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Unformatted text preview: terry (ect328) – oldhomework 36 – Turner – (59130) 1 This printout should have 14 questions. Multiplechoice 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 > ∞ >p> f f <q < ∞ >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: . 548872. Explanation: M = q p = (11 . 3068 cm) (20 . 6 cm) = . 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 < ∞ >p> f <q < <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 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) = . 399516 ....
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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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