Question

Coherent light with wavelength 603 nm passes through two very narrow slits, and the interference pattern is observed on a screen a distance of 3.00 m from the slits. The first-order bright fringe is a distance of 4.84 mm from the center of the central bright fringe.
Required:
For what wavelength of light will the first-order dark fringe (the first dark fringe next to a central maximum) be observed at this same point on the screen?

Answer 1

Answer:

λ  = 4.023 10⁻⁷ m

Explanation:

The double-slit interference phenomenon is described by

          d sin θ = (m + ½) λ          destructive interference

          d sin θ = m λ                  constructive interference

we can use trigonometry

         tan θ = y / L

how these experiments occur for small angles

         tan θ = sin θ/cos θ = sin θ

         sin θ = y / L

we substitute

        d y / L = (m + ½) λ        destructive interference

        d y / L = m λ                 constructive interference

with the expression for constructive interference we look for the separation of the slits

        d = m λ L / y

        d = 1 603 10⁻⁹ 3 /4.84 10⁻³

        d = 3.738 10⁻⁴ m

Now let's analyze the case where the distance for constructive and destructive interference occurs at the same point y = 4.84 mm = 4.84 10⁻³m

          d y / L = (m + ½)  λ

          λ = $\frac{ d \ y}{L\ (m+ 1/2) }$

the first strip is for m = 1  

let's calculate

         λ  = $\frac{3.738 \ 10^{-4} 4.84 \ 10^{-3} }{ 3 \ ( 1 + 0.5) }$

         λ  = 4.023 10⁻⁷ m