A thin film of polystyrene is used as an antireflective coating for fabulite (known as the substrate). the index of refraction of the polystyrene is 1.49, and the index of refraction of the fabulite is 2.409. what is the minimum thickness of film required? assume that the wavelength of the light in air is 500 nanometers.

For this case we only need the refractive index of the polystyrene. For an antireflective coating, we need a quarter of wave thickness at the wavelength in the air. Which means that the antireflective coating needs to be as thick as 1/4 of the wavelength, divided by the coating’s refractive index. This is expressed mathematically in the form:

x = λ / (4 * n)

where,

x = thickness

λ = wavelength of light

n = index of refraction of polystyrene

Substituting:

x = 500 nm / (4 * 1.49)
x = 500 nm / 5.96
x = 83.90 nm

AFOKE88 AFOKE88 · Answer 2 · 2022-02-21T17:09:11+01:00

The minimum thickness of the film required assuming that the wavelength of the light in air is 500 nanometers is; 83.9 nm

Calculations about Destructive Interference

We are given;

Index of refraction of polystyrene; η_p = 1.49

Index of refraction of Fabulite; η_f = 2.409

Wavelength of the light in air; λ = 500 nm

In this question, it will be discovered that light will first of all be reflected two times, first from the upper layer of coating and then from the interface between the coating and the polystyrene material.

The additional path travelled by the light in coating is 2t while the path difference is 2tη_p.

Thus, minimum thickness of film required will be gotten from the formula used in destructive interference which is;

2tη_p = λ/2

t = λ/2 * 1/2η_p

t = λ/(4η_p)

t = 500/(4 * 1.49)

t = 83.9 nm

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