# Coating Eyeglass Lenses(destructive interference)

• Joe Butler
In summary, the minimum thickness of film needed on the eyeglass lenses to cancel light of wavelength 550nm reflected towards the eye at normal incidence is 96.01nm. To determine if any other wavelengths of visible light will be canceled or enhanced in the reflected light, you will need to solve for λ in the equations 2nt = (m+.5)λ and 2nt = mλ, using the known value of t from part (a). This will allow you to investigate both constructive and destructive interference.
Joe Butler

## Homework Statement

Eyeglass lenses can be coated on the inner surfaces to reduce the reflection of stray light to the eye. If the lenses are medium flint glass of refractive index 1.62 and the coating is fluorite of refractive index 1.432, (a) what minimum thickness of film is needed on the lenses to cancel light of wavelength 550nm reflected towards the eye at normal incidence, and (b) will any other wavelengths of visible light be canceled or enhanced in the reflected light?

## Homework Equations

2nt = (m+.5)λ
n = refractive index of film
t = thickness of film
m = 1, 2, 3, …
λ = light wavelength in vacuum (air)

## The Attempt at a Solution

I have already solved the first part to find that t=96.01nm but i don't know how to solve for part b. I was thinking that you would have to plug the thickness 96.01nm into 2nt = mλ to solve for any possible constructive interference wavelengths but i am not sure.

Yes, you are thinking correctly.
From part (a), you now know the thickness t. So it is no longer an unknown.
For part (b), λ is the unknown in the equations 2nt = (m+.5)λ and 2nt = mλ. (You want to investigate both constructive and destructive interference.)

## 1. What is destructive interference in the context of coating eyeglass lenses?

Destructive interference refers to the phenomenon that occurs when two waves of light with opposite phases meet and cancel each other out, resulting in a decrease in the overall intensity of light. This is used in coating eyeglass lenses to reduce glare and improve visual clarity.

## 2. How does destructive interference help in coating eyeglass lenses?

By coating eyeglass lenses with a thin layer of material that has a different refractive index than the lens itself, destructive interference can be achieved. This causes any reflected light to be out of phase and therefore cancelled, resulting in reduced glare and improved vision.

## 3. What materials are commonly used in coating eyeglass lenses for destructive interference?

The most commonly used materials for coating eyeglass lenses are magnesium fluoride and titanium dioxide. These materials have a lower refractive index than the lens material and are able to create destructive interference when applied in a thin layer on the surface of the lens.

## 4. How thick does the coating need to be for destructive interference to occur on eyeglass lenses?

The thickness of the coating needed for destructive interference to occur depends on the wavelength of light and the refractive index of the coating material. Typically, the coating is only a few micrometers thick, which is much thinner than a human hair.

## 5. Is destructive interference the only method for coating eyeglass lenses?

No, there are other methods for coating eyeglass lenses such as anti-reflective coating and mirror coating. However, destructive interference is a popular method because it is effective in reducing glare and improving vision without altering the appearance of the lens.

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