I have a thin film question that I can't seem to figure out. Can one of you show me what I'm doing wrong? The problem is: A thin film 450 nm thick is suspended in air, and white light is shone perpendicular to the surface. The index of refraction of the film is 1.37. At what wavelength will the visible light reflected off the film undergo fully constructive interference? This is what I tried doing: the light reflecting off the top of the film will undergo a phase change of pi. The light reflecting from the bottom surface will have no phase change from reflection, but it will have a phase change of 2*n*L (which is the optical path difference). L is the thickness of the film = 450 nm. This means the total phase difference is 2*L*n + (1/2)lambda. Since I want this to be constructive interference, I set this eqaul to m*lambda, where m is some integer. Since I want the first wavelength where this can happen, I set m = 1. Thus my equation now reads 2*450*1.37 + (1/2)lambda = lambda. Solve for lambda, and I get the wavelength being 1233 nm. This is the wrong answer. What am I overlooking here?
Ok, I've read the question more carefully now. I believe they want you to find the wavelength in the visible that interferes constructively. 2466 nm, which does interfere constructively, is nevertheless outside the visible.
So how do I find any other frequency? I thought that setting m = 1 would find the absolute lowest frequency where that could occur.
You know how to do this already. Read the question carefully, you are not told to find the minimum frequency (maximum wavelength) that exhibits constructive interference, instead you are asked to find the wavelength in the visible that will produce constructive interference.
You're right. I mistakenly assumed that a bigger wavelength was a higher frequency. I figured that the lowest value of m would produce the closest to visible light that could exist. So, I tried setting a different value for m, and I came up with 822 nm, which is still outside the visible range. So then I used m=3, and came up with 493.2 nm, which is definitely visible. I'm trying it... And it works!! Thanks for the help, Physics Monkey. I would never have thought of that.
