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Unformatted text preview: benavides (jjb2356) – 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 . 1 cm in front of a lens forms an image on a screen 9 . 2 cm behind the lens. Find the focal length of the lens. Correct answer: 7 . 15061 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 . 1 cm) (9 . 2 cm) (32 . 1 cm) + (9 . 2 cm) = 7 . 15061 cm . 002 (part 2 of 2) 10.0 points What is the magnification of the object? Correct answer: . 286604. Explanation: Magnification is M = s ′ s = (9 . 2 cm) (32 . 1 cm) = . 286604 . 003 (part 1 of 2) 10.0 points A convergent lens has a focal length of 25 . 4 cm . The object distance is 8 . 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: 13 . 0024 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 = 25 . 4 cm. Solution: Substituting these values into the lens equation q = 1 1 f 1 p = 1 1 (25 . 4 cm) 1 (8 . 6 cm) = 13 . 0024 cm  q  = 13 . 0024 cm ....
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This note was uploaded on 03/26/2012 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
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