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

Answer

D

Explanation

At the area of total darkness, in double slit apparatus, minima will occur for both the wavelength which are incident simultaneously and normally.
`((2n+1)/(2))lamda_(1)=((2m+1))/(2)lamda_(2)implies(2n+1)/(2m+1)=(lamda_(2))/(lamda_(1))`
`implies(2n+1)/(2m+1)=(560)/(400)=(7)/(5)` or `10n=14m+2`
By inspection the two solutions are
(i). if `m_(1)=2,n_(1)=3`
(ii). if `m_(2)=7,n_(2)=10`
`therefore` Distance between are as correspond to these points.
`therefore` Distance `DeltaS=(Dlamda_(1))/(d)[((2n_(2)+1)-(2n_(1)+1))/(2)]`
Now putting `n_(2)=10` and `n_(1)=3`
`DeltaS=4xx7xx10^(-3)mimpliesDeltaS=28mm`