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(a) An object is placed 29.0 cm from the lens. Construct a ray diagram, find the image distance, and describe the image. SOLUTION Conceptualize Because the lens is diverging, the focal length is negative ✓ Categorize Because the lens is diverging, we expect it to form an upright, reduced Analyze Find the image distance (in cm) by using the following equation: 9 q= M = - Find the magnification of the image from the following equation: -음. f P (b) An object is placed 10.0 cm from the lens. Construct a ray diagram, find the image distance, and describe the image. SOLUTION The ray diagram for this situation is shown in figure (b). Analyze Find the image distance (in cm) by using the following equation: M = cm 1 97 q= Find the magnification of the image from the following equation: 1 9 q= (c) An object is placed 6.00 cm from the lens. Construct a ray diagram, find the image distance, and describe the image. SOLUTION Analyze Find the image distance (in cm) by using the following equation: M = cm . The ray diagram for this situation is shown in figure (a). , virtual image for any object position. cm Find the magnification of the image from the following equation:

(a) An object is placed 29.0 cm from the lens. Construct a ray diagram, find the image distance, and describe the image. SOLUTION Conceptualize Because the lens is diverging, the focal length is negative ✓ Categorize Because the lens is diverging, we expect it to form an upright, reduced Analyze Find the image distance (in cm) by using the following equation: 9 q= M = - Find the magnification of the image from the following equation: -음. f P (b) An object is placed 10.0 cm from the lens. Construct a ray diagram, find the image distance, and describe the image. SOLUTION The ray diagram for this situation is shown in figure (b). Analyze Find the image distance (in cm) by using the following equation: M = cm 1 97 q= Find the magnification of the image from the following equation: 1 9 q= (c) An object is placed 6.00 cm from the lens. Construct a ray diagram, find the image distance, and describe the image. SOLUTION Analyze Find the image distance (in cm) by using the following equation: M = cm . The ray diagram for this situation is shown in figure (a). , virtual image for any object position. cm Find the magnification of the image from the following equation:

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Images Formed by a Diverging Lens A diverging lens has a focal length of 10.0 cm. An image is formed by a diverging lens. The object is farther from the lens than the focal point. 0 9 10.0 cm/ The object is at the focal point. O, F₁ 10.0 cm -9 Fo The object is closer to the lens than the focal point. OI 10.0 cm
Transcribed Image Text:Images Formed by a Diverging Lens A diverging lens has a focal length of 10.0 cm. An image is formed by a diverging lens. The object is farther from the lens than the focal point. 0 9 10.0 cm/ The object is at the focal point. O, F₁ 10.0 cm -9 Fo The object is closer to the lens than the focal point. OI 10.0 cm
(a) An object is placed 29.0 cm from the lens. Construct a ray diagram, find the image distance, and describe the image. SOLUTION Conceptualize Because the lens is diverging, the focal length is negative Categorize Because the lens is diverging, we expect it to form an upright, reduced Analyze Find the image distance (in cm) by using the following equation: 1 = 9 = Find the magnification of the image from the following equation: M = - = P EXERCISE 1 P (b) An object is placed 10.0 cm from the lens. Construct a ray diagram, find the image distance, and describe the image. SOLUTION The ray diagram for this situation is shown in figure (b). Analyze Find the image distance (in cm) by using the following equation: 1 1 1 9 P 9 = = Find the magnification of the image from the following equation: 9 M = - P 1 9 q= cm (c) An object is placed 6.00 cm from the lens. Construct a ray diagram, find the image distance, and describe the image. SOLUTION Analyze Find the image distance (in cm) by using the following equation: M = object distance focal length cm 1 1 P Find the magnification of the image from the following equation: cm Finalize For all three object positions, the image position is negative and the magnification is a positive number smaller than 1, which confirms that the image is virtual, smaller than the object, and upright. . The ray diagram for this situation is shown in figure (a). , virtual image for any object position. An object is placed in front of a diverging lens. If the image is -6.00 cm from the lens, and the magnified image is 25% of the object, what are the object distance (in cm) and focal length (in cm) of the lens? cm cm
Transcribed Image Text:(a) An object is placed 29.0 cm from the lens. Construct a ray diagram, find the image distance, and describe the image. SOLUTION Conceptualize Because the lens is diverging, the focal length is negative Categorize Because the lens is diverging, we expect it to form an upright, reduced Analyze Find the image distance (in cm) by using the following equation: 1 = 9 = Find the magnification of the image from the following equation: M = - = P EXERCISE 1 P (b) An object is placed 10.0 cm from the lens. Construct a ray diagram, find the image distance, and describe the image. SOLUTION The ray diagram for this situation is shown in figure (b). Analyze Find the image distance (in cm) by using the following equation: 1 1 1 9 P 9 = = Find the magnification of the image from the following equation: 9 M = - P 1 9 q= cm (c) An object is placed 6.00 cm from the lens. Construct a ray diagram, find the image distance, and describe the image. SOLUTION Analyze Find the image distance (in cm) by using the following equation: M = object distance focal length cm 1 1 P Find the magnification of the image from the following equation: cm Finalize For all three object positions, the image position is negative and the magnification is a positive number smaller than 1, which confirms that the image is virtual, smaller than the object, and upright. . The ray diagram for this situation is shown in figure (a). , virtual image for any object position. An object is placed in front of a diverging lens. If the image is -6.00 cm from the lens, and the magnified image is 25% of the object, what are the object distance (in cm) and focal length (in cm) of the lens? cm cm
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