1. Not finding help here? Sign up for a free 30min 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!

Optics qn

  1. Apr 16, 2009 #1
    1. The problem statement, all variables and given/known data
    Light falls at normal incidence onto a transparent film on a substrate as shown
    2007examqn.jpg

    i. what is the optical path length difference in the case of normal incidence?
    ii. if nf>no and nf>ns, what is the condition for constructive interference in the case of normal incidence? explain your answer.


    2. Relevant equations
    OPL=nd


    3. The attempt at a solution
    ok im studying for an exam and this qn has been on the last 2 exams, and i have absolutely no idea how to do it, it cant be that, just an understanding issue. ok i know that the OPL=nd, i got no idea what the questions is asking me, or how to do it, plz can anyone give me some info?
     
  2. jcsd
  3. Apr 16, 2009 #2
    anyone is it something like OPLn(AB+BA)=2nt?
     
  4. Apr 18, 2009 #3
    anyone
     
  5. Apr 18, 2009 #4
    Your condition for constructive interference is such that twice the thickness of the glass is equal to an integer number of wavelength (the wavelength is taken from inside the film) MINUS Pi (because the rays reflected off the air-film boundary will be given a Pi phase boost because nf>no, however the rays reflecting off the film-substrate boundary will not, since ns<nf)
     
  6. Apr 18, 2009 #5
    so it it just for constructive, 2ntCos(theta)=m*lambda and for destructive 2ntCos(theta)=(m+0.5)*lambda, when m is an integer?
     
  7. Apr 18, 2009 #6
    so it it just for constructive, 2ntCos(theta)=m*lambda and for destructive 2ntCos(theta)=(m+0.5)*lambda, when m is an integer?
     
  8. Apr 18, 2009 #7

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    No, but you almost have the right idea. As Maverick said, you need to account for the extra π phase shift that happens at A, but not at B. This is essentially equivalent to adding half a wavelength to the path difference.
     
  9. Apr 18, 2009 #8
    ahh ok... so then for constructive interference it is the opposite? 2nt=(0.5+m)lambda...... its the A and B part im confused about, becasue some gets reflected and some goes straight through.. or something.

    i understand but still lost lol, for part a. the phase difference is pi?
     
    Last edited: Apr 18, 2009
  10. Apr 18, 2009 #9
    or is it (m-0.5)lambda?
     
  11. Apr 18, 2009 #10
    sorry for some reason my computer is laggy and saying the same thing over and over again every time i make a post...
     
  12. Apr 20, 2009 #11

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Yes, either one of those would be fine.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Optics qn
  1. Optics qn (Replies: 7)

  2. Moment of inertia qn (Replies: 1)

  3. Modern Optics (Replies: 1)

Loading...