In a Young's double slit experiment, the intensity at some point on the screen is found to be 3 4 times of the maximum of the interference pattern. The path difference between the interfering waves at this point is x where is wavelength of the incident light. The value of x is _______.

Step-by-step solution

The intensity at a point in Young's double slit experiment is given by I = I_ max ^2 ( 2 ) Given I = 3 4 I_ max 3 4 I_ max = I_ max ^2 ( 2 ) ( 2 ) = 3 2 2 = 6 = 3 The phase difference is related to the path difference x by = 2 x 2 x = 3 x = 6 Comparing with x = x , we get x = 6 Answer: 6