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Today 1/22

Today 1/22. Light Interference: read Text 27.1,2 HW: 1/22 Handout “Interference (more than one frequency)” due Friday 1/24 Today: Questions? Example Problem Young’s Double Slit experiment Peer Guidance Center Begins Wed afternoon Wit 209. A little review. Waves

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Today 1/22

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  1. Today 1/22 • Light Interference: read Text 27.1,2 • HW: 1/22 Handout “Interference (more than one frequency)” due Friday 1/24 • Today:Questions? Example ProblemYoung’s Double Slit experiment • Peer Guidance CenterBegins Wed afternoon Wit 209

  2. A little review • Waves • Two types, transverse and longitudinal • Wave speed depends only on the medium • Period, Frequency, Amplitude--just like SHM • v = f • Interference • Superposition-- adding waves • It’s all about Path Length Difference, , and.. • Sources “in” or “out” of “Phase”

  3. Example: What is the lowest frequency of sound that will produce destructive interference here? PL1 = 2.0m Two sources emit in phase PL2 = 2.2m

  4. 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

  5. Young’s Double Slit (like two speakers) Wave troughs Does the pattern expand or contract when:-the slits move closer together?-the wavelength increases? Wave crests Single frequency source  In phase at the slits

  6. 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)

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

  8. 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) are “in phase”

  9. 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”

  10. 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)

  11. 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

  12. 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°

  13. 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

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