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Interactive single-slit diffraction: adjust slit width and wavelength to see the intensity pattern and central maximum width change.

Variables

SymbolNameSIDimensionRange
aaSlit width
Width of the single slit aperture
mL0.00001 – 0.001
thetathetaAngle to minimumoutput
Angle from the central axis to the m-th dark fringe
rad10 – 0.5
mmOrder number
Integer order of the dark fringe (1, 2, 3, ...)
dimensionless11 – 10
lambdalambdaWavelength
Wavelength of the incident light
mL3.8e-7 – 7.5e-7

Deep dive

Derivation
Divide the slit of width a into infinitesimal strips. Each strip acts as a Huygens wavelet source. At angle θ, the path difference between the top and bottom of the slit is a·sin(θ). Dark fringes occur when this equals mλ (m = ±1, ±2, ...) because the slit can be divided into m pairs of strips that cancel pairwise. Formally, integrating the electric field contributions: E(θ) = E₀·sinc(πa·sinθ/λ), giving minima at a·sin(θ) = mλ.
Experimental verification
Grimaldi first observed (1665). Fraunhofer provided precise measurements using his diffraction grating. Modern: single-slit patterns are reproduced routinely with lasers and adjustable slits in undergraduate labs.
Common misconceptions
  • The formula gives bright fringes — it gives dark fringes (minima). Bright maxima are approximately halfway between.
  • Narrower slit means narrower pattern — the opposite: narrower slit produces wider diffraction spread.
  • Diffraction and interference are different phenomena — diffraction IS interference from a continuous aperture.
Real-world applications
  • Telescope resolution: the Airy disk is the 2D analog, setting the diffraction limit.
  • CD/DVD reading: laser diffraction from pit tracks encodes data.
  • X-ray crystallography: diffraction from atomic-scale 'slits' reveals molecular structure.
  • Acoustic diffraction: sound bending around doorways and obstacles.

Worked examples

First dark fringe of a laser through a slit

Given:
a:
0.00005
lambda:
6.33e-7
m:
1
Find: theta
Solution

θ = arcsin(mλ/a) = arcsin(1 × 6.33×10⁻⁷ / 5×10⁻⁵) = arcsin(0.01266) = 0.01266 rad ≈ 0.725°

Central maximum width on a screen

Given:
a:
0.0001
lambda:
5e-7
L:
2
Find: width of central maximum
Solution

Width = 2λL/a = 2 × 5×10⁻⁷ × 2.0 / 1×10⁻⁴ = 0.02 m = 20 mm

Scenarios

What if…
  • scenario:
    What if a >> λ?
    answer:
    Diffraction becomes negligible (θ → 0). Light passes in a straight beam — the geometric optics limit.
  • scenario:
    What if a ≈ λ?
    answer:
    First minimum at θ ≈ 90°. Light spreads into a hemisphere — maximum diffraction.
  • scenario:
    What if you use a circular aperture instead?
    answer:
    The pattern becomes an Airy disk: concentric bright and dark rings. The first dark ring is at sin(θ) = 1.22λ/D.
Limiting cases
  • condition:
    a → ∞
    result:
    theta → 0
    explanation:
    Wide slit means negligible diffraction — geometric optics limit.
  • condition:
    a → lambda
    result:
    theta → 90°
    explanation:
    Slit width equals wavelength: light spreads into a hemisphere.
  • condition:
    lambda → 0
    result:
    theta → 0
    explanation:
    Short wavelength means sharp shadows — ray optics regime.

Context

Francesco Maria Grimaldi · 1665

Grimaldi first observed diffraction fringes. Fraunhofer later provided the mathematical treatment for far-field single-slit patterns.

Hook

Why does a laser beam spread out after passing through a narrow slit?

A 633 nm laser shines through a 0.05 mm slit. Find the angle to the first dark fringe using a*sin(theta) = m*lambda.

Dimensions:
lhs:
a·sin(θ) → [L]·[1] = [L]
rhs:
m·λ → [1]·[L] = [L]
check:
Both sides are [L] = meters. ✓
Validity: Valid in the Fraunhofer (far-field) regime where screen distance L >> a²/lambda. For near-field, use Fresnel diffraction theory.

Related formulas