In a double-slit interference experiment, the slit separation is 0.40 mm and the screen is 2.0 m away. Light of wavelength 550 nm is used. What is the distance between the central bright fringe and the third-order bright fringe on the screen? (Assume small angles.)
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Correct answer
C. 8.25 mm
Principle or equation
For double-slit interference, the position of the m-th bright fringe from the central maximum is y_m = m λ L / d, where λ is the wavelength, L is the distance to the screen, and d is the slit separation.
Why this answer is correct
Given d = 0.40 mm = 4.0 × 10^-4 m, L = 2.0 m, λ = 550 nm = 5.50 × 10^-7 m. For m = 3, y = 3 × (5.50 × 10^-7)(2.0) / (4.0 × 10^-4) = 8.25 × 10^-3 m = 8.25 mm.
Example
If the slit separation is 0.20 mm and the screen is 1.5 m away with λ = 500 nm, the first-order fringe is at y = (5.00 × 10^-7)(1.5)/(2.0 × 10^-4) = 3.75 mm.
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