A converging lens has a focal length of f- 18.0 cm. (a) An object is placed 54.0 cm from the lens. Construct a ray diagram, find the image distance, and describe the image. SOLUTION Conceptualize Because the lens is converging, the focal length is positive Categorize Because the object distance is larger than the focal length, we expect the image to be real Analyze Find the image distance in cm by using the following equation (If you need to use co or -00, enter INFINITY or -INFINITY, respectively.): 111 9 M Find the magnification of the image from the following equation: q= Р cm Finalize the positive sign for the image distance tells us that the image is indeed real and on the back side of the lens. The magnification of the image tells us that the image is reduced in height by one half, and the negative sign for M tells us that the image is inverted (b) An object is placed 18.0 cm from the lens. Find the image distance and describe the image. Categorize Because the object is at the focal point, we expect the image to be infinitely far away. Analyze Find the image distance in cm (If you need to use co or -00, enter INFINITY or -INFINITY, respectively.): 1 1 q= cm Finalize This result means that rays originating from an object positioned at the focal point of a lens are refracted so that the image is formed at an infinite distance from the lens; that is, the rays travel parallel M- (c) An object is placed 9.0 cm from the lens. Construct a ray diagram, find the image distance, and describe the image. SOLUTION Categorize Because the object distance is smaller than the focal length, we expect the image to be virtual. The ray diagram for this situation is shown in figure (b). Analyze Find the image distance in cm using this equation (If you need to use co or-co, enter INFINITY or -INFINITY, respectively.): ✔ (see this table). We expect the possibilities of both real and virtual images. Find the magnification of the image using this equation: 9. cm ✔The ray diagram for this situation is shown in figure (a). cm (a) Find the location of the image (in cm). cm Finalize the negative image distance tells us that the image is virtual and formed on the side of the lens from which the light is incident, the front side. The image is [enlarged, and the positive sign for M tells us that the image is upright. EXERCISE Suppose the image of an object is upright and magnified 1.50 times when the object is placed 13.0 cm from a particular converging lens. (b) Find the focal length of the lens (in cm). to one another after refraction.

College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter23: Mirrors And Lenses
Section: Chapter Questions
Problem 7CQ: Suppose you want to use a converging lens to project the image of two trees onto a screen. One tree...
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Images Formed by a Converging Lens
An image is formed by a converging lens.
The object is farther from the
lens than the focal point.
a
F₁
f
9
F₂
The object is closer to
the lens than the focal
point.
I, F, O
b
F₂
Transcribed Image Text:Images Formed by a Converging Lens An image is formed by a converging lens. The object is farther from the lens than the focal point. a F₁ f 9 F₂ The object is closer to the lens than the focal point. I, F, O b F₂
A converging lens has a focal length of f = 18.0 cm.
(a) An object is placed 54.0 cm from the lens. Construct a ray diagram, find the image distance, and describe the image.
SOLUTION
Conceptualize Because the lens is converging, the focal length is positive ✓✓ (see this table). We expect the possibilities of both real and virtual images.
Categorize Because the object distance is larger than the focal length, we expect the image to be real
The ray diagram for this situation is shown in figure (a).
Analyze Find the image distance in cm by using the following equation (If you need to use ∞ or -co, enter INFINITY or -INFINITY, respectively.):
1 _1
f
Р
1
9
=
q=
Find the magnification of the image from the following equation:
M =
Finalize The positive sign for the image distance tells us that the image is indeed real and on the back side of the lens. The magnification of the image tells us that the image is reduced in height by one half, and the negative sign for M tells us that the image is inverted
(b) An object is placed 18.0 cm from the lens. Find the image distance and describe the image.
Categorize Because the object is at the focal point, we expect the image to be infinitely far away.
Analyze Find the image distance in cm (If you need to use ∞ or -00, enter INFINITY or -INFINITY, respectively.):
1 1
9
q=
P
1
q
1
f P
Finalize This result means that rays originating from an object positioned at the focal point of a lens are refracted so that the image is formed at an infinite distance from the lens; that is, the rays travel parallel
q=
cm
(c) An object is placed 9.0 cm from the lens. Construct a ray diagram, find the image distance, and describe the image.
SOLUTION
Categorize Because the object distance is smaller than the focal length, we expect the image to be virtual. The ray diagram for this situation is shown in figure (b).
Analyze Find the image distance in cm using this equation (If you need to use ∞ or -00, enter INFINITY or -INFINITY, respectively.):
1 1
f P
M = -
cm
9
P
Find the magnification of the image using this equation:
cm
Finalize The negative image distance tells us that the image is virtual and formed on the side of the lens from which the light is incident, the front side. The image is enlarged
cm
EXERCISE
Suppose the image of an object is upright and magnified 1.50 times when the object is placed 13.0 cm from a particular converging lens.
(a) Find the location of the image (in cm).
cm
(b) Find the focal length of the lens (in cm).
to one another after refraction.
, and the positive sign for M tells us that the image is upright.
Transcribed Image Text:A converging lens has a focal length of f = 18.0 cm. (a) An object is placed 54.0 cm from the lens. Construct a ray diagram, find the image distance, and describe the image. SOLUTION Conceptualize Because the lens is converging, the focal length is positive ✓✓ (see this table). We expect the possibilities of both real and virtual images. Categorize Because the object distance is larger than the focal length, we expect the image to be real The ray diagram for this situation is shown in figure (a). Analyze Find the image distance in cm by using the following equation (If you need to use ∞ or -co, enter INFINITY or -INFINITY, respectively.): 1 _1 f Р 1 9 = q= Find the magnification of the image from the following equation: M = Finalize The positive sign for the image distance tells us that the image is indeed real and on the back side of the lens. The magnification of the image tells us that the image is reduced in height by one half, and the negative sign for M tells us that the image is inverted (b) An object is placed 18.0 cm from the lens. Find the image distance and describe the image. Categorize Because the object is at the focal point, we expect the image to be infinitely far away. Analyze Find the image distance in cm (If you need to use ∞ or -00, enter INFINITY or -INFINITY, respectively.): 1 1 9 q= P 1 q 1 f P Finalize This result means that rays originating from an object positioned at the focal point of a lens are refracted so that the image is formed at an infinite distance from the lens; that is, the rays travel parallel q= cm (c) An object is placed 9.0 cm from the lens. Construct a ray diagram, find the image distance, and describe the image. SOLUTION Categorize Because the object distance is smaller than the focal length, we expect the image to be virtual. The ray diagram for this situation is shown in figure (b). Analyze Find the image distance in cm using this equation (If you need to use ∞ or -00, enter INFINITY or -INFINITY, respectively.): 1 1 f P M = - cm 9 P Find the magnification of the image using this equation: cm Finalize The negative image distance tells us that the image is virtual and formed on the side of the lens from which the light is incident, the front side. The image is enlarged cm EXERCISE Suppose the image of an object is upright and magnified 1.50 times when the object is placed 13.0 cm from a particular converging lens. (a) Find the location of the image (in cm). cm (b) Find the focal length of the lens (in cm). to one another after refraction. , and the positive sign for M tells us that the image is upright.
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