2. Review For Waves Test

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4. v = f f = 1/T = 720. x10 -9 m, v = c = 3.00 x 10 8 m/s f = 4.17E+14 Hz T = 1/f = 2.4E-15 s 4.17E+14 Hz, 2.4E-15 s W What is the frequency of 720. nm (1 nm = 1x10 -9 m) light? What is its period(Speed = 3.00 x 10 8 m/s)

5. n = c / v n = 1.33, c = 3.00 x 10 8 m/s 2.25 x 10 8 m/s W What is the speed of light in water? n = 1.33

6. f air = f water n = c / v v = f find f, then v, then f = v/ = 4.16667E+14 Hz v water = c/n = 225563909.8 m/s = v/f = (225563909.8 m/s)/(4.16667E+14 Hz) = 541. nm or 720/1.33 = 541 nm 541 nm W What is the wavelength of 720. nm light in water? n = 1.33 (720. nm is its wavelength in a vacuum, the frequency remains the same) (4)

7. Draw the red rays, be able to label the angle of incidence (θ 1 in this picture) and the refracted angle (θ 2 )

8. n 1 sin  1 = n 2 sin  2 n 1 = 1.33,  c = ??,  2 = 90 o, n 2 = 1.00  c = sin -1 (1.00xsin(90 o )/1.33) 48.8 o W What is the Critical angle for an air-water interface? θcθc

9. n 1 sin  1 = n 2 sin  2 n 1 = 2.42,  c = ??,  2 = 90 o, n 2 = 1.33  c = sin -1 (1.33xsin(90 o )/2.42) 33.3 o in the diamond W What is the Critical angle for an water-diamond interface? Where does the critical angle occur? θcθc

10. More than one polarizer: I = I o cos 2  I o – incident intensity of polarized light I – transmitted intensity (W/m 2 )  – angle twixt polarizer and incident angle of polarization Demo two polarizers IoIo ½I o ( ½ I o )cos 2 

11. Two polarizers are at an angle of 37 o with each other. If there is a 235 W/m 2 beam of light incident on the first filter, what is the intensity between the filters, and after the second? I = I o cos 2  After the first polarizer, we have half the intensity: I = 235/2 = 117.5 W/m 2 and then that polarized light hits the second filter at an angle of 37 o : I = (117.5 W/m 2 ) cos 2 (37 o ) = 74.94 = 75 W/m 2

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13. .45 m = 5 / 4 = 4 / 5 (.45 m) =.36 m.36 m 36 cm W The waveform is 45 cm long. What is the ?

14. .62 = 2 / 4 = 4 / 2 (.62 m) = 1.24 m v = f, f = v/ = (343 m/s)/(1.24 m) = 277 Hz 1.24 m, 277 Hz W The waveform is 62 cm long. What is the ? If it is a sound wave (v = 343 m/s), what is its frequency (v = f )

15. A string is 32.0 cm long, and has a wave speed of 281.6 m/s. Draw the first three modes of resonance. Find for each mode 1. The wavelength, 2. The frequency. Hint v = f.32 m = 2/4, =.64 m, v = f, f = 281.6/.64 = 440 Hz.32 m = 4/4, =.32 m, v = f, f = 281.6/.32 = 880 Hz.32 m = 6/4, =.2133 m v = f, f = 281.6/.2133 m = 1320 Hz

16. A pipe with both ends open is 1.715 m long, sound travels at 343 m/s along the pipe. Draw the first three modes of resonance. Find for each mode 1. The wavelength, 2. The frequency. Hint v = f 1.715 m = 2/4, = 3.43 m, v = f, f = 343/3.43 = 100. Hz 1.715 m = 4/4, = 1.715 m, v = f, f = 343/1.715 = 200. Hz 1.715 m = 6/4, = 1.14333 m v = f, f = 343/ 1.14333 = 300. Hz

17. A pipe with one end closed, one end open is also 1.715 m long, sound travels at 343 m/s along the pipe. Draw the first three modes of resonance. Find for each mode 1. The wavelength, 2. The frequency. Hint v = f 1.715 m = 1/4, = 6.86 m, v = f, f = 343/6.86 = 50. Hz 1.715 m = 3/4, = 2.28666 m, v = f, f = 343/2.28666 = 150. Hz 1.715 m = 5/4, = 1.372 m v = f, f = 343/1.372 = 250. Hz

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19. Moving source higher frequency f’ = f{ v } {v + u s } f = 256 hz, u s = 40.0 m/s, v = 343 m/s, and - 290. Hz W A car with a 256 Hz horn approaches you at 40.0 m/s. What frequency do you hear? (3) (use v sound = 343 m/s)

20. Moving source lower frequency f’ = f{ 1 } {1 + v s /v } f’ = 213 Hz, f = 256 Hz, v = 343 m/s, and + 69 m/s away from you W What speed in what direction is the same car (f = 256 Hz) moving if you hear 213 Hz (use v sound = 343 m/s)

21. Moving observer higher frequency f’ = f{v ± u o } { v } f = 440.0 Hz, f’ = 463 Hz, v = 343 m/s, and + 17.9 m/s W A running person who is late for a concert hears the concertmaster who is playing an A 440. Hz. How fast and in what direction are they running if they hear a frequency of 463 Hz. (use v sound = 343 m/s)

22. Question D on this page is a tricky little one about wavelength and Doppler effect. What you need to know is this: 1.v = fλ 2.That the wavelength gets shorter by the same amount in front of a moving object, that it gets longer in back e.g. – suppose the wavelength of a car horn is 2.0 m when the car is sitting still, if it moves so that the wavelength is 1.8 m in front of the car, it will be 2.2 m long behind the car. The problem can be solved without knowing this through the use of some fairly difficult algebra.

23. What is the velocity of a 1.12 m wave with a frequency of 32 Hz? v = f = (32 Hz)(1.12 m) = 35.84 m/s = 36 m/s 36 m/s W

24. What is the frequency of a sound wave that has a wavelength of 45 cm, where the speed of sound is 335 m/s v = f f = v/ = (335 m/s)/(.45 m) = 744.444 = 740 Hz 740 Hz W

25. Be able to draw the circles, and know where the approaching wavelength is, and the receding wavelength is

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27. What is the velocity of a 1.12 m wave with a frequency of 32 Hz? v = f = (32 Hz)(1.12 m) = 35.84 m/s = 36 m/s 36 m/s W

28. TOC If the difference in distance from the sources is an integer number of wavelengths, you get constructive interference A B

29. TOC If the difference in distance from the sources is an integer number of wavelengths, you get constructive interference A B

30. TOC If the difference in distance from the sources is an integer number of wavelengths, you get constructive interference Difference is: 0, 1, 2, 3 … A B

31. If the difference in distance from the sources has a remainder of a half wavelength, you get destructive interference: A B

32. A B If the difference in distance from the sources has a remainder of a half wavelength, you get destructive interference:

33. A B If the difference in distance from the sources has a remainder of a half wavelength, you get destructive interference: Difference.5, 1.5, 2.5, 3.5 …

34. To figure out two source problems: 1.Calculate the 2.Find the difference in distance 3.Find out how many it is 4.Decide: __.0 = constructive __.5 = destructive __.1 = mostly constructive __.25 = ???

35. Two speakers 3.0 m apart are making sound with a wavelength of 48.0 cm. A. What is the frequency of this sound if v = 343 m/s? v = f, 343 m/s = f (.48 m) f = 714.5833333 715 Hz W

36. Two speakers 3.0 m apart are making sound with a wavelength of 48.0 cm. If I am 2.12 m from one speaker, and 3.80 m from the other, is it loud, or quiet, and how many wavelengths difference in distance is there? 3.80 m - 2.12 m = 1.68 m (1.68 m)/(.48 m) = 3.5 = destructive interference 3.5 wavelengths, destructive W

37. Two speakers 3.0 m apart are making sound with a wavelength of 48.0 cm. If I am 5.17 m from one speaker, and 8.05 m from the other, is it loud, or quiet, and how many wavelengths difference in distance is there? 8.05 m - 5.17 m = 2.88 m (2.88 m)/(.48 m) = 6.0 = constructive interference 6.0 wavelengths, constructive W

38.  ≈ b  = Angular Spread = Wavelength b = Size of opening b 656 nm light is incident on a single slit with a width of 0.12 mm. What is the approximate width of spread behind the slit? = 656E-9 m b = 0.12E-3 m  = (656E-9 m)/(0.12E-3 m) = 0.0055 radians or about 0.31 o

39. Try this problem: Sound waves with a frequency of 256 Hz come through a doorway that is 0.92 m wide. What is the approximate angle of diffraction into the room? Use 343 m/s as the speed of sound. Use v = f, so = 1.340 m Then use  ≈ b  ≈ 1.5 rad What if the frequency were lower? Sub Woofers

40. Rayleigh Criterion  = 1.22 b  = Angle of resolution (Rad) = Wavelength (m) b = Diameter of circular opening (m) (Telescope aperture) the bigger the aperture, the smaller the angle you can resolve. Central maximum of one is over minimum of the other

41.  = 1.22 b  = ?, = 550 x 10 -9 m, b = 2.54 m  = 2.64173E-07 2.6 x 10 -7 radians W What is the angular resolution of the 100 inch (2.54 m) diameter telescope on the top of Mt Wilson? (use 550 nm as the wavelength) (uh 550 nm = 550 x 10 -9 m)

42.  = 1.22 b  = 6.00 x 10 -7, = 550 x 10 -9 m, b = ? b = 1.12 m 1.1 m W What diameter telescope do you need to resolve two stars that are separated by 1.8 x 10 11 m, but are 3.0 × 10 17 m from us? (use 550 nm as the wavelength) (AU, 32 LY) hint  = s/r = (1.8 x 10 11 m)/(3.0 × 10 17 m)