In a Young's double-slit experiment, monochromatic light of wavelength 600 nm passes through two slits separated by 0.20 mm. What is the distance between consecutive bright interference fringes on a screen placed 1.50 meters away?

Answer and explanation

Correct answer: B. 4.50 mm (4.50 * 10^-3 m)

  1. Step 1: The fringe separation formula in double-slit interference is: Delta y = (lambda * L) / d.
  2. Step 2: Convert units to meters: lambda = 600 10^-9 m, L = 1.50 m, d = 0.20 10^-3 m.
  3. Step 3: Calculate numerator: lambda L = (600 10^-9 m) 1.50 m = 9.00 10^-7 m^2.
  4. Step 4: Divide by d: Delta y = (9.00 10^-7) / (2.0 10^-4) = 4.50 * 10^-3 m = 4.50 mm.
Answer
4.50 mm (4.50 * 10^-3 m).
Why not the others: '2.25 mm' uses d = 0.40 mm; '9.00 mm' forgets to divide by 2 in the denominator; '0.45 mm' makes a power-of-ten unit conversion error.

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