EBK PHYSICS FOR SCIENTISTS & ENGINEERS
5th Edition
ISBN: 9780134296074
Author: GIANCOLI
Publisher: VST
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Textbook Question
Chapter 33, Problem 82GP
A small object is 25.0 cm from a diverging lens as shown in Fig. 33-48. A converging lens with a focal length of 12.0 cm is 30.0 cm to the right of the diverging lens. The two-lens system forms a real inverted image 17.0 cm to the right of the converging lens. What is the focal length of the diverging lens?
FIGURE 33-48 Problem 87.
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Students have asked these similar questions
106 In Fig. 34-52, an object is placed in front of a converging lens
at a distance equal to twice the focal length fi of the lens. On the
other side of the lens is a concave mirror of focal length f2 sepa-
rated from the lens by a distance 2(f + f). Light from the object
passes rightward through the lens, reflects from the mirror, passes
leftward through the lens, and forms a final image of the object.
What are (a) the distance between the lens and that final image
and (b) the overall lateral magnification M of the object? Is the im-
age (c) real or virtual (if it is virtual, it requires someone looking
through the lens toward the mirror), (d) to the left or right of the
lens, and (e) inverted or noninverted relative to the object?
to
E26 2i + f)-
Figure 34-52 Problem 106.
An object is 23 cm from a converging lens. A diverging lens with a focal length of− 15 cm is 31 cm beyond the converging lens. The two-lens system forms a real inverted
image 52 cm past the diverging lens. What is the focal length of the converging lens?
(numbers 32-33) An object is placed 24 cm from a converging lens with a focal length of 8 cm.
32. How far is the image from the lens?
8 cm
16 cm
12 cm
18 cm
24 cm
33. The image is:
real and larger
real and smaller
virtual and smaller
virtual and larger
none of the above answers are correct
Chapter 33 Solutions
EBK PHYSICS FOR SCIENTISTS & ENGINEERS
Ch. 33 - Where must the film be placed if a camera lens is...Ch. 33 - A photographer moves closer to his subject and...Ch. 33 - Use ray diagrams to show that a real image formed...Ch. 33 - Light rays are said to be reversible. Is this...Ch. 33 - A cat with its tail in the air stands facing a...Ch. 33 - An underwater lens consists of a carefully shaped...Ch. 33 - (III) A bright object is placed on one side of a...Ch. 33 - (II) A diverging lens with a focal length of 14 cm...Ch. 33 - (II) Suppose that a correct exposure is 1250S at f...Ch. 33 - (I) A human eyeball is about 2.0 cm long and the...
Ch. 33 - (II) Reading glasses of what power are needed for...Ch. 33 - (II) The closely packed cones in the fovea of the...Ch. 33 - (II) Sherlock Holmes is using an...Ch. 33 - (II) A converging lens of focal length = 12 cm is...Ch. 33 - (III) Given two 12-cm-focal-length lenses, you...Ch. 33 - A small object is 25.0 cm from a diverging lens as...Ch. 33 - Two converging lenses are placed 30.0 cm apart....
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- Figure P38.43 shows a concave meniscus lens. If |r1| = 8.50 cm and |r2| = 6.50 cm, find the focal length and determine whether the lens is converging or diverging. The lens is made of glass with index of refraction n = 1.55. CHECK and THINK: How do your answers change if the object is placed on the right side of the lens? FIGURE P38.43arrow_forwardFigure P23.28 shows a curved surface separating a material with index of refraction n1 from a material with index n2. The surface forms an image I of object O. The ray shown in red passes through the surface along a radial line. Its angles of incidence and refraction are both zero, so its direction does not change at the surface. For the ray shown in blue, the direction changes according to n1 sin 1 = n2 sin 2. For paraxial rays, we assume 1 and 2 are small, so we may write n1 tan 1 n2 tan 2. The magnification is defined as M = h/h. Prove that the magnification is given by M = n1q/n2p. Figure P23.28arrow_forwardFigure P23.28 shows a curved surface separating a material with index of refraction n1 from a material with index n2. The surface forms an image I of object O. The ray shown in red passes through the surface along a radial line. Its angles of incidence and refraction are both zero, so its direction does not change at the surface. For the ray shown in blue, the direction changes according to n1 sin 1 = n2 sin 2. For paraxial rays, we assume 1 and 2 are small, so we may write n1 tan 1 n2 tan 2. The magnification is defined as M = h/h. Prove that the magnification is given by M = n1q/n2p. Figure P23.28arrow_forward
- The end of a solid glass rod of refractive index 1.50 is polished to have the shape of a hemispherical surface of radius 1.0 cm. A small object is placed in air (refractive index 1.00) on the axis 5.0 cm to the left of the vertex. Determine the position of the image.arrow_forwardAu object of height 3.0 cm is placed at 25 cm in front of a diverging lens of focal length 20 cm. Behind the diverging lens, there is a converging lens of focal length 20 cm. The distance between the lenses is 5.0 cm. Fluid the location and size of the final image.arrow_forwardTwo rays traveling parallel to the principal axis strike a large plano-convex lens having a refractive index of 1.60 (Fig. P35.33). If the convex face is spherical, a ray near the edge does not pass through the focal point (spherical aberration occurs). Assume this face has a radius of curvature of R = 20.0 cm and the two rays are at distances h1 = 0.500 cm and h2 = 12.0 cm from the principal axis. Find the difference x in the positions where each crosses the principal axis. Figure P35.33arrow_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