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Chapter 31 Diffraction & Polarization

Chapter 31 Diffraction & Polarization. Phenomena of diffraction. Diffraction pattern of (a) a circular hole. (b) a razor blade. (c) a circular disk. (a). (b). (c). Fresnel’s wave theory & Poisson spot. dE(P). ·. . dS. P. S. wave front. Huygens-Fresnel principle.

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Chapter 31 Diffraction & Polarization

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  1. Chapter 31 Diffraction & Polarization

  2. Phenomena of diffraction Diffraction pattern of (a) a circular hole (b) a razor blade (c) a circular disk (a) (b) (c) Fresnel’s wave theory & Poisson spot

  3. dE(P) ·  dS P S wave front Huygens-Fresnel principle Every point on wave front → source of wavelets Superposition of all the wavelets→ diffraction Light vibration at point P: Intensity at point P: Interference & diffraction

  4. P S * A  a O C B f Fraunhofer diffraction by a single slit Plane monochromatic light falls on a narrow slit a) Diffraction angle =0 All light in phase→ central maximum b) Other diffraction angle The OPD:

  5. A a C  2  2  2 B Half-wave zone method ① Divide the slit into several zones with equal width of /2. ② Two rays with path difference /2 make a destructive interference. ③ Two adjacent zones completely cancel out. (Even zones → dark) ④ Odd number of zones can make a bright fringe.

  6. Diffraction pattern Intensity as a function of angle: Minima (dark) occur at Maxima (bright) occur at a) =0 → central maximum Notice: b) Comparing with interference

  7. Number of minima Example1: A monochromic light () falls on a slit, the 5th minimum occurs at =30°, Determine: a) the width of slit; b) the total number of minima. Solution: a)Dark fringeoccur at For the 5thdark fringe,m=5, =30° b) Let =90°, mmax<10 , so there are 18 minima in total

  8. I Width of slit Example2: 633-nm light falls on a slit, the screen is 6.0m away. The distance between two first maxima is 32mm. How wide is the slit? Solution: Maxima occur at For small angle:

  9. S A  dx r x O r0 P B *Intensity (1) by Huygens-Fresnel principle Light vibration of dx : Travels to Point P: r is the opticalpath difference from dx to P Note where r0 is the OPD from O to P

  10. *Intensity (2) Superposition: integral on the whole slit: where Amplitude Intensity at Point P:

  11. *Intensity (3) Intensity at Point P: a) =0 → I=I0, central maximum b) asin=m→ sinβ=0 →I=0, minima c) For other maxima Approximately:

  12. Diffraction of circular apertures Central bright spot : Airy disk Angular radius of Airy disk D is the diameter of aperture

  13. 1.0 0.8 . . . . Rayleigh Criterion Two images are just resolvable when the center of Airy disk of one image is directly over the first minimum in diffraction pattern of the other. Separated by angle  :

  14. Resolving power Limited by the wave nature of light Telescope:

  15. Microscope Limited by the wavelength  Electron microscope: ~ 1Å Scanning tunneling microscope (STM)

  16. Great wall Homework: Can an astronaut resolve Great Wall of China (3000km long, 5 m thick, and 8 m height) ? Pupil diameter ~ 5mm,  ~ 550nm, H ~ 400km

  17. *Diffraction in double-slit (a) interference factor of intensity (b) diffraction factor of intensity (c) the total intensity

  18. reflection transmission d Diffraction grating A large number of equally spaced parallel slits Diffraction grating: transmission / reflection d Grating constant: d —— distance between slits 10 ~ 104 slits/mm small d, large 

  19. Diffraction by a single slit Multi-slit interference Intensity in grating diffraction Multi-slit interference pattern modulated by diffraction Resultant intensity: N: total number of slits

  20. Interference pattern When =mπ, I = Imax —— principal maxima —— minima ( N-1 ) Secondary intensity maxima between two adjacent minima ( N-2 )

  21. I Diffraction -2 -1 1 0 2 I Interference 0 -8 4 8 -4 I Resultant intensity 4 -8 -4 8 0 Modulation by diffraction Thinking (a): N=? Thinking (b): d/a=? Principal maxima sharp lines

  22. I Resultant intensity sharp lines: dsin=m m=4, 8, 12 are missing 4 -8 -4 8 0 Missing maxima Position of principal maxima —— interference Intensity of principal maxima —— diffraction Example: d=4a

  23. Spectrum Continuous spectrum m=0 m=-1 m=3 m=-2 m=2 m=1 m=-3 Spectrum of grating White light with a continuous distribution of 

  24. Width of spectrum Example3: White light (400~700nm) falls normally on a grating (600 slits/mm). Find the angular width of the first-order spectrum. Solution: a)First-order spectrum: where: Purple (400nm): Red (700nm): ∴angular width

  25. Lines can be seen Example4: 6000Å light falls on a grating. 2-nd lines are at angle 30°, and 3-rd lines are missing. What is d and minimum a? How many lines can be seen? Solution: 3-rd lines are missing: So lines can be seen: m=0, 1, 2.

  26. X-Rays Wilhelm C. Röntgen 1901 Nobel Prize New type of radiation: invisible & penetrating New physics: macroscopic → microscopic ! Short wavelength: ~1Å →High energy

  27. X-Ray diffraction Short wavelength:  ~1Å Diffraction grating? Crystal structure! M. von Laue, 1912 W. H. Bragg & W. L. Bragg, 1913 d 1915 Nobel prize for physics

  28. Polarization Maxwell’s theory of light →electromagnetic wave Transverse wave: Polarization: direction of electric field vector Polarization status: unpolarized/natural light, linearly polarized light, partial polarized light.

  29.    parallel oscillation        vertical oscillation Natural (unpolarized) light Each photon has an oscillation direction A light source emits a large number of photons → random mixture in all possible directions Natural light or unpolarized light

  30. Polarized light If all oscillations are in one direction: Linearly polarized light (plane polarized) Partial polarized light: Linearly polarized light “+” natural light

  31.    I0 unpolarized plane-polarized Polaroids Polarized light can be obtained by: Polaroid / reflection / birefringence, … Axis of Polaroid: direction of polarizing Polarizer: only parallel component can pass

  32. E0 E0 cos  I0=kE02 Malus’ law Plane polarized light passes a Polaroid Intensity of transmission light: where α is the angle between polarized direction I=k(E0 cos)2 =I0 cos2

  33. Pass two Polaroids Example5: Natural light with intensity I0 passes two Polaroids. Determine the transmission intensity, if the angle between axes of two Polaroids is 30°. Solution: the transmission intensity:

  34. Polaroid as an analyzer Example6: Light falls on a Polaroid. When the axis of Polaroid rotates, the transmission intensity changes between I and 7I. What is the polarization status? Solution: Partial polarized light → mixture Unpolarized I1 + plane-polarized I2 Minimum: Maximum:

  35. Polarization by reflection For most angles of incidence Reflection light: (⊥> //) Transmission light: (//>⊥) (a) All light into camera (b) Polarizer as filter

  36. Brewster’s law Polarizing angle → reflection is plane-polarized reflected ray ⊥ transmitted ray: From law of refraction: Brewster’s law

  37. Polarizing angle Example7: (a)At what incident angle is sunlight reflected from a lake plane-polarized? (b) What is the refraction angle? (nwater=1.33) Solution: (a) (b) reflected ray ⊥ transmitted ray Critical angle for total internal reflection?

  38. *Birefringence Material with different index of refraction for different polarization directions. Two polarized light in ⊥ direction Different n→ Δφ→ wave plate (λ/2, λ/4)

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