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midterm 04

# midterm 04 - Version 035/AACAD midterm 04 Turner(59070 This...

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Version 035/AACAD – midterm 04 – Turner – (59070) 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 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 mirror’s surface. 62 49 φ Figure is not drawn to scale. Calculate the angle φ . 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: 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 (62 ) = 56 . As a matter of interest, in the upper-half 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 lower-half of the figure the angles (counter-clockwise) in the triangles from left to right are 21 , 21 , and 138 ; 42 , 28 , and 110 ; 70 , 41 , and 69 ; 111 , 41 , and 28 .

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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 θ c = 45 . n p sin θ 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? μ o c = 376 . 991 Ω . 1. 0.673562 2. 0.426956 3. 0.68853 4. 5.84808 5. 0.438178 6. 32.8634 7. 1.57762 8. 0.803902 9. 4.42627 10. 1.99489 Correct answer: 0 . 673562 m. Explanation: Let : P = 16 W , E max = 46 V / m , and μ 0 c = 376 . 991 Ω . The wave intensity (power per unit area) is given by I = S av = E 2 max 2 ( μ o c ) = (46 V / m) 2 2 (376 . 991 Ω) = 2 . 80643 W / m 2 . The constant μ o c (impedance of free space) turns out to be 120 π Ω. You should verify that (V / m) 2 Ω = W / m 2 . With this we can find the intensity, which multiplied by the area of a circle of radius R must gives us the source’s power P , therefore 4 π R 2 I = P = 16 W .
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midterm 04 - Version 035/AACAD midterm 04 Turner(59070 This...

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