Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Optics, concerning a thin film

  1. Jan 26, 2004 #1
    My friend asked me earlier about this homework he had. He said, there's a thin film on a piece of glass, and when you look at it, it looks green. Is this contructive interference?

    So, I was thinking, This would be kind of an iridesence? Depending on what angle you were looking at the film, you would see different colors right?

    So I was trying to think up an equation to explain this, and the best I can come up with is [tex]\lambda = 2 cos\theta h [/tex] where [tex] \lambda [/tex] is the wavelength of the light, theta is the angle of depression from parallel to the surface being observed and h is the thickness of the film itself. So I figure if the light reflects off the surface of the glass and the film, you'll get some sort of constructive interference. And that is why you see green, right?
     
    Last edited: Jan 26, 2004
  2. jcsd
  3. Jan 26, 2004 #2
    I've decided against my original equation. I believe it is sin instead of cos, I had the triangle backwards in my head when I was thinking about it. Anyways, I also don't know how to represent wether or not the interference is constructive. I know, if it travels a full wavelength inside the film, it will construct. And it goes through then up again(off the glass).

    I'm not positive, but I think the "2" is there correctly.

    So my new equation would be

    [tex] \lambda = 2sin\theta h [/tex]

    Is this correct? Someone please correct it if it's wrong, or I'll be walking around stupid.


    *Edit: No one caught that did they?
     
    Last edited: Jan 26, 2004
  4. Jan 26, 2004 #3

    Chi Meson

    User Avatar
    Science Advisor
    Homework Helper

    I think its pretty amazing if you are figuring this out yourself. BUt tehre are a couple of factors you need to work in: first of all, there is a 1/2 wave shift (crest becomes trough)in the reflection off the film surface.

    There may or may not be a 1/2 wave shift for the reflection off the second surface (the bottom of the film); it depends on what the film is made of and what it is on.

    Assuming there is no 1/2 wave shift on the second surface, then the light that goes through the film must travel a total distance (inside the film) equal to 1/2 the wavelength of that specific color of light in order to have constructive interference with the top-surface reflected light.

    OH, and the wavelength mentioned above must be the wavelength of the light in the film subsance which is (wavelength in vacuum/index of refraction). So your formula is on the right track, but it is a couple of stations short. AND, to follow convention, use the angle found between the ray and a line perpendicular to the surface (the complementary to the angle from the surface); this is known as the angle of incidence
     
    Last edited: Jan 26, 2004
  5. Jan 26, 2004 #4
    Okay that makes sense.

    But could you elaborate on the 1/2 wavelength shift?

    Btw I do have a background in this, but only highschool level, and a year or so ago. Nothing extensive either.
     
  6. Jan 27, 2004 #5

    Chi Meson

    User Avatar
    Science Advisor
    Homework Helper

    WHen waves rebound off of an interface (that's where the two media meet) the reflected wave will be "inverted" if the second medium has a wave speed that is slower than the first.

    THis means that crests become troughs and troughs become crests when the light reflects off the top surface. THis is the same as taking any sine wave and "shifting" your position on the wave forward or backward by 1/2 the wavelength: wherever you start on the wave, you end up at the same amplitude but opposite side of the equilibrium line.

    BUt as the light hits the bottom part of the film where it meets the glass, chances are that the speed of wave propagation in the film is slower than that of the glass; there is no inversion of the wave as it reflects off a "slow-fast" interface.

    But, maybe the film is a light oil on a special type of glass with a high index of refraction (higher n = slower speed), then the bottom reflection could also be a "fast-slow" reflection; this too would cause an inversion for this reflection, and then the light would have to travel a distance equal to a full wavelength inside the film.


    ADD to the already complicated formula the fact that snells' law must be used to determin the angle that the light takes inside the film (n_1 sin theta_1 = n_2 sin theta_2) Did you do optics in high school?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Optics, concerning a thin film
Loading...