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Electromagnetic waves Interference by puw61439

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									Electromagnetic waves:
Interference

Wednesday November 6,
2002


                        1
    Haidinger’s Bands: Fringes of equal inclination
                                                               d
                                                     n1
                                     Beam splitter           n2


    P
                                            1
x
            
                                           1
                f



                                           Extended
        Focal
                                           source
        plane                                             Dielectric
                                PI    P2
                                                          slab
                                                                   2
Fizeau Fringes: fringes of equal thickness

   Now imagine we arrange to keep cos ’ constant
   We can do this if we keep ’ small
   That is, view near normal incidence
   Focus eye near plane of film
   Fringes are localized near film since rays
    diverge from this region
   Now this is still two beam interference, but
    whether we have a maximum or minimum will
    depend on the value of t
                                                 3
 Fizeau Fringes: fringes of equal thickness

               I  I1  I 2  2 I1I 2 cos

      where,
                         2kt cos '
                         2kt  

Then if film varies in thickness we will see fringes as we move our eye.

These are termed Fizeau fringes.




                                                                     4
Fizeau Fringes
    2kt cos '
    2kt  




                          Beam splitter




    Extended source           n

                              n2


                      x      n
                                          5
Wedge between two plates
                              1 2



   glass                            y      D
   glass
                                             air
                       L


   Path difference    = 2y
   Phase difference  = 2ky -    (phase change for 2, but not for 1)

  Maxima 2y = (m + ½) o/n

  Minima 2y = mo/n


                                                                6
Wedge between two plates
Maxima 2y = (m + ½) o/n

                                              y   D
Minima 2y = mo/n
                                                  air
Look at p and p + 1 maxima             L

yp+1 – yp = o/2n  Δx

where Δx = distance between adjacent maxima

Now if diameter of object = D

Then L = D

And (D/L) Δx= o/2n or D = oL/2n Δx
                                                      7
Wedge between two plates
   Can be used to test the quality of surfaces




Fringes follow contour of constant y

Thus a flat bottom plate will give straight fringes, otherwise
ripples in the fringes will be seen.


                                                                 8
Newton’s rings
        Used to test spherical surfaces




                               Beam splitter




                                               9
Newton’s rings
  R- Rcos = y                       Maxima when,
  or,                                y = (m+1/2) o/2n
  R2=(R-y)2+r2
      R2(1-2y/R) + r2               Gives rings,
                             
  i.e.   r2   = 2yR              R   Rm2=(m+1/2)oR/n
                         R



                                     y

                                 r



                                                         10
Reflection from dielectric layer:
Antireflection coatings
 Important in instruments such as cameras
  where reflections can give rise to spurious
  images
 Usually designed for particular wavelength
  in this region – i.e. where film or eye are
  most sensitive



                                            11
Anti-Reflection coatings                                     2
                                                 1
  A. Determine thickness of film

                                            n1        air

       n1 < n2 < n3                         n2       film
                                            n3       glass

  Thus both rays (1 and 2) are shifted in phase by  on reflection.

   For destructive interference (near normal incidence)

                          2n2t=(m+1/2)

   Determines the thickness of the film
   (usually use m=0 for minimum t)

                                                                 12
Anti-Reflection coatings                             1
                                                                 2
                                                           air
  B. Determine refractive index of film
                                                n1       A’          A
        Near normal incidence                   n2        film
        Amplitude at A                          n3       glass

   EA       '  ' Eo
               2n1  2n2   n3  n2 
   EA         n  n  n  n    n  n 
                                       
               1    2  2   3     3   2 

                n  n2
   EA         3
                n3  n2                   Since  ’ ~ 1



                                                                     13
Anti-reflection coating
 B. Determine refractive index of film

  Amplitude at A’
                                   n2  n1
                         E A'   
                                   n2  n1
                                                       n3  n2 n2  n1
To get perfect cancellation, we would like EA = E A’          
                                                       n3  n2 n1  n2



    n2  n1n3                            should be index of AR film



                                                                      14
Multiple Beam interference
   Thus far in looking at reflectivity from a dielectric
    layer we have assumed that the reflectivity is small
   The problem then reduces to two beam interference
   Now consider a dielectric layer of uniform thickness
    d and assume that the reflectivity is large e.g. || >
    0.8
   This is usually obtained by coating the surface of the
    layer with a thin metallic coating – or several
    dielectric coatings to give high reflectivity
   Or, one can put coatings on glass plates , then
    consider space between plates
                                                        15
Multiple beam interference
               Let 12 =  21= ’ 12=  21= ’

                                 ’’ Eo              (’)5’Eo
                       Eo                  (’)3’Eo         (’)7’Eo
      

n1

 n2
          ’

               A             B         C          D
 n1
                   
                                   (’)2’Eo                 (’)6’Eo
                        ’ Eo                   (’)4’Eo                 16
Multiple Beam Interference
   Assume a (for the time being) a monochromatic source
   , ’ small ( < 30o) usually
   Now || = |’| >> , ’
   Thus reflected beams decrease rapidly in amplitude
    (from first to second)
   But amplitude of adjacent transmitted beam is about the
    same amplitude
   Amplitude of successfully reflected beams decreases
    slowly (from the second)
   Thus treat in transmission where contrast should be
    somewhat higher
   The latter is the configuration of most applications

                                                          17

								
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