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Unformatted text preview: Phys 2214 Fall 2008  Solutions to Problem Set #9 1. (YF 36.6) a) Relating speed, frequency, and wavelength of a wave: v = λf → λ = v f Plugging in, λ = 800 km h * 1 . h = 800 km b) The location of minima from a single slit Fraunhofer di raction pattern (distant observer approximation) are given by: a sin θ = mλ → θ = arcsin mλ a so the smallest angles at which minima occur will be arcsin 1 * 800 km 3700 km = 0 . 22 rad = 13 degrees for the Australia/Antarctica gap, and arcsin 1 * 800 km 4500 km = 0 . 18 rad = 10 degrees for the Africa/Antarctica gap. 2. (YF 36.27) a) In the observations described in this problem, the bright bands corre spond to 2slit di raction pattern maxima, while the missing bands are minima of the single slit envelope to this pattern. The distance to the screen is 90 . cm so let's assume we can make distant observer approxima tions and so long as our resulting values for d (slit separation) and a (slit width) are 90 cm, we will not have to revise this assumption. Sketching our setup: 1 2slit maxima are given by d sin θ = m 1 λ → d = m 1 λ sin θ Considering the rst bright band, m 1 = 1 , λ = 500 nm, and θ is small since it is equal to arctan 1 cm 90 cm . By the small angle approximation sin θ ≈ tan θ = 1 cm 90 cm . Plugging in yields: d = 4 . 50 × 10 5 m Similarly, 1slit minima are given by a sin θ = m 2 λ → a = m 2 λ sin θ Considering the rst missing band, m 2 = 1 and sin θ ≈ tan θ = 3 cm 90 cm , which yields: a = 1 . 50 × 10 5 m Check: a and d are both 90 . cm so our distant screen approximation was reasonable. b) To nd the total number of bright bands within  θ  ≤ 90 dregrees we can solve the equation for 2slit interference maxima, d sin θ = m 1 λ , for m 1 when θ = 90 degrees to nd the highest order max in this angle range. m 1 = d sin90 degrees λ = d λ = 90 And so the highest order max for θ < 90 degrees is the m 1 = 90 band. Since a = d 3 , by similar reasoning we can show that there will also be 30 1slit minima in this range, each canceling a 2slit maximum. This leaves 60 bright bands on either side of the di raction pattern for a total of 120 bright bands in addition to the central max. Optional note: Why do we say that the pattern from single slit di rac tion provides an envelope to the pattern from 2 or more slits? Why isn't the envelope to the Nslit pattern really N overlapping single slit patterns? The answer is that it is, however we can approximate pretty will using the simpler case of a single 1slit di raction pattern modulating the Nslit interference. Here's why: suppose the distance to our screen is l and our slit spacing is d . No matter how far the screen is, the observed spacing between single slit patterns from consecutive lines in our grating will be d....
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 Fall '07
 GIAMBATTISTA,A
 Wavelength, single slit

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