1. Supplementary Material This set of slides contains material dealing with thin films and with the Michelson Interferometer. Both of these phenomena can be explained nicely by the interference of waves. Also included are slides about the telescope.

2. Interference: Thin Films Before, we had several different parts of a wide beam interfering with one another. Can we find other ways of having parts of a beam interfere with other parts?

3. Interference: Thin Films We can also use reflection and refraction to get different parts of a beam to interfere with one another by using a thin film. air film water reflected red interferes with refracted/reflected/refracted blue.

4. Interference: Thin Films Blue travels an extra distance of 2t in the film. air film water reflected red interferes with refracted/reflected/refracted blue. t

5. Interference: Thin Films And, blue undergoes two refractions and reflects off of a different surface. air film water reflected red interferes with refracted/reflected/refracted blue. t

6. Interference: Thin Films When a wave encounters a new medium: –the phase of the refracted wave is NOT affected. –the phase of the reflected wave MAY BE affected.

7. Interference: Thin Films When a wave on a string encounters a fixed end, the reflected wave must interfere with the incoming wave so as to produce cancellation. This means the reflected wave is 180 degrees (or /2) out of phase with the incoming wave.

8. Interference: Thin Films When a wave on a string encounters a fixed end, the reflected wave must interfere with the incoming wave so as to produce cancellation. This means the reflected wave is 180 degrees (or /2) out of phase with the incoming wave.

9. Interference: Thin Films When a wave on a string encounters a free end, the reflected wave does NOT have to destructively interfere with the incoming wave. There is NO phase shift on this reflection.

10. Interference: Thin Films When a wave on a string encounters a free end, the reflected wave does NOT have to destructively interfere with the incoming wave. There is NO phase shift on this reflection.

11. Interference: Thin Films When light is incident on a SLOWER medium (one of index of refraction higher than the one it is in), the reflected wave is 180 degrees out of phase with the incident wave. When light is incident on a FASTER medium, the reflected wave does NOT undergo a 180 degree phase shift.

12. Interference: Thin Films If n a < n f < n w, BOTH red and blue reflected rays will be going from fast to slow, and no difference in phase will be due to reflection. air film water reflected red interferes with refracted/reflected/refracted blue. t

13. Interference: Thin Films If n a n w, there WILL be a 180 degree phase difference ( /2) due to reflection. air film water reflected red interferes with refracted/reflected/refracted blue. t

14. Interference: Thin Films There will ALWAYS be a phase difference due to the extra distance of 2t/. air film water reflected red interferes with refracted/reflected/refracted blue. t

15. Interference: Thin Films When t= /2 the phase difference due to path is 360 degrees (equivalent to no difference) air film water reflected red interferes with refracted/reflected/refracted blue. t

16. Interference: Thin Films When t= /4 the phase difference due to path is 180 degrees. air film water reflected red interferes with refracted/reflected/refracted blue. t

17. Interference: Thin Films Recall that the light is in the FILM, so the wavelength is not that in AIR: f = a /n f. air film water reflected red interferes with refracted/reflected/refracted blue. t

18. Interference: Thin Films reflection: no difference if n f < n w ; 180 degree difference if n f > n w. distance: no difference if t = a /2n f 180 degree difference if t = a /4n f Total phase difference is sum of the above two effects.

19. Interference: Thin Films Total phase difference is sum of the two effects of distance and reflection For minimum reflection, need total to be 180 degrees. –anti-reflective coating on lens For maximum reflection, need total to be 0 degrees. –colors on oil slick

20. Thin Films: an example An oil slick preferentially reflects green light. The index of refraction of the oil is 1.65, that of water is 1.33, and of course that of air is 1.00. What is the thickness of the oil slick?

21. Thin Films: an example Since we have preferentially reflected green light, the TOTAL phase difference must be 0 degrees (or 360 degrees).

22. Thin Films: an example Since we have preferentially reflected green light, the TOTAL phase difference must be 0 degrees (or 360 degrees). Since we have n f > n w, we have 180 degrees due to reflection.

23. Thin Films: an example Since we have preferentially reflected green light, the TOTAL phase difference must be 0 degrees (or 360 degrees). Since we have n f > n w, we have 180 degrees due to reflection. Therefore, we need 180 degrees due to extra distance, so need t = a /4n f where a = 500 nm, n f = 1.65, and so: t = 500 nm / 4(1.65) = 76 nm.

24. Michelson Interferometer Split a beam with a Half Mirror, the use mirrors to recombine the two beams. Mirror Half Mirror Screen Light sourc e

25. Michelson Interferometer If the red beam goes the same length as the blue beam, then the two beams will constructively interfere and a bright spot will appear on screen. Mirror Half Mirror Screen Light sourc e

26. Michelson Interferometer If the blue beam goes a little extra distance,  s, the the screen will show a different interference pattern. Mirror Half Mirror Screen Light sourc e ss

27. Michelson Interferometer If  s = /4, then the interference pattern changes from bright to dark. Mirror Half Mirror Screen Light sourc e ss

28. Michelson Interferometer If  s = /2, then the interference pattern changes from bright to dark back to bright (a fringe shift). Mirror Half Mirror Screen Light sourc e ss

29. Michelson Interferometer By counting the number of fringe shifts, we can determine how far  s is! Mirror Half Mirror Screen Light sourc e ss

30. Michelson Interferometer If we use the red laser ( =632 nm), then each fringe shift corresponds to a distance the mirror moves of 316 nm (about 1/3 of a micron)! Mirror Half Mirror Screen Light sourc e ss

31. Limits on Resolution: further examples: telescope Light from far away is almost parallel. objective lens eyepiece fofo fefe

32. Limits on Resolution: further examples: telescope The telescope collects and concentrates light. objective lens eyepiece fofo fefe

33. Limits on Resolution: further examples: telescope Light coming in at an angle,  in is magnified to  out. objective lens eyepiece fofo fefe x

34. Limits on Resolution: further examples: telescope  in = x/f o,  out = x/f e ; M =  out /  in = f o /f e objective lens eyepiece fofo fefe x

35. Limits on Resolution: further examples: telescopes telescopes –magnification: M =  out /  in = f o /f e –light gathering: Amt  D 2 –resolution: 1.22 = D sin(  limit ) so  in =  limit and  out = 5 arc minutes  so  limit  1/D implies M useful = 60 * D where D is in inches –surface must be smooth on order of

36. Limits on Resolution: Telescope M max useful =  out /  in =  eye /  limit = 5 arc min / (1.22 * / D) radians = (5/60)*(  /180) / (1.22 * 5.5 x 10 -7 m / D) = (2167 / m) * D * (1 m / 100 cm) * (2.54 cm / 1 in) = (55 / in) * D

37. Limits on Resolution: Telescope What diameter telescope would you need to read letters the size of license plate numbers from a spy satellite?

38. Limits on Resolution: Telescope need to resolve an “x” size of about 1 cm “s” is on order of 100 miles or 150 km  limit then must be (in radians) = 1 cm / 150 km = 7 x 10 -8  limit = 1.22 x 5.5 x 10 -7 m / D = 7 x 10 -8 so D = 10 m (Hubble has a 2.4 m diameter)