In a Young's double slit experiment using identical slits, the intensity of light at a certain point on the screen is observed to be 75 % of the maximum possible intensity. If the minimum path difference at this point is expressed as n , where is the wavelength of the monochromatic light used, then the value of n is ________.

Step-by-step solution

Let the maximum intensity in the interference pattern be I_ max . The resultant intensity I at a point with phase difference is given by: I = I_ max ^2 ( 2 ) Given that I = 0.75 I_ max = 3 4 I_ max , we have: 3 4 I_ max = I_ max ^2 ( 2 ) ( 2 ) = 3 2 For the minimum path difference, we take the smallest positive angle: 2 = 6 = 3 The relationship between path difference x and phase difference is: x = 2 x = 2 3 = 6 Comparing this with n , we get n = 6 . Answer: 6