1. Part 1 Interference of light
Chapter Wave Optics
Part 1 Interference of light

2. Colorful interfering phenomena
Soap film under sunlight

3. Oil film under sunlight

4. Interference fringes produced by wedge-shaped air film between two flat glass plates

5. Newton’s rings
(Equal thickness fringes)
Equal inclination fringes

6. §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
Stimulating
pumping
Stimulation
Radiation
The persistent of light is

7. · ·  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)

8. Coherent light 2. Laser (from stimulated radiation 受激辐射)
f= (E2-E1)/h E1 E2 f
Identical wave train (frequency, vibrating direction, phase)
Coherent light

9. Good monochromaticity
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
Bad monochromaticity I  I 

10. 4. Coherency
Coherent lights： Identical frequency, vibrating direction, phase
Can produce coherent superposition

11. The intensity distribution of resultant light in crossing space：
If I1=I2,
The intensity of light varies between bright and dark alternately.
If I1 and I2 are not coherent light, then
No varying

12. 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.：

13. The way of division of wavefront：

14. The way of division of amplitude.：

15. §17-2 Two beams interference
Study by yourself
§ 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：
The distribution characteristics of the fringes and the distance between them.
3. Analyze the distribution characteristics of the fringes of the white light.

16. II. Fresnel’s double-mirror experiment
III. Lloyd’s mirror experiment S S’ M M’ d E E’ A B C S
Half-wave loss

17. §17-3 Optical path and optical path difference, Property of thin lens
I. Optical path
--speed of light in medium
=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.

18. Let in vacuum：
, c
In the medium with refractive index n： ,
Then
II. Optical path difference
The phase difference：
：the wavelength in vacuum

19. III. The half-wave loss of reflecting light
Phase shift 
n1<n2
Has half-wave loss
n1>n2 
No half-wave loss
 The transmitting ( refracting ) light never have half-wave loss.

20. 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 is inverse.
AF=  CF=  BF ，they are converging on point F and forming a bright image point.

21. §17-4 Interference by division of amplitude
I. Equal-inclination interference
1. The interference of reflecting light
The optical path difference is

22. And

23. The conditions that occur bright and dark fringes are
--Bright fringes
--Dark fringes

24. Discussions：
 ，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

25.  The interference fringes are series of concentric circles
 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,
k smaller
R smaller,
i smaller,
k larger

26. film
Reflecting plate
lens
screen

27. film
Reflecting plate
lens
screen

28. 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
----Dark

29. 2. The interference of transmitting light
The optical path difference is
----bright
----dark
The reflecting light and transmitting light are compensative each other.

30. [Example] A thin oil film (n= 1. 30) is illuminated by the white light
[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?

31. Solution : According to
k=1  e=emin

32. The film is observed at the normal direction of the film, i=0:
For k =1,
--green
For k =2,
--Ultraviolet

33. n0 < n < n MgF2 glass II. Application
1.Transmission enhanced film  
MgF2
(anti-reflecting film)
glass
(增透膜)
n0 < n < n
The two beams reflected by the upper and bottom interface of the MgF2 film all have half-wave loss.
The optical path difference between  and  is

34. MgF2 glass --reflecting beams are destructive. 
--reflecting beams are destructive.
--transmitting beams are constructive.
The minimum thickness of MgF2： or
--optical thickness

35. 2. Reflection enhanced film(增反膜)
Considering the reflected beams by each interface.
For the first film,
k=1 
For the second film,
k=1 

36. III. Equal-thickness interference
1. Wedged interference
Assume the incident beams are perpendicular
to the interfaces of the film.
medium wedge
At the contact edge(e=0) , k =0, appears dark fringe.
--bright
--dark

37. For air wedge,
air wedge
Discussion
 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.

38.  The thickness difference between two adjacent bright(or dark) fringes:
For air wedge,

39.  The distance between two bright (or dark) fringes:
--identical distance

40.  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.

41. [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.) Si
SiO2

42. Solution：
SiO2 Si
The optical difference between two rays reflected by the upper and lower surface of SiO2 is
At the contact edge, k=0，the bright fringe at d should correspond to k=6

43. At points with thickness e ,
Plane glass
Plane-convex
2. Newton’s ring
At points with thickness e ,
--bright
--dark
At the center(e =0): corresponding to a dark point with k =0.

44. Discussion:
  is determined by e
--equal-thickness interference
 The fringes are series of concentric circles.
-- Newton’s ring
 the radius r of the circle: as

45. Bright circle,
Dark circle,
 the distance between two adjacent circles :
-- alienation(稀疏)near the center, dense near the edge.

46. When the Plane-convex lens moves upwards, the circle fringes are swallowed at the center.

47. [Example] A drop of oil is on a plane glass
[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）

48. Solution
Because n1<n2，n2>n3，
there is half-wave loss in  .
i.e.,
At the edge of oil, e=0, we have
i.e.,  satisfies dark fringes condition, So
There is a dark circle fringe at the oil edge, corresponding to k=0.

49.  The center dark point corresponds to k=4.
 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.

50. §17-5 Michelson’s interferometer
I. Instrument
M1 is fixed, M2 can be moved
Compensating plate
Beam spliter

51. Discussion
M1and M2are perpendicular to each other--equal inclination interference produced by air film.
-- concentric circles
Assume the thickness of the air film between M1and M2 is d
Then, at the center of the fringes
M2 moves ：
--The number of the fringes out of or in the center.

52. M1and M2 are not perpendicular to each other -- equal thickness interference produced by air wedge.
II. Application
Measure length d—known  , read out n，
Measure ----read out n and d，
Measure refractive index n—known ，read out d
Look for “ether（以太）”
-- “ zero”result

53. Michelson’s interferometer be used to measure the refractive index of gas.

54. Michelson’s interferometer be used to measure flow field.