Question

A material having an index of refraction of 1.35 is used as an antireflective coating on a piece of glass (n = 1.50). What should be the minimum thickness of this film in order to minimize reflection of 420 nm light?

Answer 1

Thickness = 77.7778 nm

Explanation:

Given that:

The refractive index of the glass = 1.50

The refractive index of the non reflecting coating = 1.35

The wavelength of the light = 420 nm

The thickness can be calculated by using the formula shown below as:

`Thickness=\frac {\lambda}{4* n}`

Where, n is the refractive index of the non reflecting coating = 1.35

`{\lambda}` is the wavelength

So, thickness is:

`Thickness=\frac {420\ nm}{4* 1.35}`

Thickness = 77.7778 nm