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FINAL - Version 012/AAADA nal 01 Turner(59130 This...

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Version 012/AAADA – final 01 – Turner – (59130) 1 This print-out should have 29 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A light ray is traveling in air with an index of refraction of unity and it is reflected at the boundary of a second medium with an index of refraction n . Consider the optical coating on a glass lens where the index of refraction of the coating is n , where n is greater than the index of refraction of the air. air lens t n θ θ 0 1 2 Assume: The index of refraction of the lens is greater than that of the coating. To minimize the reflection of a ray with a wavelength 500 nm and incident angle θ 0, what is the minimum nonzero thickness of the coating? 1. t = n λ 2. t = 3 λ 4 n 3. t = λ 4 n correct 4. t = n λ 2 5. t = n λ 8 6. t = λ n 7. t = 3 n λ 4 8. t = λ 2 n 9. t = λ 8 n 10. t = n λ 4 Explanation: Destructive interference occurs when the difference between the phase angle of the inci- dent ray reflected from the outer surface of the coating (ray 1) and the phase angle of the ray reflected from the inner surface of the coating (ray 2) are at π, 3 π, 5 π etc. The phase differ- ences are due to the path difference of the two rays and the change of their relative phases due to reflections. Putting them together, it gives π, 3 π, 5 π... = (2 t ) k n + | π π | = 2 t k n . With k n = 2 π λ n , it implies that the minimum thickness is given by t = λ 4 n . 002 10.0 points A charged particle beam (shot horizontally) moves into a region where there is a constant magnetic field of magnitude 0 . 00169 T that points straight down. The charged particles in the beam move in a circular path of radius 1 . 27 cm. If the charged particles in the beam were accelerated through a potential difference of 203 V, determine the charge to mass ratio of the charged particles in the beam. 1. 683515000000.0 2. 881343000000.0 3. 1211960000000.0 4. 319542000000.0 5. 59045400000.0 6. 1262090000000.0 7. 31081400000.0 8. 92455400000.0 9. 259099000000.0 10. 161118000000.0 Correct answer: 8 . 81343 × 10 11 C / kg. Explanation: Let : B = 0 . 00169 T , r = 1 . 27 cm = 0 . 0127 m , and V = 203 V .
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Version 012/AAADA – final 01 – Turner – (59130) 2 The kinetic energy of the electron after accel- erated through the potential difference is 1 2 m v 2 = q V . And the centrifugal force experienced by the charge particle in the magnetic field is q v B sin(90 ) = m v 2 r then we have v = q b r m then 1 2 m q 2 B 2 r 2 m 2 = q V . Thus , the charge-to-mass ratio is R = q m = 2 V B 2 r 2 = 2(203 V) (0 . 00169 T) 2 (0 . 0127 m) 2 = 8 . 81343 × 10 11 C / kg . 003 10.0 points A circular arc has a uniform linear charge density of 7 nC / m. The value of the Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . 257 2 . 5 m x y What is the magnitude of the electric field at the center of the circle along which the arc lies? 1. 52.5474 2. 44.527 3. 38.9437 4. 12.5546 5. 33.4949 6. 27.0988 7. 51.917 8. 39.3889 9. 60.6649 10. 19.4604 Correct answer: 39 . 3889 N / C. Explanation: Let : λ = 7 nC / m = 7 × 10 9 C / m , Δ θ = 257 , and r = 2 . 5 m .
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