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976_PartUniversity Physics Solution

# 976_PartUniversity Physics Solution - 34-14 Chapter 34...

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34-14 Chapter 34 E XECUTE : The image is behind the lens, so 0 s ′ < . The thin-lens equation gives 1 1 1 1 1 1 4.17 cm 5.00 cm 25.0 cm s s s f s + = = = , on the same side of the lens as the ant. E VALUATE : Since 0 s ′ < , the image will be erect. 34.52. I DENTIFY : Apply Eq.(34.24). S ET U P : 1 160 mm 5.0 mm 165 mm s ′ = + = E XECUTE : (a) 1 1 2 (250 mm) (250 mm)(165 mm) 317. (5.00 mm)(26.0 mm) s M f f = = = (b) The minimum separation is 4 0.10 mm 0.10 mm 3.15 10 mm. 317 M = = × E VALUATE : The angular size of the image viewed by the eye when looking through the microscope is 317 times larger than if the object is viewed at the near-point of the unaided eye. 34.53. (a) I DENTIFY and S ET U P : Figure 34.53 Final image is at so the object for the eyepiece is at its focal point. But the object for the eyepiece is the image of the objective so the image formed by the objective is 19.7 cm ° 1.80 cm = 17.9 cm to the right of the lens. Apply Eq.(34.16) to the image formation by the objective, solve for the object distance s . 0.800 cm; 17.9 cm; ? f s s = = = 1 1 1 1 1 1 , so s f s s f s f s s f ′ − + = = = E XECUTE : (17.9 cm)( 0.800 cm) 8.37 mm 17.9 cm 0.800 cm s f s s f + = = = + ′ − (b) S ET U P : Use Eq.(34.17). E XECUTE : 1 17.9 cm 21.4 0.837 cm s m s = − = − = − The linear magnification of the objective is 21.4. (c) S ET U P : Use Eq.(34.23): 1 2 M m M = E XECUTE : 2 2 25 cm 25 cm 13.9 1.80 cm M f = = = 1 2 ( 21.4)(13.9) 297 M m M = = − = − E VALUATE : M is not accurately given by 1 1 2 (25 cm) / 311, s f f = because the object is not quite at the focal point of the objective 1 1 ( 0.837 cm and 0.800 cm). s f = = 34.54. I DENTIFY : Eq.(34.24) can be written 1 1 2 2 1 s M m M M f = = . S ET U P : 1 1 120 mm s f ′ = + E XECUTE : 16 mm : 120 mm 16 mm 136 mm; 16 mm f s s = = + = = . 1 136 mm 8.5 16 mm s m s = = = . 1 124 mm 4 mm : 120 mm 4 mm 124 mm; 4 mm 31 4 mm s f s s m s = = + = = = = = . 1 122 mm 1.9 mm : 120 mm 1.9 mm 122 mm; 1.9 mm 64 1.9 mm s f s s m s = = + = = = = = . The eyepiece magnifies by either 5 or 10, so: (a) The maximum magnification occurs for the 1.9-mm objective and 10x eyepiece: 1 e (64)(10) 640. M m M = = = (b) The minimum magnification occurs for the 16-mm objective and 5x eyepiece: 1 e (8.5)(5) 43. M m M = = = E VALUATE : The smaller the focal length of the objective, the greater the overall magnification.

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Geometric Optics 34-15 34.55. I DENTIFY : -number / f f D = S ET U P : 1.02 m D = E XECUTE : 19.0 (19.0) (19.0)(1.02 m) 19.4 m. f f D D = = = = E VALUATE : Camera lenses can also have an f -number of 19.0. For a camera lens, both the focal length and lens diameter are much smaller, but the f -number is a measure of their ratio. 34.56. I DENTIFY : For a telescope, 1 2 f M f = − .
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976_PartUniversity Physics Solution - 34-14 Chapter 34...

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