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!

Interference by amplitude division (double reflection) for wedges

  1. Aug 30, 2012 #1
    1. The problem statement, all variables and given/known data

    Transparent wax of refractive index n=1.3 is deposited on top of a glass plate of width 1cm and refractive index n=1.5. The thickness of the wax is 0.01mm at one end of the plate and tapers uniformly to zero at the other end of the plate, which is defined to be at x=0. At this end the surface of the wax and of the glass form a small angle α. Light is incident on the plate from above, i.e. it goes through the wax, and is normal to the surface of the glass plate. Fringes are formed due to interference of light reflected from the top surface of the wax and from the glass.

    i) Find the value of α.
    ii) At what values of x do bright fringes occur if λ=520nm?

    2. Relevant equations

    Thickness of wedge, d = x tan(α) ≈ αx

    3. The attempt at a solution

    i) tan(α)=0.01/10 = 1×10-3 radians

    ii) I'm not sure whether bright fringes occur at xn = ((ρ+0.5)λ)/2nα) or xn = (ρλ/2nα)
     
  2. jcsd
  3. Aug 31, 2012 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    You see the wedge from above. The bright fringes occur at those places where the phase difference between the ray reflected from the wax and that reflected from the glass makes integer times 2pi. The phase can change upon reflection and during travelling through a medium. The phase change upon reflection is pi when the wave arrives from a lower index medium to the surface of a higher index material, and it is zero in the opposite case. What is the phase change when the wave reflects from the wax and what is it when it reflects from the glass?

    ehild
     
  4. Sep 2, 2012 #3
    Does that mean the phase change when light reflects from wax = pi and the phase change when light reflects from glass = pi?
    How would I use this to find the values of x for which bright fringes occur?
     
    Last edited: Sep 2, 2012
  5. Sep 2, 2012 #4

    ehild

    User Avatar
    Homework Helper
    Gold Member

    As the phase changes at the interfaces cancel, the total phase difference is the phase shift inside the layer. It is
    (4pi/λ) nd =2kπ,
    where k is integer and d is the thickness of the layer. That means that bright fringes occur where the thickness is d=kλ /2n.

    ehild
     
  6. Sep 2, 2012 #5
    Thank you so much!
     
  7. Sep 2, 2012 #6

    ehild

    User Avatar
    Homework Helper
    Gold Member

    You are welcome.


    ehild
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Interference by amplitude division (double reflection) for wedges
Loading...