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Reviewed CSCA Physics question · Standard

In a double-slit experiment, the distance between the slits is 0.25 mm and the screen is 2.0 m away. The second-order bright fringe is observed at a distance of 8.0 mm from the central bright fringe. What is the wavelength of the light used?

  1. 500 nm
  2. 400 nm
  3. 600 nm
  4. 800 nm
Show the answer and explanation

Correct answer

A. 500 nm

Principle or equation

For constructive interference in Young's double-slit experiment, the position of the m-th bright fringe from the central maximum is given by y_m = m λ L / d, where d is slit separation, L is distance to screen, and λ is wavelength.

Why this answer is correct

Given d = 0.25 mm = 2.5 × 10^-4 m, L = 2.0 m, m = 2, y_2 = 8.0 mm = 8.0 × 10^-3 m. Solve for λ: λ = y_2 d / (m L) = (8.0 × 10^-3 × 2.5 × 10^-4) / (2 × 2.0) = 2.0 × 10^-6 / 4 = 5.0 × 10^-7 m = 500 nm.

Example

If the second bright fringe is 8.0 mm from center with d=0.25 mm and L=2.0 m, then λ=500 nm.

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