1. Physics 361 Principles of Modern Physics Lecture 5

2. Photons and Electromagnetic Waves Light has a dual nature. It exhibits both wave and particle characteristics –Applies to all electromagnetic radiation –Different frequencies allow one or the other characteristic to be more easily observed The photoelectric effect and Compton scattering offer evidence for the particle nature of light –When light and matter interact, light behaves as if it were composed of particles Interference and diffraction offer evidence of the wave nature of light

3. Wave Properties of Particles In 1924, Louis de Broglie postulated that because photons have wave and particle characteristics, perhaps all forms of matter have both properties Furthermore, the frequency and wavelength of matter waves can be determined

4. de Broglie Wavelength and Frequency The de Broglie wavelength of a particle is The frequency of matter waves is

5. Dual Nature of Matter The de Broglie equations show the dual nature of matter Each contains matter concepts –Energy and momentum Each contains wave concepts –Wavelength and frequency

6. Suggested in 1924, Measured in 1926 and 1927 Both experiments performed electron diffraction off a crystal lattice (Davisson, Germer, and Thompson)

7. The Electron Microscope The electron microscope depends on the wave characteristics of electrons Microscopes can only resolve details that are slightly smaller than the wavelength of the radiation used to illuminate the object The electrons can be accelerated to high energies and have small wavelengths

8. Example: Find the de Broglie wavelength of a 1500-kg car whose speed is 30 m/s. Solution: The car’s wavelength is =h/mv=6.63x10 -34 Js/(1.5x10 3 )(30m/s) = 1.5x10 -38 m

9. Example: Find the de Broglie wavelength of a 1500-kg car whose speed is 30 m/s. Solution: The car’s wavelength is =h/mv=6.63x10 -34 Js/(1.5x10 3 )(30m/s) = 1.5x10 -38 m The wavelength is so small compared to the car’s dimension that no wave behavior is to be expected.

10. Example: Compare the de Broglie wavelength of 54-eV electrons with spacing of atomic planes in a crystal, which is 0.91x10 -10 m. Solution: KE of a 54-eV electron is KE=(54eV)(1.6x10 -19 J/eV)=8.6x10 -18 J KE=1/2 mv 2, mv=(2mKE) 1/2 =h/mv=h/(2mKE) 1/2 =1.7x10 -10 m Comparable to the spacing of the atomic planes, so diffraction occurs m=9x10 -31 kg h=6.63x10 -34 Js

11. What are the matter waves made up of? (Waves of what?) Compare to electromagnetic waves. In the case of photons, electricity and magnetism tell us the waves are spatial variations in the electric and magnetic fields. Detectors will measure an intensity proportional to the |amplitude| 2 of the E&M wave. Since we know energy is delivered by discrete photons, the detector must just be measuring the number of photons arriving per unit time. Since photons all travel the same speed, this is the same as saying the number per unit length.

12. Intensity of matter waves measure number per unit length. Matter waves vary in space and are some function of position. Their |amplitude| 2 is also proportional to number of particles per unit length measured by a detector, as with photons. This also applies to infinitesimal slices in space. But for a finite-magnitude particle wave, this would imply zero particles in the slice! The is telling us the probability to find a particle in that slice of space. These wave magnitudes are just proportional to probability to find a particle in that region. This is actually the same for photons as well.

13. To understand matter waves and the basics of quantum mechanics, we will explore simple interference effects (two slit interference)

14. Let’s first understand a single slit pattern A single slit placed between a distant light source and a screen produces a diffraction pattern –It will have a broad, intense central band –The central band will be flanked by a series of narrower, less intense secondary bands Called secondary maxima –The central band will also be flanked by a series of dark bands Called minima

15. Single Slit Diffraction According to Huygen’s principle, each portion of the slit acts as a source of waves The light from one portion of the slit can interfere with light from another portion The resultant intensity on the screen depends on the direction θ

16. Single Slit Diffraction, 2 All the waves that originate at the slit are in phase Wave 1 travels farther than wave 3 by an amount equal to the path difference (a/2) sin θ If this path difference is exactly half of a wavelength, the two waves cancel each other and destructive interference results

17. Single Slit Diffraction, 3 In general, by extending this trick, we can find that destructive interference occurs for a single slit of width a when sin θ dark = mλ / a –m =  1,  2,  3, … Doesn’t give any information about the variations in intensity along the screen

18. Single Slit Diffraction, 4 The general features of the intensity distribution are shown A broad central bright fringe is flanked by much weaker bright fringes alternating with dark fringes The points of constructive interference lie approximately halfway between the dark fringes

19. 19 The dotted blue line is due to single slit diffraction The red solid line is due to the interference between slits sin θ dark = mλ / a m =  1,  2,  3, … sin θ bright = mλ / d m =0  1,  2,  3, … Example of LASER light double-slit diffraction Single slit Double slit d>>a