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
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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Can you match using this image please?
![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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F63df6dc9-e0c2-460f-ad25-c0b34955e7ea%2F2d039457-1fd7-4222-9d8e-fe72087648a3%2Fiyh2tjm_processed.jpeg&w=3840&q=75)
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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F63df6dc9-e0c2-460f-ad25-c0b34955e7ea%2F2d039457-1fd7-4222-9d8e-fe72087648a3%2Fju82ihe_processed.jpeg&w=3840&q=75)
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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