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

1.The emitting process of light §17-1 Monochromaticity and Coherence 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)

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

Good monochromaticity Bad monochromaticity I I   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 )

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

The intensity distribution of resultant light in crossing space： If I1=I2, The intensity of light varies between bright and dark alternately. If I1andI2 are not coherent light, then No varying

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. 5. The methods to obtaining coherent light: • 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.：

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.

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

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

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

Phase shift Has half-wave loss No half-wave loss III. The half-wave loss of reflecting light  n1<n2 n1>n2   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.

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

And

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

，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 levelk Discussions： --Equal inclination interference

R larger, i larger, k smaller R smaller, i smaller, k larger  The interference fringes are series of concentric circles. They are alienation(稀疏)near the center, and dense near the edge.

screen lens Reflecting plate film

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

----bright ----dark 2. The interference of transmitting light The optical path difference is • 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 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

  MgF2 glass II. Application 1.Transmission enhanced film (anti-reflecting film) (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 --optical thickness --reflecting beams are destructive. --transmitting beams are constructive. • The minimum thickness of MgF2： or

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 

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

air wedge For 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

Si SiO2 Solution： 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

Plane-convex --bright --dark Plane glass 2. Newton’s ring At points with thickness e , • 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. 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 tok=4.  Whenthe 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.

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

Part 1 Interference of light