1. Phy2005 Applied Physics II Spring 2016 Announcements: Final exam: 50% Chs. 19-25 50% Chs 26-29 Don’t forget course evaluations Go over posted practice final for Wednesday No class 4/22

2. Principles of relativity 1.Neither an object nor any form of energy can be accelerated to a speed as fast as or faster than c. 2. An observer observes that the clock moving with a speed relative to the observer is ticking slower than his. 3. When an object passes an observer, the observer will measure the length of the object as shorter than if it was at rest relative to the observer. 4. The mass of an object increases when its speed increases. 5. Mass is a form of energy in the sense that mass and energy can be interchanged. Last time

3. Time dilation: “moving clocks run slow” Time interval from the point of view of the stationary observer Time interval from the point of view of the observer moving with the clock = Length contraction: “moving meter sticks are contracted” tt  t’  Length interval from the point of view of the stationary observer Length interval from the point of view of the observer moving with the clock = LL  L’  Lorentz factor Last time

4. Mass enhancement “moving objects get heavy” Mass from the point of view of the stationary observer Kinetic E Kinetic energy is the part of E that isn’t rest energy! m 0 = rest “mass”. An object at rest in its own frame has energy m 0 c 2 Total energy: Last time

5. Light shone on metals produces electrical current. Classical physics, should only depend on light intensity Experiment: depended not on intensity, but frequency Only if frequency was high enough do electrons get kicked out Einstein: light “particle” has energy hf, where h=Planck’s constant h=6.63 × 10 -34 J-s Photoelectric effect

6. How do we understand the intensity of light? Intensity of light (EM radiation) ≈ number of photons with same wavelength or frequency In this case, light can be considered as a massless particle, photon with energy solely determined by its wavelength (frequency). There is no conflict with Einstein’s relativistic mechanics: massless particle can have speed of light! E = hf = hc/ h = 6.6 x 10 -34 Js (Planck’s constant)

7. Q. Light of wavelength 650 nm is required to cause electrons to be ejected barely from the surface of a particular metal. What is the kinetic energy of the ejected electrons if the surface is bombarded with light of wavelength 450 nm? “…ejected barely…” means that ejected electron has ≈ 0 kinetic energy. Kinetic Energy of an electron = E photon - E threshold hc/ = E threshold (work function) for = 650 nm E threshold = (6.6 x 10 -34 Js)(3 x 10 8 m/s)/(659 x 10 -9 m) = 3.06 x 10 -18 J E k = (6.6 x 10 -34 Js)(3 x 10 8 m/s)/(450 x 10 -9 m) – 3.06 x 10 -18 = (4.42 – 3.06) x 10 -18 = 1.36 x 10 -19 J

8. A light source produces 400 nm wavelength. When the light strikes a metal surface, a stream of electrons emerges from the metal. If the intensity of the light is doubled, 1. the electrons emitted are more energetic. 2. more electrons are emitted in a given time interval 3. both of the above 4. none of the above

9. Atomic Spectra

10. Balmer Series (hydrogen spectrum) Predicts the visible emission wavelengths of Hydrogen. IR and UV wavelengths are predicted by very similar rules. strictly empirical formula invented by high school teacher -- where does it come from?

11. Bohr’s Hydrogen Atom and the Electron as a Wave Niels Bohr (1885 – 1962) Louis de Broglie (1892 – 1987) In 1913, Rutherford’s atom received a quantitative description from Niels Bohr who explained experimentally observed discrete nature of the atomic spectrum of Hydrogen. A better understanding of Bohr’s atom came after de Broglie’s conjecture (1923) that electrons should display wave properties. + - electron wave

12. The Bohr Atom from a classical perspective + - centripetal forceCoulomb force k = 9 x 10 9 Nm 2 /C 2 - NOTE: classically, this doesn’t work!!! In classical electrodynamics, an electric charge that accelerates must radiate energy, which means the electron will eventually spiral into the nucleus. Quantum mechanics keeps the atom stable.

13. It would seem that the basic idea of the quantum theory is the impossibility of imagining an isolated quantity of energy without associating with it a certain frequency de Broglie in 1923 as a graduate student light quantum (photon) E = hf = hc/ massless E = pc = hc/ = h/p wave quantity particle quantity Relativistic energy for m o = 0

14. Q. What is the de Broglie wavelength of an electron that has a kinetic energy of 100 eV? After an electron is accelerated in 100 V potential difference, its kinetic energy is 100 eV. eV unit has to be converted into SI unit, Joule. 1 eV = 1.6 x 10 -19 J E k = (1/2)m o v 2 = 1.6 x 10 -17 J v 2 = 2E k /m o = 2(1.6 x 10 -17 J)/(9.1 x 10 -31 kg) = 3.52 x 10 13 m 2 /s 2 v = 5.93 x 10 6 m/s low speed: no need to use relativistic = h/p = h/m o v = (6.6 x 10 -34 Js)/(9.1 x 10 -31 kg x 5.93 x 10 6 m/s) = 1.23 x 10 -10 m = 0.123 nm

15. Review of Chs. 26-28

16. ii rr  i =  r Law of reflection object image focus p is position of object q is position of image f is focal point image is real if putting a screen at the pt. would form an actual image otherwise, it’s virtual image is either erect or inverted (as in this fig.) p > q Spherical mirror

17. : Real vs. virtual image

18. Mirror Equation 1/p + 1/q = 1/f For a small object, f = R/2 (spherical mirror) 1/p + 1/q = 2/R Alert!! Be careful with the sign!! Negative means that it is inside the mirror!! p can never be negative (why?) negative q means the image is formed inside the mirror Magnification, M = -q/p

19. Refraction: Snell’s Law 11 11 22 v: speed of light in a medium n = c/v : index of refraction v 1 = c/n 1, v 2 = c/n 2 n 1 sin  1 = n 2 sin  2 All three beams (incident, reflected, and refracted) are in one plane. n > 1 Light always travels slower in a medium than in a vacuum, v=c/n

20. Total internal reflection n 1 (> n 2 ) n2n2 11 22 n 1 sin(  1 ) = n 2 sin(  2 ) Total internal reflection when  2 = 90  sin(  c ) = n 2 /n 1  c : critical angle < 1

21. virtual image virtual image Magnifying glass object

22. hihi hoho f D screen x D-x object distance p = D-x image distance q = x magnification M = h i /h o we found: New definitions: Rewrite: Or:

23. Lens equation and magnification 1/p + 1/q = 1/f M = -q/p Exactly the same as the mirror eq.!!! Now let’s think about the sign. positivenegative p real object virtual object (multiple lenses) q real image (opposite side of object) virtual image (same side of object) f for converging lens for diverging lens M erect image inverted image

24. Nearsighted A Farsighted

25. We know light is a wave because of interference phenomena: Interference patterns with 2 coherent sources Maxima or minima of intensity depend on difference in the path length from the 2 sources to the observation point

26. s1s1 s2s2 P r1r1 r2r2 |r 1 – r 2 | = 0,, 2, 3, …, m (m: integer)  Constructive Int. |r 1 – r 2 | = /2, 3 /2, 5 /2, …, (2m+1) /2  Destructive Int.

27. dsin  = m (0, ±, ± 2, …) Constructive (m+1) /2 (± /2, ±3 /2,…) Destructive d h  x 0 1 1 2 2 3 3 2 nd -order bright fringe 1 st order bright fringe xn Difference in two paths  s = 2x= mconstructive f Thin films

28. xn Q1: A thin film of polymer used in an antireflective coating has an index of refraction of 1.5. If light of frequency 5.5 x 10 14 Hz is incident on the film, but is found to be entirely transmitted and not reflected, what is a possible thickness x of the film (in m)? (1)9.1 x 10 -8 (2) 1.8 x 10 -7 (3) 1.3 x 10 -7 (4) 0 (5) 3.6 x 10 -7 =c/f=(3.0 x 10 8 m/s)/(5.5 x 10 14 Hz)=5.45 x 10 -7 m in film f  =(c/n)/f= 5.45 x 10 -7 m/1.5 = 3.63 x 10 -7 m for destructive interference, 2x=  f /2, … so x=  f /4 = 3.63 x 10 -7 m/4 = 9.1 x 10 -8 m path diff = 2x