1. Today 1/23  Today: Young’s Double Slit experiment, Text 27.2 Beats, 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. Young’s Double Slit (like two speakers) Single frequency source Wave crests Wave troughs Dark and bright “fringes” on a screen In phase at the slits c c c c c d d d d

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

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

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

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

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

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

9. Example: m = 0 m = 1 m = 2 m = 1 m = 2 m = 1 m = 0 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. tan  = 5mm / 2m  = 0.14° d sin  = m = 1(500 nm) d = 0.2 mm 5mm 2m

10. 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? v wave  =  c  =  3  10 8 m/s PLD = 8 m Does this equal some m or some (m + 1 / 2 ) ? v = f = 3.2m Make two lists 0 3.2m 6.4m 9.6m 12.8m 1.6m 4.8m 8.0m 11.2m 14.4m m (m + 1 / 2 ) m 0 1 2 3 4 The condition is met for destructive interference. Reception at that location is bad.

11. Your Homework Screen very far away (L) Two slits very close together (d)  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  f b “beat frequency” depends on source frequency difference

13. 0.5 s 10 Hz 12 Hz 2 Hz

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

31. Beats f b =  f 1 - f 2  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.