Double-Slit Interference
The double-slit formula gives the directions of the interference maxima: reinforcement occurs when the path difference d·sinα is an integer multiple of the wavelength.
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Formula
d \cdot \sin(\alpha_k) = k \cdot \lambdaVariables & units – Double-Slit Interference
| Symbol | Meaning | Unit |
|---|---|---|
| d | Separation of the two slits | m |
| α_k | Angle to the maximum of order k | ° |
| k | Order of the maximum (0, 1, 2, ...) | dimensionless |
| λ | Wavelength of the light | m |
Derivation & background – Double-Slit Interference
Thomas Young demonstrated the wave nature of light with the double-slit experiment in 1801. Elementary waves emerge from both slits; if their path difference is k·λ they reinforce (maximum), at (k + ½)·λ they cancel (minimum). For small angles the approximation sin α ≈ tan α = x/L holds, with screen distance L and distance x of the maximum from the centre. The same formula describes the maxima of an optical grating, there with the grating constant d.
Exam blueprint
Validity range
Holds in the far field (screen distance large compared with the slit separation) for coherent, monochromatic light. Since sin α ≤ 1 there are only finitely many orders: k ≤ d/λ.
Derivation steps
Two elementary waves reinforce each other when their path difference is a multiple of the wavelength.
- 1For a distant point at angle α the path difference of the two rays is Δs = d·sin α.
- 2Constructive interference requires Δs = k·λ, hence d·sin α_k = k·λ; minima lie at (k + ½)·λ.
Rearrangements
Wavelength
This determines the wavelength of light by measuring lengths.
Maximum position on the screen
Small-angle approximation sin α ≈ tan α = x/L.
Slit separation
Calibrating an unknown double slit or grating.
Task variant
d = 0.2 mm, screen at L = 2 m: the 1st maximum lies 6.4 mm from the centre. Find λ.
λ = d·x/(k·L) = 2×10⁻⁴ × 6.4×10⁻³/(1 × 2) = 6.4×10⁻⁷ m = 640 nm (red light).
At what angle does the 2nd maximum appear for d = 4 µm and λ = 500 nm?
sin α = k·λ/d = 2 × 5×10⁻⁷/4×10⁻⁶ = 0.25, so α = arcsin(0.25) ≈ 14.5°.
Common mistakes
Using the small-angle approximation at large angles.
Beyond about 10° compute with sin α exactly; tan α then deviates visibly.
Swapping the maxima and minima conditions.
Maxima at k·λ, minima at (k + ½)·λ path difference.
Confusing slit separation d with slit width.
d is the centre-to-centre separation; the slit width only shapes the brightness envelope.
Mixing nm, µm and mm.
Convert all lengths to metres and only convert back at the end.
Exam context
- Standard in exams: wavelength determination from screen patterns, switching between double slit and grating and transfer to electron diffraction (de Broglie).
These mistakes cost points in real exams. The set drills them until they stick.
Formula cluster
Wave optics
The classic proof of wave nature, connected to diffraction and matter waves.
Worked example
Double slit d = 0.1 mm, laser λ = 600 nm: 1st maximum at sin α = λ/d = 6×10⁻³, so α ≈ 0.34°. On a screen at L = 2 m it lies x ≈ 12 mm from the centre.
Applications
Wavelength measurement with grating or double slit, spectroscopy, structure analysis (X-ray diffraction), demonstrating the wave properties of electrons
Quanta exam set
Curated exam set for "Double-Slit Interference":
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Which formula describes Double-Slit Interference?
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Question (front)
How do you rearrange d·sin(α_k) = k·λ for Wavelength?
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Which common mistake happens with Double-Slit Interference?
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Scientific sources
Common notations & search queries
Related formulas
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Frequently asked questions about Double-Slit Interference
How do you calculate the positions of the maxima in the double slit?+
The angles of the maxima follow from d·sin(α_k) = k·λ: sin(α_k) = k·λ/d with order k = 0, 1, 2, ... The zeroth maximum always lies straight ahead in the centre. Example: d = 0.1 mm and λ = 600 nm give for k = 1 the value sin α = 6×10⁻⁷/10⁻⁴ = 0.006, so α ≈ 0.34°. The position on a screen at distance L follows from geometry: x = L·tan α, for small angles x ≈ k·λ·L/d, here 12 mm. The maxima are then equally spaced. Convert all lengths consistently to metres; mixing nm, µm and mm is the most common source of error.
What is the path difference and why does it decide bright and dark?+
The path difference Δs is the difference in distance the waves travel from the two slits to a point on the screen; for distant points Δs = d·sin α. It determines the relative timing of the two waves there. If Δs is a whole multiple of the wavelength (0, λ, 2λ, ...), crest meets crest: constructive interference, a bright maximum. If Δs is a half-integer multiple (λ/2, 3λ/2, ...), crest meets trough and the waves cancel: a dark minimum. Everything in between produces intermediate brightness. That light plus light yields darkness at the dark places is the decisive wave argument; particle trajectories alone could never explain it.
How do you determine the wavelength of light with a double slit?+
You measure lengths and convert back to nanometres; that is the core of the Young experiment. Illuminate the double slit with the light to be measured, ideally a laser, and measure three quantities: the slit separation d, the screen distance L and the distance x of the first maximum from the centre. For small angles λ = d·x/(k·L). Example: d = 0.2 mm, L = 2 m, first maximum at x = 6.4 mm: λ = 2×10⁻⁴ × 6.4×10⁻³/2 = 6.4×10⁻⁷ m = 640 nm, red light. It becomes more accurate if you average over several orders or measure the distance across many maxima and divide.
What changes when you vary the slit separation or the wavelength?+
From sin α = k·λ/d both trends can be read off directly. A larger wavelength spreads the pattern: red light (about 650 nm) produces maxima farther apart than blue (about 450 nm); with white light this creates coloured fringes, blue on the inside, red on the outside. A smaller slit separation also spreads the pattern, so a finer double slit separates better. Conversely, for large d the maxima crowd together until they merge, which is why you never observe interference from two windows. In addition sin α ≤ 1 limits the number of orders to k ≤ d/λ: for d = 4 µm and λ = 500 nm only orders up to k = 8 exist.
Does the double-slit experiment also work with electrons?+
Yes, and it is one of the most important experiments of quantum physics. Sending electrons through a double slit produces the same fringe pattern on the detector as light, described by the same formula d·sin α = k·λ, only with the de Broglie wavelength λ = h/p of the electron. The astonishing part: the pattern builds up even when the electrons fly one at a time; each electron in a sense interferes with itself. If, however, you measure which slit the electron passes, the pattern disappears. The experiment thus demonstrates wave-particle duality and the role of measurement, and in final exams it is the standard bridge from optics to quantum physics.
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How do you calculate with Double-Slit Interference?
Here is how to work through a typical Double-Slit Interference (d·sin(α_k) = k·λ) task step by step:
- 1
Task
d = 0.2 mm, screen at L = 2 m: the 1st maximum lies 6.4 mm from the centre. Find λ.
Solution path
λ = d·x/(k·L) = 2×10⁻⁴ × 6.4×10⁻³/(1 × 2) = 6.4×10⁻⁷ m = 640 nm (red light).
- 2
Task
At what angle does the 2nd maximum appear for d = 4 µm and λ = 500 nm?
Solution path
sin α = k·λ/d = 2 × 5×10⁻⁷/4×10⁻⁶ = 0.25, so α = arcsin(0.25) ≈ 14.5°.