So the image is real so the image is inverted reflect

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so the image is real. so the image is inverted. Reflect: The lens is converging and has a very short focal length. As long as the object is farther than 7.0 mm from the eye, the lens forms a real image. 24.40. Set Up: Solve: (a) and (b) (c) so the image is real. so the image is inverted. 24.41. Set Up: If the lens is reversed, and m 5 y r y 5 2 s r s . 1 s 1 1 s r 5 1 f . R 2 S ` . R 1 5 1 13.0 cm R 2 5 2 13.0 cm. R 1 S ` . 1 f 5 1 n 2 1 2 1 1 R 1 2 1 R 2 2 . m , 0 s r . 0 y r 5 my 5 1 2 0.076 21 8.0 mm 2 5 2 0.61 mm. m 5 2 s r s 5 2 1.9 cm 25 cm 5 2 0.076. s r 5 sf s 2 f 5 1 25 cm 21 1.8 cm 2 25 cm 2 1.8 cm 5 1.9 cm 5 19 mm. 1 s r 5 1 f 2 1 s 5 s 2 f sf . R 2 5 18.6 mm. 1 R 2 5 1 R 1 2 1 f 1 n 2 1 2 5 1 1 5.0 mm 2 1 1 18.0 mm 21 0.38 2 1 f 1 n 2 1 2 5 1 R 1 2 1 R 2 . m 5 y r y 5 2 s r s . 1 s 1 1 s r 5 1 f . R 1 5 1 5.0 mm. 1 f 5 1 n 2 1 2 1 1 R 1 2 1 R 2 2 . m , 0 s r . 0 0.44 cm 5 4.4 mm. y r 5 my 5 1 2 0.0273 21 16 cm 2 5 m 5 2 s r s 5 2 0.82 cm 30.0 cm 5 2 0.0273. s r 5 sf s 2 f 5 1 30.0 cm 21 0.80 cm 2 30.0 cm 2 0.80 cm 5 0.82 cm 5 8.2 mm. 1 s r 5 1 f 2 1 s 5 s 2 f sf . R 5 2 1 n 2 1 2 f 5 2 1 0.44 21 8.0 mm 2 5 7.0 mm. 1 f 5 1 n 2 1 2 1 1 R 1 2 1 R 2 2 5 1 n 2 1 2 1 1 R 2 1 2 R 2 5 2 1 n 2 1 2 R . m 5 y r y 5 2 s r s . 1 s 1 1 s r 5 1 f . R 2 5 2 R . R 1 5 R 1 f 5 1 n 2 1 2 1 1 R 1 2 1 R 2 2 . s r 5 sf s 2 f 5 1 18.0 cm 21 2 60.0 cm 2 18.0 cm 1 60.0 cm 5 2 13.8 cm. f 5 2 60.0 cm. 1 f 5 1 0.5 2 1 1 2 10.0 cm 2 1 2 15.0 cm 2 s r 5 sf s 2 f 5 1 18.0 cm 21 2 12.0 cm 2 18.0 cm 1 12.0 cm 5 2 7.2 cm. f 5 2 12.0 cm. 1 f 5 1 0.5 2 1 1 2 15.0 cm 2 1 10.0 cm 2 s r 5 sf s 2 f 5 1 18.0 cm 21 20.0 cm 2 18.0 cm 2 20.0 cm 5 2 180 cm. f 5 1 20.0 cm. 1 f 5 1 0.5 2 1 1 10.0 cm 2 1 ` 2 24-12 Chapter 24
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Solve: (a) and The image is 107 cm to the right of the lens and is 17.8 mm tall. The image is real and inverted. (b) and The image is the same as in part (a). Reflect: Reversing a lens does not change the focal length of the lens (Problem 24.44). 24.42. Set Up: For a distant object the image is at the focal point of the lens. Therefore, For the double-convex lens, and where Solve: 24.43. Set Up: Each surface could have a positive or a negative radius of curvature, so there are four distinct shapes. The four possible shapes are sketched in Figure 24.43. Figure 24.43 Solve: lens a : lens b : lens c : lens d : Each lens could also be reversed, but that doesn’t change its focal length. 24.44. Set Up: When we reverse the lens, and Solve: If is the focal length with the lens reversed, as was to be shown. 1 f r 5 1 n 2 1 2 1 1 2 R 2 2 1 1 2 R 1 22 5 1 n 2 1 2 1 1 R 1 2 1 R 2 2 5 1 f , f r 1 f 5 1 n 2 1 2 1 1 R 1 2 1 R 2 2 . R 2 S 2 R 1 . R 1 S 2 R 2 f d 5 2 13.3 cm. R 2 5 2 8.00 cm. R 1 5 2 4.00 cm, f c 5 2 4.44 cm. R 2 5 1 8.00 cm. R 1 5 2 4.00 cm, f b 5 1 4.44 cm. R 2 5 2 8.00 cm. R 1 5 1 4.00 cm, f a 5 1 13.3 cm. R 2 5 1 8.00 cm. R 1 5 1 4.00 cm, (a) (b) (c) (d) 1 f 5 1 n 2 1 2 1 1 R 1 2 1 R 2 2 5 1 0.60 2 1 1 6 4.00 cm 2 1 6 8.00 cm 2 . n 5 R 2 f 1 1 5 2.50 cm 2 1 1.87 cm 2 1 1 5 1.67. 1 f 5 1 n 2 1 2 1 1 R 1 2 1 R 2 2 5 1 n 2 1 2 1 1 R 2 1 2 R 2 5 2 1 n 2 1 2 R . R 5 2.50 cm. R 2 5 2 R , R 1 5 1 R f 5 1.87 cm. f 5 18.6 cm. 1 f 5 1 n 2 1 2 1 1 13.0 cm 2 1 ` 2 y r 5 my 5 1 2 4.76 21 3.75 mm 2 5 2 17.8 mm. m 5 2 s r s 5 2 107 cm 22.5 cm 5 2 4.76.
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  • Fall '09
  • RODRIGUEZ
  • Physics, virtual image, GEOMETRIC OPTICS

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