In Young's double slit experiment, the intensity on screen at a point, where path difference is ' /4 ', is ' K/4 '. The intensity at a point when path difference is ' ' will be [ ( /4) = 1/ 2 , ( /2) = 0, (2 ) = 1]

Step-by-step solution

The phase difference is related to the path difference x by the formula: = 2 x For a path difference of x = 4 , the phase difference is: = 2 4 = 2 The intensity I at any point on the screen in Young's double slit experiment is given by: I = I_ max ^2 ( 2 ) Substituting = 2 : I = I_ max ^2 ( 4 ) = I_ max ( 1 2 )^2 = I_ max 2 We are given that the intensity at this point is K 4 . Therefore: I_ max 2 = K 4 I_ max = K 2 Now, for a point where the path difference is x = , the phase difference is: ' = 2 = 2 The intensity