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# OHW 36 - terry(ect328 oldhomework 36 Turner(59130 This...

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terry (ect328) – oldhomework 36 – Turner – (59130) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A convergent lens has a focal length of 7 . 3 cm . The object distance is 20 . 6 cm . q h p h f f Scale: 10 cm = Find the distance of the image from the center of the lens. Correct answer: 11 . 3068 cm. Explanation: 1 p + 1 q = 1 f M = h h = - q p Convergent Lens f > 0 >p> f f <q < 0 >M > -∞ Note: The focal length for a convergent lens is positive, f = 7 . 3 cm. Solution: Substituting these values into the lens equation q = 1 1 f - 1 p = 1 1 (7 . 3 cm) - 1 (20 . 6 cm) = 11 . 3068 cm . 002 (part 2 of 2) 10.0 points Find the magnification. Correct answer: - 0 . 548872. Explanation: M = - q p = - (11 . 3068 cm) (20 . 6 cm) = - 0 . 548872 . 003 (part 1 of 2) 10.0 points A divergent lens has a focal length of 16 . 5 cm . The object distance is 24 . 8 cm . f f q h p h Scale: 10 cm = Find the distance of the image from the center of the lens. Correct answer: 9 . 90799 cm. Explanation: 1 p + 1 q = 1 f M = h h = - q p Divergent Lens f < 0 >p> 0 f <q < 0 0 <M < 1 Note: The focal length for a divergent lens is negative, f = - 16 . 5 cm. Solution: Substituting these values into the lens equation q = 1 1 f - 1 p

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terry (ect328) – oldhomework 36 – Turner – (59130) 2 = 1 1 ( - 16 . 5 cm) - 1 (24 . 8 cm) = - 9 . 90799 cm | q | = 9 . 90799 cm . 004 (part 2 of 2) 10.0 points Find the magnification. Correct answer: 0 . 399516. Explanation: M = - q p = - ( - 9 . 90799 cm) (24 . 8 cm) = 0 . 399516 . 005 10.0 points Hint: Construct a ray diagram. Given: A real object is located at “ p = 13 7 f to the left of a convergent lens with a focal length f as shown in the figure below. 0
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