In a Young's double slit experiment using monochromatic light of wavelength , the intensity of light at a specific point on the screen is observed to be 25 % of the maximum possible intensity. The minimum path difference between the interfering waves at that point is:

Step-by-step solution

Let the maximum intensity be I_ max . The intensity I at any point on the screen is given by: I = I_ max ^2 ( 2 ) where is the phase difference between the two interfering waves. We are given that I = 0.25 I_ max = 1 4 I_ max . 1 4 I_ max = I_ max ^2 ( 2 ) ( 2 ) = 1 2 For the minimum path difference, we take the smallest positive angle: 2 = 3 = 2 3 The relationship between phase difference and path difference x is: = 2 x Equating the two expressions for : 2 x = 2 3 x = 3 Answer: 3