Transcript Dark fringes

Chapter 17 Wave Optics
Part 1 Interference of light
Colorful interfering phenomena
Soap film under sunlight
Oil film under sunlight
Interference fringes produced by wedgeshaped air film between two flat glass plates
Newton’s rings
(Equal thickness fringes)
Equal inclination fringes
§17-1
Monochromaticity and
Coherence of Light
1.The emitting process of light
The life time of
electrons on the
excited state (激发态) is
11
8
~ 10  10 s
Stimulating
pumping
Stimulation Radiation
The persistent of light is
8
~ 10 s
 Lighting is intermittent(间歇）
Optical wave train
 Different wave trains are not coherent light.
·
·
independent(from
different atoms)
independent(from different
time of same atom)
2. Laser (from stimulated radiation 受激辐射)
E2
f= (E2-E1)/h
f
f
Identical wave train
f
E1
(frequency,
vibrating
direction, phase)
Coherent light
3. Monochromaticity(单色性)
•Monochromatic light：has single frequency
•Frequency width(频宽) f
•wavelength width(线宽)
Good monochromaticity: f,  are smaller.
•Spectrum curve (光谱曲线) ：The intensity
distribution of light with wavelength ( or
frequency )
Good monochromaticity
I
Bad monochromaticity

I

4. Coherency
Coherent lights： Identical frequency,
vibrating direction, phase
Can produce coherent superposition
The intensity distribution of resultant light
in crossing space：
I  I1  I 2  2 I1 I 2 cos 


If I1=I2, I  4 I1 cos
2
2
The intensity of light varies between bright
and dark alternately.
If I1 and I2 are not coherent light, then
I  I1  I 2
No varying
5. The methods to obtaining coherent light:
•principle：By using some arrangements, divide
the light wave emitted by an identical point from
monochromatic source into two beams. And
superposing these two beams in the space.
•methods：
The way of division of wavefront：
The way of division of amplitude.：
The way of division of wavefront：
S1
S
d
S2
The way of division of amplitude.：
Study by yourself
§17-2 Two beams interference
I. Young’s double-silt experiment
1. The experiment arrangement and principle.
2.The positions of bright fringes and dark fringes.
The path difference：   r  ?
The distribution characteristics of the fringes
and the distance between them.
3. Analyze the distribution
characteristics of the
fringes of the white light.
II. Fresnel’s double-mirror experiment
III. Lloyd’s mirror experiment
A
SS
C
d
M
M’
B
S
S’
E’
E
Half-wave loss
§17-3 Optical path and optical path
difference, Property of thin lens
S  nr
 nvt v--speed of light in medium
I. Optical path
c
 n t  ct
n
=Distance of light traveling through vacuum
at the same time
For a monochromatic light, f is same in
different medium, but  and v are different.
Let in vacuum： , c
In the medium with refractive index n：v ,

c
Then  '  , v 
n
n
II. Optical path difference
  n2r2  n1r1
The phase difference：
 
2


：the wavelength in vacuum
'
III. The half-wave loss of reflecting light

n1<n2
Phase shift
Has half-wave loss
n1>n2

No half-wave loss
 The transmitting ( refracting ) light never
have half-wave loss.
IV. A lens does not cause
any additional optical
path difference or
phase shift.
AF,
CF travel a larger
distance in the air and
a shorter distance in
the lens.
BF
A
B
C
is inverse.
 AF=  CF=
 BF ，they are converging on
point F and forming a bright image point.
F
§17-4 Interference by division of amplitude
I. Equal-inclination interference
1. The interference of reflecting light
The optical path
difference is

  2n2 AB  n1 AD 
2
e
 AB 
cos 
e
D
i
A
C

B
n1
n2  n1
n1
AD  AC sin i
 2e tg sin i
And
n1 sin i  n2 sin
e
D
i
A
C

B
e

  2n2
 2n1etg sin i 
cos 
2
2n2e

2

(1  sin  ) 
cos 
2
n1
n2  n1
n1
  2n2e cos  

2
 n2 cos   n2 1  sin   n2  n1 sin i
2
2
2

  2e n  n sin i 
2
2
2
2
1
2
 The
conditions that occur bright and dark
fringes are
k  1,2, --Bright fringes
k

 
 2k  1 
2
k  0,1,2, --Dark fringes
Discussions：
   (i ) ，As long
as the lights coming
from different area of
the source have same
incident angle i, they
have same  , and
belong to same
interference level k
--Equal inclination
interference
S
ii
i
i
 The interference fringes are series of
concentric circles. They are alienation(稀
疏)near the center, and dense near the
edge.
R larger,
i larger,
R smaller,
k smaller
i smaller,
k larger
screen
lens
S
Reflecting plate
i
film
i
screen
lens
S
Reflecting plate
i
film
i
When e ，we have k , the circle fringes
are produced from the center.
When e ，we have k ，the circle fringes are
swallowed at the center.
At the center,
----Bright
k


 
  2n2e   ( 2k  1) 
----Dark
2 
2
2. The interference of
transmitting light
The optical path
difference is
 '  2e n22  n12 sin2 i

e
C

B
 k
 (2k  1) 

2
n1
n2  n1
n1
----bright
----dark
k  0,1,2
The reflecting light and transmitting light are
compensative each other.
[Example] A thin oil film (n= 1.30) is illuminated
by the white light. Someone observes the
reflected light by the film. When the observing
direction has the angle 300 with respect to the
normal direction of the film, the film appears
blue ( 4800Å). Find the minimum thickness of
the oil film. If the film is observed at the
normal direction of the film, what color does
the film appear?
Solution : According to

  2e n  n sin i   k
2
k=1  e=emin
2
2
 emin 
2
1
2
( 2  1)
4 n  sin i
2
2

( 2  1)  4.8  10
4 1.3  0.5
2
7
 1.0  10 m
2
7
The film is observed at the normal direction of
the film, i=0: 2ne   2  k 
4ne
5.20  10

 
2k   1
2k   1
7
For k =1,
  5.20  10 m --green
For k =2,
  1.733  10 m
7
7
--Ultraviolet
II. Application
1.Transmission
enhanced film
(anti-reflecting film)
(增透膜)
n0 < n < n

e

n0  1
n  1.38
MgF2
n' 1.5
glass
The two beams reflected by the upper and
bottom interface of the MgF2 film all have halfwave loss.
The optical path difference between  and 
is
  2ne

  ( 2k  1)
2

--reflecting beams
are destructive.
--transmitting
beams are
constructive.
The
or
e

n0  1
n  1.38
MgF2
n' 1.5
glass
minimum thickness of MgF2：e
nemin 

4
min


4n
ne --optical thickness
2. Reflection enhanced film(增反膜)
 Considering the reflected
beams by each interface.
For the first film,
nH  2.40
nL  1.38
nL
nH
nL
  2nH e1   2  k
k  1,2, 
k=1 
nH e1   4
For the second film,
ZnS MgF2
  2nLe2   2  k
nH
nL
k  1,2, 
k=1 
nLe2   4
nH
n' 1.5
III. Equal-thickness interference
1. Wedged interference
Assume the incident beams are perpendicular
to the interfaces of the film.
  2ne   2
--bright
k



 ( 2k  1)  --dark
2

n
medium wedge
At the contact
edge(e=0) , k =0,
appears dark fringe.
For air wedge,
  2e   2
Discussion
air wedge
 The points with identical thickness e have the
same interference level k.
--Equal thickness interference
The fringes are the straight lines parallel to the
edge. They have same distance. And there is a
dark fringe at the contact edge(e =0) because of
half-wave loss.
 The thickness difference between two
adjacent bright(or dark) fringes:
l

 2nek   2  k
ek 1
2nek 1   2  ( k  1)
 e  ek 1  ek 
For air wedge,
ek

2n
e 

2
e
 The distance between two bright (or dark)
fringes:

l
ek
e
e


l

 2n
sin
ek 1
e
--identical distance
 When the upper interface of the wedge film
is moved upwards, the distance of the
fringes is constant, but all the fringes move
toward the contact edge.
A
e
B
A
e
[Example]To determine the thickness d of the
SiO2 over the Si precisely, it is usually corroded
to a wedge shape. The light with =5893Å is
incident normally from above. There are 7 bright
fringes over the length of the film. Calculate d=?
(Si: n1=3.42，SiO2: n2=1.50 are known.)
n2 d
n1
Si
SiO2
Solution：
The optical difference
between two rays
reflected by the upper
and lower surface of
SiO2 is
n2 d
n1
Si
SiO2
k

  2n2e  k k  0,1,  ek 
2n2
At the contact edge, k=0，the bright fringe
at d should correspond to k=6
6
6
 d  e6 
 1.1786  10 m
2n2
2. Newton’s ring
R
At points with
thickness e ,
  2ne   2
--bright
k




( 2k  1) --dark
Plane-convex
n
r
0
e
Plane glass
2
At
the center(e =0): corresponding to a dark
point with k =0.
Discussion:
  is determined by e
--equal-thickness interference
 The fringes are series of concentric circles.
-- Newton’s ring
 the radius r of the circle:
 r  R  R  e   2eR  e
2
as
2
e  R
2
2
R
r
2
r
e 
2R
n
0
Bright circle,
rk 
2k  1R
2n
k  1,2
kR
k  0,1,2
rk 
n
 the distance between two
adjacent circles :
Dark circle,
( k  1) R
kR
r 

n
n
1

k 1  k
R -- alienation(稀疏)near the
n center, dense near the edge.
When the Plane-convex lens moves upwards,
the circle fringes are swallowed at the center.
A
B
[Example] A drop of oil is on a plane glass. When
a monochromatic light with =5760Å is incident
on it normally, the interference fringes produced
by the reflected lights are shown in figure. The
center point of the oil is dark.
Find  Is the bright or dark
fringe at the edge of the oil?
 The maximum thickness
of the oil film =?  If the oil
spreads gradually, How do
the fringes change? (oil:
n2=1.60, glass: n3=1.50）
n2
n3
Solution
Because n1<n2，n2>n3，
there is half-wave loss in  .
i.e.,   2n e  
2
2
At the edge of oil, e=0, we have  

2
i.e.,  satisfies dark fringes condition,
  ( 2k  1)

2
So There is a dark circle fringe at the oil edge,
corresponding to k=0.
 The center dark point
corresponds to k=4.
k
 emax 
2n2
10
4  5700  10
7

 7.2  10 m
2  1.6
 When the oil spreads gradually, the dark
circle fringe located in the edge spreads
gradually, the center point changes from
dark to bright, and to dark alternately, the
level k of the fringe becomes less and less.
§17-5
Michelson’s interferometer
I. Instrument
M
'
2
M1 is fixed,
M2 can be moved M 1
d
G1
Beam spliter
G2 M 2
Compensating
plate
Discussion
M1and M2are perpendicular to each other-equal inclination interference produced by air
-- concentric circles
film.
Assume the thickness of the air film between
M1and M2 is d
Then, at the center of the fringes   2d  k
M2 moves d ：  '  2( d  d )  k' 


 d  ( k' k )  n
2
2
n--The number of the fringes out of
or in the center.
M1and M2 are not perpendicular to each
other -- equal thickness interference produced
by air wedge.
II. Application
d  n

2
Measure length d—known  , read out n，
Measure ----read out n and d，
Measure
out d
Look
refractive index n—known ，read
d  n  1e
for “ether（以太）” -- “ zero”result
Michelson’s interferometer be used to measure
the refractive index of gas.
Michelson’s interferometer be used to measure
flow field.