# Visualize refer to figure 2355 1 1 1 fs s s s f s f

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Visualize: Refer to Figure 23.55. 1 1 1 fs s s s f s f + = = We are given 60 cm, 20 cm, f s = − = and 1.0 cm. h = Solve: ( 60 cm)(20 cm) 15 cm 20 cm 60 cm fs s s f ′ = = = − + The negative sign means the image is behind the mirror; it is a virtual image. The magnification is 15 cm 20 cm 0.75. m s s = − = = This means the image is upright and has a height of (0.75)(1.0 cm) 0.75 cm. h mh ′ = = = Assess: Ray tracing will confirm these results.

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23.69. Model: Assume the lens is a thin lens and the thin-lens formula applies. Solve: Because we want to form an image of the spider on the wall, the image is real and we need a converging lens. That is, both s and s are positive. This also implies that the spider’s image is inverted, so 1 2 M s s = − = − . Using the thin-lens formula with 1 2 , s s ′ = 1 2 1 1 1 1 1 1 s s f s s f + = + = 3 1 3 s f s f = = We also know that the spider is 2.0 m from the wall, so s + s = 2.0 m = s + 1 2 s ( ) 1 3 4.0 m 133.3 cm s = = Thus, 1 3 44 cm f s = = and 2.0 m 1.33 m 0.67 m 67 cm. s ′ = = = We need a 44 cm focal length lens placed 67 cm from the wall.
23.70. Model: Assume the lens to be a thin lens. Solve: Because we want to form an image of the candle on the wall, we need a converging lens. We have 200 cm. s s + = Using the thin-lens formula, 1 1 1 1 1 1 200 cm 32 cm s s f s s + = + = ( ) 2 2 200 cm 6400 cm 0 s s + = The two solutions to this equation are s = 160 cm and 40 cm. When s = 160 cm, then 200 cm 160 cm 40 cm. s ′ = = The magnification is 40 cm 0.25 160 cm s M s = − = − = − so the image is inverted and its height is (2.0 cm)(0.25) = 0.50 cm. When s = 40 cm, then s = 200 cm 40 cm = 160 cm. The magnification is 160 cm 4 40 cm s M s = − = − = − so the image is again inverted and its height is (2.0 cm)(4) = 8.0 cm.

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23.71. Model: The eye is a converging lens and assume it is a thin lens. Solve: (a) The diameter of an adult eyeball is typically 4.0 cm. (b) The near point distance is approximately 10 inches 25 cm. (c) Using the thin-lens formula, 1 1 1 s s f + = 1 1 1 25 cm 4.0 cm f + = 1 29 100 cm f = f 3.4 cm
23.72. Model: Assume the projector lens is a thin lens. Solve: (a) The absolute value of the magnification of the lens is 98 cm 49 2 cm h M h = = = Because the projector forms a real image of a real object, the image will be inverted. Thus, 49 s M s = − = − 49 s s ′ = We also have s + s = 300 cm s + 49 s = 300 cm s = 6.0 cm s = 294 cm Using these values of s and s , we can find the focal length of the lens: 1 1 1 1 1 5.9 cm 6.0 cm 294 cm f f s s = + = + = (b) From part (a) the lens should be 6.0 cm from the slide.

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23.73. Visualize: The lens must be a converging lens for this scenario to happen, so we expect f to be positive. In the first case the upright image is virtual 2 ( 0) s ′ < and the object must be closer to the lens than the focal point. The lens is then moved backward past the focal point and the image becomes real 2 ( 0). s ′ > 1 1 1 ss f s s f s s + = = + We are given 1 10cm s = and 1 2. m = Solve: Since the first image is virtual, 0. s ′ < We are told the first magnification is 1 1 1 1 2 20 cm. m s s s = = − = − We can now find the focal length of the lens.
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