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Daily_003.SOLN

# Daily_003.SOLN - the speakers in which case the...

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© 2011 E. Szarmes P HYS 274 D AILY A SSIGNMENT 3 — DUE M ONDAY , J AN 24/11 P ROF . S ZARMES Q2B.2 According to Eq. Q2.1, the angles of constructive interference ϑ n from two sources depend only on the wavelength λ and the slit separation d between the sources (and not the width of the sources). The formula is d sin ϑ n = n λ , or , where d = 3.5 m in this problem. We aren’t given the wavelength of the sound waves, but we are given their frequency f = 320 Hz, and thus we calculate λ = v / f = 343 m/s / 320 Hz = 1.07185 m. Therefore, we calculate the following angles of constructive interference (note that λ / d = 0.3062): For n = 0 ϑ 0 = 0° n = ±1 ϑ ±1 = arcsin [(±1) 0.3062] = ±17.8° n = ±2 ϑ ±2 = arcsin [(±2) 0.3062] = ±37.8° n = ±3 ϑ ±3 = arcsin [(±3) 0.3062] = ±66.7° n = ±4 ϑ ±4 = arcsin [(±4) 0.3062] = undefined. Note : I assumed that the phrasing of the problem, “two speakers ... 3.5 m apart”, referred to the center- to-center separation. Some of you assumed that it referred to the distance between the inner edges of
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Unformatted text preview: the speakers, in which case the center-to-center separation is d = [speaker separation]+[speaker width] = 3.5 m + 0.44 m = 3.94 m, and the corresponding angles are 0, 15.8, 33.0, and 54.7. Either case received full marks. Q2B.12 When light of wavelength is diffracted by a single slit of width 'a', the nulls in the diffraction pattern occur at angles n (measured from the central maximum) given by (These n clearly have a different meaning than the n in the previous problem!) If the angle between the first two nulls on either side of the central maximum is 1.2, then each null makes an angle of 0.6 with respect to the central maximum. Therefore, 1 = 0.6, and if = 633 nm = 0.633 m we calculate the slit width to be a = 60.4 m. n = sin-1 n d n = sin 1 n a , and 1 = sin 1 a for the first null....
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