In the Young's double slit experiment, the intensity of light at a point on the screen is 'K', being the wavelength of light used. The intensity at a point where path difference is 6 will be [ ( 6 ) = 3 2 ]

Step-by-step solution

The maximum intensity in Young's double slit experiment is given as I_ max = K . The phase difference is related to the path difference x by the formula: = 2 x Given the path difference x = 6 , the phase difference is: = 2 6 = 3 The intensity I at any point on the screen is given by: I = I_ max ^2 ( 2 ) Substituting the values of I_ max and : I = K ^2 ( 6 ) Since ( 6 ) = 3 2 , we get: I = K ( 3 2 )^2 = 3K 4 Answer: 3K 4