MIDTERM 4

# MIDTERM 4 - Version 005/AAABB midterm 04 Turner(59130 This...

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Version 005/AAABB – midterm 04 – Turner – (59130) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points The magnitude of the radius of a spherical mirror is | R | = a . An object is placed in front of this spherical mirror. It forms a virtual image. The magnification is M = 1 2 . The mirror is 1. convergent 2. divergent correct Explanation: Since there is a reduced vertical image, it must be a convex (divergent) mirror. 002 (part 2 of 2) 10.0 points The object distance is 1. 1 4 a 2. 1 2 a correct 3. 3 2 a 4. 5 2 a 5. 1 3 a 6. a 7. 3 a 8. 2 3 a 9. 2 a 10. 3 4 a Explanation: Basic Concepts: 1 p + 1 q = 1 f = 2 R m = h h = q p Concave Mirror f > 0 > p > f f < q < 0 > m > −∞ f > p > 0 −∞ < q < 0 > m > 1 Convex Mirror 0 > f > p > 0 f < q < 0 0 < m < 1 Solution: Since there is a reduced virtual image, it must correspond to convex mirror, or R = −| R | . Also, virtual image implies p q = p | q | = 1 | M | = 2 . Hence, 1 p + 1 q = 2 R 1 p parenleftbigg 1 1 | M | parenrightbigg = 2 | R | 1 p (1 2) = 2 a 1 p = 2 a . Notice: The last equality confirms the fact that our choice of R = −| R | is correct. p = a 2 003 10.0 points The reflecting surfaces of two intersecting flat mirrors are at an angle of 65 , as shown in the figure. A light ray strikes the horizontal mirror at an angle of 48 with respect to the mirror’s surface. 65 48 φ Figure is not drawn to scale. Calculate the angle φ .

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Version 005/AAABB – midterm 04 – Turner – (59130) 2 1. 68.0 2. 66.0 3. 62.0 4. 70.0 5. 52.0 6. 50.0 7. 48.0 8. 76.0 9. 56.0 10. 64.0 Correct answer: 50 . Explanation: Basic Concept: θ incident = θ reflected Solution: θ 1 θ 2 φ θ Figure is to scale. The sum of the angles in a triangle is 180 . In the triangle on the left we have angles θ , 180 θ 1 2 , and 180 θ 2 2 , so 180 = θ + 180 θ 1 2 + 180 θ 2 2 , or θ 1 + θ 2 = 2 θ . (1) In the triangle on the right we have angles θ 1 , θ 2 , and φ . 180 = θ 1 + θ 2 + φ , so θ 1 + θ 2 = 180 φ . (2) Combining Eq. 1 and 2, we have φ = 180 2 θ = 180 2 (65 ) = 50 . As a matter of interest, in the upper-half of the figure the angles (clockwise) in the triangles from left to right are 42 , 42 , and 96 ; 84 , 25 , and 71 ; 109 , 23 , and 48 ; 132 , 23 , and 25 ; and in the lower-half of the figure the angles (counter-clockwise) in the triangles from left to right are 23 , 23 , and 134 ; 46 , 25 , and 109 ; 71 , 42 , and 67 ; 113 , 42 , and 25 . 004 10.0 points At what distance from a 47 W electromag- netic wave point source (like a light bulb) is the amplitude of the electric field 40 V / m? μ o c = 376 . 991 Ω . 1. 1.93649 2. 2.2964 3. 9.0312 4. 3.30544 5. 1.19487 6. 1.72763 7. 1.32759 8. 2.00798 9. 1.1134 10. 30.249 Correct answer: 1 . 32759 m. Explanation: Let : P = 47 W , E max = 40 V / m , and μ 0 c = 376 . 991 Ω .
Version 005/AAABB – midterm 04 – Turner – (59130) 3 The wave intensity (power per unit area) is given by I = S av = E 2 max 2 ( μ o c ) = (40 V / m) 2 2 (376 . 991 Ω) = 2 . 12207 W / m 2 .

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