Explanation Given info: The focal length of first lens is 15.0 cm and the focal length of second lens is 10.0 cm . The distance of the object from the first lens is 50.0 cm . The distance between the first and second lens is 35.0 cm . The expression for the focal length of lens 1 is, 1 a 1 + 1 b 1 = 1 f 1 Here, a 1 is the distance of the object from the lens 1. b 1 is the distance of the image from lens 1. Rearrange the above equation for the value of b 1 . b 1 = f 1 a 1 a 1 − f 1 Substitute 15.0 cm for f 1 and 50.0 cm for a 1 in the above expression. b 1 = ( 15.0 cm ) ( 50.0 cm ) 50.0 cm − 15.0 cm = 21.4 cm The expression for magnification of the image is, m 1 = − b 1 a 1 Substitute 21.4 cm for b 1 and 50.0 cm for a 1 in the above expression. m 1 = − 21.4 cm 50.0 cm = − 0.43 The image of the first lens acts as the object for the second lens. The expression for the distance of the object from second lens is, a 2 = 35.0 cm − b 1 Substitute 21.4 cm for b 1 in the above expression. a 2 = 35.0 cm − 21.4 cm = 13.6 cm The expression for the focal length of lens 2 is, 1 a 2 + 1 b 2 = 1 f 2 Here, b 2 is the distance of the image from lens 2

Physics for Scientists and Engineers, Technology Update (No access codes included)

9th Edition

ISBN: 9781305116399

Author: Raymond A. Serway, John W. Jewett

Publisher: Cengage Learning

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Chapter 36, Problem 36.7OQ

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The factor of difference between the size of final image and the size of the object.

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Chapter 36, Problem 36.6OQ

Chapter 36, Problem 36.8OQ

Chapter 36 Solutions

Physics for Scientists and Engineers, Technology Update (No access codes included)

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The factor of difference between the size of final image and the size of the object.

Solution Summary: The author explains that the factor of difference between the size of final image and the object is 1.20.

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The factor of difference between the size of final image and the size of the object.

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Chapter 36 Solutions