(II) A lens appears greenish yellow (λ = 570nm is strongest) w...

Question

(II) A lens appears greenish yellow (λ = 570nm is strongest) when white light reflects from it. What minimum thickness of coating (n = 1.25) do you think is used on such a glass lens (n = 1.56), and why?

Accepted Answer

Hi everyone. Let's take a look at this practice problem dealing with thin films. So in this problem, a thin film of soapy water with a refractive index of N equal to 1.33 covers the surface of a plastic toy with a refractive index of N equal to 1.50. When white light hits the surface, the reflected light appears blue with a wavelength of lambda equal to 460 nanometers being the most prominent. What is the minimum thickness of the soap film that creates this reflection? We're given four possible choices as our answers. For choice A we have 1.7 multiplied by 10 to the negative 7 m. For choice B, we have 3.4 multiplied by 10 to the negative 7 m. For choice C we have 4.5 multiplied by 10 to the negative 7 m. And for choice D, we have 5.6 multiplied by 10 to the negative 7 m. The first thing we want to do is draw a diagram of our problem through and draw a horizontal line indicating the surface of the soapy film. And we're gonna draw a line below that, another horizontal line below that indicating the surface of the toy. So we're gonna have light coming in, it's gonna hit the surface of the film and it's gonna do two things. It's going to reflect part of the light which is going to go to your eye. And then another part is going to be refracted through the film and it's gonna reflect off the surface of the toy and that's gonna come back up, refract back through the film and back into air. And the two rays, um the one that is reflected at the surface of the film and the other that has been reflected off the surface of the toy are both going parallel back to your eye once they reach air. Now, we're asked to calculate the minimum thickness of this film. So we'll call that thickness T. And the way we're going to do that is calculate the differences in the phase changes between these two rays. So we'll call the one that has been reflected off the um film, we'll call it ray one and the one that's been reflected off the toy, we're gonna call re two. Now, once the rays are in air, they travel the same distance to your eye. So there's not gonna be any additional phase changes. So we really only need to worry about the phase changes due to the reflections and the additional travel distance or ray two. So we're gonna calculate the difference between those two phase changes, we'll call that delta Phi and that's gonna be equal to P two minus by one or 52 is going to be the additional phase that has been picked up due to the uh additional path length as well as the reflection. And phi two is going to be um the phase change relative to the original ray coming in and striking the surface um due to the reflection at the surface of the film. So we're gonna add or we're going to calculate those quantities, we'll have delta phi is equal to. And the first thing, we're gonna have the phase change for phi two. And so that's gonna be due to the path difference between it and ray one. So it has to travel in a distal distance which is twice the thickness. So will that be two T and we'll need this actually in a phase. So we actually want an angle. So we're gonna divide this by the wavelength of light in the film, so that'll be landed the film. And so they don't give me the fraction of the phase that I've gone through. I will need to multiply that by two pi to get the actual amount of the phase change that has occurred, but we also have to take into account the reflection. So since I'm going from a um lower index of refraction, in this case, the 1.33 and it's being refracted or being reflected off of a surface material has an index of refraction that is higher than that, which in this case is 1.5 we're gonna pick up an additional pi phase change. We have to add to our path difference phase change. A pi we look now at ray one and look at its phase change. And here we're just gonna have the face changed due to the surface. And again, it's gonna be the similar case as our other ray. So here we're, we have light coming in in air which has an index of refraction of one and it's being reflected off a material that has an index of refraction of 1.33. So the index of fraction of air is lower than the index of refraction of the film. So we're gonna pick up an additional pi phase change. So we'll have minus pie. Now, this phase difference here has to be an integer multiple of two pi because we're looking for constructive interference here, if we had destructive interference, the uh raised would cancel out and we'd see nothing. But since we want to see the light, we want constructive interference. So it has to be an integer multiple of two pi. But here we're asked for the minimum thickness. So we want the minimum phase change or minimum phase difference. That means that's just gonna be equal to two pi. So we're basically setting our end value equal to one. So now I can solve this equation for the thickness. I noticed that the pies will cancel out, just have a positive and a negative one. And I can divide both sides by two pi. And so that'll leave me with two T divided by a Lambda film is equal to one. So we can solve this for T and that would just be T is equal to Lambda film divided by two. Now, the problem I wasn't given the wavelength of light in the film, I was given the wavelength of light in air. So I'm gonna have to relate the wavelength in air to the wavelength of the film. So recall your relation for that, that is going to be in air multiplied by Lambda. Air is equal to in film multiplied by Lambda film where in air is gonna be index of refraction of air. Lambda air is could be the wavelength of light in air in film is our index of refraction of film. And LAMBDA film is our wavelength of light in the film. We're gonna simplify this a little bit we're gonna set in air equal to one since the index of a fraction of air is one. And we're gonna solve this for Lambda film. So we have lamb A film is equal to LAMBDA error divided by in film. And here we just um divide both sides by LAMBDA film and set in air equal to one. So I can take this expression plug it in for Lambda film in our equation for the thickness. And so I'll have tea is equal to Lambda air divided by the quantity of two in film. We have values for both the wavelength and the index of refraction. So we'll plug this in. So we'll have T is equal to the wavelength in air was 460 nanometers. But I'll need to convert that into meters. I'll take the 460 multiply it by 10 to the negative nine converted into meters. I'll then divide that by the quantity of two multiplied by the index or a fraction of the film which is 1.33. And so I can plug those values to my calculator. I'll get T it's equal to 1.7 multiplied by 10 to the negative 7 m. And here just kept two significant figures. And so if I look at my answer choices, this corresponds to answer choice. A so just a quick recap of what we've done here, we started off by looking at the differences in the phase changes of our two reflected rays, one off the surface of the film and the one off the surface of the toy be taking care to include the effects of those reflections. This allowed us to write our thickness in terms of the wavelength of light in the film. We then had to relate that wavelength of light in the film to the wavelength and light of air. And once we made that um connection, we could plug in for the wavelength of light in the film in terms of the wavelength of light and air into our formula and calculate a value for our thickness. So I hope that this has been useful and I'll see you in the next video.

(II) A lens appears greenish yellow (λ = 570nm is strongest) when white light reflects from it. What minimum thickness of coating (n = 1.25) do you think is used on such a glass lens (n = 1.56), and why?

Key Concepts

Thin Film Interference

Wavelength and Refractive Index

Constructive Interference Condition

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