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Object Image 100 50 50 cm Focal peint f The image, which can be seen by looking through the lens, is one-third the size of the object and upright.

Object Image 100 50 50 cm Focal peint f The image, which can be seen by looking through the lens, is one-third the size of the object and upright.

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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A diverging lens with a focal length of 50 cm is placed 100 cm from a flower. Where is the image? What is its magnification? PREPARE The flower is in the object plane. We use ray tracing to locate the image. Then we use Equation 18.9 O to find the magnification. SOLVE Figure 18.40 O shows the ray-tracing diagram. The three special rays (labeled a, b, and c to match Tactics Box 18.3 D do not converge. However, they can be traced backward to an intersection = 33 cm to the left of the lens. Because the rays appear to diverge from the image, this is a virtual image and s'< 0. The magnification is -33 cm m = = 0.33 100 cm FIGURE 18.40 Ray-tracing diagram for a diverging lens. a Object Image 100 50 50 cm Focal point f The image, which can be seen by looking through the lens, is one-third the size of the object and upright. ASSESS The three special rays appear diverge from a single point, and the image orientation and size in our ray-tracing diagram match the computed magnification -the image is upright and smaller than the object-so we have confidence in our results.
Transcribed Image Text:A diverging lens with a focal length of 50 cm is placed 100 cm from a flower. Where is the image? What is its magnification? PREPARE The flower is in the object plane. We use ray tracing to locate the image. Then we use Equation 18.9 O to find the magnification. SOLVE Figure 18.40 O shows the ray-tracing diagram. The three special rays (labeled a, b, and c to match Tactics Box 18.3 D do not converge. However, they can be traced backward to an intersection = 33 cm to the left of the lens. Because the rays appear to diverge from the image, this is a virtual image and s'< 0. The magnification is -33 cm m = = 0.33 100 cm FIGURE 18.40 Ray-tracing diagram for a diverging lens. a Object Image 100 50 50 cm Focal point f The image, which can be seen by looking through the lens, is one-third the size of the object and upright. ASSESS The three special rays appear diverge from a single point, and the image orientation and size in our ray-tracing diagram match the computed magnification -the image is upright and smaller than the object-so we have confidence in our results.
Match the answers with questions based on Example 18.9. (Hint: Be careful with signs.) v What is the object distance? A. -50 cm v What is the image distance? B. -100 cm v What is the focal length of the lens? C. 100 cm v What is the magnification? D.0 E. 50 cm F. -0.33 G.0.33 H.33 cm I. -33 cm
Transcribed Image Text:Match the answers with questions based on Example 18.9. (Hint: Be careful with signs.) v What is the object distance? A. -50 cm v What is the image distance? B. -100 cm v What is the focal length of the lens? C. 100 cm v What is the magnification? D.0 E. 50 cm F. -0.33 G.0.33 H.33 cm I. -33 cm
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