Download K = 1

light interference
1) Young’s Double slit experiment
The positions of bright fringes:
D
x
K
nd
K = 0，1，2，...
The positions of dark fringes:
D

2 K  1
x
nd
2
K = 1，2，...
The distance between adjacent bright/dark fringes:
D
x  xk 1  xk 

nd
2) Interference of equal inclination
The conditions of bright fringes:
1

2n2e cos    k   
2

The conditions of dark fringes:
2n2e cos   k
For one incident angle, if the reflected lights form bright
fringe then the transmitted lights for dark fringe.
3) Interference of equal thickness
a) Interference in a wedge-shaped film
 2k  1
1
  2n2e   
2

2
k  0,1,2...
k  k  1,2...
Bright fringe
The thickness of film for the bright fringes:
1 

k  1,2...
ek   k  
2  2n2

The thickness of film for the dark fringes:
ek  k

2n2
k  0,1,2...
 ek
l
ek 1 h
The thickness difference of the film between two adjacent
bright/dark fringes:
e  ek 1  ek 

2n2
The difference on the surface
between two adjacent bright/dark
fringes:
e


l 


sin  2n2 sin  2n2
l
A
e h
 ek e k  1 B
l
b) Newton’s rings
The radii of bright fringes in Newton’s rings:
(2k  1) R
r
2n2
k  1, 2, 3,
The radii of dark fringes in Newton’s rings:
kR
r
n2
k  0, 1, 2, 3, 
4) Michelson Interferometer
M1
1
G1
S
M1 M
2
G2
A
2
The relationship between the
distance of mirror M1 Δd and the
number of moved fringes m:
d  m / 2
2ˊ
1ˊ
B
§3-4 light diffraction
1) Classification of light diffraction
A. Fraunhofer Single Slit Diffraction
Diffraction of parallel lights
B. Fresnel Single Slit Diffraction
2) Comparison of diffraction & interference
Yellow light
Intensity distribution
Pattern of diffraction
Pattern of interference
white light
Pattern of diffraction
Pattern of interference
3) Fraunhofer Single Slit Diffraction
A. Experimental set-up
B. results
C. Discussion
Fresnel-zone half band Method
菲涅耳“半波带法”
The width of the slit is a.
Assume light A, A1, A2, A3 and B pass
through the slit.
The optical difference between light
A and light B is:
  BC  a sin

A
A1
a A2
A3
P
B

C
2
Divided BC with λ/2.
Draw lines from the divided points parallel with AC.
Divided the wave surface in the slit into several half bands.
The number of the half bands in
the slit:
  AB sin  a sin  N  2
a sin 
N 
 2
It depends on the diffraction angle θ.

A
A1
a A2
A3
P
B

2
C
The phase difference of two adjacent
half bands is: π
The superposition of these two half bands
results in destructive interference.
For a given diffraction angle θ, if the wave
surface is divided into even half bands, the
corresponding point on the viewing screen
is the center of dark fringe.

A
A1
a A2
A3
P
B

2
C
D. Conditions of bright/dark fringes in light diffraction
1) Condition of dark fringes:

a sin   2k  k k  1, 2, 3, 
2
2) Condition of bright fringes:

a sin  (2k  1)
2
k  1, 2, 3, 
3)When θ=0,
a sin  0
Middle bright fringe
Θis variable.
E. location of bright/dark fringes in light diffraction
x
The distance between k fringe
and the O axis is:
xk  ftg  f sin 
Small angle approximation
a
xk

o
f
f: focal length of the lens
x
x
1) If the fringe is dark:
a sin   k k  1, 2, 3, 
a
k
sin    
a
f
xk   k 
a
xk

o
f
k  1,2,3...
x
x
2) If the fringe is bright:

a sin  (2k  1)
k  1, 2, 3, 
2
a
2k  1
sin   

2a
f
xk    2k  1
2a
xk

o
f
x
F. The width of middle bright fringe
The width of the middle bright
fringe equals the location
difference between k=1 and k=-1
dark fringes.
The location of k=1 dark fringe:
f
x1 
a
a
x
xk

O
f
L0 in this figure
f
The width of middle bright fringe: l0  2 x1  2
a
l
l0
l
f
The width of middle bright fringe: l0  2 x1  2
a
For a given lens and used light wavelength, the smaller the slit, the
more obvious the diffraction is.
a  l0 
If a   , l , l0 , xk  0 all the fringes located in the middle of the
viewing screen, no diffraction is observed. The light travels in
straight line.
Example 3-1
In Fraunhofer single slit diffraction, the width of the slit a=100λ, the
focal length of the lens f=40 cm. Find: the width of middle and k=1
bright fringes.
Solution: The width of middle bright fringe is:
f 2  40
l0  2

 0.8cm
a
100
The width of k=1 bright fringe is the location difference between
k=2 and k=1 dark fringes:
2 f f f
x2  x2  x1 


 0.4cm
a
a
a
Example 3-2
In Fraunhofer single slit diffraction, the wave surface in the slit
corresponding to k=3 dark fringe can be divided into 2  3  6 half
bands. If the width of the slit decreases into its half, the k=3 dark
fringe will turn to be bright fringe.
(1)  a sin   3
( 2)  a sin   3

3
N

6
/2 /2
a
1

   sin   3  3
2
2
2
4) Circle Aperture Fraunhofer Diffraction
A. Experimental set-up
爱里斑
L2
L1
Ariy Spot
R
S
f
E
 — half
angle of
Ariy spot
B. Discuss
D sin  1.22
L2
L1
The first dark fringe:
R
)θ
S
f
E
D: The diameter of the circle aperture
Θ: the half angle of Ariy spot
Θis very small.
sin     1.22

D
5) Resolution of optical systems
Once you are able to see two separate headlights, you describe the
light sources as being resolved.
A. Resolution of single slit
S1
S1
S2
(
S2
Slit
The diffraction pattern is
distinguished. The two light
sources are resolved.
(
Slit
The diffraction pattern
overlapped. The two light
sources are not well resolved.
Rayleigh’s criterion 瑞利判据
When the central maximum of the diffraction pattern of one
source falls on the first minimum of the diffraction pattern of
another source, the sources are said to be just resolved. The
limiting condition of resolution is known as Rayleigh’s criterion.
点物 S1 的衍射中心的最大恰好与另一个点物 S2 的第一衍
射的极小相重合时，恰可分辨两物点。
For the single slit, the first minimum
diffraction (dark fringe) satisfies the
following relationship:
S1
(
S2
a sin   k k  1
a sin   

That is: sin    
Slit
a
Therefore, the limiting angle of resolution for a slit with width a is:
 min 

a
B. Resolution of circle aperture
S1
resolved
S2
S1
S2 
min
S1
S2
100%
73.6%
f
80%
Just resolved
Can not resolved
The limiting angle of resolution for a circle aperture with diameter D
is:
 min  1.22

D
Compare with limiting angle of single slit
 min 

a
6) Resolving power 分辨本领
R
1
 min
D

1.22
Increasing D or decreasing λcan enhance the resolving power.
§3-5 the diffraction grating
Planar grating
concave grating
1) Kinds and construction of grating
Transmission grating
透
射
光
栅
Reflection grating
反
射
光
栅
刻痕处不透光
2) Grating constant 光栅常数
d  ab
b
a: width of transmission part
b: width of reflection part
For example, a grating ruled with 5000 lines/cm has a
grating constant / slit spacing of
1
4
d
cm  2  10 cm
5000
a
3) The function of diffraction grating
Single slit diffraction


  ( 2k  1)
a sin   
2

 k
明
暗
Decreasing the width of the slit a can enhance the resolution of the
diffraction, but it will decrease the energy of middle fringe, which
will make the diffraction fringe unclear.
Grating: many many narrow slits
4) Schematic diagram of diffraction
grating
 A E

G

F
O
a
P
b
f
5) Intensity distribution in diffraction grating
I
u
v
u: intensity distribution of diffraction grating
v: intensity distribution of single slit diffraction
Diffraction grating: the superposition of many single slit-diffraction
光栅是单缝衍射和多缝干涉的共同作用效果，是
多个单缝衍射叠加（干涉）的结构