# chapt34 - Chapter 34: Images Problem 1 A moth at about eye...

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Chapter 34: Images Problem 1 A moth at about eye level is 10 cm in front of a plane mirror; you are behind the moth, 30 cm from the mirror. For what distance must you focus your eyes to see the image of the moth in the mirror: that is, what is the distance between your eyes and the apparent position of the image? Answer 40 cm. Solution The image of the moth will be as far behind the mirror as the moth "object" is in front of the mirror. Therefore the image will be 10 cm behind the mirror and your eye will be 40 cm from that image. Problem 2 You look through a camera toward an image of a hummingbird in a plane mirror. The camera is 4.30 m in front of the mirror. The bird is at camera level, 5.00 m to your right and 3.30 m from the mirror. What is the distance between the camera and the apparent position of the bird’s image in the mirror? Answer 9.10 m Solution The situation is illustrated in Diagram 34-2. The reflected image of the hummingbird is as far behind the mirror as the hummingbird itself is in front of the mirror. The distance from the camera to the reflected image is the length of the hypotenuse of a right triangle with sides 7.60 m and 5.00 m. Therefore the distance is d = 7.60 m ( ) 2 + 5.00 m ( ) 2 = 9.10 m 4.30 m 5.00 m 3.30 m 3.30 m Camera Hummingbird Reflected Image of Hummingbird Diagram 34-2

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34-2 Chapter 34: Images Problem 4 In Fig. 34-32, an isotropic point source of light S is positioned at distance d from a viewing screen A and the light intensity I P at point P (level with S) is measured. Then a plane mirror M is placed bethind S at distance d. By how much is I P multiplied by the presence of the mirror? Answer 10/9 S P M A d d Figure 34-32 Solution As a result of the light reflecting from the mirror there will be a reflected image of the source located distance d behind the mirror or distance 3d from point P. The intensity of the light from the reflected image will be 1/9 the intensity from S itself since the intensity of an isotropic point source of light is inversely proportional to the square of the distance from the source. Therefore the total intensity is I P + I P /9 = 10I P /9. Problem 5 Figure 34-33 shows a small lightbulb suspended at distance d 1 = 250 cm above the surface of the water in a swimming pool where the water depth is d 2 = 200 cm. The bottom of the pool is a large mirror. How far below the mirror’s surface is the image of the bulb? (Hint: Construct a diagram of two rays like that of Fig. 34-3, but take into account the bending of light rays by refraction. Assume that the rays are close to a vertical axis through the bulb, and use the small-angle approximation in which sin θ tan θ θ .) Answer 351 cm d 1 x w θ 1 d 2 θ 1 θ 1 θ 2 θ 2 I Figure 34-33 Solution The image is located a distance x below the mirror. Referring to Figure 34-33, we see that d 2 + x = w tan θ 1 ( ) and w = d 1 tan 1 ( ) + 2 d 2 tan 2 ( ) = tan 1 ( ) d 1 + 2 d 2 tan 2 ( ) tan 1 ( ) For small angles,
Chapter 34: Images 34-3 tan θ 2 ( ) tan 1 ( ) sin 2 ( ) sin 1 ( ) = n 1 n 2 Combining these three equations yields x = d 1 + 2 d 2 n 1 n 2 d 2 = 250 cm + 2 200 cm ( ) 1 1.33 200 cm = 351 cm

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## This note was uploaded on 08/03/2009 for the course PHYSICS 2049 taught by Professor C during the Spring '09 term at Vanderbilt.

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chapt34 - Chapter 34: Images Problem 1 A moth at about eye...

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