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Reviewed CSCA Physics question · Standard

In a double-slit interference experiment, the slit separation is 0.30 mm and the screen is 2.5 m away. Light of wavelength 500 nm is used. What is the distance between the central bright fringe and the first-order bright fringe on the screen?

  1. 2.1 mm
  2. 4.2 mm
  3. 8.3 mm
  4. 0.42 mm
Show the answer and explanation

Correct answer

B. 4.2 mm

Principle or equation

For small angles in double-slit interference, the position of the m-th bright fringe is y_m = m λ L / d.

Why this answer is correct

For m = 1: y = (1)(500×10^-9 m)(2.5 m) / (0.30×10^-3 m) = 4.17×10^-3 m = 4.2 mm.

Example

With d = 0.50 mm, L = 3.0 m, λ = 450 nm, y_2 = 2×450×10^-9×3.0 / 0.50×10^-3 = 5.4 mm.

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