In a Young's double slit interference experiment, the fringe pattern is observed on a screen placed at a distance D from the slits. The slits are separated by a distance d and are illuminated by monochromatic light of wavelength . Determine the distance from the central point where the intensity falls to half the maximum.

Step-by-step solution

The intensity at a point on the screen in Young's double slit experiment is given by I = I_ max ^2 ( 2 ) , where is the phase difference. Given that the intensity falls to half the maximum, I = I_ max 2 . I_ max 2 = I_ max ^2 ( 2 ) ^2 ( 2 ) = 1 2 ( 2 ) = 1 2 2 = 4 = 2 The phase difference is related to the path difference x by = 2 x . 2 x = 2 x = 4 For a point at a distance y from the central maximum, the path difference is x = yd D . yd D = 4 y = D 4d