In a Young's double slit experiment, the maximum intensity of the interference pattern is I₀ . What is the minimum non-zero path difference between the two waves at a point on the screen where the intensity is I₀ 2 ?

Step-by-step solution

The intensity I at any point in the interference pattern is given by I = I₀ ^2 ( 2 ) , where is the phase difference. Given I = I₀ 2 , we have: I₀ 2 = I₀ ^2 ( 2 ) ( 2 ) = 1 2 For the minimum non-zero path difference, we take the smallest positive angle: 2 = 4 = 2 The relationship between phase difference and path difference x is = 2 x . Substituting the value of : 2 = 2 x x = 4 Answer: 4