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Index of Refraction and Thickness

  1. Dec 13, 2011 #1
    1. The problem statement, all variables and given/known data
    What is the thinnest soap film (excluding the case of zero thickness) that appears black when illuminated with light with a wavelength of 520 ? The index of refraction of the film is 1.34, and there is air on both sides of the film.

    2. Relevant equations
    Im wondering do I use the equation for constructive 2nt=m*lamda OR destructive 2nt=(m+1/2)*lambda??<--- this is mainly where I am confused.

    3. The attempt at a solution
    (0.5*520)/(2*1.34)=t
     
  2. jcsd
  3. Dec 13, 2011 #2

    ehild

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    Do you get any light reflected from the bulb when it is black? What happens in case of constructive and destructive interference? When will you get zero light intensity?

    ehild
     
  4. Dec 13, 2011 #3
    ahh i understand..i think. So anytime no light is reflected for refraction indexes...or black for slits etc...then its destructive. And you only "swap" construct. and destructive formulas when you have an air gap aka "wedge"?
     
  5. Dec 13, 2011 #4

    ehild

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    So it is destructive interference, but you need to us the formula 2nd=m*lambda. (d is the layer thickness and n is the refractive index.)
    You see the soap bubble when light waves, reflected from it, reach your eyes. The light wave is reflected from the wall of the bubble, but partly from the front surface of the wall and partly from the back surface. The reflected waves interfere, and the resultant wave is either stronger (constructive interference) or weaker (destructive interference) than the original one. The intensity of the resultant wave depends on the phase difference between the constituent waves. The phase of the wave changes by Δφ=(2π/λ) n d when it travels a distance d in a medium of refractive index n.
    The wave reflected from the back surface of the wall travels through the wall twice, so its phase changes by (4π/λ) n d. But the phase can change upon reflection, too. If the wave reflects from an optically denser medium, (with higher reflective index than the one of he medium the wave arrived from) the phase changes by pi.
    The soap bubble has higher refractive index than air.
    So the part of the incident wave that is reflected from the front surface of the wall changes its phase by pi. The other part of the wave enters into the soap layer, traverses through it, and reflects from the back surface. No phase change at that surface, as the wave reflects from a lower refractive index material. After reflection, the wave travels through the layer back and joins to the other part.
    So the phase of the first wave changes by pi, the phase of the other one changes by 4π/λ nd and the phase difference between them is Δφ=4π/λ nd-π.
    Destructive interference means that the phase difference of odd number of pi. (4π/λ)nd-π=(2m-1)π, that is 4/λ nd=2m, 2nd=mλ.

    ehild
     
  6. Dec 13, 2011 #5
    We were only taught about the phase shift being either pi or 0. I understand that its destructive (black light). Im not sure about your way of going about the phase shift...I dont think I've learned that yet. I do know, that the first reflected ray has a phase shift pi and the one from the back (high n--->low n) has a value of 0...therefore phase shift is pi. What is the rule for "swapping equations" between constructive and destructive...?
     
  7. Dec 13, 2011 #6

    ehild

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    Well, so you were not taught about the mathematical representation of waves.

    Both things, the layer thickness and the phase change at the interface(s) determine if the interference is constructive or destructive.
    When the phase change at reflection is the same for both waves, the interference is constructive if 2nt=m*lambda and destructive if 2nt=(m+1/2)*lambda.
    If one wave changes phase by pi, and the other does not, the interference is constructive if 2nt=(m+1/2)*lambda and destructive if 2nt=m*lambda.

    ehild
     
    Last edited: Dec 13, 2011
  8. Dec 13, 2011 #7
    Light reflected at the air/film boundary experiences a ∏ (λ/2) phase change.;
    Light at the film/air boundary experiences 0 phase change.
    If the film is very thin (much less than λ) then the 2 reflected waves have a phase change of λ/2 and destructive interference occurs..... the film will look black. Soap bubbles look black when they are very thin.
    If the film has a thickness of λ/2 then a path difference of λ is introduced between the front ray and the back (reflected )ray.... an overall phase difference of 3λ/2 so destructive interference occurs for the λ that satisfies this condition.
    Can you see how to find whether max or min occur.... take into account the phase difference at the reflecting surfaces then take account of the path difference as a result of the film thickness.
     
  9. Dec 13, 2011 #8

    ehild

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    @Technician: 3λ/2 is not phase difference. (You can see it from the dimension: it is length, and the phase must be dimensionless.) I think you meant 3pi instead.


    ehild
     
  10. Dec 13, 2011 #9
    You are absolutely correct ehild..... I just like to use ∏ = λ/2 (2∏ =λ) as a phase difference.
    I prefer thinking in wavelengths (path differences) than in phase differences.
    I don't think it makes any difference to the logic??????
    It may be crude but I think of the reflection at the air/film boundary as a 'kick' of λ/2 to the wave.
    I agree that 3pi is the correct term for a phase difference
    Perhaps it would be better if I said 'a phase difference equivalent to λ/2'
     
    Last edited: Dec 13, 2011
  11. Dec 13, 2011 #10

    ehild

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    It would be better. Or you can say "equivalent path difference corresponding to pi".
    I just do not understand why not the phase difference is used instead of the path difference + phase change mixed.

    ehild
     
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