A Young's double slit apparatus is illuminated simultaneously by a light source containing two wavelengths, ₁ = 420 nm and ₂ = 560 nm . The distance between the two slits is 1.0 mm and the screen is placed at a distance of 1.5 m from the slits. What is the minimum non-zero distance from the central maximum where the bright fringes of both wavelengths coincide?

Step-by-step solution

Let the n₁ -th bright fringe of ₁ coincide with the n₂ -th bright fringe of ₂ . The position of the n -th bright fringe from the central maximum is given by y = n D d . For the fringes to coincide, y₁ = y₂ : n₁ ₁ D d = n₂ ₂ D d n₁ ₁ = n₂ ₂ Substituting the given wavelengths: n₁ (420) = n₂ (560) n₁ n₂ = 560 420 = 4 3 For the minimum non-zero distance, n₁ and n₂ must be the smallest possible integers. Thus, n₁ = 4 and n₂ = 3 . The minimum distance y is: y = n₁ ₁ D d y = 4 420 10⁻⁹ 1.5 1.0 10⁻³ y = 2520 10⁻⁶ m y = 2.5