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Inteference of light ray in thin flim

  1. Aug 31, 2014 #1
    1. The problem statement, all variables and given/known data
    according to the notes in the photo, the light ray undergo of phase change of 0.5 λ when it strike and reflected from glass surface (denser) at point A . so the OPD of ray 1 &2 = 2nt-0.5λ..
    but in the 2nd photo , the OPD of ray 1 & 2 is given by 2nt+0.5λ . i am not sure which one is correct. can someone enlighten me on this?



    2. Relevant equations



    3. The attempt at a solution
     

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  3. Aug 31, 2014 #2

    BvU

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    Well well, you've really gone out of your way to make life easy for us poor helpers...

    Now all we need is some relevant equations and your attempt at a solution, right ?
     
  4. Aug 31, 2014 #3
    Sorry. I dont umderstand the situation so I hope someone can help me out on this.
     
  5. Aug 31, 2014 #4
    Or should I post this problem in other section? Which section should I post?
     
  6. Aug 31, 2014 #5

    ehild

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    They are equivalent. The electric field changes to the opposite when reflected from the surface of a higher index material. The phase change can be taken either pi or -pi, all the same. (If you add pi or - pi to the argument of a sine function, you get the same: sin(x+pi) = sin(x) cos (pi) = -sin(x) and sin(x-pi)= sin(x)cos (-pi) = -sin(x) as cos(pi)=cos(-pi)=-1.) They correspond to optical path differences +λ/2 or -λ/2.
    I would prefer 2nt+0.5 λ for the optical path difference, so as it is positive even with very thin layers.

    ehild
     
    Last edited: Aug 31, 2014
  7. Sep 1, 2014 #6
    hi, please refer to the photo 2 . It is shown that the n start from 1 , but not 0 .
    If the n start form 0 , the thickness surely will become negative value. which is indeed wrong.( from mathematical treatment)
    I cant understand why the n cant be start from 1 from the physics theory. can you explain on this?
     
  8. Sep 1, 2014 #7

    ehild

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    You asked if the OPD (optical path difference) can be written as 2nt-0.5λ or 2nt+0.5λ. In that context, n is the refractive index and t is the thickness of the layer.
    If you use μ to denote the refractive index, you should write the OPD as 2μt-0.5λ.

    If you mean the condition of constructive interference, the optical path difference should be integer multiple of λ, 2μt-0.5λ=kλ or 2μt=(k+0.5)λ, k=0, 1,2,....

    In case the OPD is taken 2μt+0.5λ, the condition of constructive interference is 2μt=(k-0.5)λ, k=1,2,....


    ehild
     
  9. Sep 1, 2014 #8
    why for
    2μt=(k-0.5)λ , the k starts from 1 but cant be 0 ? can you explain it using physics theory?
     
  10. Sep 1, 2014 #9

    ehild

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    t, the layer thickness would be negative at k=0, which is impossible.

    ehild
     
  11. Sep 2, 2014 #10
    i knew this (from maths) , but can you explain in 'physics way' ?
     
  12. Sep 2, 2014 #11

    ehild

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    Have you seen anything with negative thickness? For example how would a -0.2 m thick wall look like ?????

    ehild
     
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