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Match the answers with questions based on Example 18.8. (Hint: Be careful with signs.) v What is the object distance? v What is the image distance? v What is the focal length of the lens? v What is the magnification? A. 0 B. -4.0 cm С. -6.0 ст D. 4.0 cm E. 12 cm F. 6.0 cm G.-3 H 3

Match the answers with questions based on Example 18.8. (Hint: Be careful with signs.) v What is the object distance? v What is the image distance? v What is the focal length of the lens? v What is the magnification? A. 0 B. -4.0 cm С. -6.0 ст D. 4.0 cm E. 12 cm F. 6.0 cm G.-3 H 3

College Physics
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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18.8 Finding the Image for a Converging Lens To see a flower better, you hold a 6.0-cm-focal-length magnifying glass 4.0 cm from the flower. What is the magnification? PREPARE The flower is in the object plane. We use ray tracing to locate the image. Once the image distance is known, we can use Equation 18.9 O to find the magnification. SOLVE Figure 18.38 O shows the ray-tracing diagram. The three special rays diverge from the lens, but we can use a straightedge to extend the rays backward to the point from which they diverge. This point, the image point, s'= -12 cm. From Equation 18.9 O the magnification is seen to be 12 cm to the left of the lens. Because this is a virtual image, the image distance is s' 12 cm 3.0 4.0 cm FIGURE 18.38 Ray-tracing diagram for a magnifying glass. Trace these rays back to the image location. Focal point Image 12 cm 8 8 Object f ASSESS The three special rays converge to a single point, and the image orientation and size in our ray-tracing diagram match the computed magnification-the image is upright and much larger than the object-so we have confidence in our results.
Transcribed Image Text:18.8 Finding the Image for a Converging Lens To see a flower better, you hold a 6.0-cm-focal-length magnifying glass 4.0 cm from the flower. What is the magnification? PREPARE The flower is in the object plane. We use ray tracing to locate the image. Once the image distance is known, we can use Equation 18.9 O to find the magnification. SOLVE Figure 18.38 O shows the ray-tracing diagram. The three special rays diverge from the lens, but we can use a straightedge to extend the rays backward to the point from which they diverge. This point, the image point, s'= -12 cm. From Equation 18.9 O the magnification is seen to be 12 cm to the left of the lens. Because this is a virtual image, the image distance is s' 12 cm 3.0 4.0 cm FIGURE 18.38 Ray-tracing diagram for a magnifying glass. Trace these rays back to the image location. Focal point Image 12 cm 8 8 Object f ASSESS The three special rays converge to a single point, and the image orientation and size in our ray-tracing diagram match the computed magnification-the image is upright and much larger than the object-so we have confidence in our results.
Match the answers with questions based on Example 18.8. (Hint: Be careful with signs.) v What is the object distance? A. 0 v What is the image distance? B. 4.0 cm v What is the focal length of the lens? C. -6.0 cm v What is the magnification? D. 4.0 cm Е. 12 сm F. 6.0 cm G.-3 Н.З 1. -12 cm
Transcribed Image Text:Match the answers with questions based on Example 18.8. (Hint: Be careful with signs.) v What is the object distance? A. 0 v What is the image distance? B. 4.0 cm v What is the focal length of the lens? C. -6.0 cm v What is the magnification? D. 4.0 cm Е. 12 сm F. 6.0 cm G.-3 Н.З 1. -12 cm
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