Today 1/23

1. Today 1/23 • Today:Young’s Double Slit experiment, Text 27.2Beats, Text 17.4 • HW: 1/23 Handout “Young’s Double Slit” due Monday 1/24 • Friday: Phase Shifts on Reflection, Text 27.3 (See fig 27.11) • Peer Guidance Center 1-4 Mon-Thur Wit 211 (Not 209)

2. c d c d c d c d c Young’s Double Slit (like two speakers) Wave troughs Dark and bright “fringes” on a screen Wave crests Single frequency source  In phase at the slits

3. Always true for any interference problem Sources In Phase: Constructive if PLD = m Destructive if PLD = (m + 1/2) PLD = “path length difference” Sources Out of Phase: Constructive if PLD = (m + 1/2) Destructive if PLD = m m = 0, 1, 2, 3,… (I used “n” the other day)

4. PLD = d sin (d = slit separation)  PLD (close enough)  d  Two slit geometry (screen far away) d Screen

5. PDL = d sin (d = slit separation)  Two slit geometry d Screen d sin = m constructive interference d sin = (m+ 1/2) destructive interference When the sources (slits) and “in phase”

6.  A simpler picture Two slits very close together (d) Screen very far away (L) d sin = m constructive interference d sin = (m+ 1/2) destructive interference When the sources (slits) and “in phase”

7. m = 2 m = 1 m = 1 m = 0 m = 0 m = 0 m = 1 m = 1 m = 2 The m’s 0 “zeroth order” fringe 1 “first order” fringe 2 “second order” fringe d sin = m d sin = (m+ 1/2)

8. y Distance between fringes, y tan  = y/L m = 2 m = 1 m = 1 m = 0 L  m = 0 m = 0 m = 1 m = 1 m = 2

9. Example: m = 2 m = 1 m = 1 m = 0 5mm  m = 0 2m m = 0 m = 1 m = 1 Light with a wavelength of 500 nm passes through two closely spaced slits and forms an interference pattern on a screen 2m away. The distance between the central maximum and the first order bright fringe is 5 mm. What is the slit spacing? The light is in phase at the slits. m = 2 d sin  = m = 1(500 nm) tan  = 5mm / 2m d = 0.2 mm  = 0.14°

10. m m (m + 1/2) 0 1 2 3 4 Example: Twin radio antennas broadcast in phase at a frequency of 93.7MHz. Your antenna is located 150 m from one tower and 158 m from the other. How is the reception, good or bad? vwave=c=3108 m/s PLD = 8 m Does this equal some m or some (m + 1/2) ? v = f   = 3.2m Make two lists 0 1.6m 3.2m 4.8m The condition is met for destructive interference. Reception at that location is bad. 6.4m 8.0m 9.6m 11.2m 12.8m 14.4m

11.  Your Homework Two slits very close together (d) Screen very far away (L) d sin = m constructive interference d sin = (m+ 1/2) destructive interference When the sources (slits) and “in phase”

12. Beats • Occur when the frequencies of the sources are not the same • Frequencies must be close • Locations for constructive interference move over time • Causes sound to get loud and soft • fb “beat frequency” depends on source frequency difference

13. 10 Hz 12 Hz 0.5 s 2 Hz

14. Sources emitting different frequencies. In this case they are alternately in and out of phase as time goes by. Source 1 Source 2

15. Sources emitting different frequencies. In this case they are alternately in and out of phase as time goes by. Source 1 Source 2

16. Sources emitting different frequencies. In this case they are alternately in and out of phase as time goes by. Source 1 Source 2

17. Sources emitting different frequencies. In this case they are alternately in and out of phase as time goes by. Source 1 Source 2

18. Sources emitting different frequencies. In this case they are alternately in and out of phase as time goes by. Source 1 Source 2

19. Sources emitting different frequencies. In this case they are alternately in and out of phase as time goes by. Source 1 Source 2

20. Sources emitting different frequencies. In this case they are alternately in and out of phase as time goes by. Source 1 Source 2

21. Sources emitting different frequencies. In this case they are alternately in and out of phase as time goes by. Source 1 Source 2

22. Sources emitting different frequencies. In this case they are alternately in and out of phase as time goes by. c Source 1 Source 2

23. Sources emitting different frequencies. In this case they are alternately in and out of phase as time goes by. c Source 1 Source 2

24. Sources emitting different frequencies. In this case they are alternately in and out of phase as time goes by. c Source 1 Source 2

25. Sources emitting different frequencies. In this case they are alternately in and out of phase as time goes by. c Source 1 Source 2

26. Sources emitting different frequencies. In this case they are alternately in and out of phase as time goes by. c Source 1 Source 2

27. Sources emitting different frequencies. In this case they are alternately in and out of phase as time goes by. c Source 1 Source 2

28. Sources emitting different frequencies. In this case they are alternately in and out of phase as time goes by. c Source 1 Source 2

29. Sources emitting different frequencies. In this case they are alternately in and out of phase as time goes by. c Source 1 Source 2

30. Sources emitting different frequencies. In this case they are alternately in and out of phase as time goes by. c Source 1 Source 2 Now the locations of constructive (and destructive) interference move in time. A stationary listener hears “Beats.”

31. Beats fb = f1 - f2 The beat frequency tells you the difference between the two source frequencies.

32. You want to know the frequency of a tuning fork. You test it by playing it at the same time as a tuning fork with a known frequency of 342 Hz and you hear beats at a rate of 5 per second. You then play it at the same time as one with a known frequency of 349 Hz and the beats are heard at a rate of 12 per second. What is the frequency of the tuning fork? a. 347 Hz b. 361 Hz c. 345.5 Hz d. 337 Hz e. 354 Hz.