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Unit 4 · Lesson 2b

2bInterference and Young's Double Slit

Analyze constructive and destructive interference, calculate fringe positions in Young's experiment, and explore thin film interference in everyday phenomena.

Interference is the definitive proof that light is a wave. It underlies anti-reflection coatings on camera lenses, the colors of soap bubbles, and precision measurement tools used in engineering and astronomy.

Lesson Overview

When two or more waves overlap they interfere — adding together (constructive) or canceling (destructive). In this lesson you will analyze Young's double-slit experiment, calculate fringe positions using path difference conditions, explore thin film interference, and connect these ideas to real-world applications like anti-reflection coatings and optical instruments.

Key Concepts

Young's Double-Slit Experiment

Two coherent slits separated by distance d produce alternating bright and dark fringes on a screen, demonstrating the wave nature of light.

Constructive Interference

Bright fringes occur when path difference = mλ: d sin θ = mλ (m = 0, ±1, ±2, …).

Destructive Interference

Dark fringes occur when path difference = (m + ½)λ: d sin θ = (m + ½)λ.

Fringe Spacing

y = mλL/d gives the position of the m-th bright fringe on a screen at distance L. Fringe spacing Δy = λL/d.

Thin Film Interference

Light reflecting from the top and bottom surfaces of a thin film can interfere constructively or destructively depending on film thickness and phase shifts at boundaries.

Applications

Anti-reflection coatings (destructive interference eliminates glare), soap bubbles (colorful patterns), optical flats (testing surface flatness).

Example 1

In a double-slit experiment, d = 0.20 mm, λ = 550 nm, and the screen is 1.5 m away. Find the fringe spacing Δy.

Answer:Δy = λL/d = (550 × 10⁻⁹ × 1.5)/(0.20 × 10⁻³) = 825 × 10⁻⁹/2.0 × 10⁻⁴ = 4.125 × 10⁻³ m ≈ 4.1 mm. Bright fringes are spaced 4.1 mm apart.
Example 2

Using the setup from Example 1, find the angle to the third bright fringe (m = 3).

Answer:d sin θ = mλ → sin θ = mλ/d = (3 × 550 × 10⁻⁹)/(0.20 × 10⁻³) = 1650 × 10⁻⁹/2.0 × 10⁻⁴ = 8.25 × 10⁻³ → θ = sin⁻¹(8.25 × 10⁻³) ≈ 0.47°.
Example 3

Find the angle to the second dark fringe (m = 1, counting from m = 0 as the first dark fringe) in the same setup.

Answer:d sin θ = (m + ½)λ with m = 1 → sin θ = (1.5 × 550 × 10⁻⁹)/(0.20 × 10⁻³) = 825 × 10⁻⁹/2.0 × 10⁻⁴ = 4.125 × 10⁻³ → θ ≈ 0.24°.
Example 4

A double-slit experiment produces bright fringes spaced 3.0 mm apart on a screen 2.0 m away using 600 nm light. Find the slit separation d.

Answer:Δy = λL/d → d = λL/Δy = (600 × 10⁻⁹ × 2.0)/(3.0 × 10⁻³) = 1200 × 10⁻⁹/3.0 × 10⁻³ = 4.0 × 10⁻⁴ m = 0.40 mm.
Example 5

A thin film of oil (n = 1.45) on water (n = 1.33) has thickness 200 nm. Does 580 nm light reflect constructively or destructively? (Assume one phase shift at the air–oil boundary.)

Answer:Optical path difference = 2nt = 2 × 1.45 × 200 nm = 580 nm = λ. With one phase shift (half-wavelength), the condition for constructive reflection is 2nt = (m + ½)λ. Here 2nt = λ = 1λ, which equals (½ + ½)λ = 1λ for m = 0 — this satisfies the constructive condition. The film appears bright for 580 nm light.
Guided Problem 1

In a double-slit experiment with d = 0.15 mm, λ = 480 nm, and L = 2.0 m, find the fringe spacing.

Hint: Use Δy = λL/d. Make sure all units are in meters before calculating.

Guided Problem 2

What happens to the fringe spacing if the slit separation d is doubled while everything else stays the same?

Hint: Look at the formula Δy = λL/d. How does Δy change when d → 2d?

Guided Problem 3

Two slits are separated by 0.30 mm. Light of wavelength 500 nm creates fringes on a screen. At what angle does the fourth bright fringe appear?

Hint: Use d sin θ = mλ with m = 4.

Guided Problem 4

Explain why the double-slit experiment requires coherent light sources (same wavelength, constant phase difference).

Hint: Think about what happens to the interference pattern if the phase difference between the two sources keeps changing randomly.

Guided Problem 5

A soap film in air has n = 1.35. What minimum thickness produces constructive reflection for 540 nm light? (Two phase shifts cancel.)

Hint: With two phase shifts (at both surfaces), constructive reflection requires 2nt = mλ. Use m = 1 for minimum thickness.

Key Vocabulary

Constructive Interference

Occurs when two waves are in phase (path difference = mλ); their amplitudes add, producing a bright fringe.

Example: In Young's experiment, bright fringes appear where the path difference from the two slits is a whole number of wavelengths.

Destructive Interference

Occurs when two waves are out of phase by half a wavelength (path difference = (m + ½)λ); they cancel, producing a dark fringe.

Example: Dark fringes in the double-slit pattern appear where the path difference is a half-integer multiple of the wavelength.

Path Difference

The difference in distance traveled by two waves from their sources to a given point; determines whether interference is constructive or destructive.

Example: If one wave travels 3.5λ and another travels 3.0λ to the same point, the path difference is 0.5λ — destructive interference.

Thin Film Interference

Interference between light reflected from the top and bottom surfaces of a thin transparent film; produces colorful patterns depending on film thickness and wavelength.

Example: The rainbow colors on a soap bubble result from thin film interference as different wavelengths constructively interfere at different thicknesses.

Interactive Practice — 5 Questions

1

Constructive interference in a double-slit experiment occurs when the path difference equals:

2

In a double-slit experiment, increasing the slit separation d will:

3

Dark fringes in Young's experiment occur at:

4

A double-slit setup has d = 0.25 mm, λ = 500 nm, L = 1.0 m. What is the fringe spacing?

5

Thin film interference is used in anti-reflection coatings to:

Independent Practice

1

In a double-slit experiment, d = 0.10 mm, λ = 630 nm, L = 2.5 m. Find (a) the fringe spacing and (b) the position of the third bright fringe from the center.

2

A double-slit experiment produces fringes 2.5 mm apart on a screen 1.8 m away. If λ = 450 nm, find the slit separation d.

3

Explain the difference between constructive and destructive interference using the concept of path difference.

4

A soap film (n = 1.33) in air has thickness 300 nm. For what visible wavelength does constructive reflection occur? (Assume two phase shifts cancel, so condition is 2nt = mλ, m = 1.)

5

★ In a double-slit experiment, the 4th bright fringe (m = 4) appears at y = 8.0 mm on a screen 1.2 m away. If λ = 500 nm, find d. Then find the angle to the first dark fringe and the fringe spacing.

Challenge
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Common Mistakes

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Using the single-slit minima formula (a sin θ = mλ) for double-slit bright fringes.

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Double-slit bright fringes use d sin θ = mλ (slit separation d). Single-slit minima use slit width a.

❌

Forgetting that m = 0 is the central bright fringe, not a dark fringe.

✅

The central maximum is at m = 0 (zero path difference). Dark fringes start at m = 0 in the destructive condition: d sin θ = (0 + ½)λ.

❌

Mixing up the thin film conditions when there are 0, 1, or 2 phase shifts.

✅

Each reflection at a boundary going from low-n to high-n adds a half-wavelength phase shift. Count the number of such reflections (0, 1, or 2) to determine the constructive/destructive conditions.

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Math Tips

📌

For small angles (θ < 5°), sin θ ≈ tan θ ≈ y/L, so the fringe position formula y = mλL/d is a good approximation. For larger angles, use d sin θ = mλ exactly.

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Fringe spacing Δy = λL/d is constant across the pattern (for small angles). If you know any two of {Δy, λ, L, d}, you can find the third.