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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)