COLLEGE PHYSICS
2nd Edition
ISBN: 9781464196393
Author: Freedman
Publisher: MAC HIGHER
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Chapter 24, Problem 11QAP
To determine
Discuss how someone get use to inverted world by wearing inverting glasses for time and get back to normal after it has been removed for several days.
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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.
Chapter 24 Solutions
COLLEGE PHYSICS
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- •• (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_forwardIt 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?•arrow_forwardA clerk at the local grocery store wears glasses that make her eyes look larger than they actually are. Is the clerk nearsighted or farsighted? Explain.arrow_forward
- •• A certain telescope uses a concave spherical mirror that has a radius equal to 8.0 m. Find the location and diameter of the image of the moon formed by this mirror. The moon has a diameter of 3.5 X 10°m and is 3.8 x 10* m from Earth. 31arrow_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•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_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 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_forwardExplain why an object in water always appears to be at a depth shallower than it actually is? Why do people sometimes sustain neck and spinal injuries when diving into unfamiliar ponds or waters?arrow_forward
- 34.40 • A diverging lens with a focal length of -48.0 cm forms a virtualimage 8.00 mm tall, 17.0 cm to the right of the lens. Determine theposition and size of the object. Is the image erect or inverted? Are theobject and image on the same side or opposite sides of the lens? Draw aprincipal-ray diagram for this situation.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_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