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mysMu x canvass L x Files x Client's x Secure https://edugen.wileyplus.com/edugen/student/mainfr uni Apps E tting The Java e Complete print for Ja Used 2007 Lexus ES 3 D New Tab I Help y Physics (PHYs 1303, 1304, 1307, Intro Hall, ay damentals of Physics, 10e Gradebook ORION Uu satEEN PRINTER VERSION BACK. Chapter 34, Problem 139 In the figure, a sand grain is 3.5 cm from thin lens 1, on the central axis through the two symmetric lenses. The distance between focal point and lens is 4.2 cm for both lenses; the lenses are separated by 8.4 cm. (a) What is the distance between lens 2 and the image it produces of the sand grain? Is that image (b) to the left or right of lens 2, (c) real or virtual, and (d) inverted relative to the sand grain or not inverted? (a) Number (b) (c) the tolerance is +/-2% click if you would like to show work for this question Open Show Work Question Attempts: o of 5 used SAVE FOR LATER SUBMIT ANSWER Copyright 2000-2017 by John Wiley & Sons, Inc. or related companies. All rights reserved olicy 1 2000-2017 John Wiley & Sons, Inc. All Rights Reserved A Division of John Wiley 8 Sons, Inc. Version 4.Explanation / Answer
here,
object distance, do = 3.5 cm
focal length of lens1, f1 = -4.2 cm
focal length of lens2, f2 = +4.2 cm
distance b/w the lenses. d = 8.4 cm
From thin lens equation image formed by lens1 will be at distance:
1/f = 1/do + 1/di
1/di = 1/f - 1/do
1/di = -1/4.2 - 1/3.5
image distance, di = -1.909 cm
this image will be object for lens 2, do = 8.4 - 1.909 = 6.491 cm
From thin lens equation image formed by lens2 will be at distance:
1/f = 1/do + 1/di
1/di = 1/f - 1/do
1/di = 1/4.2 - 1/(6.491)
image distance, di = 11.9 cm
A) final; image distance, di = 11.9 cm
B) since final image distance is positive, so image will be to left of lens2
C) since final image distance is positive, image will be real
D) image will be uprighted
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