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 one fourth of the maximum.

Step-by-step solution

The intensity at any point on the screen in a Young's double slit experiment is given by: I = I₀ ^2 ( 2 ) where I₀ is the maximum intensity and is the phase difference. Given that the intensity falls to one fourth of the maximum: I₀ 4 = I₀ ^2 ( 2 ) ( 2 ) = 1 2 2 = 3 = 2 3 The phase difference is related to the path difference x by: = 2 x 2 3 = 2 x x = 3 The path difference is also given by x = yd D , where y is the distance from the central maximum. yd D = 3 y = D 3d