Transcript (等倾干涉) — equal thickness interference.

30-3. Interference in Thin Films (P691)
1. Equations for Thin-Film Interference
Both rays 1 & 2 are derived
from the same incident beam
  n2 (ab  bc ) n1ad  
P
1
i
d.
a
L
   /2 (n2>n1;n1>n2)
   0 (n1<n2<n3;n1>n2>n3)
2
i

.b
.
c
n1
n2
n1
(Pay more attention on the half-wavelength loss ! )
we get
  2 L n22  n12 sin2 i      ( i , L)
which depends on the thickness L of film
and incident angle i.


Bright fringe
2
m
(
m

1
,
2
,
3
,



)
 ( i , L)  2m22 (m  1,2,3,  )


Dark
fringe
((22mm''11))  (m
'

0
,
1
,
2
,



)
(
m
'

0
,
1
,
2
,



)
2. Discussion: 
22

(1) If L= constant, optical path difference depends on
the incident angle i — equal inclination interference.
(等倾干涉)
The rays with the same incident angle i form
a fringe with the same order.
(2) If i = constant, Optical path difference depends
on the thickness — equal thickness interference.
(等厚干涉)
The same order of fringe corresponds
to the same thickness L.
(3) It’s a method dividing amplitude to get coherent light.
3. Wedge-shaped film (劈尖) (P692):
Incoming ray is perpendicular to a thin wedge
which is put in the air.


Bright
fringes
2
m
(
m

1
,
2
,
3
,



)


2

  2n 2 L 

Dark fringes
2

(2m'1)
(m'  0,1,2,  )
2

：
10 ~ 10 rad
4

Supplementary explanation:
ray1
ray2
5
air
a

(1) The zeroth dark fringe on the edge
shows the half-wavelength loss.
n2 ·
air
L
(2) The same fringe corresponds to the equal
thickness—— equal thickness interference.
(3) Separation of fringes:
L 


2n2
l
B D
l

2n2 sin 
Lm Lm+1
 ，l 
L  l sin 
(4) For multi-chromatic incident light, a series
of colorful fringes is seen next to one another.
L
L 

2n2
( n2=1 for air )
(5) Applications:
l
• measure small length:
Scale: wavelength of light
Magnification coefficient:

2n sin 

1
2n sin 
• test flatness of surface (检查表面平整度):
The pattern observed shows when
plates are optically flat.
等厚条纹
待测工件
The pattern observed shows
when plates are not so flat.
例 工件上放平板玻璃成空气劈尖，单色光垂直照射，
干涉条纹如图。工件表面上纹路凹还是凸？纹深?
解：等厚线：纹路是凹的


a
2 H 
b2
a sin   H 
 b sin  
ab
b a
平板玻璃
工件
H

H
lk 纹路处 
4. Newton’s Ring (P692):
The interference only occurs for
rays reflected by the top and button
surfaces of the air film.


m  1, 2, 3    Bright
  2m 2
  2e   
2 (2m '1)  m '  0, 1, 2    Dark
2

( 2m  1) R
Newton’s ring
rB 
2
, m  1,2,3,  
C
R
e
rD  m' R , m'  0,1,2,  
(1) It is a typical case of equal
thickness interference
(2) The dark central point is the
result of half-wavelength loss.
Unequal
separation
r
B
O A
30-4. Michelson Interferometer (P695)
M1&M2are
2 mirrors
M1 is the virtual image of M1
formed by the splitter mirror.
M2
M1
 interference in the thin air film
between M1 and M2,
d2
S
G2
2
compensator
1
E
  2 d  m  
  d  m 

2
M1
1
beam
splitter
M1//M2  equal inclination interference
M1//M2 equal thickness interference
Applications:
d1
2
G1
等厚条纹
EXAMPLE:
A thin film with index of refraction n = 1.40 is
placed in one arm of a Michelson interferometer,
perpendicular to the optical path. If this causes a
shift of 7.0 fringes of the pattern produced by
light of = 589 nm, what is the film thickness?
Solution: The change of optical path difference due
to the insertion of the film of thickness L is:
2(n  1) L  Δ m
 Δm
(589nm )(7.0)
L

 5.2 m
2(n  1)
2(1.40  1)
•Questions (思考题）
•P697 13; P69715
•Problems (练习题）
• P699 23; P69925; P69932