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

Giancoli Douglas 5th edition

Ch. 33 - Lenses and Optical Instruments

Problem 18c

Chapter 32, Problem 18c

(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 d_T = d_i + d_o 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.

Verified step by step guidance

Step 1: Begin by understanding the lens equation, which relates the object distance (dₒ), image distance (dᵢ), and focal length (f) of a converging lens:

\frac{1}{dₒ} + \frac{1}{dᵢ} = \frac{1}{f} .

This equation will be central to solving the problem.

Step 2: Recognize that the total distance between the object and the screen is fixed, denoted as d_T = dₒ + dᵢ. Use this relationship to express one variable in terms of the other. For example,

dₒ = d_T - dᵢ .

Step 3: Substitute

dₒ = d_T - dᵢ

into the lens equation. This gives:

\frac{1}{dₒ} = \frac{1}{(} .

Combine this with

\frac{1}{dᵢ} = \frac{1}{f} - \frac{1}{(} .

Step 4: Solve the resulting equation for dᵢ. This will yield two possible solutions for dᵢ, corresponding to the two positions of the lens where the image is formed. The distance between these two lens positions can be determined by finding the difference between the two values of dᵢ.

Step 5: To determine the ratio of the image sizes, recall that the magnification (M) of a lens is given by

M = \frac{dᵢ}{dₒ} .

Use the two values of dᵢ and their corresponding dₒ values to calculate the magnifications for each lens position. The ratio of the image sizes will be the ratio of these magnifications.

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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 (dₒ), the image distance (dᵢ), and the focal length (f) of a lens. It is expressed as 1/f = 1/dₒ + 1/dᵢ. This formula is essential for understanding how the position of the object and the image changes when the lens is moved, allowing for the calculation of distances and image characteristics.

Magnification

Magnification (M) is the ratio of the height of the image (hᵢ) to the height of the object (hₒ), and it can also be expressed as M = -dᵢ/dₒ. This concept is crucial for determining how the size of the image changes relative to the object when the lens is adjusted, which is necessary for solving the problem regarding the ratio of image sizes.

Converging Lens Behavior

A converging lens focuses parallel rays of light to a point known as the focal point. The behavior of light through a converging lens is fundamental to understanding image formation, including the characteristics of real and virtual images, their positions, and sizes, which are all influenced by the lens's focal length and the distances of the object and image.

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Related Practice

Textbook Question

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?

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Textbook Question

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

1/ƒ_D = (1/ƒ_T) - (1/ƒ_C)

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Textbook Question

It is desired to magnify reading material by a factor of 3.0 x when a book is placed 9.0 cm behind a lens.

(a) Draw a ray diagram and describe the type of image this would be.

(b) What type of lens is needed?

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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

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?