In YDSE, bichromatic light of wavelengths 400 nm and 560 nm
are used. The distance between the slits is 0.1 mm and the distance between the
plane of the slits and the screen is 1m. The minimum distance between two
successive regions of complete darkness is

Options

A. (a) `4mm`
B. (b) `5.6mm`
C. (c) `14mm`
D. (d) `28mm`

Answer

D

Explanation

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`