B using the thin lens formula 1 1 1 s s f 1 1 1 15 cm

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(b) Using the thin-lens formula, 1 1 1 s s f + = 1 1 1 15 cm 20 cm s + = 1 1 60 cm s = − 60 cm s ′ = − The image height is obtained from 60 cm 4 15 cm s M s = − = − = + Thus, the image is 4 times larger than the object or ( ) 4 4 2.0 cm 8.0 cm. h Mh h ′ = = = = The image is upright. These values agree with those obtained in part (a).

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23.64. Model: Use ray tracing to locate the image. Assume the converging lens is a thin lens. Solve: (a) The figure shows the ray-tracing diagram using the steps of Tactics Box 23.2. After refraction, the three special rays converge and give an image 50 cm away from the converging lens. Thus, s = + 50 cm. The image is inverted and its height is 0.65 cm. (b) Using the thin-lens formula, 1 1 1 1 1 1 75 cm 30 cm s s f s + = + = 1 1 50 cm s = 50 cm s ′ = The image height is obtained from 50 cm 2 75 cm 3 s M s = − = − = − The image height is ( )( ) 2/3 1 cm 0.67 cm. h Mh ′ = = − = − Because of the negative sign, the image is inverted. These results agree with those obtained in part (a).
23.65. Model: Use ray tracing to locate the image. Assume the diverging lens is a thin lens. Solve: (a) The figure shows the ray-tracing diagram using the steps of Tactics Box 23.3. After refraction, the three special rays do not converge. The rays, on the other hand, appear to meet at a point that is 8.5 cm on the same side of the lens as the object. So 8.5 cm. s ′ = − The image is upright and has a height of 1.1 cm. (b) Using the thin-lens formula, 1 1 1 s s f + = 1 1 1 15 cm 20 cm s + = 1 7 60 cm 8.6 cm 60 cm 7 s s = − = − = − The image height is obtained from ( ) 60/7 cm 4 0.57 15 cm 7 s M s = − = − = + = Thus, the image is 0.57 times larger than the object, or (0.57)(2.0 cm) 1.14 cm. h Mh ′ = = = The image is upright because M is positive. These values agree, within measurement accuracy, with those obtained in part (a).

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23.66. Model: Use ray tracing to locate the image. Assume the diverging lens is a thin lens. Solve: (a) The figure shows the ray-tracing diagram using the steps of Tactics Box 23.3. After refraction from the diverging lens, the three special rays do not converge. However, the rays appear to meet at a point that is 20 cm on the same side as the object. So 20 cm. s ′ = − The image is upright and has a height of 0.3 cm. (b) Using the thin-lens formula, 1 1 1 1 1 1 30 cm 60 cm 20 cm s f s = = = − 20 cm s ′ = − The image height is obtained from 20 cm 1 0.33 60 cm 3 s M s = − = − = = Thus, ( )( ) 0.33 1.0 cm 0.33 cm, h Mh ′ = = = and the image is upright because M is positive. These values for s and h agree with those obtained in part (a).
23.67. Visualize: Refer to Figure 23.57. 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) 30 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 30 cm 20 cm 1.5. m s s = − = = This means the image is upright and has a height of (1.5)(1.0 cm) 1.5 cm. h mh ′ = = = Assess: Ray tracing will confirm these results.

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23.68.
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