See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

Path difference x = yd D here d 5 y 2 2    (as d = 5) and D = 10 d = 50  so x = 5 5 2 50               = 4  corresponding phase difference will be  = 2 2 ( ) 4 2                         x or 2 4     I = I0 cos2 2        I0 cos2 4        = 0I 2 SECTION – B