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Ch. 33 - Lenses and Optical Instruments
Giancoli Douglas - Physics for Scientists and Engineers 5th edition
Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Not the one you use?Change textbook
All textbooksGiancoli Douglas 5th editionCh. 33 - Lenses and Optical InstrumentsProblem 19
Chapter 32, Problem 19

Two 28.0-cm-focal-length converging lenses are placed 16.5 cm apart. An object is placed 35.0 cm in front of one lens.
(a) Where will the final image formed by the second lens be located?
(b) What is the total magnification?

Verified step by step guidance
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Step 1: Start by analyzing the first lens. Use the lens formula \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \( f \) is the focal length, \( d_o \) is the object distance, and \( d_i \) is the image distance. For the first lens, \( f = 28.0 \, \text{cm} \) and \( d_o = 35.0 \, \text{cm} \). Solve for \( d_i \), the image distance for the first lens.
Step 2: Determine the position of the image formed by the first lens relative to the second lens. The lenses are 16.5 cm apart, so calculate the distance of the image from the second lens by subtracting the distance between the lenses from \( d_i \) of the first lens. This new distance becomes the object distance \( d_o \) for the second lens.
Step 3: Analyze the second lens using the lens formula \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \). For the second lens, \( f = 28.0 \, \text{cm} \) and \( d_o \) is the distance calculated in Step 2. Solve for \( d_i \), the image distance for the second lens, which gives the location of the final image.
Step 4: Calculate the magnification for each lens using the formula \( M = -\frac{d_i}{d_o} \). For the first lens, use the values of \( d_i \) and \( d_o \) from Step 1. For the second lens, use the values of \( d_i \) and \( d_o \) from Step 3.
Step 5: Determine the total magnification by multiplying the magnifications of the two lenses: \( M_{\text{total}} = M_1 \times M_2 \). This gives the overall magnification of the system.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Lens Formula

The lens formula relates the object distance (u), image distance (v), and focal length (f) of a lens, expressed as 1/f = 1/v - 1/u. This formula is essential for determining the position of the image formed by a lens based on the object's position and the lens's focal length.
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Lens Maker Equation

Magnification

Magnification is the ratio of the height of the image to the height of the object, and it can also be calculated using the formula M = -v/u, where v is the image distance and u is the object distance. This concept helps in understanding how much larger or smaller the image appears compared to the object.
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Mirror Equation

Converging Lenses

Converging lenses, or convex lenses, are thicker at the center than at the edges and focus parallel incoming light rays to a point known as the focal point. Understanding how these lenses work is crucial for analyzing the behavior of light as it passes through multiple lenses in a system.
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Ray Diagrams for Converging Lenses
Related Practice
Textbook Question

(II) A diverging lens is placed next to a converging lens of focal length ƒC , as in Fig. 33–14. If ƒT represents the focal length of the combination, show that the focal length of the diverging lens, ƒD , is given by


1/ƒD = (1/ƒT) - (1/ƒC)

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Textbook Question

(II) In a film projector, the film acts as the object whose image is projected on a screen (Fig. 33–46). If a 105-mm-focal-length lens is to project an image on a screen 22.5 m away, how far from the lens should the film be? If the film is 24 mm wide, how wide will the picture be on the screen?

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Textbook Question

(III) A bright object is placed on one side of a converging lens of focal length f, and a white screen for viewing the image is on the opposite side. The distance dT = di + do between the object and the screen is kept fixed, but the lens can be moved. Determine a formula for the distance between the two lens positions in part (a), and the ratio of the image sizes.

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Textbook Question

An object is placed 96.0 cm from a glass lens (n = 1.52) with one concave surface of radius 22.0 cm and one convex surface of radius 18.5 cm.

(a) Where is the final image?

(b) What is the magnification?

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Textbook Question

A diverging lens with ƒ = -36.5 cm is placed 14.0 cm behind a converging lens with ƒ = 20.0cm. Where will an object at infinity be focused?

Textbook Question

An object is located 1.35 m from an 8.0-D lens. By how much does the image move if the object is moved (a) 0.90 m closer to the lens, and (b) 0.90 m farther from the lens?