1. Physics 11 Advanced Mr. Jean May 28 th, 2012

2. The plan: Video clip of the day Wave Interference patterns Index of refraction Slit & Double Slit interference Brewster’s Angle Polarizer More interference Problems

3. Here is a really fancy drawing showing the geometry of the path lengths. We have two slits, S1 and S2. Two rays are shown - the straight lines that go from the slits to the point P.

4. r 1 = distance of path for wave 1 r 2 = distance of path for wave 2 S 1 = Slit #1 S 2 = Slit #2 d = Distance between slits P = point where interference is observed. L = Length between slits and surface. y = height change from normal.

5. m  fringe number This is the equation you will have available on the AP Physics Test. Use it wisely. You end up with a central bright fringe. This is the zeroth-order maximum (m = 0). You can see that the angle  would be zero for this fringe. The bright central fringe is bracketed by a series of smaller bright fringes for the different integer values of m. The next set of fringes is the first-order maximum, then we have the second- order maximum, and so on.

6. Demonstrations: 7 - 1 to 7 - 4 Videos worth checking out: –http://www.brightstorm.com/science/physics/http://www.brightstorm.com/science/physics/ –Topic: “Light”

7. Light Interference:

11. Example Problem:

15. Example:

17. Light Slit pattern: Lab video demonstrations –Lab 7.1, 7.2 & 7.3

18. Brewster’s Angle: When light encounters a boundary between two media with different refractive indices, some of it is usually reflected. The fraction that is reflected is dependent upon the incoming light's polarization and angle of incidence. At some point this light source will be entirely polarized thus causing no reflected light to be visible through Polaroid.

20. Brewster’s Angle Demonstration: Lab Demo 7.4 http://webphysics.davidson.edu/physlet_re sources/bu_semester2/c27_brewster.html

21. Phase Change: Reflection & Phase: –Light reflecting from a boundary can do so in phase (sort of a free end reflection) or out of phase (a fixed end reflection). (Spring vs. Spring with rope) –The thing that determines whether the reflected wave is in or out of phase is the difference in speed for light in the two media. –The wave will undergo a 180  phase change when it is reflected from a medium that has a higher index of refraction than the one it came through.

22. Phase Change:

23. There will be no phase change if the wave is reflected from a medium that has a lower refractive index. Ex: You would get the phase change for light traveling through air and reflecting off glass. Glass has a higher index of refraction than air. Ex: You would not get a phase change for light traveling through glass and being reflected off water, since water has a lower index of refraction than glass.

24. Thin Film Interference: Thin Film Interference: This occurs when light travels through a very thin layer of transparent material. Thin film interference occurs with oil films, soap bubbles, etc. Light that is incident on the film has several things happen to it. –Some of the light is reflected off the top of the film. These waves have a 180  phase change since the index of refraction for the film is greater than for air. –Next, the light that goes into the film is refracted as it travels from air into the film. Some of the light goes into the air on the other side of the film. This light is refracted (back the other way). –Finally, some of the light is reflected off the air/film surface. This light does not undergo any phase change.

25. The film has a thickness of t. We let n be the index of refraction for the film. The index of refraction for air is, of course, 1. Ray 1 reflecting off the surface of the film has a 180  phase change. Ray 2 reflecting off the opposite film surface has no phase change. The two rays are out of phase.

26. Thin Film Problems: The two waves will recombine when you look into the film and the rays enter your eyes. If the path difference is half of the wavelength, or an odd multiple of the wavelength, then the waves will end up in phase and you will see constructive interference – a bright fringe. The basic kind of problem involves finding the minimum thickness that will cause constructive or destructive interference. This minimum would be when the wave came straight down onto the film. This means that the angle of incidence is zero.

27. We can solve for the wavelength in the film! We start with the equation for the index of refraction.

28. We’ll call the wavelength in the film f. This means that the minimum thickness is given by:

29. Example:

30. Important Questions for Chapter 9: P. 443 –Questions: 24, 29, 31 & 37 P. 444 & 445 –Questions: 44, 49, 50, 61 & 64