1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Interference in Thin Films

  1. Dec 18, 2014 #1
    1. The problem statement, all variables and given/known data
    In order to decide the thickness of a SiO2 layer(n=1.5) on Si, a wedge is prepared as shown in the picture. A total of 7 DARK fringes was observed, when light of wavelength 589nm shines prependicularly on the wedge. It is known that light reflected from the Si/SiO2 surface would have a phase change of π rad.

    physexam1q3.png
    ( I don't know why the image cannot appear after posting, it does appear in edit mode. Here is the link:
    http://s3.postimg.org/cbvntdri9/physexam1q3.png)

    i) Determine whether the fringe at the thinner end of the SiO2 wedge is bright or dark. Explain briefly. (2M)
    ii) Calculate the thickness of the SiO2 layer. (3M)
    iii) If the setup is illuminated by a light of shorter wavelength, what chages would you expect for the number of fringes observed? Explain briefly. (2M)
    iv) If the setup is placed in water(n=1.33), what changes would you expect for the number of fringes observed? Explain your answer briefly. (2M)

    2. Relevant equations
    For constructive interference, 2nt = (m+1/2)λ, m = 0, 1, 2......
    For destructive interference, 2nt = mλ, m = 0, 1, 2......

    3. The attempt at a solution
    i) Dark? Just guess. What should I consider?
    ii) 2nt = mλ, m = 7
    2(1.5)t = 7*(589*10^-9)
    t = 1.96*10^-7m
    iii) Just guess: Because 2nt = mλ, where 2nt is a constant. When λ decreases, m has to be increased. So the number of fringes observed increases.
    iv) Same? But it doesn't seem correct......

    Thank you!!
     
  2. jcsd
  3. Dec 18, 2014 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    You observe reflection. Interference happens between the directly reflected ray (blue) and the one, entering into the layer and reflected back from the SiO2-Si interface (red)
    For constructive interference, the phase difference between the blue and red rays should be integer times 2pi. What is the phase change of the directly reflected light?
    What are the correct equations for constructive and destructive interference?
     
  4. Dec 18, 2014 #3
    The equations are
    "For constructive interference, path difference = mλ, m = 0, ±1, ±2......
    For destructive interference, path difference = (m + 1/2)λ, m = 0, ±1, ±2......" ?

    And both the red ray and the blue ray has undergone phase change?
    Since SiO2's n=1.5> air's n =1, the blue ray has phase change of pi rad.
    And it's given that the red ray has a phase change of pi rad.
     
  5. Dec 19, 2014 #4

    ehild

    User Avatar
    Homework Helper
    Gold Member

    They both undergo phase change at the boundary from where the reflect.
     
  6. Dec 19, 2014 #5
    But the path difference between they reflect changes continuously in part a)? How should I deal with this?
     
  7. Dec 19, 2014 #6

    ehild

    User Avatar
    Homework Helper
    Gold Member

    You see a dark strip on the wedge where the thickness of SiO2 corresponds to the condition of destructive interference.The thickness depends on x, the horizontal distance from the edge and the angle of the incline - how?
     
  8. Dec 19, 2014 #7
    tanθ=thickness/horizontal distance,
    thickness=horizontal distance*tanθ?
     
  9. Dec 19, 2014 #8

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Yes. What should be the thickness at the first dark strip?
     
  10. Dec 19, 2014 #9
    1/2λ=1/2*589=294.5nm?
     
  11. Dec 19, 2014 #10

    ehild

    User Avatar
    Homework Helper
    Gold Member

    The wavelength 589 nm is given for air. You have to multiply the thickness by the refractive index. And the ray travels the thickness twice, once forward, then backward.
     
  12. Dec 19, 2014 #11
    Oh yes. 2*1.5*thickness=1/2λ=294.5, thickness=294.5/3=589/6nm?
     
  13. Dec 19, 2014 #12

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Good. You see the first dark fringe where the thickness is 589/6 nm. Where do you see the seventh one?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Interference in Thin Films
Loading...