# S03x3aan - NAME PHYS-1200 PHYSICS II QUIZ 3 SOLUTIONS PART A 1 D 2 D 3 E 4 D PART B 1 a N1 = L 7.50 10-6 m = 5.00 10 7 m L L = n =(1.30(15 b N 2

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Unformatted text preview: NAME PHYS-1200 PHYSICS II QUIZ 3 APRIL 25, 2003 SOLUTIONS PART A. 1. D 2. D 3. E 4. D PART B. 1. a) N1 = L 7.50 10 -6 m = 5.00 10 - 7 m L L = n = (1.30)(15) b) N 2 = /n c) THEY ARE COMPLETELY OUT OF PHASE N2 N1 = 4.5, which is an integer plus . 2. a) N1 = 15 N2 = 19.5 5. C 6. D 7. A 8. C 9. B 10. A b) c) NAME PART C. 1. a) d sin = m , so d = m 1(5.0 10-8 m) = sin sin2.0 d = 1.4 10-6 m = 1.4 m m 1(5.0 10-8 m) = b) a sin = m , so a = sin sin5.0 c) Eel = a = 5.7 10-7 m = 0.57 m p2 h2 (6.63 10-34 J s) 2 = = 2m 2m2 2(9.11 10-31 kg)(5.0 10-8 m)2 Eel = 9.65 10-23 J = 6.0 10-4 eV = 0.60 meV y = 1.7 10-2 m = 1.7 cm d) y = D tan = (0.50 m) tan 2.0 2 Em 2. a) I = , so Em = 2c 0 I = 2(3.0 108 m/s)(4 10-7 H/m)(6.0 10-2 W/m 2 ) 2c 0 Em = 6.7 V/m b) I = Ps , so 4 r 2 Ps = 4 r 2 I = 4 (2.0 m)2 (6.0 10-2 W/m 2 ) Ps = 3.0 W c) Ephoton = hf = (6.63 10-34 Js)(3.0 1012 Hz) Ephoton = 2.0 10-21 J = 1.2 10-2 eV = 12 meV d) In time t, the energy that strikes the screen is U = IAt. Then, the number of photons in U IAt (6.0 10-2 W/m 2 )(2.0 10-4 m 2 )(1.0 s) = = one second is, N = E photon E photon 2.0 10 - 21 J N = 6.0 1015 ...
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## This test prep was uploaded on 04/18/2008 for the course PHYS 1200 taught by Professor Stoler during the Spring '06 term at Rensselaer Polytechnic Institute.

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S03x3aan - NAME PHYS-1200 PHYSICS II QUIZ 3 SOLUTIONS PART A 1 D 2 D 3 E 4 D PART B 1 a N1 = L 7.50 10-6 m = 5.00 10 7 m L L = n =(1.30(15 b N 2

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