In a double-slit interference experiment, the slit separation is 0.50 mm and the screen is 3.0 m away. Light of wavelength 450 nm is used. What is the distance between the central bright fringe and the second-order bright fringe? (Assume small angles.)

Correct answer

A. 5.4 mm

Principle or equation

For double-slit interference, the position of the m-th bright fringe from the central maximum is y_m = mλL/d, where λ is wavelength, L is distance to screen, and d is slit separation.

Why this answer is correct

Given λ = 450 × 10⁻⁹ m, L = 3.0 m, d = 0.50 × 10⁻³ m, and m = 2. Then y₂ = 2 × (450 × 10⁻⁹) × 3.0 / (0.50 × 10⁻³) = 5.4 × 10⁻³ m = 5.4 mm.

Example

With λ = 600 nm, L = 2 m, d = 0.2 mm, the first-order fringe is at y₁ = (600×10⁻⁹)(2)/(0.2×10⁻³) = 6.0 mm.