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. `4mm`
B. `5.6mm`
C. `14mm`
D. `28mm`

Answer

D

Explanation

Let nth minima of 400 nm coincides with mth minima of 560 nm, then
`(2n - 1) ((400)/(2)) = (2m - 1) ((560)/(2))`
or `(2n-1)/(2m-1) = (7)/(5) = (14)/(10) =` …
i.e., `4^(th)` minima of 400 nm coincides with `3^(rd)` minima of 560 nm.
Location of this minima is,
`Y_(1) = ((2xx4-1)(100)(400xx10^(-6)))/(2xx0.4) = 14 nm`
Next `11^(th)` minima of 400 nm coincides with `8^(th)` minima of 560 nm, then
`(2n-1) ((400)/(2)) = (2m - 1) ((560)/(2))`
or `(2n-1)/(2m-1) = (7)/(5) = (14)/(10)= ` ...
i.e., `11^(th)` minima of 400 nm coincides with `8^(th)` minima of 560 nm.
Location of this minima is, `+Y_(2)` =
`((2xx11-1)(100)(400xx10^(-6)))/(2xx0.1) = 42mm`
Required distance `= Y_(2)-Y_(1) = 28 mm`
Hence, the correct optical is (D).