1. 1 Chapter 9 Interference February 16 General considerations of interference 9.1 General considerations Introduction: Wave equation  Superposition principle Interference: The interaction between two or more waves produces in space a resultant irradiance which is different from the sum of the component irradiances: E = E 1 +E 2, but I  I 1 + I 2. Interferometric devices: wavefront splitting and amplitude splitting. Poynting vector: is the instantaneous power flow across an unit area whose normal is parallel to S. For a harmonic, linearly polarized plane wave: Irradiance (intensity): The time-averaged energy transport per unit area per unit time.

2. 2 Superposition of two planar, linearly polarized waves (vector model): Phase difference interference term 1)When E 01 and E 02 are perpendicular, I 12 =0, no interference. 2)When E 01 and E 02 are parallel, Note that cos  averages to 0 in space: energy is conserved.

3. 3 Total constructive interference: Total destructive interference: When I 1 =I 2 =I 0, Interference fringes between two plane-waves: Bright fringes:  Fringes are parallel planes perpendicular to k 1 -k 2. k1k1 k2k2 k1-k2k1-k2  Exercise: Prove that the distance between the fringe planes is Discussion: Fabrication photographic gratings.

4. 4 9.2 Conditions for interference 1)For producing stable patterns, the two sources must have the same frequency. 2)For producing interference patterns, the two fields must have parallel components. 3)For producing clear patterns, the two sources must have similar amplitude. 4)For producing interference patterns, coherent sources are required. Temporal coherence: Time interval  t c in which the light resembles a sinusoidal wave. (~10 ns for quasi- monochromatic light sources.) Coherence length:  l c = c  t c. Relation to bandwidth (uncertainty principle):  t c  1. Useful relations: Spatial coherence: The correlation of the phase of a light wave between different locations. =Relative phase predictable

5. 5 Read: Ch9: 1-2 Homework: Ch9: 2,3 Due: February 27

6. r1r1 r2r2 S1S1 S2S2 m=-1 0 +1 P m = M a 6 At a far field, I 1  I 2  I 0. Maximum: Minimum: Hyperboloids Superposition of two spherical waves: r1r1 r2r2 P S1S1 S2S2 Interference patterns: 1)Viewing screen perpendicular to the S 1 -S 2 axis: Concentric Rings. 2)Viewing screen parallel to the axis: Parallel fringes (Young’s experiment). 3)Maximum order of rings: -M< m <M, where M =a/. (  -a <r 1 -r 2 <a ) February 23 Interference of spherical waves

7. 7 Variation of the interference rings: When reducing a : 1)The rings swallow. (What we observe in the Michelson interferometer.) 2)The distances between neighbor rings at a fixed viewing point increases. Superposition of a plane wave and a spherical wave: Move S 1 to infinity. r1r1 r2r2 S1S1 S2S2 m=-1 0 +1 P m = M a

8. 8 (*Reading) Interference pattern and holography: 1)Two coherent waves produce an interference pattern. 2)The interference pattern is photographically recorded, which makes a hologram. 3)Now remove one of the waves (called the object wave), and illuminate the hologram with the other wave (called the reference wave). The removed wave is then magically reproduced in space! A B Record A Reconstruction

9. 9 (*Reading) Size of the interference rings: Note: 1) The size of the nth ring from the center: 2) S 1, S 2 and the viewing screen form a lens system: 3) Distance between rings: 4) For interference between a plane wave and a spherical wave: r1r1 r2r2 P S1S1 S2S2 xmxm u v (xn)(xn)

10. 10 (*Reading) Proof of the change of the rings (Method I, complicated but most accurate): r1r1 r2r2 P S1S1 S2S2 R x D  1) Calculate dR/dx for a given m: 2) Calculate dR/dm at a given R: At a given R,  The ring separation at (R, x) is given by

11. 11 (*Reading) Proof of the change of the rings (Method II, easier): r1r1 r2r2 P S1S1 S2S2 R x D  (*Reading) Proof of the change of the rings (Method III, simplified): 1) Calculate d  /dx for a given m: 2) Calculate d  /dm at a given  :  The ring separation at ( , x) is given by At a given ,  The ring separation at (m, x) is given by At a given m,  R

12. 12 Read: Ch9: 1-2 No homework

13. a s  13 February 25 Wavefront-splitting Interferometers 9.3 Wavefront-splitting interferometers 9.3.1 Young’s experiment (1801) Young’s originality: Obtaining spatially coherent light from a pinhole, and splitting the wavefront of the spatially coherent light with two pinholes. Optical path difference: r 2 – r 1 = a sin θ = ay/s ( If s>> a, s>> y ) Positions of bright fringes: m = 0, ±1, ±2, … is the order number. Distance between fringes: Intensity distribution of the fringes: ay/s

14. 14 Effects of finite coherent length: Light from each slit has a coherent length l c. For sunlight (blackbody) l c  3. The waves from two slits can only interfere if r 2 – r 1 <l c. The contrast of the fringes degrades when the amount of the overlap between uncorrelated wavegroups increases. P r 2 – r 1 B A A'A' B'B' lclc Other wavefront-splitting interferometers Fresnel’s double mirror: Slits S 1 and S 2 act as virtual coherent sources. They are images of slit S in the two mirrors.  = r 2 – r 1 P M1M1 M2M2 S1S1 S2S2 S a r1r1 r2r2 s y Shield Space between fringes:

15. 15 Fresnel’s double prism: Interference between light refracted from the upper and the lower prisms. The prisms produce two virtual coherent source S 1 and S 2. Question: Where are S 1 and S 2 ? S1S1 S2S2 S a  Lloyd’s mirror: Interference between light from source S and its image S ' in the mirror. Glancing incidence causes a phase shift of , therefore the fringes are complementary to those of Young’s. S S'S' a s y

16. 16 Read: Ch9: 3 Homework: Ch9: 6,14,17,19 Due: March 6

17. 17 9.4 Amplitude-splitting interferometers 9.4.1 Dielectric films – double-beam interference Application: Optical coating reduces or enhances the reflection of optical devices through thin-film interference effects. Fringes of equal inclination Consider the first two reflections (other reflections are weak if the reflectance is not large). Optical path length difference: February 27 Amplitude-splitting Interferometers-1 d A B C D P S ii tt nfnf n1n1 n2n2

18. 18 1) External reflection, n i <n t. There is a phase shift of  between the incident electric field and the reflected electric field. 2) Internal reflection, n i >n t. There is no phase shift between the incident electric field and the reflected electric field. Phase shift in reflections when the angle of incidence is small: E is kiki ii rr tt nini E rs krkr E ts ktkt ntnt E rp E tp E ip

19. 19 Phase difference: Assume n f >n 1, n f >n 2. There is a phase shift of  between external and internal reflections (when the incident angle is not large). Bright spot: Dark spot: Extended source: All rays inclined at the same angle arrive at the same point.  Fringes of equal inclination: Arcs centered on the perpendicular from the eye to the film. Haidinger fringes: The fringes of equal inclination viewed at nearly normal incidence.  Concentric circular bands.

20. 20 Read: Ch9: 4-5 Homework: Ch9: 27,29,32,34 Due: March 6

21. Thickness: d =  x Interference maximum: 2n f d m = (m + ½) Distance between fringes:  x = f /2  21 March 2 Amplitude-splitting Interferometers-2 Fringes of equal thickness Fizeau fringes: Contours from a non-uniform film when viewed at nearly normal incidence.  x n1n1 n2n2 nfnf d Compare: For the interference between two plane waves, the distance between the fringe planes is

22. 22 Interference colors When a white light source is used in an interference experiment, each spectral component will produce its own interference pattern in space. The overlapped interference patterns from all spectral components result in the appearance of a special sequence of colors. Calculated interference colors Crossed polarizers Parallel polarizers Examples: Soap bubbles Oil films Thin film coatings Crystals between polarizers Applications: Identify minerals Control coatings Inspect strains

23. 23 Newton’s rings: Interference pattern from an air film between two glass surfaces. x d Thickness: Bright rings: 2n f d m = (m + ½)  Dark rings: Discussions: Superposition of a plane wave and a spherical wave. Fabricating zone plates.

24. 24 Beam splitter Movable mirror (©WIU OptoLab) S S1S1 S2S2 M1M1 M2M2 2dcos   P Optical path length difference: Phase difference: Dark fringes: 9.4.2 Mirrored interferometers Michelson interferometer 1)Compensation plate: Negates dispersion from the beam splitter 2)Collimated source: Fringes of equal thickness 3)Point source: Interference of spherical waves 4)Extended source: Fringes of equal inclination Application: Accurate length measurement.   from the splitter.

25. 25 Read: Ch9: 4-5 No homework

26. 26 9.6 Multiple beam interference Interference between multiple reflection and multiple refractions: tt d S ii nfnf March 4, 6 Multiple beam interference Reduces to fringes of equal inclination when r is small.

27. 27 F = 0.2 (r 2 = 0.046) F = 1 (r 2 = 0.17) F = 200 (r 2 = 0.87) Coefficient of finesse: Airy function:

28. 28 9.6.1 The Fabry-Perot interferometer 1)High resolving power 2)Prototype of laser cavity S P d Half-width of transmission: Finesse:

29. 29 Applications: I. How a laser wavelength is selected. 1) The laser itself is a Fabry-Perot cavity. Two highly reflective mirrors select an output of a certain wavelength with an extremely narrow bandwidth, which is called a mode. 2) The mode separation in a laser cavity (  =c/2l) can be narrower than the gain profile of the laser medium, therefore output of multiple modes is possible. 3) Laser modes can be longitudinal or transverse.

30. 30 II. How a scanning spectrum analyzer works. 1)A scanning spectrum analyzer can be used to measure the modes of a laser. 2) The analyzer is a Fabry-Perot interferometer whose length can be slightly varied by piezoelectric device. 3) A saw-tooth voltage is used to repeatedly change the cavity length, and the output signal is displayed on an oscilloscope. scanning

31. 31 ii rr tt nini ntnt 1 t r r t tr' rt r2r2 tt'

32. 32 Read: Ch9: 6 Homework: Ch9: 37,40,42,45,47 Due: March 13 Hint to P9.45: You are asked for the refractive index and the thickness of the film, so two conditions should be figured out. In order to completely eliminate the overall reflection, the light reflected from the front surface and from the rear surface of the film must have exactly the same amplitude, as well as having exactly a phase difference of . Suppose, then n f can be found from the Fresnel equations if we require the two reflections exactly cancel each other. You may also prove that won’t work for this problem.

33. 33 “You can’t judge a book by its cover.”