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Chapter 24: Problem 63

An astronomical telescope provides an angular magnification of 12. The two converging lenses are \(66 \mathrm{cm}\) apart. Find the focal length of each of the lenses.

Short Answer

Expert verified
Answer: The focal length of the objective lens is 60.96 cm, and the focal length of the eyepiece lens is 5.08 cm.

Step by step solution

01

Identify the formula for angular magnification

The formula for the angular magnification (m) of a telescope is given by: $$m=\frac{f_\mathrm{o}}{f_\mathrm{e}}$$, where \(f_\mathrm{o}\) is the focal length of the objective lens, and \(f_\mathrm{e}\) is the focal length of the eyepiece lens. In this case, \(m=12\). Our goal is to find the values of \(f_\mathrm{o}\) and \(f_\mathrm{e}\).
02

Identify the formula for the lens tube length

The length of the telescope's tube is given by the distance between its objective and eyepiece lenses (\(L\)). This is related to the focal lengths of the lenses by the following formula: $$L = f_\mathrm{o} + f_\mathrm{e}$$ In this problem, the telescope's tube length is given as \(L=66\,\mathrm{cm}\).
03

Write the system of equations

We can use the two formulas obtained in steps 1 and 2 to write a system of equations: $$\begin{cases} 12 = \frac{f_\mathrm{o}}{f_\mathrm{e}} \\ 66 = f_\mathrm{o} + f_\mathrm{e} \end{cases}$$
04

Solve the system of equations

We can multiply the first equation by \(f_\mathrm{e}\) to eliminate the denominator: $$12f_\mathrm{e} = f_\mathrm{o}$$ Now, substitute this expression for \(f_\mathrm{o}\) in the second equation: $$66 = 12f_\mathrm{e} + f_\mathrm{e}$$ Combine terms and solve for \(f_\mathrm{e}\): $$66 = 13 f_\mathrm{e}$$ $$f_\mathrm{e} = \frac{66}{13}$$ $$f_\mathrm{e} = 5.08\,\mathrm{cm}$$ Now we can substitute this value of \(f_\mathrm{e}\) in the equation for \(f_\mathrm{o}\): $$f_\mathrm{o} = 12f_\mathrm{e}$$ $$f_\mathrm{o} = 12(5.08)$$ $$f_\mathrm{o} = 60.96\,\mathrm{cm}$$
05

Find the focal length of each lens

We have found the focal length of the objective lens (\(f_\mathrm{o}\)) to be \(60.96\,\mathrm{cm}\) and the focal length of the eyepiece lens (\(f_\mathrm{e}\)) to be \(5.08\,\mathrm{cm}\).

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Most popular questions from this chapter

The wing of an insect is \(1.0 \mathrm{mm}\) long. When viewed through a microscope, the image is \(1.0 \mathrm{m}\) long and is located \(5.0 \mathrm{m}\) away. Determine the angular magnification.
An insect that is 5.00 mm long is placed \(10.0 \mathrm{cm}\) from a converging lens with a focal length of \(12.0 \mathrm{cm} .\) (a) What is the position of the image? (b) What is the size of the image? (c) Is the image upright or inverted? (d) Is the image real or virtual? (e) What is the angular magnification if the lens is close to the eye?
Two converging lenses, separated by a distance of \(50.0 \mathrm{cm},\) are used in combination. The first lens, located to the left, has a focal length of \(15.0 \mathrm{cm} .\) The second lens, located to the right, has a focal length of \(12.0 \mathrm{cm} .\) An object, \(3.00 \mathrm{cm}\) high, is placed at a distance of \(20.0 \mathrm{cm}\) in front of the first lens. (a) Find the intermediate and final image distances relative to the corresponding lenses. (b) What is the total magnification? (c) What is the height of the final image?
A microscope has an objective lens of focal length \(5.00 \mathrm{mm} .\) The objective forms an image \(16.5 \mathrm{cm}\) from the lens. The focal length of the eyepiece is \(2.80 \mathrm{cm} .\) (a) What is the distance between the lenses? (b) What is the angular magnification? The near point is \(25.0 \mathrm{cm} .\) (c) How far from the objective should the object be placed?
A converging lens with a focal length of \(3.00 \mathrm{cm}\) is placed $24.00 \mathrm{cm}\( to the right of a concave mirror with a focal length of \)4.00 \mathrm{cm} .\( An object is placed between the mirror and the lens, \)6.00 \mathrm{cm}\( to the right of the mirror and \)18.00 \mathrm{cm}$ to the left of the lens. Name three places where you could find an image of this object. For each image tell whether it is inverted or upright and give the total magnification.
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