A Young's double slit experiment is performed using a light source that emits two wavelengths, 450 nm and 600 nm . The separation between the slits is 0.5 mm . If the physical distance from the central maximum to the first point where the bright fringes of the two wavelengths coincide is 3.6 mm , the distance of the screen from the slits is

Step-by-step solution

The condition for the spatial coincidence of bright fringes is: n₁ ₁ = n₂ ₂ n₁ (450) = n₂ (600) n₁ n₂ = 600 450 = 4 3 The first coincidence occurs for the 4^ th bright fringe of the 450 nm light (which overlaps with the 3^ rd bright fringe of the 600 nm light). The position of this fringe on the screen is given by: y = n₁ ₁ D d Substituting the given values: 3.6 10⁻³ = 4 450 10⁻⁹ D 0.5 10⁻³ Solving for D : D = 3.6 10⁻³ 0.5 10⁻³ 1800 10⁻⁹ D = 1.8 10⁻⁶ 1.8 10⁻⁶ = 1.0 m Answer: 1.0 m