iPad 1:20 AM 61% edugen.wileyplus com Assignment signment NEXT FULL SCREEN PRINT

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iPad 1:20 AM 61% edugen.wileyplus com Assignment signment NEXT FULL SCREEN PRINTER VERSION .BACK Chapter 25, Problem 29 An object is located 16.6 cm in front of a conve mirror, the image being 9.98 cm behind the mirror. A second object, twice as tall as the first one, is placed in front of the mirror, but at a different location. The image of this second object has the same height as the other image. How far in front of the mirror is the second object located? 20 Number units Click if you would like to show work for this question QRen Show Work By accessing this Question Assistance, you will learn while you eam points based on the Point Potential Policy set by your Question Attempts: o of 4 used SAVE OR LATER SUSMITANSwER Earn Maximum Points available only if you n four nswer this question correctly attempts or less. Copyright 2000-2017 by John Wiley & Sons, Inc. or related companies. All rights reserved

Explanation / Answer

Given,

o1 = 16.6 cm ; i1 = -9.98 cm ;

h2 = 2h1 ; h2' = h1'

We know from lens equation that,

1/f = 1/i + 1/o

f1 = i1 x o1 / ( i1 + o1) = -9.98 x 16.6 / ( -9.98 + 16.6 ) = -25.03 cm

m1 = -i1/o1 = h1'/h1 => m = -(-9.98)/16.6 = 0.601

h' = 0.601 h

Since second object, twice as tall as the first one and image of this second object has the same height as the other image

m = -i2/o2 = 0.301 => i2 = 0.301 o2

again from lens equation

1/f = 1/i + 1/o

1/-25.05 = 1/0.301 o2 + 1/o2 = 1/o2 ( 1/0.301 + 1) = 4.32 / o2

o2 = 4.32 x -25.03 = - 108.13 cm

Hence, o2 = -108.13 cm

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