Unit 4 · Lesson 2a

2aDiffraction

Discover how waves bend around obstacles and through openings, and apply single-slit and grating equations to predict diffraction patterns.

Diffraction reveals the wave nature of light and is the key principle behind spectroscopy, X-ray crystallography, and the colorful patterns on CDs — tools that underpin chemistry, biology, and materials science.

Lesson Overview

Diffraction is the bending and spreading of waves around obstacles or through openings. In this lesson you will explore single-slit diffraction, diffraction gratings, and the condition for minima. You will also see how diffraction is applied in X-ray crystallography and the colorful patterns on CDs.

Key Concepts

Diffraction

The bending and spreading of waves when they pass through a narrow opening or around an obstacle; most noticeable when the opening size is comparable to the wavelength.

Single-Slit Diffraction

A single slit of width a produces a central bright maximum flanked by dark minima. Minima occur at a sin θ = mλ (m = ±1, ±2, …).

Condition for Minima

a sin θ = mλ, where a is slit width, θ is the angle to the minimum, λ is wavelength, and m is a non-zero integer.

Diffraction Grating

Many equally spaced slits (spacing d); produces sharp, bright maxima at d sin θ = mλ. Used to separate wavelengths of light.

X-ray Crystallography

X-rays diffract off atomic planes in crystals; the pattern reveals atomic spacing and molecular structure (e.g., DNA double helix).

CDs and DVDs

The closely spaced tracks on a CD act as a diffraction grating, separating white light into a rainbow of colors.

Example 1

Light of wavelength 600 nm passes through a single slit of width 0.10 mm. Find the angle of the first dark minimum.

Answer:a sin θ = mλ → sin θ = mλ/a = (1 × 600 × 10⁻⁹)/(0.10 × 10⁻³) = 6.0 × 10⁻³ → θ = sin⁻¹(6.0 × 10⁻³) ≈ 0.34°. The first dark fringe is very close to the center because the slit is much wider than the wavelength.
Example 2

A single slit of width 0.025 mm is illuminated with 500 nm light. Find the angle of the second dark minimum.

Answer:a sin θ = mλ → sin θ = 2 × (500 × 10⁻⁹)/(0.025 × 10⁻³) = 1000 × 10⁻⁹/25 × 10⁻⁶ = 0.040 → θ = sin⁻¹(0.040) ≈ 2.3°.
Example 3

A diffraction grating has 500 lines/mm. Find the angle of the first-order maximum for 550 nm light.

Answer:Grating spacing d = 1/500 mm = 2.0 × 10⁻³ mm = 2.0 × 10⁻⁶ m. d sin θ = mλ → sin θ = (1 × 550 × 10⁻⁹)/(2.0 × 10⁻⁶) = 0.275 → θ = sin⁻¹(0.275) ≈ 15.96° ≈ 16.0°.
Example 4

Using the grating from Example 3, find the angle of the second-order maximum for 550 nm light.

Answer:sin θ = mλ/d = (2 × 550 × 10⁻⁹)/(2.0 × 10⁻⁶) = 0.550 → θ = sin⁻¹(0.550) ≈ 33.4°.
Example 5

A diffraction grating produces a first-order maximum at 20° for light of wavelength 480 nm. How many lines per mm does the grating have?

Answer:d sin θ = mλ → d = mλ/sin θ = (1 × 480 × 10⁻⁹)/sin 20° = 480 × 10⁻⁹/0.342 = 1.404 × 10⁻⁶ m. Lines/mm = 1/(1.404 × 10⁻³ mm) ≈ 712 lines/mm.
Guided Problem 1

Light of wavelength 700 nm passes through a 0.050 mm slit. Find the angle of the first dark minimum.

Hint: Use a sin θ = mλ with m = 1. Convert all lengths to the same unit (meters) before dividing.

Guided Problem 2

Why does diffraction become more noticeable as the slit width decreases?

Hint: Compare the ratio λ/a. What happens to sin θ (and therefore θ) as a gets smaller?

Guided Problem 3

A diffraction grating has 300 lines/mm. What is the grating spacing d in meters?

Hint: d = 1/(lines per meter). Convert 300 lines/mm to lines/m first.

Guided Problem 4

A grating with d = 1.5 × 10⁻⁶ m is illuminated with 600 nm light. Find the maximum order m that can be observed.

Hint: The maximum order occurs when sin θ = 1 (θ = 90°). Solve mλ/d ≤ 1 for m.

Guided Problem 5

Explain why X-rays (λ ≈ 0.1 nm) rather than visible light are used to study crystal structure.

Hint: Compare the wavelength of X-rays to the spacing between atoms in a crystal (~0.1–0.3 nm).

Key Vocabulary

Diffraction

The bending and spreading of waves as they pass through an opening or around an obstacle; most pronounced when the opening is comparable in size to the wavelength.

Example: Sound diffracts around corners easily because its wavelength (cm to m) is comparable to everyday obstacles.

Single-Slit Diffraction

The diffraction pattern produced by a single narrow slit; characterized by a wide central maximum and narrower secondary maxima separated by dark minima at a sin θ = mλ.

Example: Shining a laser through a 0.1 mm slit produces a wide bright central band with dark fringes on either side.

Diffraction Grating

An optical element with many equally spaced slits or grooves; produces sharp, bright maxima at d sin θ = mλ and is used to separate (disperse) wavelengths.

Example: Spectroscopes use diffraction gratings to separate the colors in starlight and identify chemical elements.

Order (m)

An integer (0, ±1, ±2, …) labeling each bright maximum in a diffraction pattern; m = 0 is the central maximum.

Example: The first-order maximum (m = 1) appears at a smaller angle than the second-order maximum (m = 2).

Interactive Practice — 5 Questions

1

The condition for dark minima in single-slit diffraction is:

2

A diffraction grating with 400 lines/mm has grating spacing d equal to:

3

Diffraction is most noticeable when:

4

A grating (d = 2.0 × 10⁻⁶ m) gives a first-order maximum at 15° for a certain wavelength. What is λ?

5

X-ray crystallography works because:

Independent Practice

1

Light of wavelength 450 nm passes through a single slit of width 0.030 mm. Find the angles of the first and second dark minima.

2

A diffraction grating has 600 lines/mm. Find the angle of the first-order maximum for red light (λ = 700 nm) and violet light (λ = 400 nm).

3

Explain why a narrower slit produces a wider central diffraction maximum.

4

A CD has track spacing of about 1.6 μm. For what angle does first-order diffraction occur for green light (λ = 530 nm)?

5

★ A diffraction grating produces second-order maxima for two wavelengths at 30° and 45° respectively. Find both wavelengths if d = 2.4 × 10⁻⁶ m. Then find the angle separation between their third-order maxima.

Challenge
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Common Mistakes

Using the grating equation d sin θ = mλ for single-slit minima.

Single-slit minima use a sin θ = mλ (slit width a). The grating equation uses slit spacing d for maxima.

Forgetting to convert nm to meters before substituting into the diffraction equations.

Always convert: 1 nm = 1 × 10⁻⁹ m. Mixing nm and mm in the same equation gives wrong answers.

Thinking m = 0 gives a dark minimum in single-slit diffraction.

m = 0 is excluded from the minima condition; it corresponds to the bright central maximum.

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Math Tips

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For diffraction gratings, find d by taking the reciprocal of the line density: if the grating has N lines/mm, then d = 1/N mm. Convert to meters: d (m) = 1/(N × 10³).

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The maximum possible order is m_max = floor(d/λ). Any m that gives sin θ > 1 is physically impossible — check this before reporting an answer.