In a Young's double slit experiment, the initial path difference between the two interfering waves at a point P on the screen is 4 . A thin transparent sheet of refractive index and thickness t = 3( - 1) is then placed in front of one of the slits. The ratio of the initial intensity at point P to the new intensity at the geometric center O of the screen is

Step-by-step solution

The intensity at any point is given by I = I_ ^2 ( 2 ) . Initially at point P, the path difference is x_P = 4 . The phase difference at P is _P = 2 4 = 2 . The initial intensity at P is: I_P = I_ ^2 ( 4 ) = I_ 2 When the transparent sheet is introduced, the additional optical path difference created at the geometric center O is: x_O = ( - 1)t = ( - 1) [ 3( - 1) ] = 3 The new phase difference at O is _O = 2 3 = 2 3 . The new intensity at O is: I_O = I_ ^2 ( 3 ) = I_ ( 1 2 )^2 = I_ 4 The required ratio is: I_P I_O =