Visible light travels in air, which has an index of refraction of 1.0. It is incident on plastic which is covered by a coating (such that light hits the coating before the plastic). The coating has an index of refraction of 1.6 and is 0.50 microns thick. The ray reflected off the air-coating interface and the ray reflected off the coating-plastic interface experience interference. What frequencies could the light be

  λ = 0.64 10⁻⁶ m = 640 nm

Explanation:

The two reflected rays experience constructive interference, if we can see them, so we can one condition for interference, but let's see two phenomena that occur

* A ray when it is reflected by a surface of major spare part index has a phase change of 180º, these passes at the air-plastic interface

* A ray when passing a material with a refractive index changes its wavelength

       λ= λ₀ / n

taking into account these facts the condition in constructive interference is

             2 n t = (m + ½) λ

             λ= 2 n t / (m + ½)

if we suppose that we have the first inference m = 0

            λ = 2 1.6 0.5 10⁻⁶ (0 + ½)

            λ = 3.2 10⁻⁶ m

this wavelength is in the infrared

suppose an interference of m = 1

          λ = 1.6 10⁻⁶ / (1 + 1/2)

          λ = 1.06 10-6 m

m = 2

         λ = 1.6 10⁻⁶ / (2 + 1.5)

         λ = 0.64 10⁻⁶ m

this wavelength corresponds to the visible range (640 nm) orange

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