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solution_pdf36 - tovar(jdt436 homework 36 Turner(59070 This...

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tovar (jdt436) – homework 36 – Turner – (59070) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 (part 1 oF 2) 10.0 points An object located 34 cm in Front oF a lens Forms an image on a screen 9 . 73 cm behind the lens. ±ind the Focal length oF the lens. Correct answer: 7 . 56506 cm. Explanation: Basic Concepts: 1 p + 1 q = 1 f m = h h = - q p Converging Lens f > 0 >p > f f <q < 0 >m> -∞ f >p > 0 -∞ <q < 0 1 Diverging Lens f < 0 >p > 0 0 0 <m< 1 Solution: The Formula relating the Focal length to the image distance s and the object distance s is 1 s + 1 s = 1 f so f = s s s + s = (34 cm) (9 . 73 cm) (34 cm) + (9 . 73 cm) = 7 . 56506 cm . 002 (part 2 oF 2) 10.0 points What is the magnifcation oF the object? Correct answer: - 0 . 286176. Explanation: Magnifcation is M = - s s = - (9 . 73 cm) (34 cm) = - 0 . 286176 . 003 (part 1 oF 2) 10.0 points A convergent lens has a Focal length oF 15 . 2 cm . The object distance is 7 . 6 cm . f f q h p h Scale: 10 cm = ±ind the distance oF the image From the center oF the lens. Correct answer: 15 . 2 cm. Explanation: 1 p + 1 q = 1 f M = h h = - q p Convergent Lens f > 0 f > p> 0 -∞ 0 >M > 1 Note: The Focal length For a convergent lens is positive, f = 15 . 2 cm. Solution: Substituting these values into the lens equation q = 1 1 f - 1 p = 1 1 (15 . 2 cm) - 1 (7 . 6 cm) = - 15 . 2 cm | q | = 15 . 2 cm . 004 (part 2 oF 2) 10.0 points ±ind the magnifcation. Correct answer: 2.
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tovar (jdt436) – homework 36 – Turner – (59070) 2 Explanation: M = - q p = - ( - 15 . 2 cm) (7 . 6 cm) = 2 .
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