How did the si became -10.8?

College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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How did the si became -10.8?

EXAMPLE 5.17 An object O0', 9.0 cm high, 27 cm in front of a concave lens of
focal length -18 cm. Determine the position and height of its iamge Il' (a) by
construction and (b) by computation.
SOLUTION (a) Choose the two convenient rays from O shown in Fig. 5.47..
1. A ray OP, parallel to the optical axis, is refracted outward in the direction D
as if it came from the principal focus F.
2. A ray through the optical center of the lens is drawn as a straight line OC.
Then II' is the image of 00'. Images formed by concave or divergent lenses
are virtual, erect, and smaller.
D
O'
F
I'
C
Fig.5.47. Example 5.17.
1
1
1
1
|(b)
or
So
Si
27 cm
Si
18 cm
or
S; = –10.8 cm = –11 cm
|
Since s, is negative, the image is virtual, and it is 11 cm in front of the lens.
10.8 cm
Si
Linear magnification
= 0.40
%3D
27 cm
|°s|
height of image = (0.40)(9.0 cm) = 3.6 cm
or
Transcribed Image Text:EXAMPLE 5.17 An object O0', 9.0 cm high, 27 cm in front of a concave lens of focal length -18 cm. Determine the position and height of its iamge Il' (a) by construction and (b) by computation. SOLUTION (a) Choose the two convenient rays from O shown in Fig. 5.47.. 1. A ray OP, parallel to the optical axis, is refracted outward in the direction D as if it came from the principal focus F. 2. A ray through the optical center of the lens is drawn as a straight line OC. Then II' is the image of 00'. Images formed by concave or divergent lenses are virtual, erect, and smaller. D O' F I' C Fig.5.47. Example 5.17. 1 1 1 1 |(b) or So Si 27 cm Si 18 cm or S; = –10.8 cm = –11 cm | Since s, is negative, the image is virtual, and it is 11 cm in front of the lens. 10.8 cm Si Linear magnification = 0.40 %3D 27 cm |°s| height of image = (0.40)(9.0 cm) = 3.6 cm or
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