COLLEGE PHYSICS
2nd Edition
ISBN: 9781464196393
Author: Freedman
Publisher: MAC HIGHER
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Chapter 24, Problem 15QAP
To determine
To Explain:
why the lens in a digital camera must move away from the light sensor in order to focus on a nearby object.
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It has become common to replace the cataract-clouded lens of the eye with an internal lens. This intraocular lens can be chosen so that the person has perfect distant vision. Will the person be able to read without glasses? If the person was nearsighted, is the power of the intraocular lens greater or less than the removed lens?•
•• A certain telescope uses a concave spherical mirror that
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31
• (a) Use a ray diagram to determine the approximate location ofthe image produced by a concave lens when the object is at a distance 12 ∙ ƒ ∙ from the lens. (b) Is the image upright or inverted? (c) Isthe image real or virtual? Explain
Chapter 24 Solutions
COLLEGE PHYSICS
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- • An object is a distance 2ƒ from a convex lens. (a) Use a ray diagram to find the approximate location of the image. (b) Is theimage upright or inverted? (c) Is the image real or virtual? Explain.arrow_forward• A mirrored-glass gazing globe in a garden is 31.9 cm in diameter.What is the focal length of the globe?arrow_forward•• An object is 15.0 cm in front of a converging lens that has a focal length equal to 15.0 cm. A diverging lens that has a focal length whose magnitude is equal to 15.0 cm is located 20.0 cm in back of the first. (a) Find the location of the final image and describe its properties (for example, real and inverted) and (b) draw a ray diagram to corroborate your answers to Part (a). SSM 55arrow_forward
- •• (a) An object that is 3.00 cm high is placed 25.0 cm in front of a thin lens that has a power equal to 10.0 D. Draw a ray di- agram to find the position and the size of the image and check your results using the thin-lens equation. (b) Repeat Part (a) if the object is placed 20.0 cm in front of the lens. (c) Repeat Part (a) for an object placed 20.0 cm in front of a thin lens that has a power equal to -10.0 D. SSM 45arrow_forward• A concave mirror produces a virtual image that is three times as tallas the object. (a) If the object is 16 cm in front of the mirror, what is theimage distance? (b) What is the focal length of this mirror?arrow_forward•48 An object is moved along the central axis of a thin lens while the lateral magnification m is measured. Figure 34-43 gives m ver- sus object distance p out to p, = 8.0 cm. What is the magnification of the object when the object is 14.0 cm from the lens? p (cm)arrow_forward
- •• You view a nearby tree in a concave mirror. The inverted imageof the tree is 4.8 cm high and is located 7.0 cm in front of the mirror.If the tree is 28 m from the mirror, what is its height?arrow_forwardA camera with a 50.0 mm focal length lens is being used to photograph a person standing 8.00 m away. • How far from the lens must the film be? Provide the solution: cmarrow_forward• The angle between the Sun and a rescue aircraft is 54°. Whatshould be the angle of incidence for sunlight on a plane mirror sothat the rescue pilot sees the reflected light?arrow_forward
- Bifocal lenses are prescribed for a patient, the components having focal length of 40cm & -300cm. • What far points of the patients eye?arrow_forward•42 Figure 34-40 gives the lateral magnification m of an object versus the object distance p from a lens as the object is moved along the cen- tral axis of the lens through a range of values for p out to p, = 20.0 cm. What is the magnification of the ob- ject when the object is 35 cm from the lens? E 0.5 p (cm) Figure 34-40 Problem 42.arrow_forwardSupposed your physics instructor wears a pair of glasses with diverging lenses with focal length -60 cm. While looking at a distant tree, the image of the tree is only 58 cm from the glasses. . Find the actual location of the tree. • • Sketch a diagram showing the eye, the glasses, the image of the tree, and the tree itself. Identify whether the image of the tree is real or virtual.arrow_forward
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Convex and Concave Lenses; Author: Manocha Academy;https://www.youtube.com/watch?v=CJ6aB5ULqa0;License: Standard YouTube License, CC-BY