2. Physics 1402: Lecture 35 Today’s Agenda Announcements: –Midterm 2: graded soon … »solutions –Homework 09: Wednesday December 9 Optics –Diffraction »Introduction to diffraction »Diffraction from narrow slits »Intensity of single-slit and two-slits diffraction patterns »The diffraction grating

3. Diffraction

4. Fraunhofer Diffraction (or far-field) Incoming wave Lens  Screen

5. Fresnel Diffraction (or near-field) Incoming wave Lens Screen P (more complicated: not covered in this course)

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

7. First Destructive Interference: (a/2) sin  = ± /2 sin  = ± /a m th Destructive Interference: (a/4) sin  = ± /2 sin  = ± 2  /a Second Destructive Interference: sin  = ± m /a m=±1, ±2, … How do we understand this pattern ? See Huygen’s Principle

8. 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 ?

9. Phasor Description of Diffraction Can we calculate the intensity anywhere on diffraction pattern ?  =  =  N   / 2  =  y sin (  ) /  = N  = N 2   y sin (  ) / = 2  a sin (  ) / Let’s define phase difference (  ) between first and last ray (phasor) 1st min. 2nd max. central max. (a/  sin  = 1: 1st min.

10. Yes, using Phasors ! Let take some arbitrary point on the diffraction pattern This point can be defined by angle  or by phase difference between first and last ray (phasor)  The arc length E o is given by : E o = R  sin (  /2) = E R / 2R The resultant electric field magnitude E R is given (from the figure) by : E R = 2R sin (  /2) = 2 (E o /  ) sin (  /2) = E o [ sin (  /2) / (  /2) ] I = I max [ sin (  /2) / (  /2) ] 2 So, the intensity anywhere on the pattern :  = 2  a sin (  ) /

11. Other Examples What type of an object would create a diffraction pattern shown on the left, when positioned midway between screen and light source ? A penny, … Note the bright spot at the center. Light from a small source passes by the edge of an opaque object and continues on to a screen. A diffraction pattern consisting of bright and dark fringes appears on the screen in the region above the edge of the object.

12. Resolution (single-slit aperture) Rayleigh’s criterion: two images are just resolved WHEN: When central maximum of one image falls on the first minimum of another image sin  =  / a  min ~  / a

13. 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:

14. EXAMPLE A ruby laser beam ( = 694.3 nm) is sent outwards from a 2.7- m diameter telescope to the moon, 384 000 km away. What is the radius of the big red spot on the moon? a. 500 m b. 250 m c. 120 m d. 1.0 km e. 2.7 km  min = 1.22 (  / a) R / 3.84 10 8 = 1.22 [ 6.943 10 -7 / 2.7 ] R = 120 m ! Earth Moon

15. 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

16. Example The centers of two slits of width a are a distance d apart. Is it possible that the first minimum of the interference pattern occurs at the location of the first minimum of the diffraction pattern for light of wavelength  ? d a a 1st minimum interference: d sin  = /2 1st minimum diffraction: a sin  = The same place (same  ) : /2d = /a a /d =  No!

17. 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 !

18. 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