2. Physics 1202: Lecture 28 Today’s Agenda Announcements: –Midterm 2: solutions HW 8 this FridayHW 8 this Friday Diffraction –Review Polarization –Reflection by surface

3. Diffraction

4. Experimental Observations: (pattern produced by a single slit ?)

5. So we can calculate where the minima will be ! sin  = ± m /a m=±1, ±2, … Why is the central maximum so much stronger than the others ? So, when the slit becomes smaller the central maximum becomes ?

6. Diffraction patterns of two point sources for various angular separation of the sources Resolution (circular aperture)  min = 1.22 (  / a) Rayleigh’s criterion for circular aperture:

7. Two-Slit Interference Pattern with a Finite Slit Size I diff = I max [ sin (  /2) / (  /2) ] 2 Diffraction ( “ envelope ” function):  = 2  a sin (  ) / I tot = I inter. I diff Interference (interference fringes): I inter = I max [cos (  d sin  /  ] 2 smaller separation between slits => ? smaller slit size => ? The combined effects of two-slit and single-slit interference. This is the pattern produced when 650-nm light waves pass through two 3.0- mm slits that are 18 mm apart. Animation

8. Application X-ray Diffraction by crystals Can we determine the atomic structure of the crystals, like proteins, by analyzing X-ray diffraction patters like one shown ? A Laue pattern of the enzyme Rubisco, produced with a wide-band x-ray spectrum. This enzyme is present in plants and takes part in the process of photosynthesis. Yes in principle: this is like the problem of determining the slit separation (d) and slit size (a) from the observed pattern, but much much more complicated !

9. Determining the atomic structure of crystals With X-ray Diffraction (basic principle) 2 d sin  = m  m = 1, 2,.. Crystalline structure of sodium chloride (NaCl). length of the cube edge is a = 0.562 nm. Crystals are made of regular arrays of atoms that effectively scatter X-ray Bragg ’ s Law Scattering (or interference) of two X-rays from the crystal planes made-up of atoms

10. Polarization of light Recall E&M wave This is an example of linearly polarized light –Electric field along a fixed axis ( here y ) y x z Most light source are nonpolarized –Electric field along random axis 22_all_imgs_in_ppt E

11. Polarizers Made of long molecules (organic polymers) –Block electric field along their length –Electric field perpendicular passes through So E after =E cos  Recall that I ~ E 2 E E I = I 0 cos 2  Malus’s law

12. Polarization of Electromagnetic Waves (light) as view along direction of propagation linearly polarized  unpolarized 

13. I = I max cos 2  Polarization by Absorption E = E max cos 

14. Two polarizers have polarization angles set at  1 =15 o and  3 = 90 o with respect to vertical axis, as show on the Figure below. (a)By what factor is the intensity of unpolarized light attenuated going through both polarizers (I f /I i = ? ). EXAMPLE: I f /I i = cos 2 (90-15) = 0.07 I f /I i = cos 2 (45-15) = 0.75 I f /I i = cos 2 (90-45) = 0.55 I f /I i = 0.5x0.75 x 0.55 = 0.21 ! I = I max cos 2  unpolarized polarized (b)Does the attenuation factor increase or decrease if a third polarizer is inserted in-between the two polarizers, with polarization angle  2 =45 o ? Unpolarized light through linear polarizer: I f /I i = = 0.50 I f /I i = 0.5 I f /I i = 0.50 x 0.07 = 0.04 !

15. Polarization by Reflection Brewster ’ s angle: n 1 sin  p = n 2 sin  2 note :  2 = 90 -  p using : n 1 = 1 and n 2 = n n = sin  p / cos  p Or: n = tan  p for n =1.55  p = 57 o When unpolarized light is incident on a reflecting surface, the reflected and refracted beams are partially polarized. The reflected beam is completely polarized when the angle of incidence equals the polarizing angle (Brewster’s angle)

16. EXAMPLE a.22.2° b. 7.7° c. 16.6° d. 36.9° e. 53.1° How far above the horizon is the moon when its image, reflected in a calm lake, is completely polarized ? (n water = 1.33) n = tan  p  p = tan -1 (1.33) = 53.1 o

17. Polarization by Double Refraction Unpolarized light incident on a calcite crystal splits into an ordinary (O) ray and an extraordinary (E) ray which are polarized in mutually perpendicular directions. A calcite crystal produces a double image because it is a birefringent (n o =1.658, n E =1.486) material. There are materials where the speed of light is not the same in all directions. Such (birefingent) materials thus have two indexes of refraction And light beam splits into two beam, ordinary (O) and extraordinary (E) ray E and O rays are also polarized in mutually perpendicular directions.

18. Application A plastic model of an arch structure under load conditions observed between perpendicular polarizers. Such patterns are useful in the optimal design of architectural components. Bright area ==> max stress Dark area => minimum stress Some materials become birefringent when stressed (glass, plastic,..) Since the birefringence effectively rotates the polarization of the light when such an object is placed between the “ polarizer ” and “ analyzer ” oriented at 90 o, only the light passing thorough stressed portion of the material will be observed. This provides a way to “ image ” stress in model plastic structures.