Transcript Interference in Dielectric films (amplitude division):

Interference in Dielectric films (amplitude division):
Colours on surface of soap films are associated
with interference of light in single or multiple
(mother-of-pearl) thin surface layers of
transparent materials
Nature of interference pattern depends on
variables such as size and spectral width of
source and shape & reflectance of film
Double-beam interference from a film
P
Rays reflected from the top and bottom
plane surfaces of the film are brought
together at P by a lens
n0
nf
A
C
t
Transparent
film
B
ns
Substrate
Thin transparent film –
oil slick, metal oxide
layer, evaporated coating
on a flat, glass substrate
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Interference in Dielectric films (amplitude division):
Incident ray (1) hits A,
splits into reflected (2) and
refracted (3) rays
P
2
5
1
n0
nf
A
3
C
Transparent
film
t
B
ns
4
Refracted (3) ray is then
reflected at B
Reflected (4) ray gets
refracted into n0 at C,
emerging as (5)
Substrate
Multiple reflections may occur within film layer (dashed blue lines) until
intensity is lost; corresponding multiple rays emerge from film
But assuming reflectance is small, only 1st and 2nd emerging rays (2) &
(5) are considered
The two rays are brought together by converging lens (e.g. eye), to
superimpose and interfere at P
The two rays traverse different optical paths, have relative phase
difference, and constructive & destructive interference occur at P
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Interference in Dielectric films (amplitude division) :
D
(1)
(2)
i
i
n0
G
A
nf
i
t
t
E
(3)
(5)
C
F
(4)
t
t
B
ns
Optical path difference between emerging beams (2) & (5) is:
 = nf (AB + BC) – n0 (AD)
Let’s split AB = AE + EB
Thus,
and
BC = BF + FC
 = [ nf (AE + FC) – n0 (AD) ] + nf (EB + BF)
This shall be shown to be equal to zero
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Interference in Dielectric films (amplitude division) :
From Snell’s law:
From the figure:
and
Therefore,
n0 sin i  n f sin t
AE  AG sin t   AC 2sin t
AD  AC sin i
 sin t
2 AE  AC sin t  AD
 sin i
(11-2)
(11-3)
 n0

  AD
 nf


n0 AD  2n f AE  n f  AE  FC
So that
(11-1)




(11-4)
resulting in optical path difference of   n f ( EB  BF)  2n f EB (11-5)
However,
EB  t cost , hence,
  2n f t cost
(11-6)
(in terms of angle of refraction)
n0
From Snell’s law, sin  t 
and thus,
sin  i
nf
2
 n0 
2
1/ 2
  2n f t[1  sin  t ]  2n f t[1    sin 2  i ]1/ 2  2t[n 2f  n02 sin 2  i ]1/ 2 (11-7)
n 
 f 
(in terms of angle of incidence)
[What about the phase
shift due to reflection?]
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Interference in Dielectric films (amplitude division) :
Taking into account possible additional relative phase shift (as beam
(2) is externally reflected - nf > n0; while beam (5) is internally
reflected - nf > ns resulting in phase shift of  between them), the
following conditions apply:
Constructive interference:
 p   r  m
(11-8)
 p   r  (m  12 )
(11-9)
and
Destructive interference:
where p  optical path difference
r  equivalent path difference arising from phase change on
reflection
 i   t  0 ;   2n f t
Corresponding phase difference   k  (2 0 ) 
For normal incidence
Taking into account relative phase shift
  (2 0 )   
(11-10a)
(11-10b)
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Application of Interference in Dielectric films
(antireflecting coating) :
P
Application: production of antireflecting
coatings on optical surfaces
(2)
How does it work?
(5)
n0 = 1 (air)
nf
A
C
n0 = 1 (light of wavelength 0 in air)
t
B
ns
If film thickness
ns > nf (no relative phase shift between
beams (2) & (5))
[Consider normal incidence]
t  f 4
(f = wavelength of light in the film)
the optical path difference becomes   2n f t  n f  f / 2  0 2
Destructive interference occurs at this wavelength and to some extent at
neighbouring wavelengths  light reflected from film is the incident
spectrum minus the wavelength region around 0
Extinction is more efficient if amplitudes of two reflected beams are equal
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(total destruction)
Application of Interference in Dielectric films (antireflecting coating) :
In the case of normal incidence, the reflection coefficient (ratio of
reflected to incident electric field amplitudes) is given by:
1 n
r
1 n
n2
where n 
 relative index
n1
(11-11)
Condition for amplitudes of electric field reflected internally and
externally to be equal are (assume non-absorbing film):
their relative indices are equal, that is,
Since n0 = 1, we choose
n f  ns
nf
ns

or n f  n0 ns
n0 n f
(11-12)
Thus, to reduce reflectance in optical instruments handling white
light, we choose film thickness /4 with  in which detection system
is most sensitive (e.g. for the eye,  = 550 nm). Assuming n = 1.50 for
glass lens, ideally nf = 1.5 = 1.22. The nearest practical film
material is MgF2 with n = 1.38. Loss of reflected light near the middle
of VIS spectrum results in predominantly blue and red ends, thus,
coatings appear purple in reflected light.
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Fringes of equal thickness :
Air wedge formed with 2 microscopic
slides, t increases to the right
Fizeau fringes
(extended)
(contours reveal
nonuniformities in
thickness of film)
Arrangement of apparatus to view
interference from a wedge-shaped film
Interference by irregular film. Variation in film
thickness and angle of incidence determine the
wavelength regions reinforced by interference
For film of varying thickness t, optical path difference will vary as in
  2n f t cost
Thus, at normal incidence, bright & dark fringes will be seen because of
varying  due to varying t  producing fringes of equal thickness
Constructive and destructive interferences occur according to (11-8) &
(11-9)
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Fringes of equal thickness :
Beam splitter (in arrangement of apparatus) is set at angle 45 to
incident light (to enable light to strike film at normal incidence and
to provide transmission of part of reflected light into detector)
At normal incidence,
cos t  1 and   2n f t
Therefore, condition for bright and dark fringes (11-8) & (11-9) is:
2n f t   r 
m
bright
(m  12 ) dark
(11-13)
where r = /2 or 0 (depends on whether there is relative phase shift
 between rays reflected from top and bottom surfaces of film or
not)
In this case, r = /2
As t increases from 0 to d, (1113) is satisfied for consecutive
orders of m; a series of equally
spaced, alternating bright &
dark fringes (virtual) appear
internal
air reflection
glass
air
glass
external
reflection
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Newton’s rings :
Newton’s rings
apparatus.
Geometry for
production of Newton’s
rings
R 2  rm2  ( R  t m ) 2
Interference fringes of equal thickness produced by air wedge between lens and optical flat.
Equal-thickness contours for a perfectly spherical surface results in
concentric circles around point of contact with optical flat
At point of contact t = 0 (for m = 0 in (11-13)) appears dark (destructive
interference) because r = /2 due to reflection
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Newton’s rings :
(a)
(b)
Newton’s rings in (a) reflected light and
(b) transmitted light are complimentary
Newton’s rings can also be observed
through light transmitted through optical
flat, but, has poor contrast (incomplete
cancellation in destructive interference)
and bright centre.
Concentric ring pattern will be
distorted if there are
irregularities in the surface of
the lens
Radius rm of mth-order dark
fringe, air-film thickness tm,
and radius of curvature R of
lens surface is related
through:
rm2  t m2
R
(11-14)
2t m
rm is measured, tm of air
determined through condition
(11-13), and R can then be
found
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Newton’s rings (quantitative example):
Problem:
Plano-convex lens (n = 1.523) of 0.125 diopter power is placed, convex surface
down, on an optically flat surface. Using a travelling microscope and sodium light
( = 589.3 nm), interference fringes are observed. Determine the radii of 1st and
10th dark rings.
Solution:
relative phase difference: r = /2
Condition (11-13): 2nftm + ½ = (m + ½)  
tm = m  / 2nf
tm  m 2
Film is in air, nf = 1,
thus,
2
2
Ring radius given by: R  rm  t m
(neglecting very small term tm2, we have
2t m
rm2  2 Rt m )
Radius of curvature from lensmaker’s equation: 1  (n  1) 1  1 
R R 
f
2 
 1
f = 1/P = 1/0.125 = 8 m, n = 1.523, R2  , thus, R1 = R = 4.184 m.
Then,
rm2  2 Rt m  2 Rm 2   mR
r12  (1)(4.184)(589.3 109 )  2.466 106 m2 ;
r1  1.57 mm
r102  (10)( 4.184)(589.3 10 9 )  2.466 10 5 m2 ;
r10  4.97 mm
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Interference method to measure film-thickness :
(uses fringes of equal
thickness)
Arrangement:
Monochromatic light from
LS channeled to beam
splitting prism BS through
fiber optics cable LP
At BS, one beam
transmitted to flat mirror M
and other to film surface
After reflection at M and film
surface F, two beams are
transmitted by BS into
microscope MS and
interference occurs
Beam reflected from M can
be considered to come from
virtual image M’
Interference fringes
produced by light
reflected from the
film surface and
substrate used to
determine film
thickness d.
Thus, interference patterns results from
interference due to air film between
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reflecting plane at M’ and film F
Interference method to measure film-thickness :
Mirror M can be moved toward or away from beam splitter to equalize
optical path lengths.
Mirror M can also be tilted to make M’ more or less parallel to film surface
When M’ and film surface are not precisely parallel, Fizeau fringes due to
a wedge will be observed through microscope
Light beam striking film will cover edge of film F, so that two fringe
systems are seen side by side
1
2
3
Sketch of one side of the valley
shown in photo.
Valley-like channel in
interference pattern due to
depression in film surface
1
2
3
Fringe pattern shifts by x at film
edge
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Interference method to measure film-thickness :
The shift in the two interference pattern corresponds to the difference in
thickness in the air films at their juncture (joining part).
M’
tF
At film F : 2ntF  r  mF 
tS
d
F
S
juncture
At substrate S : 2nt S   r  mS 
Difference : 2n(t S  t F )  (mS  mF )
2n(t )  (m)
Thickness of film d can be determined from the shift (or translation) in
the two fringe systems through (consider normal incidence):
Bright fringes satisfy:  p   r  2nt   r  m
(t = thickness of air film
at some point)
If air-film thickness changes by t = tS  tF = d, order of interference m
changes accordingly, and we have
(11-15)
2nt  2d  (m)
(n = 1 for air film)
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Interference method to measure film-thickness :
For e.g., increasing t by /2 changes order of any fringe by m =1, that
is, fringe pattern translates by one whole fringe
Therefore, for a shift of fringes of magnitude x, the change in m is
given by m  x / x resulting in
d  (x / x)( / 2)
(11-16)
We measure fringe spacing x and fringe shift x; and film thickness d is
then calculated.
Problem: when using monochromatic light, net shift of fringe systems is
ambiguous as x = 0.5x will look exactly like x = 1.5x
To solve: edge not sharp such that we can trace the one-to-one
translation of the each fringe in the fringe patterns as seen in the photo
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Fizeau interferometer
The basic use of a Fizeau interferometer for measuring the surface height differences
between two surfaces was described earlier. If the reference surface is flat, then
surface flatness is being measured.
Typical fringe pattern
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Fizeau interferometer(continued)
Surface height error and fringe deviation
Surface height error =
    
  
 2  S 
For a given fringe the separation
between the two surfaces is constant
 can be measured to within 0.1 (i.e. 1/10). Therefore the surface height error
S
can be measured to /20 (~30 nm)
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