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Thin Film Interference and reflection

  1. Dec 8, 2009 #1
    1. The problem statement, all variables and given/known data
    What is the thinnest film of a coating with n = 1.43 on glass (n=1.52) for which destructive interference of the red component (660 nm) of an incident white light beam in air can take place by reflection?

    2. Relevant equations
    2t = m*lambda?

    3. The attempt at a solution
    There really is none. Honestly I have been reading through the book for like an hour and trying to figure out what everything means but I just don't understand what to do at all. Any help at all is really appreciated. This just doesn't make sense to me :confused:
  2. jcsd
  3. Dec 8, 2009 #2

    Andrew Mason

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    The purpose of the coating is to create two reflecting surfaces separated by a small distance (some light will always reflect when it passes from one medium into another with a different index of refraction). If this distance is 1/4 of a wavelength of the incident light, the light reflecting off the coating/lens surface destructively interferes with the light reflecting off the air/coating surface (ie the wave reflected from the coating/glass boundary is a 1/2 wavelength out of phase with the light reflecting from the air/coating boundary). This reduces light reflection and increases the amount of light passing through the lens.

    So you have to create a layer of thickness equal to 1/4 of the wavelength of the light.

  4. Dec 8, 2009 #3


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    The thickness of the coating should be 1/4 of the wavelength inside the layer. The wavelength changes when the light enters from air to an other material with refractive index n. If the wavelength of the incident light is [itex] \lambda_0 [/itex], that in the material is [itex] \lambda=\lambda_0/n [/itex]. So the thickness of the antireflecting coating should be [itex] d=\lambda_0/(4n)[/itex].

  5. Dec 8, 2009 #4
    Thank you very much. That helped me out a lot. I couldn't not figure that out at all.

    Thanks again.
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