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Chapter 24: Problem 49

You have a set of converging lenses with focal lengths $1.00 \mathrm{cm}, 10.0 \mathrm{cm}, 50.0 \mathrm{cm},\( and \)80.0 \mathrm{cm} .$ (a) Which two lenses would you select to make a telescope with the largest magnifying power? What is the angular magnification of the telescope when viewing a distant object? (b) Which lens is used as objective and which as eyepiece? (c) What should be the distance between the objective and eyepiece?

Short Answer

Expert verified
Answer: The largest magnifying power for the telescope is 80, with an 80.0 cm lens as the objective and a 1.00 cm lens as the eyepiece. The distance between the objective and the eyepiece should be 81.0 cm for maximum angular magnification.

Step by step solution

01

Select lenses for the telescope with the largest magnifying power

To maximize the magnifying power, we need the lens with the longest focal length for the objective lens and the lens with the shortest focal length for the eyepiece. From the given lenses with focal lengths \(1.00 \mathrm{cm}, 10.0 \mathrm{cm}, 50.0 \mathrm{cm},\) and \(80.0 \mathrm{cm}\), we'll use the \(80.0 \mathrm{cm}\) lens for the objective and \(1.00 \mathrm{cm}\) lens for the eyepiece.
02

Calculate the angular magnification

The angular magnification \((M)\) of a telescope is given by the ratio of the focal lengths of the objective (\(f_o\)) and the eyepiece (\(f_e\)): $$ M = \frac{f_o}{f_e} $$ Using the selected lenses with focal lengths \(80.0 \mathrm{cm}\) (objective) and \(1.00 \mathrm{cm}\) (eyepiece), we can calculate the magnification as follows: $$ M = \frac{80.0}{1.00} = 80 $$ The telescope's angular magnification when viewing a distant object is 80.
03

Identify the objective and eyepiece lenses

As we have selected the lenses in step 1, the lens with focal length \(80.0 \mathrm{cm}\) is the objective, and the lens with focal length \(1.00 \mathrm{cm}\) is the eyepiece.
04

Calculate the distance between the objective and eyepiece

When viewing a distant object, the angular magnification of the telescope is maximum. In this case, the distance between the objective and eyepiece is equal to the sum of their focal lengths: $$ D = f_o + f_e $$ Using the selected objective (\(80.0 \mathrm{cm}\)) and eyepiece (\(1.00 \mathrm{cm}\)) focal lengths, we can find the distance as: $$ D = 80.0 + 1.00 = 81.0 \mathrm{cm} $$ The distance between the objective and eyepiece should be 81.0 cm for maximum angular magnification.

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Most popular questions from this chapter

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