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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
Chapter 32, Problem 9

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?

Verified step by step guidance
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Step 1: Recall the lens formula: 1f = 1do + 1di, where f is the focal length of the lens, do is the object distance, and di is the image distance. The focal length f can be calculated from the lens power P using the formula f = 1P. Substitute P = 8.0 diopters to find f in meters.
Step 2: Using the lens formula, calculate the initial image distance di when the object is located at do = 1.35 m. Rearrange the lens formula to solve for di: di = 11f - 1do. Substitute the values of f and do to find di.
Step 3: For part (a), calculate the new object distance when the object is moved 0.90 m closer to the lens. The new object distance is do = 1.35 - 0.90 m. Use the lens formula again to calculate the new image distance di for this new object distance.
Step 4: For part (b), calculate the new object distance when the object is moved 0.90 m farther from the lens. The new object distance is do = 1.35 + 0.90 m. Use the lens formula again to calculate the new image distance di for this new object distance.
Step 5: To find the image movement in each case, subtract the initial image distance from the new image distance for both part (a) and part (b). The magnitude of the difference gives the amount by which the image moves in each scenario.

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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 through the equation 1/f = 1/v - 1/u. This formula is essential for determining the position of the image formed by a lens based on the position of the object.
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Lens Maker Equation

Magnification

Magnification (M) is the ratio of the height of the image (h') to the height of the object (h), and it can also be expressed as M = -v/u. Understanding magnification helps in analyzing how the size and position of the image change as the object is moved relative to the lens.
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Mirror Equation

Sign Convention for Lenses

The sign convention for lenses dictates that distances measured in the direction of the incoming light (toward the lens) are negative, while distances measured in the direction of the outgoing light (away from the lens) are positive. This convention is crucial for correctly applying the lens formula and interpreting the results.
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Ray Diagrams for Converging Lenses
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

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?

(c) What is the power of the lens in diopters?

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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

A 105-mm-focal-length lens is used to focus an image on the sensor of a camera. The maximum distance allowed between the lens and the sensor plane is 132 mm.

(a) How far in front of the sensor should the lens (assumed thin) be positioned if the object to be photographed is 10.0 m away? (b) 3.0 m away? (c) 1.0 m away?

(d) What is the closest object this lens could photograph sharply?

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Textbook Question

A sharp image is located 373 mm behind a 235-mm-focal-length converging lens. Find the object distance by calculation.

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