Title: Electromagnetic waves: Interference
1Electromagnetic waves Interference
- Wednesday November 6, 2002
2Haidingers Bands Fringes of equal inclination
d
n1
n2
Beam splitter
P
?1
x
?
?1
f
Extended source
Focal plane
Dielectric slab
PI
P2
3Fizeau 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
4Fizeau Fringes fringes of equal thickness
where,
Then if film varies in thickness we will see
fringes as we move our eye. These are termed
Fizeau fringes.
5Fizeau Fringes
Beam splitter
Extended source
n
n2
n
x
6Wedge between two plates
1
2
glass
D
y
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
7Wedge between two plates
Maxima 2y (m ½) ?o/n Minima 2y m?o/n
D
y
air
Look at p and p 1 maxima yp1 yp ?o/2n ?
?x? where ?x distance between adjacent
maxima Now if diameter of object D Then L?
D And (D/L) ?x ?o/2n or D ?oL/2n ?x
L
8Wedge 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.
9Newtons rings
Used to test spherical surfaces
Beam splitter
10Newtons rings
Maxima when, y (m1/2) ?o/2n Gives
rings, Rm2(m1/2)?oR/n
R- Rcos? y or, R2(R-y)2r2 ? R2(1-2y/R)
r2
?
i.e. r2 2yR
R
R
y
r
11Reflection 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
12Anti-Reflection coatings
2
1
A. Determine thickness of film
air
n1
n2
film
n1 lt n2 lt n3
n3
glass
Thus both rays (1 and 2) are shifted in phase by
? on reflection.
For destructive interference (near normal
incidence) 2n2t(m1/2)? Determines the
thickness of the film (usually use m0 for
minimum t)
13Anti-Reflection coatings
2
1
air
B. Determine refractive index of film
n1
A
A
Near normal incidence Amplitude at A
n2
film
n3
glass
Since ? ? 1
14Anti-reflection coating
B. Determine refractive index of film
Amplitude at A
To get perfect cancellation, we would like EA E
A
should be index of AR film
15Multiple 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. ? gt 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
16Multiple 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
17Multiple Beam Interference
- Assume a (for the time being) a monochromatic
source - ?, ? small ( lt 30o) usually
- Now ? ? gtgt ?, ?
- 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