1. Chapter 24: Thin Films Diffraction Diffraction Grating

2. Hint: Be able to do the homework (both the problems to turn in AND the recommended ones) you’ll do fine on the exam! Wednesday, February 3, 1999 in class Ch. 22, 23, 24 (Sect. 1-6) You may bring one 3”X5” index card (hand-written on both sides), a pencil or pen, and a scientific calculator with you.

3. DUE TODAY 11 pm

4. For the case of a thin film of index of refraction n surrounded by air, the maxima and minima occur under the following conditions Constructive interference Destructive interference

5. Air, n = 1 Water, n = 1.33 Glass, n = 1.5 t In this case, the blue and red rays suffer a 180 o phase change, so the conditions for maxima and minima are exactly the opposite of what appears on the last slide!

6. When working problems with thin films, you must first evaluate what happens to the reflected ray at each boundary. For rays travelling from low n to high n, a 180 o phase shift will occur. Then, simply calculate the difference in path length based upon the thickness of the film.

7. For cases in which both reflected rays suffer the same fate (either phase shifted or not), path length differences of 1/2 a wavelength lead to destructive interference. For cases in which one reflected ray suffers a phase shift of half a wavelength, path length differences of 1 full wavelength lead to destructive interference.

9. In studying the two slit interference pattern, we’ve made some simplifying assumptions and neglected some important physical processes. Perhaps the most important of these is known as

10. The way we’ve drawn our diagrams for the two slit interference, we’ve included the process already without really commenting on its existence. Let’s take a close-up view of one of the slits...

11. If light is travelling in a straight path, why don’t we observe the following:

12. Geometric optics cannot explain the diffraction-- the bending of light around edges--that is actually observed in our experiments with the slits) When we project the light passing through a single slit onto a distant screen, guess what we see….

13. Central maximum Secondary maxima 1122 Secondary minima

14. Right about this point, you should be asking yourself... Why in the world should light from a single slit produce what looks like an interference pattern on the screen???? It is useful to employ Huygens’ principle to explain our observations...

15. Recall that Huygens’ principle advised us to treat each portion of a wave front of light as a source of light waves... screen Thinking about the problem this way allows us to see the single slit as containing a large number of sources, each of which can interfere.

16. screen  a A dark spot will appear on the screen at locations where the path length difference  is half a wavelength. minima

17. screen a minima This is true for any pair of “sources” separated by half the slit width...

18. Destructive interference will occur as long as the path length difference is some integral multiple of the quantity /a Therefore, angles at which destructive interference occurs are given by Where m = +/- 1, +/- 2, +/- 3, ….

19. As with the two-slit interference pattern, we can relate the distance above the central maximum of each dark fringe with a little geometry... screen a L y

20. Notice that when the slit width, a, is less than the wavelength, the quantity on the right side is greater than 1! Therefore, for a <, we will not observe the diffraction interference pattern.

21. The secondary maxima can be found on either side of the central maximum approximately half-way between the locations of successive minima. The central maximum is found at the angle  = 0 and the secondary maxima at angles given by: Note that the central maximum is twice as wide as the secondary maxima.

22. The formulae for the diffraction pattern look a lot like those we saw for the two-slit interference pattern, but they are very different! diffraction minimaTwo-slit maxima

24. Conceptually, it’a quite simple to extend our discussion of diffraction and Young’s experiment to the case of the interference pattern produced from an object known as a Instead of the large number of “sources” that we considered in the diffraction pattern using Huygen’s principle, the diffraction grating consists of a very large number of slits, each one of which acts as its own source. You can think of it as an n-slit interference pattern (where n is the number of slits).

25. Diffraction Grating Diffraction gratings are usually described by the number of slits per unit length. L N slits As in Young’s experiment, the observed diffraction pattern depends upon the slit separation, the wavelength of the incident light, and the angle to the screen.

26. observed interference pattern 01-22 central maximum order of the secondary maxima We can proceed as before (using geometry) to locate the maxima. Notice that these lines tend to be very narrow.

27. N-slit maxima Where m is the order number and d the slit separation, given by L / N for the diffraction grating. As you will see in the laboratory, diffraction gratings provide a great way to separate the constituent wavelengths of incident light, since each wavelength will have a unique angle for the appearance of bright lines when examined through the grating. All wavelengths will have a bright at angle  = 0, so a bright white line will appear at this angle (usually).

28. Geometric Optics Mirrors Lenses Images Interference Diffraction

29. The wavelength ( ) and frequency (f) of electromagnetic waves are related to their propagation speeds by 10 -9 10 -6 10 -3 110 3 TV Microwaves Infrared VISIBLE Ultra-Violet X-ray c = 3 X 10 8 m/s

30. ii rr (a.k.a. Snell’s Law) ii rr air water rr n1n1 n2n2

31. The frequency with which waves leave one medium must equal the frequency with which they enter another (otherwise we’d have waves piling up at a boundary, and we never observe that to occur). c v Since the frequency of the light must be the same in both media, it’s the wavelength that is changing

32. For light rays travelling into a material with a smaller index of refraction, there exists a critical angle greater than which the light rays cannot enter the second medium. These light rays are totally reflected by the boundary back into the original medium. The law of reflection applies in this case.

33. Let’s start by studying what happens when light strikes a plane mirror (a 2D surface).. p the distance of the object from the mirror q the distance of the image from the mirror h = h’p = q h object height object image h’ image height Magnification

34. Concave: Convex:

35. For spherical mirrors, f and R are related by geometry: f = R / 2 M > 0 means the image is erect. M < 0 means the image is inverted. p is the object distance q is the image distance

36. n1n1 n2n2 object image p is the distance of the object from the surface q is the distance of the image from the surface

37. A plane has a radius of curvature equal to infinity, so... Diverging (f < 0) Thinnest in the middle. Parallel light rays from infinity diverge as they pass through the lens. Converging (f > 0) Thickest in the middle. Bring light rays from infinity to a focus on the opposite side of the lens.

38. Light can pass through a lens in either direction. So it’s not surprising that a lens will have two foci. focal length, f The thin lens equation relates the object and image distances to the focal length. R2R2 R1R1 front back

39. front back “focus of back surface” “focus of front surface” IMAGE OBJECT Converging lens

40. Whether you’re dealing with mirrors, refracting surfaces, or lenses, p and q are defined to be positive where the light rays actually are. Real objects will have p > 0. Real images will have q > 0. Virtual objects will have p < 0. Virtual images will have q < 0.

41. To determine the sign of R correctly, examine where the light ray goes AFTER encountering the surface... For a lens, the light passes through to the back side... q > 0q < 0 R < 0 For a mirror, the light reflects back on the same side... q < 0R > 0q > 0 f > 0

42. 1) To determine whether p > 0 or p < 0, ask yourself, “Where is the light coming from?” On the side of the boundary from which light comes, p > 0! 2) To determine whether q > 0 or q < 0, ask yourself, “Where does the light go?” On the side of the boundary that light travels after encountering the surface, q > 0.

43. monochromatic light source crests troughs bright dark Two-slit interference pattern Observed interference pattern

44. d screen L y r1r1 r2r2    For small  So, bright fringes dark fringes

45. When light reflects off the boundary of a material with a higher index of refraction than that in which the light is incident on the surface, this phase shift of 180 o will occur. mirror, n > 1 incident crest trough reflected Air, n = 1 If the reflected rays lie on top of one another (as will be the case if the light is normally incident on the top surface), the reflected rays can interfere with one another! Air, n = 1 Water, n = 1.33 Air, n = 1 t

46. The path length difference is 2 t constructive destructive When working problems with thin films, you must first evaluate what happens to the reflected ray at each boundary. For rays travelling from low n to high n, a 180 o phase shift will occur. Then, simply calculate the difference in path length based upon the thickness of the film. For cases in which both reflected rays suffer the same fate (either phase shifted or not), path length differences of 1/2 a wavelength lead to destructive interference. For cases in which one reflected ray suffers a phase shift of half a wavelength, path length differences of 1 full wavelength lead to destructive interference.