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Tutorial Exercise A man standing 1.61 m in front of a shaving mirror produces an inverted image 16.9 cm in front of it. How close to the mirror should he stand if he wants to form an upright image of his chin that is twice the chin's actual size? Step 1 When the man stands 1.61 m in front of the mirror, the object is located at a distance p = +1.61 m 161 cm. We are given that a real image is formed at the image distance of q = +16.9 cm. We find the focal length f of the mirror from the mirror equation, 1/f 1/p+1/q. Solving for the focal length, we obtain the following. pq p+q For the product in the numerator, pg= cm², and for the sum in the denominator, cm, and thus for the focal length, we obtain Submit pq p+q Skip (you cannot come back) cm.

Tutorial Exercise A man standing 1.61 m in front of a shaving mirror produces an inverted image 16.9 cm in front of it. How close to the mirror should he stand if he wants to form an upright image of his chin that is twice the chin's actual size? Step 1 When the man stands 1.61 m in front of the mirror, the object is located at a distance p = +1.61 m 161 cm. We are given that a real image is formed at the image distance of q = +16.9 cm. We find the focal length f of the mirror from the mirror equation, 1/f 1/p+1/q. Solving for the focal length, we obtain the following. pq p+q For the product in the numerator, pg= cm², and for the sum in the denominator, cm, and thus for the focal length, we obtain Submit pq p+q Skip (you cannot come back) cm.

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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Tutorial Exercise A man standing 1.61 m in front of a shaving mirror produces an inverted image 16.9 cm in front of it. How close to the mirror should he stand if he wants to form an upright image of his chin that is twice the chin's actual size? Step 1 When the man stands 1.61 m in front of the mirror, the object is located at a distance p = +1.61 m 161 cm. We are given that a real image is formed at the image distance of q = +16.9 cm. We find the focal length f of the mirror from the mirror equation, 1/f 1/p+1/q. Solving for the focal length, we obtain the following. pq p+q For the product in the numerator, pg= cm², and for the sum in the denominator, cm, and thus for the focal length, we obtain Submit pq p+q Skip (you cannot come back) cm.
Transcribed Image Text:Tutorial Exercise A man standing 1.61 m in front of a shaving mirror produces an inverted image 16.9 cm in front of it. How close to the mirror should he stand if he wants to form an upright image of his chin that is twice the chin's actual size? Step 1 When the man stands 1.61 m in front of the mirror, the object is located at a distance p = +1.61 m 161 cm. We are given that a real image is formed at the image distance of q = +16.9 cm. We find the focal length f of the mirror from the mirror equation, 1/f 1/p+1/q. Solving for the focal length, we obtain the following. pq p+q For the product in the numerator, pg= cm², and for the sum in the denominator, cm, and thus for the focal length, we obtain Submit pq p+q Skip (you cannot come back) cm.
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