In a young's double slit experiment, two wavelength of 500 nm and 700 nm were used. What is the minimum distance from the central maximum where their maximas coincide again ? Take `D//d = 10^(3)` . Symbols have their usual meanings.

Answer

3.5 mm

Explanation

let the `n_(1)^(th)` maxima of `lamda_(1)` coincide with `n_(2)^(th)` maxima of `lamda_(2)`.
`n_(1)(lamda_(1)D)/(d)=n_(2)(lamda_(2)D)/(d)implies(n_(1))/(n_(2))=(lamda_(2))/(lamda_(1))`
`implies(n_(1))/(n_(2))=(700)/(500)implies(n_(1))/(n_(2))=(7)/(5)`
Minimum integral value permitted for `n_(1)` is 7
`therefore` minimum distance `=n_(1)(lamda_(1)D)/(d)` where `(D)/(d)=10^(3)`
`=(7(500xx10^(-9))xx10^(3))/(1)=3.5mm`