# A thin film of a material with n=1.3 is used to coat a piece of glass (n=1.5). For what minimum film thickness will there be no glare from yellow-green light (λ=570 nm in air)? (Please show all...

A thin film of a material with n=1.3 is used to coat a piece of glass (n=1.5). For what minimum film thickness will there be no glare from yellow-green light (λ=570 nm in air)?

(Please show all your steps and a diagram to support your method.)

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**Answer: 110nm**

For a thin film, light reflects off of the surfaces involved and interfere with each other. The most common example is the soap bubble. When white light hits it you see patterns of different colors, this is due to interference via thin films.

There are four basic ideas you need to consider:

1. Interference - when two or more waves arrive simultaneously in the same point in space.

2. Constructive interference occurs when waves arrive in phase, meaning a crest of one wave lines up perfectly with a crest of another. The path difference will be an integer multiple of the wavelength: `mlambda`

3. Destructive interference occurs when waves arrive out of phase by an "half-integral number of wavelengths"* in other words, when a crest meats up with a trough, the amplitudes cancel. `(m+1/2lambda)`

4. Reflection Phase Shift "an extra half-wavelength phase shift occurs when light reflects from an optically more dense medium"* In other words from air to glass, the phase shift happens. From glass to air, the phase shift does not happen.

Let's apply these four steps to the problem.

1. Interference. Some of the light will reflect off of the front of the thin film, and some of the light will reflect off the back of the thin film. If we consider these two rays and that they are perpendicular to the film, then interference will occur. The ray that goes through the film and bounces back has to travel twice the thickness of the film before it starts to interfere with the front-reflected ray.

2. We want the green glare to be removed, so we do not want constructive interference.

3. We want both reflected rays to eliminate each other. The phase shift required will be due to the extra distance the back-reflected ray must travel, in addition to any surface reflection phase shift.

4. Reflection phase shift. Since both rays are reflecting off of a more optically dense medium (e.g. air to film, film to glass), they both shift by half a wavelength. Since this is the same for both, the effect is no net phase change between the two rays. This means that all of the phase change is due to the thickness of the film.

Let t be the thickness of the film. The ray must travel 2t to exit back out. We want 2t to equal `(m+1/2)lambda_n` so that it will be out of phase. If t is going to be the thinnest, then we select m=0.

Thus: `2t=lambda_n/2` , where `lambda_n` is the wavelength of the wave in the material. It is defined as

`lambda_n = lambda/n `

Substituting and solving for t,

`t=lambda/(4n)rArr(570text(nm))/(4*(1.3))~~110text(nm)`

*p683 Physics 6ed, Giancolli.