Calculate the wavelength of light that has its third minimum at an angle of 30.0º when falling on double slits separated by 3.00~\mu\text{m}3.00 μm.

Given values:

As we know,

→

By putting the values, we get

By converting it to "nm", we get

Thus the response above is correct.  

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λ = 428.57 nm

Explanation:

Young's double slit experiment is described by the expression

          d sin θ = m λ                    constructive interference

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

where d is the distance between the slits, λ is the wavelength, m is an integer that describes the order of interference

In this case they give us the angle 30º, the distance between the slits d = 3.00 μm = 3.00 10⁻⁶ m and the order of interference m = 3; we use the destructive interference equation

         λ = d sin θ / (m + ½)

let's calculate

        λ = 3 10⁻⁶ sin 30 / (3 +1/2)

         λ = 4.2857 10⁻⁷ m

we reduce to nm

        λ = 4.2857 10⁻⁷ (10⁹ nm / 1 m)

         λ = 428.57 nm

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