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# Mid4 - Answers - Arb Kellen Midterm 4 Due 10:00 pm Inst...

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Arb, Kellen – Midterm 4 – Due: Nov 30 2004, 10:00 pm – Inst: Turner 1 This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points A screen is illuminated by monochromatic light as shown in the figure below. The distance from the slits to the screen is 6 . 4 m . 3 . 6 cm 6 . 4 m 0 . 72 mm S 1 S 2 θ viewing screen What is the wave length if the distance from the central bright region to the seventh bright fringe is 3 . 6 cm . Correct answer: 578 . 562 nm. Explanation: Basic Concepts: For bright fringes, we have d sin θ = m λ , and for dark fringes, we have d sin θ = m + 1 2 λ , where m = 0 , ± 1 , ± 2 , ± 3 , · · · . From geometry, we have y = L tan θ . Let : y = 3 . 6 cm = 0 . 036 m , L = 6 . 4 m , and d = 0 . 72 mm = 0 . 00072 m . r 2 r 1 y L d S 1 S 2 θ = tan - 1 y L · viewing screen δ d sin θ r 2 - r 1 P O 6 S 2 Q S 1 90 Q r 2 r 1 d S 1 S 2 θ = tan - 1 y L · θ δ d sin θ r 2 - r 1 6 S 2 Q S 1 90 Q Solution: The angle θ from the slits’ mid- point to the y position on the screen is θ = arctan h y L i = arctan (0 . 036 m) (6 . 4 m) = 0 . 00562494 rad . The wavelength of the light for the seventh bright fringe, m = 7, is λ = d sin θ m = (0 . 00072 m) sin(0 . 00562494 rad) (7) = 5 . 78562 × 10 - 7 m = 578 . 562 nm . 002 (part 1 of 4) 10 points An object is placed 10 m before a convex lens with focal length 5 . 2 m . Another concave lens is place 18 . 5 m behind the first lens with a focal length - 12 m (see the figure below). Note: Make a ray diagram sketch in order to check your numerical answer.

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Arb, Kellen – Midterm 4 – Due: Nov 30 2004, 10:00 pm – Inst: Turner 2 0 5 10 15 20 25 p 1 f 1 f 1 10 m 5 . 2 m f 1 = f 2 18 . 5 m - 12 m f 2 = At what distance is the first image from the first lens? Correct answer: 10 . 8333 m. Explanation: Basic Concept: The thin lens equation is 1 p + 1 q = 1 f . Let : f 1 = 5 . 2 m , f 2 = - 12 m , p 1 = 10 m , and = 18 . 5 m . 0 5 10 15 20 25 p 1 q 1 p 2 f 1 f 1 f 2 q 2 Note: After sketching the ray diagram above and using the positions on the scale, we have f 1 = (10 m) - (4 . 8 m) = 5 . 2 m , or = (15 . 2 m) - (10 m) = 5 . 2 m p 2 = q 1 = (20 . 8333 m) - (10 m) = 10 . 8333 m f 2 = (28 . 5 m) - (40 . 5 m) = - 12 m , or = (16 . 5 m) - (28 . 5 m) = - 12 m , and q 2 = (23 . 822 m) - (28 . 5 m) = - 4 . 67797 m . Solution: q 1 = 1 1 f 1 - 1 p 1 = 1 1 5 . 2 m - 1 10 m = 10 . 8333 m . 003 (part 2 of 4) 10 points What is the magnification of the first image? Correct answer: - 1 . 08333 . Explanation: m 1 = - q 1 p 1 = - 10 . 8333 m 10 m = - 1 . 08333 . 004 (part 3 of 4) 10 points At what distance is the second image from the second lens? Correct answer: 4 . 67797 m. Explanation: p 2 = - q 1 = (18 . 5 m) - (10 . 8333 m) = 7 . 66667 m . q 2 = 1 1 f 2 - 1 p 2 = 1 1 - 12 m - 1 7 . 66667 m = - 4 . 67797 m | q 2 | = 4 . 67797 m .
Arb, Kellen – Midterm 4 – Due: Nov 30 2004, 10:00 pm – Inst: Turner 3 005 (part 4 of 4) 0 points What is the magnification of the final image, when compared to the initial object? Correct answer: 0 . 610169 . Explanation: m 2 = - q 2 p 2 = - - 4 . 67797 m 7 . 66667 m = 0 . 610169 . m final = m 1 m 2 = ( - 1 . 08333) (0 . 610169) = - 0 . 661017 .

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Mid4 - Answers - Arb Kellen Midterm 4 Due 10:00 pm Inst...

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