Question Detail
Question
Using the wavelength calculated in the previous problem (553.3 nm), where would the second dark fringe appear in Young's experiment with the same setup (d = 0.04 mm, x = 2 m)?
Correct Answer
Option B is the correct answer.
Detailed Explanation
Given: λ = 553.3 nm = 553.3 × 10-9 m, d = 0.04 mm = 0.04 × 10-3 m, x = 2 m. For the second dark fringe, n = 3 (n=1 for the first dark fringe, n=3 for the second, etc.). The formula for dark fringes is yd/x = nλ/2. Rearranging for y: y = (nλx) / (2d) y = (3 × 553.3 × 10-9 m × 2 m) / (2 × 0.04 × 10-3 m) y = (3319.8 × 10-9) / (0.08 × 10-3) y = 41497.5 × 10-6 m = 0.0414975 m ≈ 4.15 cm.
Hint
For dark fringes, remember that n takes odd integer values (1, 3, 5, ...) and the formula is yd/x = nλ/2. Be careful with the 'n' value for the 'second' dark fringe.