Physics 2 - Lecture 5: Diffraction - Huynh Quang Linh

DIFFRACTION
• Diffraction is the deviation of light from a straight-line path
when the light passes through an aperture or around an
obstacle.
• Diffraction is due to the wave nature of light 
This principle states that we can consider every
point of a wave front as a source of secondary
wavelets.
The position of the wave front at any later time is
the envelope of the secondary wavelets at that time.
To find the resultant displacement at any point, we
combine all the individual displacements produced
by these secondary waves, using the superposition
principle and taking into account their amplitudes
and relative phases 
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  1. DIFFRACTION Tran Thi Ngoc Dung – Huynh Quang Linh – Physics A2 HCMUT 2016
  2. DIFFRACTION • Diffraction is the deviation of light from a straight-line path when the light passes through an aperture or around an obstacle. • Diffraction is due to the wave nature of light.
  3. 2 Huygens’s Principle This principle states that we can consider every point of a wave front as a source of secondary wavelets. M . r The position of the wave front at any later time is O the envelope of the secondary wavelets at that time. To find the resultant displacement at any point, we combine all the individual displacements produced by these secondary waves, using the superposition principle and taking into account their amplitudes and relative phases M d O
  4. The optical path difference between the rays of 2 adjacent strips is /2. There is destructive interference. M =>The combined light from 2 adjacent strips is slit completely cancels at M. => If the number of strips is EVEN, M is DARK F If the number of trips is ODD, M is BRIGHT /2 sin o 1 /2 MQ S 2asin λ Dark Fringe : N 2m sin m m 1; 2; 3 λ a 2asin λ Bright Fringe : N 2m 1 sin (2m 1) (2m 1) 3; 5 λ 2a
  5. Central Bright Fringe : sin 0 λ Dark Fringe : sin m m 1, 2, 3 a λ Bright Fringe : sin (2m 1) 2m 1 3; 5 2a If a λ :thecentralfringeis verynarrow no diffraction.
  6. 38.3 Resolution of Single-Slit and Circular Apertures
  7. Limiting angle of resolution for a circular aperture   1.22 D: the diameter of the aperture min D Limiting angle of resolution for a slit of width a   min a
  8. MULTIPLE-SLIT DIFFRACTION + N slits + Width of a slit : a N slits + Distance between 2 adjacent slits: d  M There is diffraction by every slit and a d interference by slits . dsin F  Principal minima sin  m m 1, 2, 3 a o  Principal maxima sin  m m 0, 1, 2, 3 d screen There are N slits, between 2 adjacent principal maxima, there are N-1 secondary minima, N-2 secondary maxima
  9. + The width of maxima is equal. + If N is large, the width of maxima is small, we have sharp spectral lines. + Intensity of principal maxima is proportinal to N^2. Three slit Diffraction
  10. DIFFRACTION GRATING An array of a large number of parallel slits, all with the same width and spaced equal distances is called a diffraction grating. Glass Metal Studying Visible light 2 types Ultraviolet light (Glass absobs ultraviolet - transmission grating light) - reflection grating Gratings can be made by using a diamond point to scratch many equally spaced grooves on a glass or metal surface d= grating spacing n=1/d : number of slits per unit length transmission grating reflection grating -reflection grating
  11. Example 36.4 Width of a grating spectrum The wavelengths of the visible spectrum are approximately 380 nm (violet) to 750 nm (red). (a) Find the angular limits of the first order visible spectrum produced by a plane grating with 600 slits per millimeter when white light falls normally on the grating. (b) Do the first-order and second-order spectra overlap? What about the second-order and third-order spectra? Do your answers depend on the grating spacing?
  12. X RAY DIFFRACTION The arrangement of atoms in a crystal of sodium chloride (NaCl) is shown in the Figure, A careful examination of the NaCl structure shows that the ions lie in discrete planes. 2dsin=m m=1,2 Bragg’s Law suppose an incident x-ray beam makes an angle  with one of the planes as in the Figure. The beam can be reflected from both the upper plane and the lower one, but the beam reflected from the lower plane travels farther than the beam reflected from the upper plane. The optical path difference is 2dsin . The two beams reinforce each other (constructive interference) when this path difference equals some integer multiple of . The same is true for reflection from the entire family of parallel planes. Hence, the condition for constructive interference (maxima in the reflected beam) is 2dsin=m
  13. Diffraction by diffraction grating. Light source is He-Ne laser Diffraction on CD
  14. X-ray diffraction for a quasicrystal. diffraction by X-ray and by electron X-ray diffraction for a hexahedral crystal.
  15. X-ray Crystallography The Braggs made so many discoveries that Lawrence described the first few years as „like looking for gold and finding nuggets lying around everywhere‟: • showed that the sodium and chloride ions were not bonded into molecules, but arranged in a lattice • could distinguish different cubic lattices • discovered the crystal structure of diamond • Lawrence Bragg was the youngest Laureate ever (25) to receive a Nobel Prize (shared with his father in 1915) • now standardly used for all kinds of materials analysis, even biological samples! • The same multi-layer interference phenomenon is now used to make highly wavelength-specific mirrors for lasers (“distributed Bragg feedback” [DBF])