Assuming that the light in air has a wavelength of 500 nanometers, the minimum thickness of the film needed is 83.9 nm.
Given; Index of refraction of polystyrene; η_p = 1.49
Index of refraction of Fabulite; η_f = 2.409
Light's in-air wavelength is 500 nm.
In order to answer this question, it will be found that light will initially be reflected twice: once from the top coating layer and once from the contact between the coating and the polystyrene material.
The light's additional path in the coating is 2t, whereas the path difference is 2t 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
Therefore The minimum thickness of the film required assuming that the wavelength of the light in air is 500 nanometers is; 83.9 nm
The complete question is : 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.
To learn more about Destructive Interference refer to :
https://brainly.com/question/23594941
#SPJ4
