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Chapter 33: Problem 69

(II) An astronomical telescope longer than about \(50 \mathrm{~cm}\) is not easy to hold by hand. Based on this fact, estimate the maximum angular magnification achievable for a telescope designed to be handheld. Assume its eyepiece lens, if used as a magnifying glass, provides a magnification of \(5 \times\) for a relaxed eye with near point \(N=25 \mathrm{~cm}\).

Short Answer

Expert verified
Maximum angular magnification is approximately 9x.

Step by step solution

01

Understanding Telescope Length and Effectiveness

A handheld telescope should be easy to hold, limiting its focal length to at most 50 cm. The total length of a telescope is the sum of the focal lengths of the objective (large) lens and the eyepiece (small) lens.
02

Defining Angular Magnification

Angular magnification \(M\) of a telescope is given by \( M = \frac{f_o}{f_e} \), where \(f_o\) is the focal length of the objective lens and \(f_e\) is the focal length of the eyepiece lens. For maximum magnification, \(f_o\) should be near the maximum length constraint.
03

Calculating Eyepiece Focal Length

Given that the eyepiece provides a magnification of \(5\times\) as a magnifying glass for relaxed eye, \( M_e = \frac{N}{f_e} = 5\). Thus, \( f_e = \frac{N}{5} = \frac{25 \, cm}{5} = 5 \, cm \).
04

Estimating Maximum Angular Magnification

To achieve maximum angular magnification with \(f_o + f_e \leq 50 \, cm\), and knowing \(f_e = 5 \, cm\), the maximum \(f_o = 50 - 5 = 45 \, cm\). Thus, \( M = \frac{f_o}{f_e} = \frac{45 \, cm}{5 \, cm} = 9 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Telescope Focal Length
The focal length of a telescope is the combined distance of how light is focused through its lenses. It is often the sum of the focal lengths of the objective lens and the eyepiece lens.

For a handheld telescope, this distance is crucial because it must not be too long to comfortably hold and use. Typically, a handheld telescope reaches the maximum usable focal length of about 50 cm.

This limitation means the combined focal lengths of both crucial lenses must fit within this boundary. Since the focal length impacts both how much a telescope can zoom and its overall length, balancing these is key.

By choosing focal lengths wisely, a user maximizes magnification while keeping the telescope compact.
  • A shorter focal length in the eyepiece means more magnification.
  • A longer focal length in the objective lens gathers more light, providing clearer images.
Objective Lens
The objective lens is the larger lens at the front of a telescope. Its main job is to gather light from distant objects and bring it to focus. This lens is essential for a clear and bright image, as more light gathering means better visual details.

The focal length of the objective lens directly impacts the angular magnification of the telescope.
For maximum magnification, extending the focal length of the objective lens is beneficial鈥攈ence the original target of around 45 cm when keeping a total length limit of 50 cm.

Remember:
  • The larger the focal length of the objective lens, the larger the telescope, but more light is collected.
  • A balance is crucial in handheld models to ensure usability and comfort.
Eyepiece Lens
The eyepiece lens is the smaller lens located at the back end of a telescope, through which you look. It serves to magnify the image formed by the objective lens.

In this exercise, the eyepiece's focal length was found using its known magnification as a standalone magnifying glass. Given it provides a magnification of 5x, and the near point of a relaxed eye is 25 cm:
  • The formula used is \( M_e = \frac{N}{f_e} \).
  • Thus, the focal length \(f_e\) equals 5 cm, derived from \( \frac{25 \, \text{cm}}{5} \).

The eyepiece is crucial for achieving the final magnified view, and its focal length determines its power.

Key points:
  • Shorter eyepiece focal lengths lead to higher magnification power.
  • Balancing this with the objective lens is crucial to staying within practical design constraints, such as keeping the telescope easy to handle and operate.

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Most popular questions from this chapter

(III) Given two 12 -cm-focal-length lenses, you attempt to make a crude microscope using them. While holding these lenses a distance \(55 \mathrm{~cm}\) apart, you position your microscope so that its objective lens is distance \(d_{0}\) from a small object. Assume your eye's near point \(N=25 \mathrm{~cm}\). (a) For your microscope to function properly, what should \(d_{\mathrm{o}}\) be? (b) Assuming your eye is relaxed when using it, what magnification \(M\) does your microscope achieve? (c) Since the length of your microscope is not much greater than the focal lengths of its lenses, the approximation \(M \approx N \ell / f_{\mathrm{c}} f_{\mathrm{o}}\) is not valid. If you apply this approximation to your microscope, what \% error do you make in your microscope's true magnification?

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

(a) Show that if two thin lenses of focal lengths \(f_{1}\) and \(f_{2}\) are placed in contact with each other, the focal length of the combination is given by \(f_{\mathrm{T}}=f_{1} f_{2} /\left(f_{1}+f_{2}\right)\). (b) Show that the power \(P\) of the combination of two lenses is the sum of their separate powers, \(P=P_{1}+P_{2}\)

(I) A double concave lens has surface radii of \(33.4 \mathrm{~cm}\) and \(28.8 \mathrm{~cm} .\) What is the focal length if \(n=1.58 ?\)

How large is the image of the Sun on film used in a camera with \((a)\) a 28 -mm-focal-length lens, \((b)\) a 50 -mm-focal-length lens, and (c) a 135 -mm- focal-length lens? (d) If the 50 -mm lens is considered normal for this camera, what relative magnification does cach of the other two lenses provide? The Sun has diameter \(1.4 \times 10^{6} \mathrm{~km},\) and it is \(1.5 \times 10^{8} \mathrm{~km}\) away.
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