02Wave Optics and Interference
Analyze double-slit interference patterns, single-slit diffraction, and thin-film interference using the wave model of light.
Wave optics reveals the true nature of light. Interference and diffraction are used in anti-reflective coatings, holograms, and precision measurement instruments.
How does treating light as a wave explain the bright and dark bands seen when light passes through two narrow slits?
Lesson Overview
Geometric optics treats light as rays, but many phenomena — interference fringes, diffraction around obstacles, iridescent soap bubbles, and polarizing sunglasses — can only be explained by the wave model of light. In this lesson you will apply Huygens' principle to understand how wavefronts propagate, derive the conditions for constructive and destructive interference, calculate fringe spacing in Young's double-slit experiment, analyse single-slit diffraction minima, use diffraction gratings, predict thin-film interference colours, and apply Malus's law and Brewster's angle for polarized light.
Interference Conditions
Constructive Interference
- Path difference: Δ = mλ (m = 0, ±1, ±2, …)
- Waves arrive in phase → amplitudes add
- Bright fringe (double-slit): d sinθ = mλ
- Thin film (one phase flip): 2t = (m + ½)λ/n
Destructive Interference
- Path difference: Δ = (m + ½)λ
- Waves arrive out of phase → amplitudes cancel
- Dark fringe (double-slit): d sinθ = (m + ½)λ
- Phase flip: reflection off a denser medium adds ½λ to path difference
Key Equations
d sinθ = mλDouble-slit bright fringes (m = 0, ±1, ±2, …)Δy = λL / dFringe spacing (small-angle approximation)a sinθ = mλSingle-slit diffraction minima (m = ±1, ±2, …)2t = (m + ½)λ / nThin-film constructive (one phase flip)I = I₀ cos²θMalus's law — transmitted intensitytan θ_B = n₂ / n₁Brewster's angle for complete polarizationWorked Examples
A double-slit setup has slit separation d = 0.25 mm, wavelength λ = 550 nm, and screen distance L = 1.2 m. Calculate the fringe spacing Δy.
Write the fringe-spacing formula: Δy = λL / d
Convert units: λ = 550 × 10⁻⁹ m, d = 0.25 × 10⁻³ m, L = 1.2 m
Substitute: Δy = (550 × 10⁻⁹ × 1.2) / (0.25 × 10⁻³)
Numerator: 550 × 10⁻⁹ × 1.2 = 6.60 × 10⁻⁷ m²
Divide: Δy = 6.60 × 10⁻⁷ / 2.50 × 10⁻⁴ = 2.64 × 10⁻³ m
In a double-slit experiment, d = 0.50 mm and λ = 600 nm. Find the angle θ for the m = 2 bright fringe.
Bright-fringe condition: d sinθ = mλ
Solve for sinθ: sinθ = mλ / d
Substitute: sinθ = (2 × 600 × 10⁻⁹) / (0.50 × 10⁻³)
sinθ = 1.20 × 10⁻⁶ / 5.00 × 10⁻⁴ = 2.40 × 10⁻³
θ = arcsin(2.40 × 10⁻³) ≈ 0.138°
A diffraction grating has 600 lines/mm. Find the first-order (m = 1) diffraction angle for λ = 500 nm.
Find slit spacing: d = 1 / 600 mm = 1.667 × 10⁻³ mm = 1.667 × 10⁻⁶ m
Grating equation: d sinθ = mλ → sinθ = mλ / d
Substitute: sinθ = (1 × 500 × 10⁻⁹) / (1.667 × 10⁻⁶)
sinθ = 5.00 × 10⁻⁷ / 1.667 × 10⁻⁶ = 0.300
θ = arcsin(0.300) ≈ 17.5°
A soap film (n = 1.33) is illuminated with λ = 550 nm light. One phase flip occurs at the top surface. Find the minimum thickness for constructive reflection.
One phase flip → constructive condition: 2nt = (m + ½)λ
For minimum thickness use m = 0: 2nt = λ / 2 → t = λ / (4n)
Wait — re-derive: 2t = (m + ½)λ / n, so 2t = λ / (2n) when m = 0
t = λ / (4n) = 550 nm / (4 × 1.33)
t = 550 / 5.32 ≈ 103.4 nm
Check with formula 2t = (0 + ½) × 550 / 1.33 = 275 / 1.33 ≈ 206.8 nm → t ≈ 103.4 nm
Polarized light of intensity I₀ = 100 W/m² passes through an analyzer rotated 60° from the polarization axis. Find the transmitted intensity.
Apply Malus's law: I = I₀ cos²θ
θ = 60°, cos 60° = 0.500
cos²60° = (0.500)² = 0.250
I = 100 × 0.250 = 25 W/m²
Guided Problems
A double-slit experiment uses λ = 480 nm, d = 0.30 mm, and L = 2.0 m. What is the fringe spacing?
Hint: Use Δy = λL / d. Convert λ to metres and d to metres before substituting.
Light of wavelength 700 nm passes through a single slit of width a = 0.14 mm. At what angle does the first diffraction minimum occur?
Hint: Single-slit minimum: a sinθ = mλ with m = 1. Solve for sinθ then take arcsin.
A diffraction grating has 400 lines/mm. For λ = 632 nm, find the second-order diffraction angle.
Hint: Find d = 1/400 mm first. Then use d sinθ = mλ with m = 2.
A glass lens (n = 1.50) is coated with MgF₂ (n = 1.38). Two phase flips occur. What minimum thickness gives destructive reflection at λ = 550 nm?
Hint: Two phase flips cancel, so destructive condition is 2nt = (m + ½)λ. Use m = 0 for minimum thickness.
Unpolarized light of intensity 80 W/m² passes through a polarizer then an analyzer at 45°. What is the final intensity?
Hint: A polarizer halves unpolarized light: I₁ = I₀/2. Then apply Malus's law with θ = 45°.
Key Vocabulary
Huygens' Principle
Every point on a wavefront acts as a source of secondary spherical wavelets; the new wavefront is the envelope of these wavelets.
Example: Explains why waves bend around corners (diffraction) and why refraction obeys Snell's law.
Coherence
Two sources are coherent if they maintain a constant phase relationship over time, producing stable interference patterns.
Example: A laser is highly coherent; sunlight is not — hence lasers produce sharp fringes.
Path Difference
The difference in distance travelled by two waves from their sources to a common point (Δ = d sinθ for double-slit geometry).
Example: When Δ = mλ the waves interfere constructively; when Δ = (m + ½)λ they interfere destructively.
Constructive Interference
Superposition of two waves that are in phase, producing a resultant amplitude equal to the sum of the individual amplitudes.
Example: Bright fringes in Young's experiment occur where path difference is a whole number of wavelengths.
Destructive Interference
Superposition of two waves that are exactly out of phase (180°), producing a resultant amplitude of zero.
Example: Dark fringes in Young's experiment occur where path difference is a half-integer number of wavelengths.
Diffraction
The spreading of waves as they pass through a narrow aperture or around an obstacle, explained by Huygens' principle.
Example: Single-slit diffraction produces a central bright maximum flanked by weaker secondary maxima.
Thin-Film Interference
Interference between light reflected from the top and bottom surfaces of a thin transparent film, producing colour effects.
Example: The rainbow colours of a soap bubble result from thin-film interference of white light.
Polarization
The restriction of a transverse wave's oscillation to a single plane; light can be polarized by transmission, reflection, or scattering.
Example: Polarizing sunglasses block horizontally polarized glare reflected from roads and water.
Check Your Understanding
Interactive Practice — 5 Questions
In Young's double-slit experiment, the fringe spacing Δy is proportional to which of the following?
A soap film has one phase flip. Which condition gives constructive interference?
Malus's law states that when polarized light of intensity I₀ passes through an analyzer at angle θ, the transmitted intensity is:
For a diffraction grating with 500 lines/mm, what is the slit spacing d?
The first single-slit diffraction minimum occurs when:
Independent Practice
In a double-slit experiment, the slit separation is 0.25 mm and the screen is 1.5 m away. Light of wavelength 600 nm is used. Calculate the fringe spacing.
Light of wavelength 500 nm passes through a single slit of width 0.10 mm. Find the angle of the first dark fringe.
A thin film of oil (n = 1.45) floats on water (n = 1.33). For light of wavelength 560 nm incident normally, find the minimum film thickness for constructive interference in reflection.
In a double-slit experiment, the 4th bright fringe is observed at 1.2 cm from the center. The slit separation is 0.30 mm and the screen is 1.0 m away. Find the wavelength of light used.
★ A diffraction grating has 500 lines/mm. Light of wavelength 589 nm (sodium D line) is incident normally. (a) Find the grating spacing d. (b) Calculate the angles of the first, second, and third order maxima. (c) What is the highest order maximum that can be observed? (d) If white light (400–700 nm) is used, find the angular range of the first-order spectrum.
ChallengeCommon Mistakes
Using the small-angle approximation (sinθ ≈ tanθ ≈ θ) when the angle is large
The approximation Δy = λL/d is valid only for small angles (θ < ~10°). For larger angles, use sinθ = mλ/d directly
Forgetting to account for the phase flip when light reflects off a more optically dense medium
Reflection from a denser medium (higher n) causes a 180° phase flip (equivalent to λ/2 path difference). Reflection from a less dense medium has no phase flip
Confusing the order m in double-slit (m = 0, ±1, ±2...) with the order in single-slit diffraction minima
Double-slit bright fringes: d sinθ = mλ (m = 0, ±1, ±2...). Single-slit dark fringes: a sinθ = mλ (m = ±1, ±2...). The central maximum of single-slit is at m = 0
Applying Malus's law starting from unpolarized light
Malus's law I = I₀cos²θ applies only when the INCIDENT light is already polarized. Unpolarized light through a polarizer gives I = I₀/2 regardless of orientation
Math Tips
Fringe spacing shortcut: Δy = λL/d. To get more widely spaced fringes, use longer wavelength (red > violet), longer screen distance L, or narrower slit separation d
Diffraction grating: d sinθ = mλ where d = 1/(lines per meter). For 600 lines/mm: d = 1.667×10⁻⁶ m. Higher order m gives larger angle — check that sinθ ≤ 1
Thin-film color: the color you SEE is the one that constructively interferes. White light minus the destructively interfering wavelength gives the complementary color (e.g., no blue → you see yellow)
Brewster's angle: tanθ_B = n₂/n₁. At this angle, reflected light is completely polarized parallel to the surface. For glass (n=1.5) in air: θ_B = arctan(1.5) ≈ 56.3°