In a Young's double slit experiment, the intensities at two points, for the path difference 4 and 3 ( being the wavelength of light used) are I₁ and I₂ respectively. If I₀ denotes the intensity produced by each one of the individual slits, then I₁ + I₂ I₀ = ( 45^ = 1 2 , 60 = 1 2 )

Step-by-step solution

The phase difference is related to the path difference x by the formula = 2 x . The resultant intensity in a Young's double slit experiment when both slits have intensity I₀ is given by: I = 4I₀ ^2 ( 2 ) For a path difference of x₁ = 4 : ₁ = 2 4 = 2 ₁ 2 = 4 = 45^ I₁ = 4I₀ ^2(45^ ) = 4I₀ ( 1 2 )^2 = 4I₀ 1 2 = 2I₀ For a path difference of x₂ = 3 : ₂ = 2 3 = 2 3 ₂ 2 = 3 = 60^ I₂ = 4I₀ ^2(60^ ) = 4I₀ ( 1 2 )^2 = 4I₀ 1 4 = I₀ Adding the two intensities: I₁ + I₂ = 2I₀ + I₀ = 3I₀ Therefore, the required ratio is: I₁ + I₂