Quantum Physics

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Quantum Physics

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Many physical phenomena of great practical interest to engineers chemists, biologists, physicists, etc. were not in Gen. Phys. I & II Quantum Physics: The development of experimental equiment and techniques  modern physics can go inside the microscopic world (atoms, electrons, nucleus, etc.)  New principles, new laws for the microscopic (subatomic) world were discoverved

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  1. GENERAL PHYSICS III Optics & Quantum Physics
  2. What does we learn in Gen. Phys. III? Many physical phenomena of great practical interest to engineers, chemists, biologists, physicists, etc. were not in Gen. Phys. I & II  Wave phenomena of light:  Interference: what happens when two or more waves overlap? Interference! (Light passing through two slits give such kind of picture)
  3. • Diffraction: The image of an object is not exact in fine details. For example, the image of a circular disk is diffused Interference & diffraction can be analyzed if we regard light as a wave  Quantum Physics: The development of experimental equiment and techniques  modern physics can go inside the microscopic world (atoms, electrons, nucleus, etc.)  New principles, new laws for the microscopic (subatomic) world were discoverved
  4.  The basis concept: wave  particle duality Examples: light wave  photon electron  electron wave etc… • Subatomic objects obey new mechanics: quantum mechanics On the basis of quatum mechanics we study structure and properties of atoms, nucleus, solids, laser rays, etc…
  5. Chapter XVII Interference of light §1. Interference of coherent sources of light §2. Interference in thin films §3. Interferometer
  6. §1. Interference of light from coherent sources: 1.1 Coherent sources of light: We consider an overlap of light that comes from two sources. A remarkable phenomen takes place, if two sources satisfy some following conditions: The sources are monochromatic. It means that they emit light of a single color. A monochromatic light corresponds to a sinusoidal electromagnetic wave with a single frequency f and wave length  Two sources have the same frequency f (the same wave length  ) Two souces are permanently in phase, or , at least, have any definite constant phase difference Then, two sources are called coherent sources.
  7.  Notes: Recall the formula for a sinusoidal e-m wave:  2  y , t  A cos     A cos   ft  A cos   t  x  x vt  kx 2  kx    v speed A amplitude 2 wavelength k  wavenumber or wavevector  f frequency 2f  angular frequency Common sources of light do not emit monochromatic light (single-frequency light) → However one can produce approximately monochromatic light: • by using filters which block all but a very narrow range of wave length • by using light from a laser
  8. 1.2 Interference of light through narrow slits: Young’s experiment on double-slit interference double- (Thomas Young performed in 1800) Monochromatic light source at a great distance, Observation or a laser. Slit pattern screen
  9. Light (wavelength  is incident on a two-slit (two narrow,  rectangular openings) apparatus: I1  If either one of the slits is closed, a diffuse image of the other slit will appear S1 on the screen. (The image will be “diffuse” due to diffraction. We will discuss this effect in more detail later.) S2 Diffraction profile Monochromatic light screen  If both slits are now open, we see (wavelength ) interference “fringes” (light and dark bands), corresponding to constructive S1 and destructive interference of the electric-field amplitudes from both slits. S2 Light fringes Dark fringes I
  10. Observer  r1 S1 Light r2 d S2  Important quantity: path difference  r2 - r1 = The light density at the location of observer depends on the path difference  A path difference corresponds to a phase difference  of two waves at the observer’s point       
  11.  One has a simple formula for the path difference, , when the observer is far from sources. sources. (Assume 2 sources radiating in phase) Observer  d r Normal to d When observer distance >> slit spacing (r >> d) : = dsin   d The corresponding phase difference  at the observer’s point:       sin      d      
  12. At observer’s points that satisfy the condition     = 0, ±1, ±2,... → two waves  m   m  are in phase and reinforce each other, we say  /d  that there is constructive interference at these points y r /d Constructive = dsin= m Interference d  0 If   m + 1)→ two waves   I cancell each other → there is -/d destructive interference Destructive L = dsin= (m + 1/ ) 2 Interference m=2 Usually we care about the linear “lines” of (as opposed to angular) constructive m=1 displacement y of the pattern interference: (because our screens are often  m=0 = sin-1(md  flat): m=-1 y = L tan m=-2
  13. The slit-spacing d is often large compared to  so that  small. , is Then we can use the small angle approximations to simplify our results: For small angles: (
  14.  Example: S1 A laser of wavelength 633 nm is y incident on two slits separated by 0.125 mm. S2 I 1. What is the spacing  between fringe maxima on a screen 2m away? y a. 1 m b. 1 mm c. 1 cm 2. If we increase the spacing between the slits, what will happen to  y? a. decrease b. stay the same c. increase 3. If we instead use a green laser (smaller   will? ), y a. decrease b. stay the same c. increase
  15. Example (continue) S1 A laser of wavelength 633 nm is y incident on two slits separated by 0.125 mm. S2 1. What is the spacing  between fringe maxima on a screen 2m away? y a. 1  m b. 1 mm c. 1 cm First question: can we use the small angle approximation? d = 125 m; = 0.633   m d >>  is small d sin = m i  ~ d   i i  mi ( i d)  y L( –  L(2 – 1) ( = L = (2 m)(0.663  2 1) d)  /d m)/125  = 0.01 m m 2. If we increase the spacing between the slits, what will happen to  y? a. decrease b. stay the same c. increase 3. If we instead use a green laser (smaller   will? ), y a. decrease b. stay the same c. increase
  16. Example (continue) S1 y A laser of wavelength 633 nm is incident on two slits separated by 0.125 mm. S2 1. What is the spacing  between fringe maxima on a screen 2m away? y a. 1  m b. 1 mm c. 1 cm 2. If we increase the spacing between the slits, what will happen to  y? a. decrease b. stay the same c. increase Since  ~ 1/d, the spacing decreases. Note: This is a general phenomenon y – the “far-field” interference pattern varies inversely with slit dimensions. 3. If we instead use a green laser (smaller   will? ), y a. decrease b. stay the same c. increase Since  ~  the spacing decreases. y ,
  17.  Note: If a monochromatic source is replaced by a white light one → how is the interference picture? We have known that the location of (light or dark) fringers depends on the wavelength  therfore • At y = 0 (the center light fringer) the maxima for all wavelengths coincide → there is a white light fringer • The nearby fringers have spectrum colors (like in rainbow) • Far fringers are not visible (are diffused)
  18. §2. Interference in thin films: This kind of interference takes place when light reflects from a thin film (for example, a soap bubble, a thin layer of oil floating on water) In this case there is an overlap eye of light waves: one is reflected from the upper, the other from the lower surface. 2.1 Calculation of path difference of two light waves: First we consider the case that the incident light is monochromatic with the wavelength  
  19. b: the thickness of the film n: the index of refraction  path difference between  the the reflected waves 1 & 2 By elementary geometry it is not difficult to obtain Using the formula one can eliminate i1 or i2 We must take into account one more effect: the phase shift of a wave after reflection
  20. We use the follwing theoretical results from Maxwell’s theory of electromagetism: • If na > nb → the phase shift of reflected wave relative to the incident na wave is zero nb • If na < nb → the phase shift of reflected wave relative to the incident wave is  radian ( a half cycle) We can take into account this phase shift by introducing a complementary term in the formula of path difference  Remark: The path difference of two reflected waves depends on b, i1 (or i2) At the points that satisfy   m (m = 0, ±1, ±2,...) → we have  cnstructive interference  the points that satisfy = (m+1/2) destructive interference At  →
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