College Physics (10th Edition)
10th Edition
ISBN: 9780321902788
Author: Hugh D. Young, Philip W. Adams, Raymond Joseph Chastain
Publisher: PEARSON
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Chapter 24, Problem 27P
A converging lens with a focal length of 90.0 cm forms an image of a 3.20-cm-tall real object that is to the left of the lens. The image is 4.50 cm tall and inverted. Where are the object and image located in relation to the lens? Is the image real or virtual?
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Chapter 24 Solutions
College Physics (10th Edition)
Ch. 24 - If a spherical mirror is immersed in water, does...Ch. 24 - For what range of object positions does a concave...Ch. 24 - If a screen is placed at the location of a real...Ch. 24 - Is it possible to view a virtual image directly...Ch. 24 - Prob. 5CQCh. 24 - On a sunny day, you can use the suns rays and a...Ch. 24 - A person looks at her reflection in the concave...Ch. 24 - What happens to the image produced by a converging...Ch. 24 - Without measuring its radius of curvature (which...Ch. 24 - Without measuring its radii of curvature (which is...
Ch. 24 - A spherical air bubble in water can function as a...Ch. 24 - Optical telescopes having a principal mirror only...Ch. 24 - A ray from an object passes through a thin lens,...Ch. 24 - If a single lens forms a real image, we can...Ch. 24 - If a single lens forms a virtual image, we can...Ch. 24 - An object lies outside the focal port of a...Ch. 24 - An object lies outside the focal point of a...Ch. 24 - Prob. 6MCPCh. 24 - An object is placed a distance 2f away from a...Ch. 24 - In order to form an image with a converging lens...Ch. 24 - A ray from an object passes through a thin lens,...Ch. 24 - As you move an object from just outside to just...Ch. 24 - As you move an object from just outside to just...Ch. 24 - You have a shiny salad bowl with a spherical shape...Ch. 24 - A candle 4.85 cm tall is 39.2 cm to the left of a...Ch. 24 - Two plane mirrors form a 60 wedge as shown in...Ch. 24 - An object is placed between two plane mirrors...Ch. 24 - If you run away from a plane mirror at 2.40 m/s,...Ch. 24 - A concave spherical mirror has a radius of...Ch. 24 - A concave spherical mirror has a radius of...Ch. 24 - The diameter of Mars is 6794 km. and its minimum...Ch. 24 - A concave mirror has a radius of curvature of 34.0...Ch. 24 - Rearview mirror. A mirror on the passenger side of...Ch. 24 - Examining your image in a convex mirror whose...Ch. 24 - A coin is placed next to the convex side of a thin...Ch. 24 - Consider a concave mirror that has a focal length...Ch. 24 - A spherical, concave shaving mirror has a radius...Ch. 24 - An object 0.600 cm tall is placed 16.5 cm to the...Ch. 24 - Repeat the previous problem for the case in which...Ch. 24 - The thin glass shell shown in Figure 24.43 has a...Ch. 24 - Dental mirror. A dentist uses a curved mirror to...Ch. 24 - The left end of a long glass rod 6.00 cm in...Ch. 24 - Prob. 19PCh. 24 - The left end of a long glass rod 8.00 cm in...Ch. 24 - A large aquarium has portholes of thin transparent...Ch. 24 - Focus of the eye. The cornea of the eye has a...Ch. 24 - A speck of dirt is embedded 3.50 cm below the...Ch. 24 - A skin diver is 2.0 m below the surface of a lake....Ch. 24 - A person is swimming 1.0 m beneath the surface of...Ch. 24 - A converging lens with a focal length of 7.00 cm...Ch. 24 - A converging lens with a focal length of 90.0 cm...Ch. 24 - You are standing 0.50 m in front of a lens that...Ch. 24 - Figure 24.44 shows an object and its image formed...Ch. 24 - Set up: 1s+1s=1f. The type of lens determines the...Ch. 24 - Figure 24.46 shows an object and its image formed...Ch. 24 - The two surfaces of a plastic converging lens have...Ch. 24 - A lens has an index of refraction of 1.7 and a...Ch. 24 - Set Up: Use 1f=(n1)(1R11R2) to calculate f and...Ch. 24 - The lens of the eye. The crystalline lens of the...Ch. 24 - The cornea as a simple lens. The cornea behaves as...Ch. 24 - An insect 3.75 mm tall is placed 22.5 cm to the...Ch. 24 - Two double-convex thin lenses each have surfaces...Ch. 24 - A converging meniscus lens (see Figure 24.30) with...Ch. 24 - A converging lens with a focal length of 12.0 cm...Ch. 24 - Combination of lenses, I. When two lenses are used...Ch. 24 - Set Up: Apply 1s+1s=1f with f = 35.0 cm. We know...Ch. 24 - Combination of lenses, II. Two thin lenses with a...Ch. 24 - A lens forms a real image that is 214 cm away from...Ch. 24 - A converging lens has a focal length of 14.0 cm...Ch. 24 - A converging lens forms an image of an...Ch. 24 - A diverging lens with a focal length of 48.0 cm...Ch. 24 - When an object is 16.0 cm from a lens, an image is...Ch. 24 - Figure 24.48 shows a small plant near a thin lens....Ch. 24 - Figure 24.49 shows a small plant near a thin lens....Ch. 24 - Figure 24.50 shows a small plant near a thin lens....Ch. 24 - Prob. 52GPCh. 24 - Where must you place an object in front of a...Ch. 24 - Set Up: Use 1s+1s=1f. A plot of 1f versus 1s...Ch. 24 - A concave mirror is to form an image of the...Ch. 24 - A lens has one convex surface of radius 6.00 cm...Ch. 24 - A 3 80-nm-tall object 24.0 cm from the center of...Ch. 24 - A lensmaker wants to make a magnifying glass from...Ch. 24 - An object is placed 18.0 cm from a screen, (a) At...Ch. 24 - In the text, Equations 24.4 and 24.7 were derived...Ch. 24 - A lens in a liquid. A lens obeys Snell s law,...Ch. 24 - Refraction of liquids. The focal length of a...Ch. 24 - Refraction of liquids. The focal length of a...Ch. 24 - If you place a concave mirror with a focal length...Ch. 24 - Refraction of liquids. The focal length of a...
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- The left face of a biconvex lens has a radius of curvature of magnitude 12.0 cm, and the right face has a radius of curvature of magnitude 18.0 cm. The index of refraction of the glass is 1.44. (a) Calculate the focal length of the lens for light incident from the left. (b) What If? After the lens is turned around to interchange the radii of curvature of the two faces, calculate the focal length of the lens for light incident from the left.arrow_forwardIn Figure P35.30, a thin converging lens of focal length 14.0 cm forms an image of the square abed, which is he = hb = 10.0 cm high and lies between distances of pd = 20.0 cm and pa = 30.0 cm from the lens. Let a, b, c. and d represent the respective corners of the image. Let qa represent the image distance for points a and b, qd represent the image distance for points c and d, hb, represent the distance from point b to the axis, and hc represent the height of c. (a) Find qa, qd, hb, and hc. (b) Make a sketch of the image. (c) The area of the object is 100 cm2. By carrying out the following steps, you will evaluate the area of the image. Let q represent the image distance of any point between a and d, for which the object distance is p. Let h represent the distance from the axis to the point at the edge of the image between b and c at image distance q. Demonstrate that h=10.0q(114.01q) where h and q are in centimeters. (d) Explain why the geometric area of the image is given by qaqdhdq (e) Carry out the integration to find the area of the image. Figure P35.30arrow_forwardTwo converging lenses having focal lengths of f1 = 10.0 cm and f2 = 20.0 cm are placed a distance d = 50.0 cm apart as shown in Figure P35.48. The image due to light passing through both lenses is to be located between the lenses at the position x = 31.0 cm indicated. (a) At what value of p should the object be positioned to the left of the first lens? (b) What is the magnification of the final image? (c) Is the final image upright or inverted? (d) Is the final image 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