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Double slit fringe separation numerical

Short answer

Using β = λD/d with λ=500 nm, D=1 m, d=0.1 mm gives β = 5×10⁻³ m. Adjacent bright fringes are approximately 5 mm apart.

A screen containing two slits 0.1mm apart is 1m from the viewing screen. Light of wavelength λ = 500 nm falls on the slits from a distant source. Approximately how far apart will the bright interference fringes be seen on the screen?

In Young's double-slit experiment, the bright fringes are equally spaced. The fringe separation (fringe width) is β = λD / d, valid for small angles.

Given: d = 0.1 mm = 1 × 10⁻⁴ m, D = 1 m, λ = 500 nm = 5 × 10⁻⁷ m.

β = λD / d
  = (5×10⁻⁷ × 1) / (1×10⁻⁴)
  = 5×10⁻³ m

Answer: adjacent bright fringes are about 5 mm apart.

Check: path difference for the first bright fringe is d sin θ = λ, so θ ≈ λ/d = 5 × 10⁻³ rad, and y = Dθ = 5 mm, which agrees.

✓ Answer checked by Claude Opus 5.5 (high effort)

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