Download Interference

Chapter 37 Interference
Soap Film Interference
Structural 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)!!
Interference of light
phenomena
Interference
condition
Interference method
Double slit
Thin film interference
Wedge
interference
Newton’s ring
Michelson
interferometer
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
1.phenomena of interference and coherent Light
1) Phenomena of interference
2) the condition to produce interference phenomena
a) why we can’t see the interference from common light
source
The many atoms ordinarily radiate in an
unsynchronized and random phase relationship.
b) Coherent light sources?
Same frequency
Constant phase difference
Same polarization (oscillation direction)
Two common way to get coherent light
p
S
Chose two points
on the same wave
front
2. Constructive and Destructive Interference
S1 : E1  E10 cos(t  10 
S2 : E2  E20 cos(t  20 
2
1
2
2
r1 )  E10 cos(t  10 
r2 )  cos(t  20 
2
  (20  10 ) 
( n2r2  n1r1 )
0
constructi ve
  2k
  
k  0,1...
destructive
 ( 2k  1)
2
0
2
0
n1r1 )
n2r2 )
if : 10  20
  k , constructi ve
n2r2  n1r1  
1
  ( k  2 ) , destructive
  nr
Optical path length
bringht
  k , constructive

 2  1  
1
dark
 ( k  2 ) , destructive
Example: find light path difference
s1
n1
n2
s2
p
r1
p
r2
= n1r1- n2r2
s1
S1p= r1
s2
S2p= r2
= (r1-e1 +n1e1) - (r2-e2 +n2e2)
  n2 L  n1 L
caution:
No optical path length difference through lens
1
s
2
s
The various light pass through the lens would
introduce no additional optical path difference or
phase shift.
3. Young’s Double-Slit Experiment
Thomas Young, 1773 –
1829 English physicist,
medical doctor, and
Egyptologist Also the
inventor of Young’s
modulus (strength of
materials)
Young’s double-slit experiment
The positions of the fringes observed in Young’s experiment can easily be
computed with the help of the following diagram.
 = d sin  = m, m = 0, 1, 2, . . .
for bright fringes
 = d sin  = (m+ ½) , m = 0, 1, 2, . . . for dark fringes
P 1143
dy
 
R
 k
bright
1
 ( k  )
2
dark
(k  0,1,2,......)
Discussion:
R
y   m
d
Bright fringe，k =0,1,2,…...
1) 0th bright fringe in y=0. If white light go through the
double slits,there is a white fringe in x=0
2) the space between adjacent bright fringes or dark
fringes
R
y 
d
D
xk
2a
(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.
R
ym
d
Example: Young’s experiment is performed with sodium
light( =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?
Solution:
d sin  m
y
d
 m
R
R
1  589  109
d  m
 20 
m
3
y
11.78  10
 10 3 m  1mm
Example: Young’s experiment is performed with white
light(4000 Å ~7000 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?
Solution:
d sin  m
R
ym
,m  2
d
R
 y  m (red   pur )
d
5
 10
 2
(
7000

4000
)

10
m
3
0.25  10
 1.2  10 3 m  1.2mm
k=2
k=1
k=0
k=1
k=2
Example: The wavelength of the incident light is
6000 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.
Solution:
   ( r  e  en )  r  e( n  1 )
   e( n  1 )  5 
5
e 
n1
5  6  10 7

m  10  5 m
1.3  1
n, e
O
Lioyd’s mirror
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.
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  . a at what value of x is the signal at
the receiver maximum?
1
  l2  l1  (m  )
2
1  D
x  ( m  ) 
2  2a 
The figure shows the plot of light intensity versus
the path difference d·sin 
4. Thin film interference
Why do some lens have
a purple color? Why
colorful bulbs?
Thin film interference
Geometrical OPD
  n2 ( AB  BC )  n1 DC
s
b1
b
a
t
i
a1
D
A
C
B
n1
n2
n3
  2t n2  n1 sin2 i
2
2
When incident angle i=0
  2t n22  n12 sin2 i
Half-Cycle Phase Shift
180o 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
180 0 phase change
No phase change
Interference in thin film
 

2tn2 (  )
2
bright
m


1
( m  ) dark


2
(m=0,1,2……)
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)?
r1
Air
n0 = 1.00
i
r2
Solution:


1
2
  2n1d   m   
dmin

5500


 996 A
4n1 4  1.38
d
MgF2 n1 = 1.38
Glass
n2 = 1.50
The point where a soap bubble bursts will be black
just as it bursts.
(a) True
(b) False
Example: White light is incident on a soap bubble ,the
thickness t=3800Å, n2=1.33 ，find the color of bubble in
reflecting light.
solution:
  2tn2 

2
 k
7600Å×1.33
2en2
 
1 =
1
k
k
2
2
In the extent of visible light (7700Å~3900Å)
k=1,…
k=2, =6739Å red
k=3, =4043Å violet
k=4,...
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 = 500nm.
Solution: The fringe at the contact line is dark

1
   2 t   ( k  ) k  0 ,1... dark
2
2
t  k / 2  
2
t t
l 

x  1.25mm
tg D
x
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。
Solution: from interference in thin
film,there are concave pitch in
worpiece
l
  2nt 
a
Standard level

workpiece
t 
H


2
 m
2
From similitude triangle
H a

t l
a 
H 
l 2
Newton’s rings
k
k  1 ,2...bright
 

  2 nt   

k  0 ,1...dark
2  ( 2k  1 )
2

r2
t 
2R
Example: A lens of radius of curvature R resting on
a plane glass plate is illuminated from above by
light of wavelength . 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.
  2nt 

2
 k
2
r
t 
2R
r  (m 
1
2)R
m = 0, 1, 2, 3, . . .
d
Example: two convex lens positioned in the following
way,find the radius of bright fringes。

  2e
k
1
=
( k  )
2

2
o2
o1 R2
R1
r
bright (k=1,2…)
dark(k=0，1,2...)
2
2
r
r
e  e2  e1 

2 R1 2 R2
e
e1
e2
(k  1 / 2) R1R2
rk 
R2  R1
5. Interferometer
2( L2  L1 )  2 t
bright
k



( 2k  1 )
dark

2

measure wavelength of light
measure displacements
An airtight chamber 5.cm long with glass window is placed
in one arm of a michelson interferometer.   500nm
Evaluating the air from the chamber causes a shift of 60
fringes.find the index of refraction of air atmospheric
pressure