1. The iridescent colors of birds, butterflies, and moths arise from structural interference effects in their feathers and scales rather than pigments. 2. Interference occurs when two light waves meet and interact, producing a pattern of bright and dark bands called interference fringes. The document discusses key terms and concepts related to interference including constructive and destructive interference. 3. Young's double-slit experiment demonstrated the wave nature of light by producing an interference pattern of alternating bright and dark fringes on a screen. The experiment led to an understanding of how interference causes the colors seen in thin films like oil films and soap bubbles.

1. Networn Rings

2. Soap Film Interference

3. Structural Color in Plants and AnimalsStructural Color in Plants and Animals
The iridescent sheen of a bluebird wing or the
extravagant colors of many butterflies and moths
arise - not from chemistry (pigments) but from
physics (interference effects)!!

4. Interference
condition
Interference
condition Double slitDouble slit Thin film interferenceThin film interference
Wedge
interference
Wedge
interference Newton’s ringNewton’s ring Michelson
interferometer
Michelson
interferometer
Interference methodInterference methodphenomenaphenomena
Interference of lightInterference of light

5. Key terms:Key terms:
Monochromatic light
Coherent waves/lights
Constructive interference
destructive interference
Antinodal /nodal curves
Interference fringes
Newton’s rings
Nonreflective/reflective coating
Michelson interferometer
Michelson-Morley experiment
Ether
photon
Plank’s constant

6. Same frequency
Constant phase difference
Same polarization (oscillation direction)
2) the condition to produce interference phenomena
a) why we can’t see the interference from common light
source
1.phenomena of interference and coherent Light1.phenomena of interference and coherent Light
The many atoms ordinarily radiate in an
unsynchronized and random phase relationship.
b) Coherent light sources?
1) Phenomena of interference

7. p
S
Chose two points
on the same wave
front
Two common way to get coherent light

8. )(
2
)( 1122
0
1020 rnrn −−−=∆
λ
π
φφφ
edestructiv
veconstructi
k
k
k
...1,0
)12(
2
=



+±
±
=∆
π
π
φ
2. Constructive and Destructive Interference2. Constructive and Destructive Interference
)
2
cos()
2
cos(: 11
0
10101
1
101011 rntErtEES
λ
π
φω
λ
π
φω −+=−+=
)
2
cos()
2
cos(: 22
0
202
2
202022 rntrtEES
λ
π
φω
λ
π
φω −+=−+=

9. 



λ+±
λ±
=δ−δ
darkedestructiv,)
2
1
k(
bringhtveconstructi,k
12
nr=δ Optical path length
2010: φφ =if




+±
±
=−
edestructivk
veconstructik
rnrn
,)
2
1
(
,
1122
λ
λ

10. δ=n1r1- n2r2 δ= (r1-e1 +n1e1) - (r2-e2 +n2e2)
s2
s1
r2
r1
pn1
n2
s2
S1p= r1
s1
pe1
e2
n1
n2
S2p= r2
LnLn 12 −=∆δ
Example: find light path difference

11. The various light pass through the lens would
introduce no additional optical path difference or
phase shift.
s
1
2
s′
No optical path length difference through lens
caution:caution:

12. Thomas Young, 1773 –
1829 English physicist,
medical doctor, and
Egyptologist Also the
inventor of Young’s
modulus (strength of
materials)
3. Young’s Double-Slit Experiment3. Young’s Double-Slit Experiment

13. Young’s double-slit experimentYoung’s double-slit experiment

14. The positions of the fringes observed in Young’s experiment can easily be
computed with the help of the following diagram.
δ = d sin θ = mλ, m = 0, ±1, ±2, . . . for bright fringes
δ = d sin θ = (m+ ½) λ, m = 0, ±1, ±2, . . . for dark fringes
P 1143

16. 1) 0th bright fringe in y=0. If white light go through the
double slits,there is a white fringe in x=0
λm
d
R
y ±= Bright fringe ， k =0,1,2,…...
bright
λ)
2
1
( +± k
λk±
dark
,......)2,1,0( =k==
R
dy
δ
Discussion:
2) the space between adjacent bright fringes or dark
fringes
d
R
y
λ
=∆ a
D
kx
2
λ
=

17. d
R
my
λ
=
(3)Experiment with white light,a colorful fringes will be
seen,to specific mth bright fringes,the violent frienge
close to center and red far away.

18. Solution:
mmm
m
y
R
md
110
1078.11
105891
20
3
3
9
==
×
××
×==∴
−
−
−
λ
Example: Young’s experiment is performed with sodium
light(λλ =589nm=589nm). Fringes are measured carefully on a
screen 100cm away from the double slit, and the center of
the twentieth fringe is found to be 11.78mm from the
center of the zeroth fringe. What is the separation of the
two slits?
λθ md =sin λm
R
y
d =

19. k=0
k=1
k=1
k=2
k=2
Solution:
2, == m
d
R
my
λ

mmm
m
d
R
my purred
2.1102.1
10)40007000(
1025.0
5
2
)(
3
10
3
=×=
×−
×
×=
−=∆∴
−
−
−
λλ
Example: Young’s experiment is performed with white
light(40004000 Å ~70007000 Å). The screen is 500cm away from
the double slit. The separation of the two slits is 0.25mm.
What is the width of the second colorful bands?
λθ md =sin

20. Solution:
n, e
O
r)ener( −+−=δ )1n(e −=
λ=−=δ 5)1n(e
m10m
13.1
1065
1n
5
e
5
7
−
−
=
−
××
=
−
λ
=∴
Example: The wavelength of the incident light is
60006000 Å.The index of refraction of the transparent
film is 1.3. The fifth bright fringe is at the center.
Find the thickness of the film.

21. An interference pattern is
produced at point P on the
screen as a result of the
combination of the direct ray
and the reflected ray. The
interference pattern has a dark
fringe at P’, and the dark and
bright fringes are reversed
relative to the pattern created
by two real coherent sources.
Lioyd’s mirror

22. Microwave transmitter at height a above the water level of
a wide lake transmits microwave of length toward a
receiver on the opposite shore.suppose d is much greater
than a and x,and . at what value of x is the signal at
the receiver maximum?
λ
a≥λ

24. λδ )
2
1
(12 +=−= mll
λ





+=
a
D
mx
2
)
2
1
(

25. The figure shows the plot of light intensity versus
the path difference d·sin θ

26. 4. Thin film interference
Why do some lens have
a purple color? Why
colorful bulbs?

27. Thin film interference

28. 1a
a
1bb
s
n3
n2
n1
t
B
C
D
i
A
DCnBCABn 12 )( −+=δ
innt 22
1
2
2 sin2 −=δ
When incident angle i=0
innt 22
1
2
2 sin2 −=δ
Geometrical OPD

29. 180o
phase change
180 0
phase change
Phase shift: when
light is incident from
a lower-index
medium and reflects
from a higher-index
medium
No phase shift: when
light is incident from
a higher-index
medium and reflects
from a lower-index
medium
Half-Cycle Phase Shift
No phase change

30. Interference in thin filmInterference in thin film
22tn=δ




+
=
darkm
brightm
λ
λ
)
2
1
(
)
2
(
λ
+
(m=0,1,2……)

31. Air n0 = 1.00
MgF2 n1 = 1.38
Glass n2 = 1.50
i
r1
r2
d
λδ 





+==
2
1
2 1 mdn
996
38.14
5500
4 1
min =
×
==
n
d
λ A
Example: Lenses are often coated with films of transparent
substances like MgF2 (n = 1.38) in order to reduce the
reflection from the glass surface. How thick a coating is
needed to produce a minimum reflection at the center of
the visible spectrum (5500 A)?
Solution:

32. The point where a soap bubble bursts will be black
just as it bursts.
(a) True (b) False

33. solution:
22tn=δ
2
λ
+ λk=
2
1
2 2
−
=∴
k
en
λ
7600Å×1.33
2
1
−k
=
In the extent of visible light (7700Å~3900Å)
k=1,…
k=2, λ=6739Å red
k=3, λ=4043Å violet
k=4,...
Example: White light is incident on a soap bubble ,the
thickness t=3800Å, n2=1.33 ， find the color of bubble
in reflecting light.

34. Solution: The fringe at the contact line is dark
dark...1,0k)
2
1
k(
2
t2 =+=+=∴ λ
λ
δ
Example: Suppose the two glass plates are 10cm long. At
one end they are in contact; at the other end they are
separated by a wire 0.020mm thick. What is the spacing
of the interference fringes seen by reflection? Is the
fringe at the line of contact bright of dark? Assume
monochromatic light with a wavelength in air of λ=
500nm.
2
2/ λλ =∆=∆ kt
mmx
D
t
tg
t
l 25.1=
∆
=
∆
=∆
θ
x

35. Solution: from interference in thin
film,there are concave pitch in
worpiece
H
2
λ
⋅=
l
a
H
l
a
θ
workpiece
Standard level
Example: check the level of smoth of the workpiece ,a
wedge is used here ， and interference fringes as
follows.judge the concave or protruding case in the
workpiece? Find the depth of it H 。
2
2
2
λ
λλδ
=∆
=+=
t
mnt
From similitude triangle
l
a
t
H
=
∆

36. Newton’s ringsNewton’s rings




=
λ
+
=λ
=
λ
+=δ
dark...1,0k
2
)1k2(
bright...2,1kk
2
nt2
R2
r
t
2
=∴

37. Rmr λ+≈ )( 2
1 m = 0, 1, 2, 3, . . .
d
Example: A lens of radius of curvature R resting on
a plane glass plate is illuminated from above by
light of wavelength λ. Circular interference fringes
(Newton’s rings) appear, associated with the
variable thickness air film between the lens and the
plate. Determine the the radii of the circular
interference maxima.
λ
λ
δ knt =+=
2
2
R
r
t
2
2
=

38. e2=δ
2
2
1
2
12
22 R
r
R
r
eee −=−=
e
o1
o2
R2
R1
e1
e2
r
12
21)2/1(
RR
RRk
rk
−
−
=
λ
Example: two convex lens positioned in the following
way,find the radius of bright fringes 。
λ)
2
1
( +k
λk bright (k=1,2…)
dark(k=0 ， 1,2...)
=2
λ
+

39. 5. Interferometer5. Interferometer




λ
+
λ
=
=−
dark
2
)1k2(
brightk
t2)LL(2 12
measure wavelength of light
measure displacements

41. An airtight chamber 5.cm long with glass window is placed
in one arm of a michelson interferometer.
Evaluating the air from the chamber causes a shift of 60
fringes.find the index of refraction of air atmospheric
pressure
nm500=λ