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Homework Help: Reflection or not with two layers with different indices of refraction

  1. Aug 13, 2004 #1
    The packet doesn't explain this concept very well, and the book doesn't seem to explain it at all.

    A glass surface (n = 1.5) is coated with a film with an index of refraction n=1.3. If light of frequency 6x10^14 cps is incident almost normally on the film, find the minimum thickness that will cause (a) reflection, and (b) non-reflection.

    Any tips on the concept or any formulas woluld be very helpful. Thanks!
  2. jcsd
  3. Aug 13, 2004 #2
    I think you have to calculate the angle of the first refracted ray entering the film, and then use this ray to calculate if the angle is at the critical angle of the glass.
    For the first part, I am assuming the ray comes from air and goes into the film.
    [tex] {n_1}{sin\theta_1} = {n_2}{sin\theta_2} [/tex]
    now sub in the values and solve for the angle of refraction.
    after doing this, you can solve to see what the critical angle of the glass will be:
    [tex] sin\theta = \frac {n_1}{n_2} [/tex]
    and see if 90 - the angle that you got in the first part is more or less than this critical angle. If it is more, then it vill cause reflection, but if it is leass, it will refract.
  4. Aug 13, 2004 #3
    when u say non-reflection, do u mean complete refraction or partial reflection-refraction? (It may not prolly matter here but i just wanted to know cuz i have not really seen the usage of non-reflection anywhere)

    -- AI
  5. Aug 13, 2004 #4

    Doc Al

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    Staff: Mentor

    thin film interference/anti-reflective coatings

    The basic idea is that when light hits a boundary some of it will reflect back. Since you have two boundaries (between air and the film and between the film and the glass) you will have two reflected waves. If those waves are out of phase, they will cancel thus minimizing reflection; on the other hand, if they are in phase, they will constructively interfere and thus maximize the reflection.

    So the trick is to figure out the phase difference between the first reflection and the second reflection. That depends on two things:
    (1) the thickness of the film. Remember that the second reflection passes through the film twice.
    (2) the phase changes at the boundaries. When light reflects from a low n to higher n boundary, the reflection gets a [itex]\pi/2[/itex] phase inversion. (So... will the two reflections have a net phase inversion in your problem?)​
    Your job is to figure out the film thickness that will give a [itex]\pi/2[/itex] (destructive) or a [itex]\pi[/itex] (constructive) phase difference between the two reflections. (Hint: a [itex]\pi/2[/itex] phase difference equals half a wavelength. What's the wavelength of the light when inside the film?)
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