1、Two beam interference,Two plane waves meet at P,Assume the observation point P is far enough away from the sources so that the wavefronts at P can be regarded as planes. And we limit our discussion to linearly polarized waves.,Superposition of two beams,Obviously,Superposition of two beams,where pre

2、sents the crossing angle between the linear polarization directions of the two beams， and,Interference Pattern,Relation between Polarization Directions,parallel,perpendicular,Coherence Conditions for Interference,In order to get stable intensity-modulated interference pattern by using two beams, two

3、 conditions must be satisfied The two waves must maintain a constant phase with respect to each other, i.e. have the same frequency and constant initial phase The two electric fields have the non-zero projective components onto each other, i.e. the polarization directions cannot be perpendicular to

4、each other,Youngs Double Slit Experiment,In 1801, Thomas Young demonstrated for the first time interference of light The coherent sources was gotten through: Light from a monochromatic source goes through a narrow slit. The narrow width of the silt make sure transmitted light comes from a tiny regio

5、n of the source which is coherent. The transmitted light is incident onto a screen containing two narrow slits,Youngs Double Slit Experiment,The symmetric narrow slits, S1 and S2 act as the two light sources The waves from the two slits come from the same source S0 and therefore are always in phase.

6、,Other Coherent Sources,Currently, it is much more common to use a laser as a coherent source The laser produces an intense, coherent, monochromatic parallel beam over a width of several millimeters,Laser pointer,Youngs double-slit interference,Side view of the interference field,Interference Patter

7、ns,Constructive interference occurs at the center point The two waves travel the same distance Therefore, they arrive in phase,Interference Patterns,The upper wave has to travel farther than the lower wave The upper wave travels one wavelength farther Therefore, the waves arrive in phase A bright fr

8、inge occurs,Interference Patterns,The upper wave travels one-half of a wavelength farther than the lower wave The trough of the bottom wave overlaps the crest of the upper wave (180 phase shift) This is destructive interference A dark fringe occurs,Youngs Experiment(Double-Slit Interference),Interfe

9、rence Equations,The path difference, , is found from the tan triangle Optical Path Difference = L = r2 r1 = d sin ,Interference Equations,This assumes the paths are parallel Not exactly, but a very good approximation (Ld),Interference Equations,For a bright fringe where total constructive interferen

10、ce happens m = 0, 1, 2, m is called the order number When m = 0, it is the zeroth order maximum When m = 1, it is called the first order maximum,Interference Equations,When total destructive interference occurs, a dark fringe is observed This needs a path difference of an odd half wavelength = d sin

11、 dark = (m + ) m = 0, 1, 2, ,Interference Equations,Y:measured vertically from the zeroth order maximum Assumptions L d, d y =LtanLsin,Interference Equations,For bright fringes For dark fringes,Usage of Youngs Double Slit Experiment,Youngs Double Slit Experiment provides a method for measuring wavel

12、ength of the light This experiment gave the wave model of light a great deal of credibility It is inconceivable that particles of light could cancel each other,A thin flake of mica (n = 1.58) is used to cover one slit of a double-slit arrangement. The central point on the screen is then occupied by

13、what used to be the seventh bright fringe. If = 550 nm, what is the thickness of the mica?,The 7th bright fringe with no mica corresponds to n = 7 The mica therefore introduce a path difference of 7 or a phase difference of 7x2 at the screen center,d,Phase with meca,The mica thickness is :,Phase wit

14、hout meca,Lloyds Mirror,Produce an interference pattern with a single light source Wave reach point P either by a direct path or by reflection The reflected ray can be treated as a ray from the image S below the mirror,Interference Pattern from Lloyds Mirror,The positions of the dark and bright frin

15、ges are reversed relative to pattern of two real sources, i.e. at point P the intensity is zero This is because there is a 180 phase change produced by the reflection,ni nt,0 30 60 90,Incidence angle,Incidence angle,p 0,p 0,S-polarization,P-polarization, phase change in grazing incidence,Fresnels bi

16、prism,L1,L2,Diagrammatic Sketch,Because the apex angle of the biprism is very small, the distance between the two images a can be written as:,So the fringe space on the screen is:,Fresnels biprism,Fresnels mirror,Diagrammatic Sketch,Billets split lens,Diagrammatic Sketch,If white light is used in do

17、uble-slit interference, position of the 0th maximum for light with different wavelength is coincidence. So white fringe appears at the screen center. Higher order maximums have spatial shift. Therefore color instead of white fringes are observed at other positions.,Interference fringe by white light

18、,Interference fringe by white light,Green light,White light,Interference in Thin Films Two beams model,第二章第4節(jié)，第三章第1，2，3節(jié),Soap bubbles,Soap solution is transparent, but soap bubbles are colorful!,Problem Solving Strategy with Thin Films,Identify the two interfering beams. Calculate the intensity of t

19、he two beams according to Fresnels formula Identify the polarization direction of the beams Calculate the phase difference between the two beams, i.e. calculate the optical path length difference and determine whether the phase change exists or not. Submit all the items into the intensity formula. D

20、iscussion.,Interference in Thin Films,Assume incident light is in the air nearly normal to the two film surfaces and is linearly polarized. Assume the film is surrounded by the air. What we observe is the interference between two weak reflected beam from the thick film.,1,Interference in Thin Films,

21、The first two beams are the stronger reflected light Let r and t present the reflectance and the transmittance E1=r E0 E2=t E0 r t = rt2E0,Because r0 and t1, E1E2E0 Two reflected beams have the same polarization,Interference fringe,Intensity Distribution,Here,A,B,Optical path length difference in th

22、in film,Phase Changes Due To Reflection,Ray 1 undergoes a phase change of 180 with respect to the incident ray. No phase shift happens in Ray 2 with respect to the incident wave.,1,Low incident angle,k: wave vector in the air, phase change in grazing incidence,S-polarization,P-polarization,ni nt,0 30 60 90,Incidence angle,p 0,p 0,S-polarization,P-polarization, phase change in grazing incidence,nt ni,Incidence angle,0 30 60 90,p 0,p 0,b,c,S-polarization,P-polarization,If you slowly turn up a laser i