Correct Answer - D
Let `n^(th)` minima of `400nm` coincides with `m^(th)` minima of `560nm` then
`(2n-1)400=(2m-1)560`
`implies(2n-1)/(2m-1)=7/5=14/10=21/15`
i.e., 4th minima of `400nm` coincides with 3rd minima of `560nm`.
The location of this minima is
`=(7(1000)(400xx10^-6))/(2xx0.1)=14mm`
Next, `11^(th)` minima of `400nm` will coincide with `8^(th)` minima of `560nm`
Location of this minima is
`=(21(1000)(400xx10^-6))/(2xx0.1)=42mm`
`:.` Required distance `=28mm`