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Physics 1502: 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. Physics 1502: Lecture 35 Today’s Agenda.

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Physics 1502: Lecture 35 Today’s Agenda

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  1. 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 Physics 1502: Lecture 35Today’s Agenda

  2. Diffraction

  3. q Lens Incoming wave Screen Fraunhofer Diffraction(or far-field)

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

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

  6. How do we understand this pattern ? First Destructive Interference: (a/2) sin Q = ± l/2 sin Q = ± l/a Second Destructive Interference: (a/4) sin Q = ± l/2 sin Q = ± 2 l/a mth Destructive Interference: sin Q = ± m l/a m=±1, ±2, … See Huygen’s Principle

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

  8. central max. 1st min. 2nd max. Phasor Description of Diffraction Let’s define phase difference (b) between first and last ray (phasor) b = S (Db) = N Db (a/l sin Q = 1: 1st min. Db / 2p = Dy sin (Q) / l • = N Db = N 2pDy sin (Q) / l = 2pa sin (Q) / l Can we calculate the intensity anywhere on diffraction pattern ?

  9. So, the intensity anywhere on the pattern : I = Imax [ sin (b/2) / (b/2) ]2 • = 2p a sin (Q) / l Yes, using Phasors ! Let take some arbitrary point on the diffraction pattern This point can be defined by angle Qor by phase difference between first and last ray (phasor) b The resultant electric field magnitude ER is given (from the figure) by : sin (b/2) = ER / 2R The arc length Eo is given by : Eo = R b ER = 2R sin (b/2) = 2 (Eo/ b) sin (b/2) = Eo [ sin (b/2) / (b/2) ]

  10. Other Examples 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. • 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.

  11. sin Q = l / a Qmin ~ l / a 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

  12. Rayleigh’s criterion for circular aperture: Qmin = 1.22 ( l / a) Resolution (circular aperture) Diffraction patterns of two point sources for various angular separation of the sources

  13. Moon Earth EXAMPLE A ruby laser beam (l = 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 Qmin = 1.22 ( l / a) R / 3.84 108 = 1.22 [ 6.943 10-7 / 2.7 ] R = 120 m !

  14. Diffraction (“envelope” function): Idiff = Imax [ sin (b/2) / (b/2) ]2 • = 2pa sin (Q) / l Interference (interference fringes): Iinter = Imax [cos (pd sin Q / l)]2 Two-Slit Interference Pattern with a Finite Slit Size Itot = Iinter . Idiff smaller separation between slits => ? 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. smaller slit size => ? Animation

  15. a d a 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 l ? 1st minimum interference: d sin Q = l /2 1st minimum diffraction: a sin Q = l No! The same place (same Q) : l /2d = l /a a /d = 2

  16. 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 !

  17. 2 d sin Q = m l m = 1, 2, .. Determining the atomic structure of crystals With X-ray Diffraction (basic principle) Crystals are made of regular arrays of atoms that effectively scatter X-ray Scattering (or interference) of two X-rays from the crystal planes made-up of atoms Bragg’s Law Crystalline structure of sodium chloride (NaCl). length of the cube edge is a = 0.562 nm.

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