A pair of glasses uses a nonreflective coating of index of refraction 1.4 to minimize reflection of light with wavelength 500nm. If the index of refraction of the glass is 1.5, what is the minimum non-zero thickness of the coating

 d = 178.57 10⁻⁹ m

Explanation:

For this exercise we must find the thickness to minimize the reflection, so the interference for the reflection must be destructive.

To find the expression we must take into account, two things:

* When the light goes from an index mordant medium to one with a higher refractive incoe, it undergoes a phase change of 180 (pi radians)

* within the film the wavelength of light is modulated by the index of refraction

           λₙ = λ₀/ n

In this case the light passes from the air to the reflective layer and undergoes a phase change of ∫π rad,  then it is reflected in the film-glass layer where it undergoes another phase change of π rad, therefore the total change of phase is 2π radians, this change is the or changes its value

period of the trigonometric functions, therefore its value does not change

  the expression for destructive interference is

           d sin θ = (me + ½) λₙ

           d sin θ = (m + ½) λ₀ / n

the minimum thickness occurs for m = 0 and if we take perpendicular incidence the sine = 1

          d = λ₀ /2 n

l

et's calculate

          d = 500 10⁻⁹ /( 2  1.4)

         d = 178.57 10⁻⁹ m

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