Monochromatic light of walelength 400nm and 560nm are incident simultaneously and normally on double slits apparatus whose slit sepation is 0.1 mm and screen distance is 1m. Distance between areas of total darkness will be

Options

A. 4 mm
B. 5.6 mm
C. 14 mm
D. 25 mm

Answer

d

Explanation

At the area of total darkness, minima will occur for both the wavelengths.
`:. ((2n - 1))/(2) lambda_(1) = ((2m + 1))/(2) lambda_(2)`
`implies (2n - 1) lambda_(1) = (2 m+ 1) lambda_(2)`
or ` ((2n + 1))/((2 m + 1)) = (560)/(400) = (7)/(5)`
or `10 n = 14 m + 2`
By inspection: For `m= 2, n = 3`. For `m = 7, n = 10`. The distance between then will be the distance between such point, i.e.,
`Delta s = (D lambda_(1))/(d) {((2n_(2) + 1) - (2n_(1) + 1))/(2)}`
Putting `n_(2) = 10` and `n_(1) - 3` and solving, we get `Delta x = 28` mm.