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

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Unformatted text preview: 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 find 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. Find 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 > >p> f f <q < >m>- f >p>- <q < >m> 1 Diverging Lens f < >p> f <q < <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 magnification of the object? Correct answer:- . 286176. Explanation: Magnification is M =- s s =- (9 . 73 cm) (34 cm) =- . 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 = Find 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 > f > p>- <q < >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 Find the magnification....
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This note was uploaded on 12/11/2009 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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

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