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Reviewed CSCA Physics question · Standard

In a double-slit interference experiment, the slit separation is 0.50 mm and the screen is 2.0 m away. Light of wavelength 600 nm is used. What is the distance between the central bright fringe and the second-order bright fringe? (Assume small angles.)

  1. 2.4 mm
  2. 4.8 mm
  3. 7.2 mm
  4. 9.6 mm
Show the answer and explanation

Correct answer

B. 4.8 mm

Principle or equation

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

Why this answer is correct

Given d = 0.50 mm = 5.0e-4 m, L = 2.0 m, λ = 600 nm = 6.0e-7 m, m = 2. y_2 = 2 * (6.0e-7) * 2.0 / (5.0e-4) = 2 * 1.2e-6 / 5.0e-4 = 2.4e-6 / 5.0e-4 = 4.8e-3 m = 4.8 mm.

Example

For m = 1 with same parameters, y_1 = 2.4 mm.

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