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Unformatted text preview: nguyen (jmn727) – homework 36 – Turner – (59070) 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 34 . 3 cm in front of a lens forms an image on a screen 9 . 41 cm behind the lens. Find the focal length of the lens. Correct answer: 7 . 38419 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 . 3 cm) (9 . 41 cm) (34 . 3 cm) + (9 . 41 cm) = 7 . 38419 cm . 002 (part 2 of 2) 10.0 points What is the magnification of the object? Correct answer: . 274344. Explanation: Magnification is M = s ′ s = (9 . 41 cm) (34 . 3 cm) = . 274344 . 003 (part 1 of 2) 10.0 points A convergent lens has a focal length of 25 . 1 cm . The object distance is 11 . 7 cm . f f q h ′ p h Scale: 10 cm = Find the distance of the image from the center of the lens. Correct answer: 21 . 9157 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 . 1 cm. Solution: Substituting these values into the lens equation q = 1 1 f 1 p = 1 1 (25 . 1 cm) 1 (11 . 7 cm) = 21 . 9157 cm  q  = 21 . 9157 cm ....
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 Spring '08
 Turner
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