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

Answer:

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:

$Thickness=\frac {\lambda}{4\times n}$

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

${\lambda}$ is the wavelength

So, thickness is:

$Thickness=\frac {420\ nm}{4\times 1.35}$

Thickness = 77.7778 nm