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Sol7 - UAHuntsville Spring 2010 PH 113 Homework Solutions...

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UAHuntsville Spring, 2010 PH 113: Homework Solutions No. 7 Solution 41: Assume that the converging lens is a thin lens. Use ray tracing to locate the image. Solve: (a) The figure shows the ray-tracing diagram using the steps of Tactics Box 23.2. The three rays after refraction converge to give an image at s ʹஒ = 40 cm. The height of the image is h ʹஒ = 2 cm. (b) Using the thin-lens formula, 1 s + 1 ʹஒ s = 1 f 1 40 cm + 1 ʹஒ s = 1 20 cm 1 ʹஒ s = 1 40 cm ʹஒ s = 40 cm The image height is obtained from M = ʹஒ s s = 40 cm 40 cm = 1 The image is inverted and as tall as the object, that is, h ʹஒ = 2.0 cm. The values for h ʹஒ and s ʹஒ obtained in parts (a) and (b) agree.
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Solution 42: 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. After refraction, the three special rays converge and give an image 50 cm away from the converging lens. Thus, s ʹஒ = + 50 cm. The image is inverted and its height is 0.65 cm. (b) Using the thin-lens formula, 1 s + 1 ʹஒ s = 1 f 1 75 cm + 1 ʹஒ s = 1 30 cm 1 ʹஒ s = 1 50 cm ʹஒ s = 50 cm The image height is obtained from M = ʹஒ s s = 50 cm 75 cm = 2 3 The image height is ʹஒ h = Mh = 2 / 3 ( ) 1 cm ( ) = 0.67 cm. Because of the negative sign, the image is inverted. These results agree with those obtained in part (a).
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Solution 43: Visualize: Refer to the figure above: 1 s + 1 ʹஒ s = 1 f ʹஒ s = fs s f We are given f = 60 cm, s = 20 cm, and h = 1.0 cm.
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