In Young's double slit experiment, the intensity of light at a point on the screen where the path difference is is K units ( is the wavelength of light used). The percentage change in intensity at a point where the path difference is 6 and the above point is

Step-by-step solution

The intensity I at any point on the screen in Young's double slit experiment is given by I = I_ max ² ( 2 ) , where is the phase difference. The phase difference is related to the path difference x by = 2 x . For the first point, the path difference is x₁ = . Thus, ₁ = 2 = 2 . The intensity at this point is I₁ = I_ max ² ( 2 2 ) = I_ max ²( ) = I_ max (1)^2 = I_ max . Given I₁ = K , we have I_ max = K . For the second point, the path difference is x₂ = 6 . Thus, ₂ = 2 6 = 3 . The intensity at this point is I₂ = I_