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In a Young’s double‑slit experiment, the distance between the slits is 0.15 mm, the wavelength of light used is 600 nm, and the screen is placed 1.2 m from the slits. What is the distance between adjacent bright fringes on the screen?

easy MCQ 2 marks Class 12 · Physics · Wave Optics
  1. 0.72 mm
  2. 0.6 mm
  3. 0.9 mm
  4. 1.08 mm
Answer
0.72 mm

The correct answer is 0.72 mm — Convert to millimetres: 7.2×10⁻⁴ m = 0.72 mm.

Step-by-step solution

  1. Compute fringe spacing using Δy = λL/d.
  2. Convert λ to meters: 600 nm = 6×10⁻⁷ m.
  3. Convert d to meters: 0.15 mm = 1.5×10⁻⁴ m.
  4. Insert values: Δy = (6×10⁻⁷ m)(1.2 m)/(1.5×10⁻⁴ m) = 7.2×10⁻⁴ m.
  5. Convert to millimetres: 7.2×10⁻⁴ m = 0.72 mm.
Common mistake: Using wavelength in nm without converting to metres or using the wrong formula for fringe spacing

Concept tested: Young’s double‑slit fringe spacing

Published · How this solution was prepared and checked

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