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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 to fill out the application form. application form It looks good on your resume, pays well, and is fun!

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Announcements Important notes on grades spreadsheets: 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). 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. Zeroes for boardwork can still lower your total points. You have already received your 8 “free” end material points. Do not take the test if that is all you need! Important: sample.sample Course grade cutoffs will not be lowered under any circumstances.

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Announcements PLC …in that case will check PLC periodically Wednesday and provide help to any students who show up. PLC will run Monday afternoon and evening as usual. There will be no Wednesday PLC unless I get an * from at least one student asking for PLC help… “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 *before Wednesday!

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Announcements Please fill out electronic course evaluations… Use the following links to complete the evaluation(s): The links should be available through Sunday, Dec 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

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

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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. Physics 2135 Final Room Assignments, Fall 2014: InstructorSectionsRoom Dr. HaleK, PSt. Pat’s Dr. KurterG, JG-31 EE Mr. LayE, H, L, QSt. Pat’s Dr. MadisonB, D120 BCH Dr. ParrisM, NG-5 H/SS Dr. PeacherF, R125 BCH Mr. UpshawA, C103/104 Centennial Hall See notes on next slide for information about room locations. Special AccommodationsTesting Center Exam is from 12:30-2:30 pm Thursday December 18! No one admitted after 12:45 pm!

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Physics 2135 Final Room Assignments, 12:30 PM, Dec. 18 InstructorSectionsRoom Dr. HaleK, PSt. Pat’s St. Pat’s Ballroom in Havener Center Dr. KurterG, JG-31 EE G-31 EE is lecture hall in EECH (Electrical Engineering) Mr. LayE, H, L, QSt. Pat’s St. Pat’s Ballroom in Havener Center Dr. MadisonB, D120 BCH BCH is Butler-Carlton Hall (Civil Engineering) Dr. ParrisM, NG-5 H/SS same as exam rooms during semester Dr. PeacherF, R125 BCH same as exam rooms during semester Mr. UpshawA, C103/104 Centennial Hall two regular classrooms with a sliding partition in between them Special AccommodationsTesting Center

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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 Exam Final (nz386.jpg)

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

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Diffraction Light is an electromagnetic wave, and like all waves, “bends” around obstacles. d <

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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

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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. a Each part of the slit acts as a source of light rays, and these different light rays interfere.

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a/2 a Divide the slit in half. Ray travels farther* than ray by (a/2)sin . Likewise for rays and . 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.

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a/2 a Destructive interference: 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

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a/2 a In general, destructive interference occurs when The above equation gives the positions of the dark fringes. The bright fringes are approximately halfway in between. AppletApplet.

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

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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. Single Slit Diffraction Intensity

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Starting equations for single-slit intensity: “Toy”

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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? y 1 = (32 mm)/2tan = y 1 /L tan sin for small 32 mm 6 m

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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. Resolution of Single Slit (and Circular Aperture)

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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.

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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! These come from the small angle approximation, and geometry.

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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

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y S2S2 S1S1 d a P L r1r1 r2r2 Double Slit DiffractionSingle Slit Diffraction Double Slit Diffraction with a a O x y

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A diffraction grating consists of a large number of equally spaced parallel slits. d = d sin The path difference between rays from any two adjacent slits is = dsin . Interference maxima occur for Diffraction Gratings 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.

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

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

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

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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. angle? 400 nm 700 nm* *Or 750 nm, or 800 nm, depending on who is observing.

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Interference Maxima: First-order violet: 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.

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First-order red: 10.9

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Application: use of diffraction to probe materials. La 0.7 Sr 0.3 Mn 0.7 Ni 0.3 O 3 La, Sr Mn, Cr

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Application: use of diffraction to probe materials. La 0.7 Sr 0.3 Mn 0.7 Ni 0.3 O 3 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.

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Application: use of diffraction to probe materials. La 0.7 Sr 0.3 Mn 0.7 Cr 0.3 O 3

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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 λ 1 and λ 2 that are nearly equal. The average wavelength is and the difference is The resolving power is defined as Diffraction Grating Resolving Power definition of resolving power mercury

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For a grating with N lines illuminated it can be shown that the resolving power in the m th order diffraction is Dispersion mercury 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. resolving power needed to resolve m th order

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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? mercury

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

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