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Unformatted text preview: Version 005/AAABB – midterm 04 – Turner – (59130) 1 This printout should have 19 questions. Multiplechoice 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 > ∞ >p> f f <q < ∞ >m> −∞ f >p> −∞ <q < ∞ >m> 1 Convex Mirror > f ∞ >p> f <q < <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 . 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 upperhalf 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 lowerhalf of the figure the angles (counterclockwise) in the triangles from left to right are 23 ◦ , 23 ◦ , and 134 ◦ ; 46 ◦ , 25 ◦ , and 109 ◦ ; 71 ◦ , 42 ◦ , and 67 ◦ ; 113 ◦ , 42 ◦ , and 25 ◦ ....
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This note was uploaded on 02/17/2010 for the course PHY 59130 taught by Professor Turner during the Fall '09 term at University of Texas.
 Fall '09
 Turner

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