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Unformatted text preview: Version 035/AACAD midterm 04 Turner (59070) 1 This printout should have 19 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points The reflecting surfaces of two intersecting flat mirrors are at an angle of 62 , as shown in the figure. A light ray strikes the horizontal mirror at an angle of 49 with respect to the mirrors surface. 62 49 Figure is not drawn to scale. Calculate the angle . 1. 54.0 2. 68.0 3. 70.0 4. 74.0 5. 48.0 6. 76.0 7. 58.0 8. 46.0 9. 62.0 10. 56.0 Correct answer: 56 . Explanation: Basic Concept: incident = reflected Solution: 1 2 Figure is to scale. The sum of the angles in a triangle is 180 . In the triangle on the left we have angles , 180 1 2 , and 180 2 2 , so 180 = + 180 1 2 + 180 2 2 , or 1 + 2 = 2 . (1) In the triangle on the right we have angles 1 , 2 , and . 180 = 1 + 2 + , so 1 + 2 = 180 . (2) Combining Eq. 1 and 2, we have = 180 2 = 180 2 (62 ) = 56 . As a matter of interest, in the upperhalf of the figure the angles (clockwise) in the triangles from left to right are 41 , 41 , and 98 ; 82 , 28 , and 70 ; 110 , 21 , and 49 ; 131 , 21 , and 28 ; and in the lowerhalf of the figure the angles (counterclockwise) in the triangles from left to right are 21 , 21 , and 138 ; 42 , 28 , and 110 ; 70 , 41 , and 69 ; 111 , 41 , and 28 . Version 035/AACAD midterm 04 Turner (59070) 2 002 10.0 points Consider a prism with the shape shown in the diagram. Its index of refraction is labeled by n p , and it is submerged in a special liquid which has an index of refraction n , where the difference n p n = 0 . 58. The light ray is perpendicularly incident from the liquid into the prism as shown in the diagram. Notice that the incident angle at both points A and B is 45 . n p A B n l 45 45 Find the index of refraction of the prism for a critical angle of 45 . 1. 1.46811 2. 1.67296 3. 1.87782 4. 1.39983 5. 1.77539 6. 1.57054 7. 1.98024 8. 1.70711 9. 1.63882 10. 1.60468 Correct answer: 1 . 98024. Explanation: Let : n = 0 . 58 c = 45 . n p sin c = ( n p n ) sin 90 n p 1 2 = n p n n p parenleftbigg 1 1 2 parenrightbigg = n n p = 2 n 2 1 = 2 (0 . 58) 2 1 = 1 . 98024 . 003 10.0 points At what distance from a 16 W electromag netic wave point source (like a light bulb) is the amplitude of the electric field 46 V / m?...
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This note was uploaded on 01/21/2010 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Physics

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