Playground
Interactive double-slit pattern: adjust slit separation and wavelength to see the interference fringes form on a simulated screen.
Variables
| Symbol | Name | SI | Dimension | Range |
|---|---|---|---|---|
| Slit separation Distance between the centers of the two slits | m | L | 0.00001 – 0.001 | |
| Angle to maximumoutput Angle from the central axis to the m-th bright fringe | rad | 1 | 0 – 0.1 | |
| Order number Integer order of the bright fringe (0, 1, 2, ...) | dimensionless | 1 | 0 – 10 | |
| Wavelength Wavelength of the incident light | m | L | 3.8e-7 – 7.5e-7 |
Deep dive
Derivation
Two coherent sources separated by distance d emit waves. At angle θ, the path difference is Δ = d·sin(θ). Constructive interference (bright fringe) occurs when Δ = mλ, giving d·sin(θ) = mλ. For small angles (sin θ ≈ tan θ = y/L), the fringe position is y_m = mλL/d.
Experimental verification
Thomas Young's 1801 experiment using sunlight through two pinholes. Modern: laser double-slit experiments reproduce the pattern precisely. Single-photon double-slit experiments confirm wave-particle duality at the quantum level.
Common misconceptions
- Dark fringes mean light is destroyed — energy is redistributed, not lost. Bright fringes get extra energy.
- Interference only works with lasers — any coherent source works; Young used filtered sunlight.
- Wider slits give sharper fringes — wider slits reduce coherence and blur the pattern.
Real-world applications
- Measuring wavelength: fringe spacing directly gives λ if d and L are known.
- Anti-reflective coatings: thin film interference cancels reflected light.
- Holography: interference patterns encode 3D information on a 2D surface.
- Quantum mechanics foundations: single-particle interference proves wave-particle duality.
Worked examples
Fringe spacing on a screen
Given:
- lambda:
- 5.5e-7
- d:
- 0.0001
- L:
- 1
- m:
- 1
Find: y_1 (first bright fringe position)
Solution
y_1 = mλL/d = 1 × 5.5×10⁻⁷ × 1.0 / 1×10⁻⁴ = 5.5×10⁻³ m = 5.5 mm
Finding wavelength from fringe pattern
Given:
- d:
- 0.0002
- L:
- 1.5
- y_spacing:
- 0.004
Find: lambda
Solution
λ = Δy × d / L = 0.004 × 2×10⁻⁴ / 1.5 = 5.33×10⁻⁷ m ≈ 533 nm (green)
Scenarios
What if…
- scenario:
- What if you use white light instead of monochromatic?
- answer:
- Each wavelength produces its own fringe pattern at different spacings. The central maximum is white, but higher orders show rainbow-colored fringes with overlapping colors.
- scenario:
- What if you close one slit?
- answer:
- The interference pattern disappears. You get a single-slit diffraction pattern instead — broader central maximum, no sharp fringes.
- scenario:
- What if d decreases?
- answer:
- Fringes spread apart (Δy = λL/d). In the limit d → 0, fringes become infinitely wide — effectively one source.
Limiting cases
- condition:
- d → ∞
- result:
- theta → 0
- explanation:
- Very wide slit separation compresses fringes to the center.
- condition:
- lambda → 0
- result:
- theta → 0
- explanation:
- Short wavelength means fringes crowd near the center — approaching ray optics.
- condition:
- m = 0
- result:
- theta = 0
- explanation:
- Central maximum is always straight ahead regardless of wavelength.
Context
Thomas Young · 1801
Young's experiment provided the first definitive proof of the wave nature of light, overturning Newton's corpuscular theory.
Hook
How can two slits in a card prove that light is a wave?
Monochromatic light (lambda = 550 nm) hits two slits separated by 0.1 mm. Find the fringe spacing on a screen 1 m away using y = m*lambda*L/d.
Dimensions:
- lhs:
- d·sin(θ) → [L]·[1] = [L]
- rhs:
- m·λ → [1]·[L] = [L]
- check:
- Both sides are [L] = meters. ✓
Validity: Valid for coherent, monochromatic light with slit width << slit separation. Assumes far-field (Fraunhofer) regime where L >> d.