intereference

1. 1
• For two beams to interfere and produce a stable
pattern, they
– Must have nearly the same frequency;
– Must have equal or nearly equal amplitudes
• the pattern will be very clear.
– Need not be in-phase; but phase difference must be
constant  coherent.
Conditions for Interference

2. 2
Wavefront-splitting
Interferometers:
Young’s Experiment
• Young was able to
produce interference
even before laser was
invented.
• In his experiment, a
single wavefront was
selected and split into
two coherent sources,
which were then used to
produce interference.

3. 3
• Consider a plane wave
illuminating a slit.
– Light will diffract out in all
directions from this primary
slit.
• This diffracted ray then
encounters two slits, S1
and S2, on another barrier.
– The wavefront arriving at the
two slits will be exactly in-
phase.
– The slits constitute two
coherent secondary sources.
– Interference occurs wherever
these two waves overlap.

4. 4

5. 5
• Consider the OPD traverse by rays from S1 and
S2 to point P, on the screen:
OPD = l2 – l1
• It can be approximated by taking a perpendicular
line from S1 to l2:
OPD = d sin 

6. 6
• For a bright fringe to exist, the OPD must be an
integral number of the wavelength:
OPD = m, m = 0, 1, 2, …
d sin  = m
• Consider sin  ≈ :
It gives the angular position of the mth order
fringe.
d
m
m

 

7. 7
• IN order to determine the vertical position of the
mth order bright fringe, on the viewing screen.
• Let y = 0 be the horizontal position midpoint
between the two slits and ym be the vertical
position of the mth order fringe along the viewing
screen.
• The position of ym can be determined from the
geometry of the experiment set-up.
L
d
m
y
d
m
L
y
m




sin

8. 8
• For a dark fringe to exist,
OPD = (m + ½), m = 0, 1, 2, …
d sin  = (m + ½) 
• Consider sin  ≈ :
• The position of the mth order dark fringe:
• The separation between two consecutive
maxima/minima can be shown as
 
d
m
m

 2
1


  L
d
m
ym
2
1


L
d
y



9. 9
Dark fringes
bright fringes
m 3
2
2
1
1
0
-1
-1
-2
-2

10. 10
• The phase difference between the two rays
 = k(l2 – l1)
• The total irradiance at an arbitrary point on the
viewing screen is
I = I1 + I2 + 2(I1I2)1/2 cos 
• In this experiment, the irradiance contribution
from both sources are equal i.e. I1 = I2 = Io
I = 2Io (1 + cos )
 










L
ydk
I
llk
I
I
o
o
o
2
cos4
2
cos4
2
cos4
2
122
2 

11. 11
• The irradiance has a maximum value when
• Consecutive maxima are separated by y.
d
Ln
y
nn
L
yd
L
ydk
n











,...2,1,0,
2
...20
d
L
d
L 

12. 12
Example
A viewing screen is separated from a double-slit
source by 1.2 m. The distance between the two
slits is 0.030 mm. The second-order bright
fringe is 4.5 cm from the center of the line.
(a) Determine the wavelength of the light.
(b) Calculate the distance between adjacent bright
fringes.

13. 13
Solution
L = 1.2 m, d = 0.030 mm, m = 2, ybright = 4.5 cm
(a) ybright = m
  560 nm
(b) y = ym+1 – ym =
 2.2 cm
d
L
d
L

14. 14
Amplitude-splitting
Interferometers:
Thin Film
• Beamsplitter is an optical device that splits a
beam of light into two – reflected beam and
transmitted beam.
• Examples: thin stretched plastic films, uncoated
glass plate, half-silvered mirror, soap bubbles.
• Both the transmitted and the reflected waves
would have lower amplitudes than the incident
wave.
• These two waves, when brought together again
at a detector, would produce interference pattern
as long as the original coherence between the
two had not been destroyed.

15. 15

16. 16
• Consider a film of uniform
thickness t and index of
refraction n.
• Assume :
i. Light rays travelling in air
are nearly normal to the two
surfaces of the film.
ii. The film is non-absorbent.
iii. The amplitude-reflection
coefficients at the interface
are so low that only the first
two reflected beams need to
be considered.

17. 17
• In practice, the amplitudes of the higher-order
reflected beams generally decreases very rapidly.
• Part of the wave is reflected, at surface A, as ray
1 with a 180o phase change and another part is
refracted into the film without phase change.
• These two rays can be consider as arising from
two coherent; they are parallel on leaving the
film.

18. 18
• The OPD between these two
rays is
OPD = nf[AB + BC] – n1(AD)
• From the diagram,
AB = BC = d/cos t,
AD = AC (nf sin f/n1)
AC = 2d tan t
then
OPD = (2nfd/cos t) – n1(AD)
= 2nfd(1 – sin2t)/cos t
= 2nfd cos t
t
t
A
B
C
D
d
n1
nf
n2
Snell’s Law at point A

19. • Case 1:
– Film is immersed in 1 medium (n1 = n2 = n)
– n < nf (soap film in air)
• The 1st reflected ray experience phase change by
180o.
• The 2nd ray, when it exits the upper surface, did
not undergo phase change.
– There exist a relative phase difference of  between
these two rays.
• If OPD = m where m = 0, 1, 2,… then a central
dark fringe is observed.
19
4
)2(
4
)2(cos
2
)2(cos2
n
f
t
tf
m
n
md
mdn







20. 20
• If OPD = where m = 0, 1, 2,… then a central
bright fringe is observed.
• A monochromatic light beam incident on a thin
film can have the reflected beam either
reinforced or diminished depending on the phase
difference and hence angle of incidence and film
thickness.
2

m
4
)12(cos
2
)12(cos2
n
t
tf
md
mdn







21. 21
• Case 2.1: Film is immersed in 2 mediums.
– n1 < nf < n2
• Both reflected rays experience phase change by
180o. There is no relative phase difference
between these two rays.
• Case 2.2: Film is immersed in 2 mediums.
– n1 > nf > n2
• Both reflected rays are in-phase with the incident
ray. There is no relative phase difference between
these two rays.
• The above equations would have to be modified
accordingly.
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22. 22
Fringes of Equal Thickness
• Fringes of equal thickness exist for which optical
thickness, nfd, is the dominant parameter rather
than i.
– Under the white-light illumination, the iridescence of thin
film (soap bubbles, oil slicks, even oxidized metal
surfaces) is the result of variations in film thickness.
• Each fringe is the locus of all point in the film for
which the optical thickness is a constant.
– In general nf does not vary.
• This is useful in determining the surface features
of optical elements.

23. 23
• Consider the surface of an
optical element (to be
examined) in contact with an
optical flat.
– Optical flat – A flat surface that
deviates by not more than
about /4 from a perfect plane.
– Individual flats of /200 can be
achieved.
• If the two surfaces are plane,
so that the film is wedge-
shaped, the fringes will be
practically straight, following
the lines of equal thickness.
– Fizeau’s fringes

24. 24
• For a thin wedge of small angle , the optical path
length difference between two reflected rays may
be approximated from the thin film case i.e.
OPD = 2nfd cos t ≈ 2nfdm
where d is the thickness at a particular point that
is
d = x

x
d

25. 25
• The condition for bright fringes becomes
2nfdm = (m + ½)
2nf (xm) = (m + ½)
• From one fringe to the next, m increases by 1, the
optical thickness of the film nd change by /2.
• If the test surface is not perfectly flat, the fringes
will be irregular in shape.
• Calculating the order of the fringes, and as the film
is air (n = 1.0), the position of the imperfect
surface can be located for further polishing.

26. 26
• A long-focus lens placed on
a test plate produces a set
of circular fringes, known
as Newton’s rings.
• Observation are usually
made at normal incidence
according to the
arrangement shown.
– The beamsplitter reflects
light from source, S down
to the plates.
– After reflection it is then
transmitted by the
beamsplitter to the low-
power microscope above
it.
S
Newton’s Rings
 Circular fringes is called
Newton’s rings.
 The radius of the nth dark fringes
 Important in the testing of optical
lenses.
 A circular pattern is obtained
only when the lens is ground to a
perfectly symmetric curvature.
n
Rm
r



27. 27
• The position of the maxima can be determined
according to that of Fizeau’s fringes.
• Assuming the lenses are just touching at the centre,
and let the radius of curvature of the lens be R, the
thickness of the air film can be calculated as
d = r2/2R
• Substituting the above expression yields
(Minima)
Fig 37-18a, p.1191
n
Rm
r


d

28. 28
• The amount of uniformity in the
concentric circular pattern is a
measure of the degree of
perfection in the shape of the lens.
• As the diameter of the fringes, r,
depends on wavelength, white
light will produce only a few
coloured rings near the point of
contact.
• Monochromatic light gives an
extensive fringe system.
• When the contact is perfect the
central spot is black due to the
relative phase change by .
 Circular fringes is called
Newton’s rings.
 The radius of the nth dark fringes
 Important in the testing of optical
lenses.
 A circular pattern is obtained
only when the lens is ground to a
perfectly symmetric curvature.
n
Rm
r



29. 29
Non-reflecting Films
• A simple and very important application of the
principles of interference in thin films is the
production of coated surfaces.
• A film of transparent substance of refractive index
n’ (= n½) is deposited on glass of a larger index n
to a thickness of one-quarter of the wavelength of
light in the film so that
• Light reflected at normal incidence is almost
completely suppressed by (destructive – rare-to-
dense reflection) interference, this corresponds to
the condition m = 0.
'4n
d



30. 30
• Light is NOT destroyed by a non-reflecting film, it
is merely redistributed such that a decrease of
reflection carries with it a corresponding increase
of transmission.
• They can be use in optical system which require
greatly reduced loss of light.

31. 31
Mirrored Interferometers
• Mirrored interferometers utilize arrangements of
mirrors and beamsplitters.
• Most important of these is Michelson
interferometer.
, M2
, M1
, O
, C

32. 32
• An extended source (as oppose to point source)
emits a wave part of which is directed at the
beamsplitter, O.
• The wave is divided into two, one segment
travels to movable mirror, M2, and the other
travels to the fixed mirror, M1.
• These are then reflected by the mirrors and
return to the beamsplitter.
– The ray from M2 passes through the beamsplitter going
downward.
– The ray from M1 deflected by the beamsplitter toward
the detector.
• They unite to produce interference pattern.

33. 33
• One ray passes through the beamsplitter trice
(reflect) while the other (transmit) only go
through it once.
• The compensator plate is used to even out the
path difference due to the beamsplitter.
• It is an exact duplicate of the beamsplitter, with
the exception of any possible silvering of thin film
coating on the beamsplitter.
• It is position at 45o so that it is parallel to the
beamsplitter.
• Any optical path difference that exist now is from
the actual path difference 

34. 34
• Fringes will not be observe by simply following
the arrangement shown.
– Mirrors M1 and M2 must be made exactly perpendicular
to each other. In physical arrangement, there are screws
at M1 used to control the tilt of the mirror.
– Light in general must be monochromatic or nearly so,
especially if the distances of OM1 and OM2 are
appreciable.
– The distances OM1 and OM2 should be roughly within a
few millimetres.
• The fringes are circular when they appear.

35. 35
• Consider the figure below:
• The fixed mirror, M1 is replaced by M1’, its virtual
reflection by the beamsplitter, O.
• These two mirrors, M1’ and M2, will have virtual
images of the extended source at L, labelled L1
and L2.
– These virtual sources are coherent – the phases of
corresponding points in the two are exactly the same at
all instants.
P’
L2
P’’
L1
2d
P
L
M2
M1'
d


36. • If d is the separation between M1’ and M2, the
virtual sources will be separated by 2d.
• When d is exactly an integral number of half-
wavelength, 2d equals to an integral number of
whole wavelength.
– All rays of light reflected normal to the mirror will be in
phase.
• Reflected rays at an angle will in general not be in
phase.
– Its path difference is 2dcos.
– The rays will reinforce each other to
produce a maxima for
2dcos = m
– For a given m, , and d, the angle  is a constant.
• the maxima will lie in the form of circles about the foot of the
perpendicular from the eye to the mirrors.
36
P’ P’’

2d
2dcos

37. 37
• Notice that the reflected rays are parallel; they do
not meet. The observer’s eyes, focus at infinity,
serve as the condensing lens for the image to form.
• As M2 moves towards M1’, d decreases. From the
expression earlier,
– cos m increases;
– m decreases;
– The mth order ring shrinks towards the centre.
– Each remaining ring broadens as more and more
fringes vanish at the centre.
• http://www.optics.arizona.edu/jcwyant/JoseDiaz/MichelsonInterferometerFr
inges.htm
• When d = 0, the central fringe would have
spread out, filling the entire field of view.

38. 38
• As M2 moves away from M1’ d increases, widely
spaced fringes reappear from the centre.
• These will gradually become more closely spaced as
the path difference increases.

39. • When M1’ and M2 are not parallel, inclined wrt
each other making small angle, it results in
wedge-shaped air film between then.
• Fizeau fringes, a pattern of straight parallel
fringes, are observed.
• The interference fringes appear to diverge from a
point behind the mirrors. The eye would have to
focus on this point to observe the localized
fringes.
39
P’
L2
L1
P
L
M2
M1'

40. 40
d reducing d = 0 d increasing
Fringes and the relative position of the movable mirror and fixed
mirror.
• The fringes are not exactly straight when d has an
appreciable value.
– They are in general curved and are always convex towards the
thin edge of the wedge.
– If d is decreased, the fringes will move to the left across the
field, a new fringe crossing the centre each time d changes by
/2.
– d ≈ 0; the fringes become straighter.
– When d increases, the fringes curve in the opposite direction.