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

i think it’s a. this is because if they measured the mass of something with a mass of 150g, then their final answer would have to be approximately 150g. though all of their answers are almost 150, rosa and tran have the closest answers (or the exact answer) to the accurate one.