HWsol09 - Phys 2214 Fall 2008 Solutions to Problem Set#9...

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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 2-slit 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 2-slit 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, 1-slit 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 2-slit 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 1-slit minima in this range, each canceling a 2-slit 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 N-slit 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 1-slit di raction pattern modulating the N-slit 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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This note was uploaded on 12/24/2008 for the course PHYS 2214 taught by Professor Giambattista,a during the Fall '07 term at Cornell.

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HWsol09 - Phys 2214 Fall 2008 Solutions to Problem Set#9...

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