Question

A piece of glass of index of refraction 1.50 is coated with a thin layer of magnesium fluoride of index of refraction 1.38. It is illuminated with light of wavelength 680 nm. Determine the minimum thickness of the coating that will result in no reflection

Answer 1

Thickness = 123.19 nm

Explanation:

Given that:

The refractive index of the glass = 1.50

The refractive index of thin layer of magnesium fluoride = 1.38

The wavelength of the light = 680 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 thin layer of magnesium fluoride = 1.38

`{\lambda}` is the wavelength

So, thickness is:

`Thickness=\frac {680\ nm}{4* 1.38}`

Thickness = 123.19 nm