Question

A jeweler wishes to show off the intrinsic brilliance of their prize diamond by giving it an antireflective coating for light of wavelength 591 nm in air. What is the minimum thickness the jeweler needs for a coating material with an index of refraction of 1.43?

Answer 1

The minimum thickness for an antireflective coating on a diamond with a refractive index of 1.43 to minimize reflection for light of wavelength 591 nm is 206.5 nanometers.

Step-by-step explanation:

The question is asking for the minimum thickness of an antireflective coating on a diamond that has a refractive index of 1.43 so that it minimizes the reflection for light of wavelength 591 nm. To achieve this, the coating should create a condition of destructive interference for light at this wavelength.

Destructive interference occurs when the path difference between two waves is an odd multiple of half wavelengths. However, when light reflects off a medium with a lower refractive index to a medium with a higher refractive index, there is a phase change of π (equivalent to half a wavelength). This means that for a single non-reflective coating, the additional path length traveled within the film must itself be an odd multiple of half the wavelength within the material, not in air:

λ_in_material = λ_in_air / n

Destructive interference condition (with a phase shift upon reflection at the diamond-coating interface):

2t = (m + ½)λ_in_material

For the first order of interference (m = 0), thus:

2t = (½)λ_in_material

2t = (½)(λ_in_air / n)

t = (½)(λ_in_air / n) / 2

Substitute the given values:

t = (½)(591 nm / 1.43) / 2

t = 206.5 nm

The minimum thickness required for an antireflective coating on the diamond for the given wavelength is 206.5 nm.