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
1st Edition
ISBN:9781938168000
Author:Paul Peter Urone, Roger Hinrichs
Publisher:Paul Peter Urone, Roger Hinrichs
Chapter25: Geometric Optics
Section: Chapter Questions
Problem 1PE: Suppose a man stands in front of a mirror as shown in Figure 25.50. His eyes are 1.65 m above the...
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