In a double-slit experiment, if the central diffraction peak contains 13 interference fringes, how many fringes are contained within each secondary diffraction peak (between m = + 1 and + 2 in Dsinθ=mλ). Assume the first diffraction minimum occurs at an interference minimum. Express your answer as an integer.

Width of central diffraction peak is given by the following expression

Width of central diffraction peak = 2 λ D / d₁

where d₁ is width of slit and D is screen distance and λ is wave length.

Width of other fringes become half, that is each of secondary diffraction fringe is equal to

λ D / d₁

Width of central interference peak is given by the following expression

Width of each of bright fringe = λ D / d₂

where d₂ is width of slit and D is screen distance and λ is wave length.

Now given that the central diffraction peak contains 13 interference fringes

so (2 λ D / d₁) / λ D / d₂ = 13

then (λ D / d₁) / λ D / d₂ = 13 / 2

= 6.5

no of fringes contained within each secondary diffraction peak = 6.5