In a Young's double slit experiment, = 500 nm , d = 1.0 mm , and D = 1.0 m . Determine the minimum distance from the central maximum at which the intensity is half of the maximum intensity.

Step-by-step solution

The intensity in Young's double slit experiment is given by I = I_ max ^2 ( 2 ) . For the intensity to be half of the maximum intensity, I = I_ max 2 . I_ max 2 = I_ max ^2 ( 2 ) ( 2 ) = 1 2 For the minimum distance, 2 = 4 = 2 . The phase difference is related to the path difference x by = 2 x . 2 = 2 x x = 4 Using the relation x = y d D , we get: y d D = 4 y = D 4 d Substituting the given values: y = 500 10⁻⁹ 1.0 4 1.0 10⁻³ y = 125 10⁻⁶ m = 1.25 10⁻⁴ m