In a Young's double slit experiment using monochromatic light of wavelength , the intensity of light at the central maximum is I₀ . Let I₁ be the intensity at a point on the screen where the phase difference between the interfering waves is 2 , and I₂ be the intensity at a point where the path difference is 6 . The ratio I₁ I₂ is :

Step-by-step solution

The intensity of light in Young's double slit experiment at a point with phase difference is given by I = I₀ ^2 ( 2 ) , where I₀ is the maximum intensity at the central maximum. For the first point, the phase difference is ₁ = 2 . I₁ = I₀ ^2 ( 4 ) = I₀ ( 1 2 )^2 = I₀ 2 For the second point, the path difference is x = 6 . The corresponding phase difference is ₂ = 2 x = 2 ( 6 ) = 3 . I₂ = I₀ ^2 ( 6 ) = I₀ ( 3 2 )^2 = 3I₀ 4 The ratio of the intensities is: I₁ I₂ = I₀ 2 3I₀ 4 = 2 3 Answer: 2 3