solutions_chapter24

# S r 5 2 1 cm 2 5 2 158 cm s 5 485 cm 1 s 1 1 2 325 s

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s r 5 2 1 3.25 21 4.85 cm 2 5 2 15.8 cm. s 5 4.85 cm. 1 s 1 1 2 3.25 s 5 1 7.00 cm 1 s 1 1 s r 5 1 f s r 5 2 3.25 s . m 5 2 s r s 5 1 3.25 m 5 y r y 5 1.30 cm 0.400 cm 5 1 3.25. 1 s 1 1 s r 5 1 f . m . 0. y r . 0 m 5 y r y 5 2 s r s . s 5 2 1 n a n b 2 s r 5 2 1 1.00 1.333 2 1 2 4.40 m 2 5 3.30 m. n a s 1 n b s r 5 0. s r 5 2 4.40 m. 5.20 m 2 0.80 m 5 4.40 m n b 5 1.333 1 water 2 . n a 5 1.00 1 air 2 . s r 5 2 1 n b n a 2 s 5 2 1 1.00 1.333 2 1 33.0 cm 2 5 2 24.8 cm. s r 5 2 1 n b n a 2 s 5 2 1 1.00 1.333 2 1 7.0 cm 2 5 2 5.25 cm. s 5 33.0 cm. 20.0 cm 1 13.0 cm 5 33.0 cm 20.0 cm 2 7.0 cm 5 13.0 cm s 5 7.0 cm. n b 5 1.00. n a 5 1.333 n a s 1 n b s r 5 0. R S ` 2.0 m 1 9.3 m 5 11.3 m s r 5 2 1 n b n a 2 s 5 2 1 1.333 1.00 2 1 7.0 m 2 5 2 9.3 m. n a s 1 n b s r 5 0. s 5 7.0 m. s r n b 5 1.333 1 water 2 . n a 5 1.00 1 air 2 . s r 5 2 1 n b n a 2 s 5 2 1 1.00 1.309 2 1 3.50 cm 2 5 2 2.67 cm. s 5 3.50 cm. n b 5 1.00. n a 5 1.309 n a s 1 n b s r 5 0. R S ` s r 5 1.93 cm 5 19.3 mm. 1.35 s r 5 1.35 2 1.00 0.50 cm . R 5 0.50 cm: s < ` s r s r 5 2.70 cm 5 27.0 mm. 1.35 s r 5 0.500 1.00 25 cm 1 1.35 s r 5 1.35 2 1.00 0.648 . s 5 25 cm: R 5 0.648 cm R 5 0.648 cm 5 6.48 mm. 1.35 2.5 cm 5 1.35 2 1.00 R s r 5 2.5 cm: s < ` 1 s < 0. s < ` R . 0. n b 5 1.35. n a 5 1.00, n a s 1 n b s r 5 n b 2 n a R . Geometric Optics 24-9

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24.30. Set Up: Since the image is inverted, and Solve: gives gives and The object is 154 cm to the left of the lens. The image is 217 cm to the right of the lens and is real. 24.31. Set Up: Since the image is projected onto a wall, the image is real and and both s and are positive, so The height of the image is three times the height of the object, so and Solve: (a) so You are 0.60 m from the lens. (b) so the image is inverted. (c) and the lens is converging. Reflect: Since the image is real the lens must be converging and your distance from the lens must be greater than the focal length. 24.32. Set Up: The type of lens determines the sign of The sign of depends on whether the image is real or virtual. is positive because the image is on the side of the lens opposite to the object. Solve: (a) and is positive so the lens is converging. (b) The image is 1.88 mm tall. and the image is real. 24.33. Set Up: The type of lens determines the sign of The sign of determines whether the image is real or virtual. is negative because the image is on the same side of the lens as the object. Solve: (a) and is negative so the lens is diverging. (b) and the image is virtual. Reflect: A converging lens can also form a virtual image, if the object distance is less than the focal length. But in that case and the image would be farther from the lens than the object is. 24.34. Set Up: The type of lens determines the sign of The sign of depends on whether the image is real or virtual. is negative because the image is on the same side of the lens as the object. Solve: (a) and is positive so the lens is converging. (b) and the image is virtual. s r , 0 y r 5 my 5 1 1.38 21 3.25 mm 2 5 4.48 mm. m 5 2 s r s 5 2 2 22.0 cm 16.0 cm 5 1.38. f f 5 ss r s 1 s r 5 1 16.0 cm 21 2 22.0 cm 2 16.0 cm 2 22.0 cm 5 1 58.7 cm. 1 f 5 s 1 s r ss r s r s r 5 2 22.0 cm; s 5 16.0 cm.
• Fall '09
• RODRIGUEZ
• Physics, virtual image, GEOMETRIC OPTICS

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