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white (rw22679) – Hw08 – Ross – (10219) 1 This print-out should have 13 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 2 . 6 for an object placed 22 cm from the lens. What is the focal length of the lens? Correct answer: 35 . 75 cm. Explanation: Let : M = 2 . 6 and p = 22 cm . 1 p + 1 q = 1 f M = h h = - q p Converging Lens f > 0 >p> f f <q < 0 >m> -∞ f >p> 0 -∞ <q < 0 >m> 1 From the magnification q = - M p so 1 f = 1 p - 1 M p = M - 1 M p = 2 . 6 - 1 (2 . 6)(22 cm) = 0 . 027972 cm 1 f = 1 0 . 027972 cm 1 = 35 . 75 cm . 002 (part 1 of 2) 10.0 points A certain LCD projector contains a single thin lens. An object 31 . 3 mm high is to be projected so that its upright image fills a screen 2 . 2 m high. The object-to-screen distance is 1 . 48 m. Find the focal length of the projection lens. Correct answer: 20 . 4698 mm. Explanation: Given : h = 31 . 3 mm = 0 . 0313 m , h = - 2 . 2 m , and L = 1 . 48 m . It is desired to form a magnified, real image on the screen using a single thin lens. To do this, a converging lens must be used and the image will be inverted. The magnification then gives M = h h = - q p q = - h h p = - - 2 . 2 m 0 . 0313 m p = 70 . 2875 p. Also we know that L = p + q = 1 . 48 m, so p + 70 . 2875 p = 1 . 48 m p = 1 . 48 m (1 + 70 . 2875) · 1000 mm 1 m = 20 . 761 mm . The lens equation gives 1 p + 1 q = 1 f 1 p + 1 70 . 2875 p = 1 f f = 70 . 2875 p 70 . 2875 + 1 = (70 . 2875) (20 . 761 mm) 70 . 2875 + 1 = 20 . 4698 mm . 003 (part 2 of 2) 10.0 points How far from the object should the lens of the projector be placed in order to form the image on the screen?
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