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. 5.6mm
C. 14mm
D. 28mm

Answer

D

Explanation

At the area of total darkness minima will occur for both the wavelengths. `((2n+1))/(2)lamda=((2m+1))/(2)lamda_(2)` `implies(2n+1)lamda_(1)=(2m+1)lamda_(2)`
or `((2n+1))/(2m+1)=(560)/(400)=(7)/(5)` or `10n=14m+2`
by inspection of m=2,n=3 and for m=7, n=10, the distance between them will be the distance between such points.
i.e, `triangles=(Dlamda_(1))/(d){((2n_(2)+1)-(2n_(1)+1))/(2)}`
put `n_(2)=10`, `n_(1)=3`
On solving we get, `triangles==28mm`.