24_InstSolManual_PDF_Part17

24_InstSolManual_PDF_Part17 - 2 16.0 cm 1 36.0 cm 5 11.1...

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24.50. Set Up: and Solve: (a) The image is 6.43 cm to the left of the lens. (b) (c) so the image is virtual. (d) so the image is erect. The principal ray diagram is sketched in Figure 24.50a. Figure 24.50 (a) The image is 4.12 cm to the left of the lens. (b) (c) so the image is virtual. (d) so the image is erect. The principal ray diagram is sketched in Figure 24.50b. 24.51. Set Up: The sign of determines whether the lens is converging or diverging. Solve: (a) and the lens is converging. (b) so the image is inverted. m , 0 0 y r 0 5 0 m 0 y 5 1 2.25 21 8.00 mm 2 5 18.0 mm. m 52 s r s 52 36.0 cm 16.0 cm 52 2.25. f . 0 f 5 ss r s 1 s r 5 1 16.0 cm 21 36.0 cm
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Unformatted text preview: 2 16.0 cm 1 36.0 cm 5 11.1 cm. s r 5 1 36.0 cm. s 5 16.0 cm. f 1 s 1 1 s r 5 1 f . m . s r , m 5 2 s r s 5 2 2 4.12 cm 7.00 cm 5 1 0.588 s r 5 sf s 2 f 5 1 7.00 cm 21 2 10.0 cm 2 7.00 cm 2 1 2 10.0 cm 2 5 2 4.12 cm. s 5 7.00 cm Image Object 2 3 1 F 1 F 2 (a) Image Object 1 3 2 F 2 F 1 (b) m . s r , m 5 2 s r s 5 2 2 6.43 cm 18.0 cm 5 1 0.357 s r 5 sf s 2 f 5 1 18.0 cm 21 2 10.0 cm 2 18.0 cm 2 1 2 10.0 cm 2 5 2 6.43 cm. s 5 18.0 cm f 5 2 10.0 cm. m 5 2 s r s . 1 s 1 1 s r 5 1 f Geometric Optics 24-17...
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