In a double-slit interference experiment, the slit separation is 0.35 mm and the screen is 2.2 m away. Light of wavelength 480 nm is used. What is the distance between the central bright fringe and the third-order bright fringe? (1 nm = 10^-9 m)
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Correct answer
A. 9.0 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 m is the order number, λ is the wavelength, L is the distance to the screen, and d is the slit separation.
Why this answer is correct
Given m = 3, λ = 480 × 10^-9 m, L = 2.2 m, d = 0.35 × 10^-3 m. y_3 = 3 × 480 × 10^-9 × 2.2 / (0.35 × 10^-3) = 3 × 480 × 2.2 / 0.35 × 10^-6 = 3168 / 0.35 × 10^-6 = 9051.43 × 10^-6 m ≈ 9.05 × 10^-3 m = 9.0 mm.
Example
If d = 0.50 mm, L = 2.0 m, λ = 600 nm, the second-order fringe is at y = 2 × 600 × 10^-9 × 2.0 / (0.50 × 10^-3) = 4.8 × 10^-3 m = 4.8 mm.
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