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Unformatted text preview: aldawsari (ma28683) – HW20 – Betancourt – (16856) 1 This printout should have 16 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 34.534.8 001 10.0 points The magnification produced by a converging lens is found to be 3 for an object placed 19 cm from the lens. What is the focal length of the lens? Correct answer: 28 . 5 cm. Explanation: Let : M = 3 and p = 19 cm . 1 p + 1 q = 1 f M = h ′ h = q p Converging Lens f > ∞ >p> f f <q < ∞ >m>∞ f >p>∞ <q < ∞ >m> 1 From the magnification q = M p so 1 f = 1 p 1 M p = M 1 M p = 3 1 (3)(19 cm) = 0 . 0350877 cm − 1 f = 1 . 0350877 cm − 1 = 28 . 5 cm . 002 (part 1 of 3) 10.0 points A real object is located 23 . 8 cm from a diverg ing lens of focal length 35 cm. What is the magnitude of the image dis tance? Correct answer: 14 . 1667 cm. Explanation: Let : p = 23 . 8 cm and f = 35 cm . 1 p + 1 q = 1 f M = h ′ h = q p Diverging Lens > f ∞ >p> f <q < <M < 1 Using the lens equation, 1 q = 1 f 1 p = p f f p q = f p p f = ( 35 cm) (23 . 8 cm) 23 . 8 cm ( 35 cm) = 14 . 1667 cm . which has a magnitude of 14 . 1667 cm . 003 (part 2 of 3) 10.0 points The image is 1. virtual, inverted, and larger. 2. virtual, erect, and larger. 3. real, inverted, and larger. 4. real, erect, and larger. 5. does not exist. 6. virtual, erect, and smaller. correct 7. real, erect, and smaller. 8. real, inverted, and smaller. 9. virtual, inverted, and smaller. Explanation: q < 0 (the image is on the same side of the lens as the object), so the image is virtual, and the magnification is M = q p = 14 . 1667 cm 23 . 8 cm = . 595238 . < M < 1, so the image is erect and smaller than the object. 004 (part 3 of 3) 10.0 points aldawsari (ma28683) – HW20 – Betancourt – (16856) 2 What is the magnitude of the magnification of the image? Correct answer: 0 . 595238. Explanation: M = 0 . 595238 . 005 (part 1 of 2) 10.0 points Hint: Construct a ray diagram. Given: A real object is located at “ p 1 = 10 7 f ” to the left of a convergent lens with a focal length f as shown in the figure below. 10 7 f f ( × f ) The image distance q 1 to the right of the lens is 1. q 1 = 13 6 f . 2. q 1 = 9 4 f . 3. q 1 = 5 2 f . 4. q 1 = 7 3 f . 5. q 1 = 7 2 f . 6. q 1 = 15 8 f . 7. q 1 = 11 5 f . 8. q 1 = 11 6 f . 9. q 1 = 10 3 f . correct 10. q 1 = 15 4 f . 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 Convergent (convex) lenses f > 0 have real images q > 0 when the object ∞ > p > f . 10 7 f f 10 3 ( × f ) Basic Concepts: 1 p + 1 q = 1 f . Solution: 1 q 1 = 1 p 1 + 1 f = 7 10 f + 1 f = 10 7 10 f = 3 10 f q 1 = 10 3 f ....
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This note was uploaded on 01/27/2011 for the course PHYS 251 taught by Professor Gavrine during the Spring '10 term at Purdue.
 Spring '10
 Gavrine

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