College Physics, Young,Geller CH24-CH30

College Physics, Young,Geller CH24-CH30 - m 5 2 s r s 5 2...

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Solve: (a) or The lens can be 14.2 cm or 3.8 cm from the object. (b) 24.65. Set Up: For the second image, the image formed by the mirror serves as the object for the lens. For the mirror, For the lens, The center of curvature of the mirror is to the right of the mirror vertex. gives Solve: (a) The principal-ray diagrams from the two images are sketched in Figures 24.65a-b. In Figure 24.65b, only the image formed by the mirror is shown. This image is at the location of the candle so the principal ray diagram that shows the image formation when the image of the mirror serves as the object for the lens is analogous to that in Fig- ure 24.65a and is not drawn. Figure 24.65 (b) Image formed by the light that passes directly through the lens: The candle is 85.0 cm to the left of the lens.
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Unformatted text preview: m 5 2 s r s 5 2 51.3 cm 85.0 cm 5 2 0.604. s r 5 sf s 2 f 5 1 85.0 cm 21 32.0 cm 2 85.0 cm 2 32.0 cm 5 1 51.3 cm. (a) Object Image F 1 F 2 1 2 3 Object (b) Image C 1 4 2 s r 5 sf s 2 f . 1 s 1 1 s r 5 1 f R 5 2 f m 5 20.0 cm f 5 32.0 cm. f m 5 1 10.0 cm. m 5 2 s r s 5 2 14.2 cm 3.8 cm 5 2 3.74. s r 5 18.0 cm 2 3.8 cm 5 14.2 cm. s 5 3.8 cm: m 5 2 s r s 5 2 3.8 cm 14.2 cm 5 2 0.268. s r 5 18.0 cm 2 14.2 cm 5 3.8 cm. s 5 14.2 cm: s 5 3.8 cm. s 5 14.2 cm s 5 1 9.00 6 5.20 2 cm. s 5 1 2 A 18.0 6 "1 18.0 2 2 2 4 1 54.0 2 B cm. s 2 2 1 18.0 cm 2 s 1 54.0 cm 2 . s 1 18.0 cm 2 s 2 5 54.0 cm 2 . 18.0 cm s 1 18.0 cm 2 s 2 5 1 3.00 cm . 1 s 1 1 18.0 cm 2 s 5 1 f . 1 s 1 1 s r 5 1 f . Geometric Optics 24-23...
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