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.

Options

A. `6.5mm`
B. `3.5mm`
C. `10.2mm`
D. `5.9mm`

Answer

B

Explanation

Let `n_(1)` bright fringe corresponding to wavelength `lambda_(1) = 500 nm` coincides with `n_(2)` bright fringe corresponding to wavelength `lambda_(2) = 700nm`.
`n_(1) (lambda_(1)D)/(d) = n_(2) (lambda_(2)D)/(d)` or `(n_(1))/(n_(2)) = (lambda_(2))/(lambda_(1)) = (7)/(5)`
This implies that `7^(th)` maxima of `lambda_(1)` coincides with `5^(th)` maxima of `lambda_(2)`. Similarly `14^(th)` maxima of `lambda_(1)` will coincide with `10^(th)` maxima of `lambda_(2)` asnd so on. Minimum distance `= (n_(1)lambda_(1)D)/(d)`
`= 7xx5xx10^(-7)xx10^(3) = 3.5xx10^(-3)m = 3.5 mm`