/(2.00 × 10⁻⁴) = 0.00250 and θ ≈ 0.143°.
- The angle is below 10°, so y₁ ≈ λL/a = 0.00500 m = 5.00 mm. The first minima are at ±5.00 mm and the central bright width is 10.0 mm.
- At that direction, edge-to-edge Δ = 500 nm. Half-width pairs have a 250 nm difference, half a cycle, and cancel. At the center all paths have equal length and reinforce.
Answer and check
The broad central maximum is bounded by first minima ±5.00 mm. Its darkness mechanism is pairwise cancellation across one opening, with a calculated angle 0.143°, distinct from the enlarged diagram angle.
Make the model respond
Focused Physics 2 model
Interference orders and standing waves
Predict first. Choose a model. Predict how wavelength changes bright orders or how harmonic number changes frequency.
Bright orders: d sin θₘ = mλ; exact screen y = L tan θₘ; small-angle estimate y ≈ Lmλ/d
- Exact finite screen positions
- Small-angle estimates
- Selected or reference point
On a narrow screen, scroll within a figure to read its labels. The table and explanation provide the same physics in text.
Order m = 1: θ = 14.48°, exact y = 0.5164 m; small-angle y ≈ 0.5 m.
Read the model data and comparisons
| Quantity | Value | Unit | Meaning or limit |
|---|---|---|---|
| Wavelength | 5.000 × 10^−7 | m | |
| Slit spacing | 2.000 × 10^−6 | m | |
| Screen distance | 2 | m | |
| Selected integer order | 1 | ||
| Required path difference mλ | 5.000 × 10^−7 | m | |
| Selected angle | 14.48 | ° | |
| Exact screen coordinate | 0.5164 | m | |
| Small-angle estimate | 0.5 | m | Only accurate when |mλ/d| ≪ 1 |
| Maximum allowed angular |m| | 4 | Equality gives grazing order, with no finite screen intersection | |
| Order -5 exact y | forbidden | m | |mλ/d| > 1 |
| Order -4 exact y | grazing | m | θ = -90° |
| Order -3 exact y | -2.268 | m | θ = -48.59° |
| Order -2 exact y | -1.155 | m | θ = -30° |
| Order -1 exact y | -0.5164 | m | θ = -14.48° |
| Order 0 exact y | 0 | m | θ = 0° |
| Order 1 exact y | 0.5164 | m | θ = 14.48° |
| Order 2 exact y | 1.155 | m | θ = 30° |
| Order 3 exact y | 2.268 | m | θ = 48.59° |
| Order 4 exact y | grazing | m | θ = 90° |
| Order 5 exact y | forbidden | m | |mλ/d| > 1 |
Two coherent slits produce constructive interference when the path difference is an integer wavelength. A bright-order number is an integer, not a continuous position. Convert the allowed angle to a flat-screen coordinate with tan θ. Using sin θ ≈ tan θ ≈ θ is a separate approximation and can noticeably underestimate positions at larger angles.
Assumptions and limits. Monochromatic coherent illumination, identical narrow slits, uniform center spacing, far-field interference, screen perpendicular to the central beam. The maximum-order condition uses exact trigonometry within this ideal far-field geometry. Finite slit width, missing orders from the diffraction envelope, brightness, peak width, and thin-film interference are not modeled.
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
With the same laser and screen distance, does halving slit width make the central peak narrower or wider?
Worked answer and reason
Twice as wide in the small-angle model, because first-minimum |y| = λL/a.
A smaller opening produces greater diffraction.
Question 2
Would a circular pinhole make the same long bright bands as a long rectangular slit?
Worked answer and reason
No. A circular aperture produces a circular diffraction pattern; aperture geometry changes how the contributions interfere.
Use the opening shape as part of the model.
JavaScript enables saved drafts and self-assessment. You can always work on paper.
Catch the tempting mistake
- The central dark minimum has m = 0.
- At zero path difference the slit contributions reinforce; the center is bright. Dark minima start at nonzero m.
Put it to work
Give the physical reason as well as the calculation. The worked solutions are on a separate page.
- Problem 13 · mixed
- Problem 14 · experimental
- Problem 20 · experimental
Scientific calculator workspace
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Scientific calculator
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For the teacher
- Have students pair the top and bottom halves of a slit at the first minimum. Distinguish single-slit minima from double-slit maxima.
Source objectives and scope
- 14.7.A — Describe the behavior of a wave and the diffraction pattern resulting from a wave passing through a single opening.
Suggested skills are teaching guidance. The course applies all science practices across its objectives.
Science references for this teaching
- AP Physics 2 Course and Exam Description, official Unit 14 — Official topic/LO/EK scope and boundaries; local pinned source pages 117–134, equations visually checked on 118,120,130,132,134.
- OpenStax University Physics 1, Energy and Power of a Wave — Verification of fixed-amplitude mechanical-wave frequency scaling and distinction from electromagnetic intensity.
- OpenStax University Physics 1, Interference of Waves — Linear superposition and pointwise addition model.
- OpenStax University Physics 3, Single-Slit Diffraction — Verification of aperture minima and small-angle conditions.
- OpenStax University Physics 3, Interference in Thin Films — Verification of reflection phase and normal-incidence thin-film paths.
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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Wavelength
Wavelength is the spatial distance between neighboring points at the same phase of a periodic wave. It connects propagation speed and frequency through speed equal to frequency times wavelength. For example, neighboring crests of a sinusoidal water wave are separated by one wavelength.
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Wavefront
A wavefront is a surface joining points of equal wave phase at a given time. Rays are perpendicular to wavefronts in the uniform isotropic ray model used here. For example, a distant source can be approximated by nearly plane wavefronts over a small region.
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Oscillation phase
Phase specifies the position within a repeating oscillation cycle. A phase difference tells how much one oscillation leads or lags another at a chosen place and time. For example, two sinusoidal waves half a cycle out of phase can cancel when their amplitudes are equal.
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Wave intensity
Wave intensity is the time-averaged energy-transfer rate per area perpendicular to propagation. It is measured in watts per square meter and differs from total power because area matters. For example, spreading the same sound power over a larger area reduces the intensity there.
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Wave interference
Wave interference is the combination of overlapping wave disturbances. In a linear medium, the signed disturbances add at each location and time without permanently destroying the passing waves. For example, two overlapping upward rope pulses temporarily make a larger upward disturbance.
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Amplitude
Amplitude is the maximum magnitude of a repeating disturbance measured from its equilibrium value. It describes the size of the oscillation rather than its frequency or propagation speed. For example, pulling a spring-mounted mass farther from equilibrium increases its oscillation amplitude when it is released from rest.