In a double-slit interference setup with L = 2.0 m, wavelength λ = 500 nm, slit separation d = 0.50 mm, what is the fringe spacing Δy for the m = 1 bright fringe?

• A. 2.0 mm
• B. 1.0 mm
• C. 0.5 mm
• D. 5.0 mm

Answer: (A)

In a double-slit setup, bright fringes satisfy d sin θ = mλ. For small angles, sin θ ≈ tan θ ≈ y/L, so the position of the m-th bright fringe is y_m ≈ mλL/d. The spacing between adjacent bright fringes is Δy = y_{m+1} − y_m ≈ λL/d, independent of m.

Plugging in L = 2.0 m, λ = 500 nm = 5.0×10^-7 m, and d = 0.50 mm = 5.0×10^-4 m gives:

Δy = (5.0×10^-7 m × 2.0 m) / (5.0×10^-4 m) = 1.0×10^-6 / 5.0×10^-4 = 2.0×10^-3 m = 2.0 mm.

So the fringe spacing is 2.0 mm. The first bright fringe would appear about 2.0 mm from the center, matching the same spacing.