Part 1 Interference of light Chapter 17 Wave Optics Part 1 Interference of light

Colorful interfering phenomena Soap film under sunlight

Oil film under sunlight

Interference fringes produced by wedge-shaped 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 Stimulating pumping Stimulation Radiation The persistent of light is

· ·  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)

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

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 

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

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

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：

The way of division of amplitude.：

§17-2 Two beams interference 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： 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 S S’ M M’ d E E’ A B C S Half-wave loss

§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.

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

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.

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.

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

And

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

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

 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

film Reflecting plate lens screen

film Reflecting plate lens screen

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

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

[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?

Solution : According to k=1  e=emin

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

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

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

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 

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

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.

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

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

[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

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

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.

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

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

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

[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）

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.

 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.

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

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.

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

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

Michelson’s interferometer be used to measure flow field.