Homework 13-solutions

# Homework 13-solutions - gonzales (pag757) – Homework 13...

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Unformatted text preview: gonzales (pag757) – Homework 13 – Hoffmann – (56745) 1 This print-out should have 26 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The magnification produced by a converging lens is found to be 3 . 2 for an object placed 18 cm from the lens. What is the focal length of the lens? Correct answer: 26 . 1818 cm. Explanation: Let : M = 3 . 2 and p = 18 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 . 2- 1 (3 . 2)(18 cm) = 0 . 0381944 cm − 1 f = 1 . 0381944 cm − 1 = 26 . 1818 cm . 002 (part 1 of 3) 10.0 points A real object is located 23 . 5 cm from a diverg- ing lens of focal length 33 . 9 cm. What is the magnitude of the image dis- tance? Correct answer: 13 . 8789 cm. Explanation: Let : p = 23 . 5 cm and f =- 33 . 9 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 = (- 33 . 9 cm) (23 . 5 cm) 23 . 5 cm- (- 33 . 9 cm) =- 13 . 8789 cm . which has a magnitude of 13 . 8789 cm . 003 (part 2 of 3) 10.0 points The image is 1. virtual, erect, and larger. 2. real, erect, and smaller. 3. virtual, erect, and smaller. correct 4. real, erect, and larger. 5. virtual, inverted, and smaller. 6. real, inverted, and smaller. 7. does not exist. 8. real, inverted, and larger. 9. virtual, inverted, and larger. 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 =-- 13 . 8789 cm 23 . 5 cm = . 590592 . < M < 1, so the image is erect and smaller than the object. gonzales (pag757) – Homework 13 – Hoffmann – (56745) 2 004 (part 3 of 3) 10.0 points What is the magnitude of the magnification of the image? Correct answer: 0 . 590592. Explanation: M = 0 . 590592 . 005 (part 1 of 2) 10.0 points An object located 33 . 9 cm in front of a lens forms an image on a screen 8 . 2 cm behind the lens. Find the focal length of the lens. Correct answer: 6 . 60285 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 = ss ′ s + s ′ = (33 . 9 cm) (8 . 2 cm) (33 . 9 cm) + (8 . 2 cm) = 6 . 60285 cm . 006 (part 2 of 2) 10.0 points What is the magnification of the object? Correct answer:- . 241888. Explanation: Magnification is M =- s ′ s =- (8 . 2 cm) (33 . 9 cm) =- . 241888 ....
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## This note was uploaded on 01/27/2012 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas.

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Homework 13-solutions - gonzales (pag757) – Homework 13...

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