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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

An object is located \(16.0 \mathrm{cm}\) in front of a converging lens with focal length \(12.0 \mathrm{cm} .\) To the right of the converging lens, separated by a distance of \(20.0 \mathrm{cm},\) is a diverging lens of focal length \(-10.0 \mathrm{cm} .\) Find the location of the final image by ray tracing and verify using the lens equations.
(a) If Harry has a near point of \(1.5 \mathrm{m},\) what focal length contact lenses does he require? (b) What is the power of these lenses in diopters?

A refracting telescope is \(45.0 \mathrm{cm}\) long and the caption states that the telescope magnifies images by a factor of \(30.0 .\) Assuming these numbers are for viewing an object an infinite distance away with minimum eyestrain, what is the focal length of each of the two lenses?
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 man requires reading glasses with \(+2.0 \mathrm{D}\) power to read a book held \(40.0 \mathrm{cm}\) away with a relaxed eye. Assume the glasses are $2.0 \mathrm{cm}$ from his eyes. (a) What is his uncorrected far point? (b) What refractive power lenses should he use for distance vision? (c) His uncorrected near point is \(1.0 \mathrm{m} .\) What should the refractive powers of the two lenses in his bifocals be to give him clear vision from \(25 \mathrm{cm}\) to infinity?
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