In Young's double slit experiment, the wavelength of light used is . The intensity on the screen at a point for path difference ' ' is 'X'. The intensity at the point for path difference ( 6 ) is ( 180^ = -1 , 30^ = 3 2 )

Step-by-step solution

The phase difference is related to the path difference x by the formula = 2 x The intensity at any point on the screen is given by I = I_ max ^2 ( 2 ) For a path difference x = , the phase difference is = 2 = 2 The intensity is I₁ = I_ max ^2 ( 2 2 ) = I_ max ^2( ) = I_ max Given that I₁ = X , we have I_ max = X For a path difference x = 6 , the phase difference is = 2 6 = 3 The intensity at this point is I₂ = I_ max ^2 ( /3 2 ) = I_ max ^2 ( 6 ) Substituting I_ max = X and ( 6 ) = 30^ = 3 2 : I₂ = X ( 3 2 )^2 = 3X