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Unformatted text preview: terry (ect328) – homework 36 – Turner – (59130) 1 This printout should have 12 questions. Multiplechoice 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 32 . 5 cm in front of a lens forms an image on a screen 10 . 5 cm behind the lens. Find the focal length of the lens. Correct answer: 7 . 93605 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 ′ = (32 . 5 cm) (10 . 5 cm) (32 . 5 cm) + (10 . 5 cm) = 7 . 93605 cm . 002 (part 2 of 2) 10.0 points What is the magnification of the object? Correct answer: . 323077. Explanation: Magnification is M = s ′ s = (10 . 5 cm) (32 . 5 cm) = . 323077 . 003 (part 1 of 2) 10.0 points A convergent lens has a focal length of 22 . 7 cm . The object distance is 11 . 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: 24 . 5743 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 = 22 . 7 cm. Solution: Substituting these values into the lens equation q = 1 1 f 1 p = 1 1 (22 . 7 cm) 1 (11 . 8 cm) = 24 . 5743 cm  q  = 24 . 5743 cm ....
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This note was uploaded on 02/17/2010 for the course PHY 59130 taught by Professor Turner during the Fall '09 term at University of Texas at Austin.
 Fall '09
 Turner

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