The maximum intensity in a Young's double slit experiment is I₀ . Distance between the slits ( d ) is 5 , where is the wavelength of light used. The intensity of the fringe, exactly opposite to one of the slits on the screen, placed at D = 10d is _______.

Step-by-step solution

The position of the point on the screen exactly opposite to one of the slits is at a distance y = d 2 from the central maximum. The path difference x at this point is given by: x = yd D Substituting y = d 2 and D = 10d : x = ( d 2 )d 10d = d^2 20d = d 20 Given that the distance between the slits is d = 5 , we have: x = 5 20 = 4 The corresponding phase difference is: = 2 x = 2 ( 4 ) = 2 The intensity at this point is given by: I = I₀ ^2 ( 2 ) Substituting = 2 : I = I₀ ^2 ( 4 ) = I₀ ( 1 2 )^2 = I₀ 2 Answer: I₀ 2