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Chapter 27: Problem 67

To construct a telescope, you are given a lens with a focal length of \(32 \mathrm{mm}\) and a lens with a focal length of \(1600 \mathrm{mm}\). (a) On the basis of focal length alone, which lens should be the objective and which the eyepiece? Explain. (b) What magnification would this telescope produce?

Short Answer

Expert verified
Objective: 1600 mm lens, Eyepiece: 32 mm lens. Magnification: 50x.

Step by step solution

01

Identify Lens Roles

In a telescope, the objective lens is the one with the longer focal length, which captures light from distant objects. The eyepiece lens has a shorter focal length and magnifies the image formed by the objective lens. Here, the lens with\[ f_o = 1600 \text{mm} \] will act as the objective lens, and the lens with\[ f_e = 32 \text{mm} \] will be the eyepiece.
02

Calculate Magnification

The magnification \( M \) of the telescope can be calculated using the formula:\[M = \frac{f_o}{f_e}.\]Substituting in the given focal lengths:\[M = \frac{1600}{32} = 50.\]Thus, the magnification of this telescope is 50 times.

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Key Concepts

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

Objective Lens
In telescope construction, the objective lens plays a crucial role by gathering light from distant objects and forming a real image. The power of a telescope to collect light is determined by the size and focal length of its objective lens.

Generally, the objective lens should have a long focal length. This extends the distance over which the light is concentrated, allowing viewers to see faraway objects in greater detail.

In your exercise, the lens with a focal length of 1600 mm is selected as the objective lens. The long focal length ensures clear and sharp images, suitable for astronomical observations or detailed terrestrial views.

Remember, a larger, longer-focal length objective lens can capture more light, improving brightness and clarity, especially in night sky observations.
Eyepiece Lens
The eyepiece lens in a telescope is designed to magnify the image formed by the objective lens. It acts as a magnifying glass, enlarging the image for your eye.

The focal length of the eyepiece lens is typically much shorter than that of the objective lens. A shorter focal length means higher magnification for the telescope.

In the exercise, the lens with a 32 mm focal length serves as the eyepiece. The compact focal length is crucial for enhancing the view of the distant object created by the objective.

When choosing an eyepiece, it is vital to ensure the right balance between image clarity and magnification. An eyepiece with too short of a focal length may provide high magnification but could compromise image quality due to distortion or reduced brightness.
Telescope Magnification
Telescope magnification determines how much larger an object appears compared to viewing with the naked eye.

This is calculated using the formula: \( M = \frac{f_o}{f_e} \), where \( f_o \) is the focal length of the objective lens and \( f_e \) is the focal length of the eyepiece lens.

The exercise provides focal lengths of 1600 mm for the objective and 32 mm for the eyepiece, yielding a magnification of 50 times.
  • This formula emphasizes the relationship between lens focal lengths and resulting magnification power.
  • The longer the objective and the shorter the eyepiece, the greater the magnification.
High magnification can enhance details of celestial bodies, but excessive magnification might reduce image quality due to atmospheric interference or limitations in lens quality.

It's important to strike a balance, choosing appropriate lenses for clear, detailed imagery while minimizing distortion.

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Most popular questions from this chapter

You are comfortably reading a book at a distance of \(24 \mathrm{cm}\). (a) What is the refractive power of your eyes? (b) Does the refractive power of your eyes increase or decrease when you move the book farther away? Explain. (For the purposes of this problem, treat the eye as a single-lens system, with the retina \(2.40 \mathrm{cm}\) from the lens.)

The medium-power objective lens in a laboratory microscope has a focal length \(f_{\text {objective }}=4.00 \mathrm{mm}\). (a) If this lens produces a lateral magnification of \(-40.0,\) what is its "working distance"; that is, what is the distance from the object to the objective lens? (b) What is the focal length of an eyepiece lens that will provide an overall magnification of \(125 ?\)

A grade school student plans to build a 35 -power telescope as a science fair project. She starts with a magnifying glass with a focal length of \(5.0 \mathrm{cm}\) as the eyepiece. What focal length is needed for her objective lens?

BIO Treating Cataracts When the lens in a person's eye becomes clouded by a cataract, the lens can be removed with a process called phacoemulsification and replaced with a man-made intraocular lens. The intraocular lens restores clear vision, but its focal length cannot be changed to allow the user to focus at different distances. In most cases, the intraocular lens is adjusted for viewing of distant objects, and corrective glasses are worn when viewing nearby objects. (a) Should the refractive power of the corrective glasses be positive or negative? (b) Choose the hest explanation from among the following: I. The refractive power should be positive -converging because the intraocular lens will make the person farsighted. II. A negative refractive power is required to bring the focal point of the intraocular lens in from infinity to a finite value.

Two thin lenses have refractive powers of +4.00 diopters and -2.35 diopters. What is the refractive power of the two if they are placed in contact? (Note that these are the same two lenses described in the previous problem.)
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