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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 108
Chapter 32, Problem 108

An astronomical telescope, Fig. 33–36, produces an inverted image. One way to make a telescope that produces an upright image is to insert a third lens between the objective and the eyepiece, Fig. 33–39b. To have the same magnification, the non-inverting telescope will be longer. Suppose lenses of focal length 150 cm, 1.5 cm, and 10 cm are available. Where should these three lenses be placed to make a non-inverting telescope with magnification 100x?

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Step 1: Understand the problem. A non-inverting telescope requires three lenses: an objective lens, an eyepiece lens, and an additional lens (erecting lens) to make the image upright. The magnification of the telescope is given as 100x, and the focal lengths of the lenses are provided: 150 cm (objective), 1.5 cm (eyepiece), and 10 cm (erecting lens). The task is to determine the placement of these lenses to achieve the desired magnification and upright image.
Step 2: Recall the formula for magnification in a telescope. The magnification (M) is given by the ratio of the focal length of the objective lens (fₒ) to the focal length of the eyepiece lens (fₑ): M = f o f e . Substitute the given magnification (M = 100) and the focal lengths of the objective and eyepiece lenses to verify the relationship: 100 = 150 1.5 . This confirms the magnification is correct with these lenses.
Step 3: Determine the placement of the erecting lens. The erecting lens is placed between the objective and the eyepiece to invert the inverted image produced by the objective lens, making the final image upright. The erecting lens should be positioned such that its focal length (10 cm) allows it to re-invert the image. The exact placement depends on the intermediate image formed by the objective lens, which is at its focal length (150 cm) from the objective. The erecting lens should be placed slightly beyond this intermediate image to ensure proper inversion.
Step 4: Calculate the total length of the telescope. The total length of the telescope is the sum of the distances between the objective lens, the erecting lens, and the eyepiece lens. The distance between the objective and the erecting lens is approximately 150 cm (focal length of the objective), and the distance between the erecting lens and the eyepiece lens is adjusted to ensure the final image is at infinity. This distance is approximately the sum of the focal lengths of the erecting lens (10 cm) and the eyepiece lens (1.5 cm).
Step 5: Summarize the lens placements. Place the objective lens at the starting point. Position the erecting lens approximately 150 cm from the objective lens. Finally, place the eyepiece lens approximately 11.5 cm (10 cm + 1.5 cm) from the erecting lens. This configuration ensures a non-inverting telescope with a magnification of 100x.

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Key Concepts

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

Magnification in Telescopes

Magnification in telescopes refers to the ratio of the angular size of the image produced by the telescope to the angular size of the object being observed. It is determined by the focal lengths of the objective and eyepiece lenses. For a non-inverting telescope, the magnification can be calculated using the formula M = f_objective / f_eyepiece, where f represents the focal length. Achieving a specific magnification, such as 100x, requires careful selection and arrangement of the lenses.
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Lens Arrangement and Focal Lengths

The arrangement of lenses in a telescope is crucial for achieving the desired image characteristics, including orientation and magnification. The focal lengths of the lenses determine how they interact with incoming light. Inserting a third lens can help correct the image orientation while maintaining the overall magnification. The placement of each lens affects the total length of the telescope and the final image produced.
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Inverted vs. Upright Images

In telescopes, the image produced by the objective lens is typically inverted due to the way light converges. To create an upright image, additional optical elements, such as a third lens, can be introduced. This lens can invert the image again, resulting in an upright final image. Understanding how lenses manipulate light and image orientation is essential for designing telescopes that meet specific observational needs.
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Related Practice
Textbook Question

The focal length f of a converging lens can be found by placing an object of known size at various locations in front of the lens and measuring the resulting real-image distances dᵢ and their associated magnifications m (minus sign indicates that image is inverted). The data taken in such an experiment are given here:



(a) Show algebraically that a graph of m vs. dᵢ should produce a straight line. What are the theoretically expected values for the slope and the y-intercept of this line? [Hint: dₒ is not constant.] (b) Using the data above, graph m vs. dᵢ and show that a straight line does indeed result. Use the slope of this line to determine the focal length of the lens. Does the y-intercept of your plot have the expected value?

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Textbook Question

As early morning passed toward midday, and the sunlight got more intense, a photographer noted that, if she kept her shutter speed constant, she had to change the f-number from f/5.6 to f/16. By what factor had the sunlight intensity increased during that time?

Textbook Question

A physicist lost in the mountains tries to make a telescope using the lenses from his reading glasses. They have powers of +2.0 D and +4.5 D, respectively.

(a) What maximum magnification telescope is possible?

(b) Which lens should be used as the eyepiece?

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Textbook Question

A 50-year-old man uses +2.5 D lenses to read a newspaper 25 cm away. Ten years later, he must hold the paper 32 cm away to see clearly with the same lenses. What power lenses does he need now in order to hold the paper 25 cm away? (Distances are measured from the lens.)

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