Two wavelengths λ₁ and λ₂ are used in Young's double slit experiment. λ₁=450 nm and λ₂=650 nm. The minimum order of fringe produced by λ₂ which overlaps with the fringe produced by λ₁ is n. The value of n is _______.

Solution

Physics sir · 3 steps

Fringes overlap at the same y, so set n₁λ₁ = n₂λ₂ and find the smallest integers.

1. Bright fringe position. The nth bright fringe is at y = (nλ D)/d. Both wavelengths use the same D and d, so two fringes coincide when nλ is the same for both.

   n₁ × 450 = n₂ × 650 ⇒ n₁/n₂ = 650/450 = 13/9

2. Smallest whole numbers. 13 and 9 share no common factor, so the smallest integer pair is n₁ = 13 and n₂ = 9. Check: 13 × 450 = 5850 and 9 × 650 = 5850, both in nm.
3. So. The 9th bright fringe of λ₂ falls on the 13th bright fringe of λ₁, so n = 9.

The common slip: 13. Reporting the order of the λ₁ fringe gives 13. The question asks for the order of the λ₂ fringe, which is 9.

The idea to keep. For overlapping fringes of two wavelengths, equate n₁λ₁ = n₂λ₂ and reduce the wavelength ratio to lowest terms. The longer wavelength always gets the smaller order.

Answer from the NTA final answer key. Solution by PrepWiser.