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Ch 34: Geometric Optics
Young & Freedman Calc - University Physics 15th Edition
Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Not the one you use?Change textbook
All textbooksYoung & Freedman Calc 15th EditionCh 34: Geometric OpticsProblem 51ac
Chapter 33, Problem 51ac

Zoom Lens. Consider the simple model of the zoom lens shown in Fig. 34.43a. The converging lens has focal length f1 = 12 cm, and the diverging lens has focal length f2 = -12 cm. The lenses are separated by 4 cm as shown in Fig. 34.43a. (a) For a distant object, where is the of the converging lens? (c) Where is the final image? Compare your answer to Fig. 34.43a.

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Step 1: Begin by analyzing part (a). For a distant object, the light rays entering the converging lens are parallel. Using the lens equation \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \( f \) is the focal length, \( d_o \) is the object distance, and \( d_i \) is the image distance, note that \( d_o \to \infty \) for a distant object. This simplifies the equation to \( \frac{1}{d_i} = \frac{1}{f} \), so the image distance \( d_i \) is equal to the focal length of the converging lens, \( f_1 = 12 \, \text{cm} \). Thus, the image of the converging lens is formed at 12 cm from the lens.
Step 2: Proceed to part (b). The image formed by the converging lens serves as the object for the diverging lens. The object distance for the diverging lens is calculated relative to its position. Since the lenses are separated by 4 cm, the object distance for the diverging lens is \( d_o = 12 \, \text{cm} - 4 \, \text{cm} = 8 \, \text{cm} \).
Step 3: For part (c), use the lens equation again for the diverging lens: \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \). Here, \( f_2 = -12 \, \text{cm} \) and \( d_o = 8 \, \text{cm} \). Substitute these values into the equation to solve for \( d_i \), the image distance for the diverging lens.
Step 4: Interpret the sign of \( d_i \) for the diverging lens. If \( d_i \) is negative, the image is virtual and located on the same side of the lens as the object. Compare the calculated position of the final image to the diagram in Fig. 34.43a to confirm its location.
Step 5: Summarize the process: (a) The image of the converging lens is at its focal length, 12 cm. (b) The object distance for the diverging lens is 8 cm. (c) Use the lens equation to find the final image position and compare it to the diagram for verification.

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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 images formed by lenses, whether converging or diverging. Understanding how to manipulate this equation allows for the calculation of image locations based on object distances and lens characteristics.
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Lens Maker Equation

Converging and Diverging Lenses

Converging lenses, such as convex lenses, focus parallel rays of light to a point, characterized by a positive focal length. In contrast, diverging lenses, like concave lenses, spread out light rays, having a negative focal length. Recognizing the behavior of these lenses is crucial for analyzing how they interact in a system, such as a zoom lens setup.
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Ray Diagrams for Diverging Lenses

Image Formation

Image formation by lenses involves the creation of an image from light rays that pass through the lens. The nature of the image (real or virtual, upright or inverted) depends on the type of lens and the relative positions of the object and lens. Understanding these principles is vital for solving problems related to multiple lens systems, as in the case of the zoom lens described.
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Refraction at Spherical Surfaces
Related Practice
Textbook Question

Contact lenses are placed right on the eyeball, so the distance from the eye to an object (or image) is the same as the distance from the lens to that object (or image). A certain person can see distant objects well, but his near point is 45.0 cm from his eyes instead of the usual 25.0 cm. Is this person nearsighted or farsighted?

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Textbook Question

Repeat Exercise 34.41 using the same lenses except for the following changes: The second lens is a diverging lens having a focal length of magnitude 60.0 cm.

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Textbook Question

Where is the near point of an eye for which a contact lens with a power of +2.75 diopters is prescribed?

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Textbook Question

A 1.20 cm tall object is 50.0 cm to the left of a converging lens of focal length 40.0 cm. A second converging lens, this one having a focal length of 60.0 cm, is located 300.0 cm to the right of the first lens along the same optic axis. Find the location and height of the image (call it I1) formed by the lens with a focal length of 40.0 cm.

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Textbook Question

You wish to project the image of a slide on a screen 9.00 m from the lens of a slide projector. If the dimensions of the picture on a 35 mm color slide are 24 mm ✖ 36 mm, what is the minimum size of the projector screen required to accommodate the image?

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Textbook Question

BIO Ordinary Glasses. Ordinary glasses are worn in front of the eye and usually 2.0 cm in front of the eyeball. Suppose that the person in Exercise 34.52 prefers ordinary glasses to contact lenses. What focal length lenses are needed to correct his vision, and what is their power in diopters?

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