If 730-nm and 640-nm light passes through two slits 0.61 mm apart, how far apart are the second-order fringes for these two wavelengths on a screen 1.1 m away?

X2 - X2'=3.24×10⁻⁴m

Explanation:

Given Data

λ1=730nm=

λ2=640nm

d=0.61 mm

D=1.1 m

Distance of the second bright fringe from the central fringe is given by

Xn = = D x n x2 λ / d

X2 = 2 D λ1 / d

X2' = 2 D λ2 / d

Separation between the second bright fringes of the two wavelengths ( the separation between the second order fringes ) is ,

X2 - X2' = 2 D ( λ1 - λ2 ) / d

X2 - X2'= {2 x 1. 1 (730×10^-9 - 640×10^-9) }/ 0.61 x 10 ^-3

X2 - X2'=3.24×10⁻⁴m

its c i just checked it

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