In a Young's double slit experiment, the intensities of the interference pattern at two points A and B on the screen are 75 % and 50 % of the maximum intensity, respectively. The ratio of the minimum path differences at points A and B is

Step-by-step solution

The intensity I at a point in a Young's double slit experiment is given by I = I_ ^2 ( 2 ) where is the phase difference between the interfering waves. For point A: I_A = 0.75 I_ = 3 4 I_ ^2 ( _A 2 ) = 3 4 ( _A 2 ) = 3 2 _A 2 = 6 _A = 3 The minimum path difference at A is x_A = 2 _A = 6 . For point B: I_B = 0.5 I_ = 1 2 I_ ^2 ( _B 2 ) = 1 2 ( _B 2 ) = 1 2 _B 2 = 4 _B = 2 The minimum path difference at B is x_B = 2 _B = 4 . The required ratio is: x_A x_B = / 6 / 4 = 4 6 = 2 3 . Answer: 2 : 3