ww.lab-initio (nz326.jpg)

1. http://ww.lab-initio.com (nz326.jpg)

2. LEAD Tutors/Peer Instructors Needed! You can tutor or be a PLC peer instructor if you have at least a 3.6 GPA and get an “A” in the course you want to tutor. Contact me or go to http://lead.mst.edu/ to fill out the application form. It looks good on your resume, pays well, and is fun!

3. Announcements Important notes on grades spreadsheets: Your scores for the end material test have been set to 8 (your “free” points) so that the spreadsheet can correctly discard low scores. I will need to exhibit incompetence somewhere in lecture to make this happen. You can enter projected scores into the spreadsheet to see how many points you need for the next higher grade (or to see if your grade is already certain). Zeroes for boardwork can still lower your total points. Important: sample. You have already received your 8 “free” end material points. Do not take the test if that is all you need! Course grade cutoffs will not be lowered under any circumstances.

4. Announcements PLC PLC will run Monday afternoon and evening as usual. There will be no Wednesday PLC unless I get an email* from at least one student asking for PLC help… …in that case will check PLC periodically Wednesday and provide help to any students who show up. *before Wednesday! “I don't know half of you half as well as I should like, and I like less than half of you half as well as you deserve.”—Bilbo Baggins

5. Announcements Please fill out electronic course evaluations…Use the following links to complete the evaluation(s):http://teacheval.mst.eduhttp://teachevalm.mst.eduThe links should be available through Sunday, Dec. 13. Note: “ready to submit” means the evaluations have not yet been submitted! For feedback or problems with the link, please contact Dr. Timothy Philpot at philpott@mst.edu

6. Announcements If anything about your grade needs fixed, fix it now (next week is too late)! See your recitation instructor! (Not me!) Don’t forget to put your used Physics 2135 textbook to good use!

11. Physics 2135 Final Room Assignments, Fall 2014: Instructor Sections Room Coming soon Special Accommodations Testing Center Exam is from 12:30-2:30 pm Thursday December 18! No one admitted after 12:45 pm! Know the exam time! Find your room ahead of time! If at 12:30 on test day you are lost, go to 104 Physics and check the exam room schedule, then go to the appropriate room and take the exam there.

12. Today’s agenda: no purple boxes. No purple boxes! I am taking a break from purple boxes. You’ve seen your last purple box agenda! Final Exam Final Final Exam http://www.nearingzero.net (nz386.jpg)

13. Two lectures ago I showed you these two plots of the intensity distribution in the double-slit experiment: Peak intensity varies with angle. Peak intensity independent of angle. Which is correct?

14. d <<d d >>d Diffraction Light is an electromagnetic wave, and like all waves, “bends” around obstacles.  This bending, which is most noticeable when the dimension of the obstacle is close to the wavelength of the light, is called “diffraction.” Only waves diffract.

15. Diffraction pattern from a penny positioned halfway between a light source and a screen. The shadow of the penny is the circular dark spot. Notice the circular bright and dark fringes. The central bright spot is not a defect in the picture. It is a result of light “bending” around the edges of the penny and interfering constructively in the exact center of the shadow. Good diffraction applets at http://ngsir.netfirms.com/englishhtm/Diffraction.htm http://micro.magnet.fsu.edu/primer/java/diffraction/basicdiffraction/ http://www.physics.uq.edu.au/people/mcintyre/applets/grating/grating.html

16.      a  Single Slit Diffraction In the previous chapter we calculated the interference pattern from a pair of slits. One of the assumptions in the calculation was that the slit width was very small compared with the wavelength of the light. Now we consider the effect of finite slit width. We start with a single slit. Each part of the slit acts as a source of light rays, and these different light rays interfere.

17. Divide the slit in half.    Ray  travels farther* than ray  by (a/2)sin. Likewise for rays  and . a/2    a a/2 If this path difference is exactly half a wavelength (corresponding to a phase difference of 180°) then the two waves will cancel each other and destructive interference results. Destructive interference: *All rays from the slit are converging at a point P very far to the right and out of the picture.

18. If you divide the slit into 4 equal parts, destructive interference occurs when If you divide the slit into 6 equal parts, destructive interference occurs when Destructive interference:    a/2    a a/2

19.    a/2    a a/2 In general, destructive interference occurs when The above equation gives the positions of the dark fringes. The bright fringes are approximately halfway in between. Applet.

20. http://www.walter-fendt.de/ph14e/singleslit.htm

21. Use this geometry for tomorrow’s single-slit homework problems. y a  If  is small,* then it is valid to use the approximation sin   . ( must be expressed in radians.) O x *The approximation is quite good for angles of 10 or less, and not bad for even larger angles.

22. Single Slit Diffraction Intensity Your text gives the intensity distribution for the single slit. The general features of that distribution are shown below. Most of the intensity is in the central maximum. It is twice the width of the other (secondary) maxima.

23. Starting equations for single-slit intensity: “Toy”

24. Example: 633 nm laser light is passed through a narrow slit and a diffraction pattern is observed on a screen 6.0 m away. The distance on the screen between the centers of the first minima outside the central bright fringe is 32 mm. What is the slit width? y1 = (32 mm)/2 tan = y1/L tan  sin   for small  32 mm 6 m

25. Resolution of Single Slit (and Circular Aperture) The ability of optical systems to distinguish closely spaced objects is limited because of the wave nature of light. If the sources are far enough apart so that their central maxima do not overlap, their images can be distinguished and they are said to be resolved.

26. When the central maximum of one image falls on the first minimum of the other image the images are said to be just resolved. This limiting condition of resolution is called Rayleigh’s criterion.

27. These come from the small angle approximation, and geometry. From Rayleigh’s criterion we can determine the minimum angular separation of the sources at the slit for which the images are resolved. For a slit of width a: For a circular aperture of diameter D: Resolution is wavelength limited!

28. If a single slit diffracts, what about a double slit? Remember the double-slit interference pattern from the chapter on interference? If the slit width (not the spacing between slits) is small (i.e., comparable to the wavelength of the light), you must account for diffraction. interference only

29. Double Slit Diffraction with a   Single Slit Diffraction Double Slit Diffraction y r1 y a a  S1 r2 O  P d S2 x L

30.  = d sin  Diffraction Gratings A diffraction grating consists of a large number of equally spaced parallel slits. The path difference between rays from any two adjacent slits is  = dsin .  If  is equal to some integer multiple of the wavelength then waves from all slits will arrive in phase at a point on a distant screen. d Interference maxima occur for

31. Ok what’s with this equation monkey business? double-slit interference constructive single-slit diffraction destructive! diffraction grating constructive d  double slit and diffraction a  a single slit but destructive

32.  d  = d sin  Diffraction Grating Intensity Distribution Interference Maxima: The intensity maxima are brighter and sharper than for the two slit case. See here and here.

33. visible light hydrogen helium mercury Application: spectroscopy You can view the atomic spectra for each of the elements here.

35. Example: the wavelengths of visible light are from approximately 400 nm (violet) to 700 nm (red). Find the angular width of the first-order visible spectrum produced by a plane grating with 600 slits per millimeter when white light falls normally on the grating. 400 nm 700 nm* angle? *Or 750 nm, or 800 nm, depending on who is observing.

36. Example: the wavelengths of visible light are from approximately 400 nm (violet) to 700 nm (red). Find the angular width of the first-order visible spectrum produced by a plane grating with 600 slits per millimeter when white light falls normally on the grating. Interference Maxima: First-order violet:

37. First-order red: 10.9

38. Application: use of diffraction to probe materials. La0.7Sr0.3Mn0.7Ni0.3O3 La,Sr Mn,Cr

39. Application: use of diffraction to probe materials. La0.7Sr0.3Mn0.7Ni0.3O3 La,Sr Mn,Cr Shoot a beam of x-rays or neutrons at an unknown material. The x-rays or neutrons diffract. Positions of peaks tell you what sets of planes exist in the material. From this you can infer the crystal structure. Intensities of peaks tell you atoms lie on the different planes, and where they are located on the planes.

40. Application: use of diffraction to probe materials. La0.7Sr0.3Mn0.7Cr0.3O3

42. mercury Diffraction Grating Resolving Power Diffraction gratings let us measure wavelengths by separating the diffraction maxima associated with different wavelengths. In order to distinguish two nearly equal wavelengths the diffraction must have sufficient resolving power, R. Consider two wavelengths λ1 and λ2 that are nearly equal. The average wavelength is and the difference is definition of resolving power The resolving power is defined as

43. mercury For a grating with N lines illuminated it can be shown that the resolving power in the mth order diffraction is resolving power needed to resolve mth order Dispersion Spectroscopic instruments need to resolve spectral lines of nearly the same wavelength. The greater the angular dispersion, the better a spectrometer is at resolving nearby lines.

44. mercury Example: Light from mercury vapor lamps contain several wavelengths in the visible region of the spectrum including two yellow lines at 577 and 579 nm. What must be the resolving power of a grating to distinguish these two lines?

45. mercury Example: how many lines of the grating must be illuminated if these two wavelengths are to be resolved in the first-order spectrum?