Young's Double Slit Experiment Calculator

Calculate fringe spacing, wavelength, and slit parameters for Young's double slit interference.

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Why Two Slits Produce Evenly Spaced Fringes

Young's double-slit pattern is created because waves from two coherent openings arrive at the screen with different path lengths. Bright fringes occur when the path difference is an integer number of wavelengths, d sinθ=mλ. Dark fringes occur halfway between those conditions, where the path difference is (m+1/2)λ.

For a distant screen and small angles, sinθ≈tanθ≈y/D. Substituting this into the bright-fringe condition gives ym≈mλD/d, so adjacent bright fringes are separated by Δy=λD/d. The equal spacing is therefore a small-angle result. If the screen is close or the fringe angle is large, use the exact angular condition rather than extending the linear approximation too far.

d sinθ=mλ,   ym≈mλD/d,   Δy≈λD/d
SymbolMeaningWhy it appears / units
λWavelengthm; longer wavelengths create wider fringe spacing.
dCenter-to-center slit separationm; smaller separation makes fringes farther apart.
DSlit-to-screen distancem; larger D spreads the angular pattern over a larger linear distance.
mFringe orderInteger for bright fringes, with m=0 at the central maximum.
ΔyAdjacent bright-fringe spacingm; approximately constant in the small-angle region.

Clear interference requires the two waves to maintain a stable phase relationship. Real slits also have finite width, so the two-slit fringes can be modulated by a broader single-slit diffraction envelope. The fringe-spacing formula describes the interference spacing, not that envelope's width.

Worked Examples

Example 1: Sodium light λ=589nm, d=0.5mm, D=2m
Δy=589e-9×2/(0.5e-3)
Result: 2.356 mm fringe spacing
Classic sodium double slit
Example 2: Measure Δy=3mm, d=0.4mm, D=1.5m
λ=3e-3×0.4e-3/1.5
Result: 800 nm — near-IR
Find unknown wavelength
Example 3: Green laser fringes
λ=532nm, d=0.25mm, D=1.20m → Δy=λD/d
Result: Δy≈2.554mm
Converting nanometers and millimeters to meters before substitution keeps the unit factors consistent.
Example 4: Position of the third bright fringe
Using Δy=2.554mm, m=3 → y3=3Δy
Result: y3≈7.66mm from the center
Fringe order counts from the central bright fringe at m=0, so the third-order maximum lies three spacings away.

Common Mistakes

⚠️
Mixing nanometers, millimeters, and meters

Convert all lengths to a consistent unit before calculating. A missed 10−3 or 10−9 factor can overwhelm the physics.

⚠️
Confusing slit width with slit separation

The double-slit interference spacing uses center-to-center separation d. Individual slit width controls the diffraction envelope and is a different geometric quantity.

⚠️
Applying Δy=λD/d at large angles without checking

The linear spacing formula uses the small-angle approximation. For larger angles, start with d sinθ=mλ and use screen geometry explicitly.

Frequently Asked Questions

Constructive vs destructive?
Constructive (bright): path difference = mλ (m=0,±1,±2). Destructive (dark): path difference = (m+½)λ. The central bright fringe (m=0) has zero path difference.
Why does decreasing d increase fringe spacing?
Δy=λD/d — inversely proportional to slit separation. Closer slits → wider fringes. This is why gratings with many fine lines produce widely spaced spectra.
Why is the central fringe bright?
At the center, the distances from the two identical slits are equal, so the path difference is zero. Zero equals mλ for m=0, which is a constructive-interference condition, producing the central bright maximum.
What happens if the wavelength increases?
For the same slit separation and screen distance, fringe spacing increases in direct proportion to wavelength. Red light therefore gives wider spacing than blue light under otherwise identical conditions.
Why must the two slits be coherent?
A stable interference pattern requires a stable phase relationship between the waves. If their relative phase changes randomly faster than the measurement can resolve, bright and dark regions average together and the visible interference contrast disappears.
Why are real double-slit fringes inside a diffraction envelope?
Each slit has finite width and therefore diffracts light by itself. The two-slit interference pattern is multiplied by this single-slit diffraction envelope, so some outer fringes become weaker and certain interference maxima can be strongly suppressed.

Formula Explorer connections

Interpretation: This relationship connects light propagation, geometry, wavelength, refraction, interference or image formation. Assumption: Use a consistent sign convention and units. Paraxial rays, thin elements, coherent light, vacuum wavelength or ideal optical components may be assumed.

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