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Chapter 26: Problem 91

An astronomical telescope has an angular magnification of -184 and uses an objective with a focal length of \(48.0 \mathrm{~cm}\). What is the focal length of the eyepiece?

Short Answer

Expert verified
The focal length of the eyepiece is approximately 0.261 cm.

Step by step solution

01

Understand the Formula for Angular Magnification

The angular magnification \(M\) of a telescope is given by 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 negative sign indicates the image is inverted.
02

Substitute Given Values

We are given that the angular magnification \(M\) is \(-184\) and the focal length of the objective \(f_o\) is \(48.0 \text{ cm} \). Substitute these values into the formula: \( -184 = - \frac{48.0 \text{ cm}}{f_e} \).
03

Solve for the Focal Length of the Eyepiece

To find \(f_e\), rearrange the equation from the previous step to get \(f_e = \frac{48.0 \text{ cm}}{184} \).
04

Calculate the Focal Length of the Eyepiece

Calculate \(f_e\) by performing the division: \(f_e = \frac{48.0}{184} \approx 0.261 \text{ cm} \).

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Key Concepts

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

Angular Magnification
Angular magnification is a measure of how much larger or smaller an object appears when viewed through a telescope compared to the naked eye. It is a critical concept in astronomy and optics as it affects the way we perceive distant objects.
  • The formula for angular magnification is given by: \( M = - \frac{f_o}{f_e} \), where:
    • \( M \) is the angular magnification
    • \( f_o \) is the focal length of the objective lens
    • \( f_e \) is the focal length of the eyepiece lens
  • The negative sign indicates that telescopes typically produce an inverted image.
In our example, the angular magnification was calculated to be -184, which means the image is magnified 184 times and inverted.By understanding this formula, you can determine how powerful a telescope is relative to the size of its lenses.
Objective Lens
The objective lens of a telescope is one of the fundamental components that plays a crucial role in gathering light.It is the main lens through which light enters the telescope and is responsible for forming the primary image.
  • The focal length of the objective lens, denoted as \( f_o \), is the distance from the lens to the point where it converges light to form an image.
  • In an astronomical telescope, longer focal lengths allow you to see further into space and with greater detail.
In the problem, the objective lens has a focal length of 48.0 cm. This value is essential for calculating the angular magnification and understanding the telescope's capability to magnify distant objects.
Eyepiece Lens
The eyepiece lens of a telescope is another essential component that magnifies the image formed by the objective lens.After the light has been gathered and converged by the objective lens, the eyepiece takes over to provide a further enlarged view to the observer.
  • The focal length of the eyepiece, \( f_e \), helps to determine the telescope's total magnification along with the objective lens.
  • A shorter focal length in the eyepiece lens results in greater magnification.
In our calculation, the eyepiece lens had a calculated focal length of approximately 0.261 cm.This short focal length relative to the objective's focal length leads to the high angular magnification of -184.It highlights how the eyepiece contributes to making distant objects appear closer.
Focal Length
Focal length is a core concept in the study of lenses and optics.It refers to the distance between a lens and its focused image.Whether for the objective or eyepiece lens, understanding focal length is crucial for manipulating and optimizing a telescope.
  • A longer focal length in a telescope's lens usually means the ability to see clearer and more detailed images.
  • It is used as a primary factor in determining the angular magnification when combined with the focal length of another lens in the telescope.
In our solution, we utilized the definition of focal length to determine the unknown variable鈥擻( f_e \)鈥攚hich stands for the focal length of the eyepiece.This calculation allowed us to find the precise magnification for the telescope setup, enhancing our understanding of how telescopes function.

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Most popular questions from this chapter

A camper is trying to start a fire by focusing sunlight onto a piece of paper. The diameter of the sun is \(1.39 \times 10^{9} \mathrm{~m},\) and its mean distance from the earth is \(1.50 \times 10^{11} \mathrm{~m} .\) The camper is using a converging lens whose focal length is \(10.0 \mathrm{~cm}\) (a) What is the area of the sun's image on the paper? (b) If \(0.530 \mathrm{~W}\) of sunlight pass through the lens, what is the intensity of the sunlight at the paper?

A converging lens \((f=25.0 \mathrm{~cm})\) is used to project an image of an object onto a screen. The object and the screen are \(125 \mathrm{~cm}\) apart, and between them the lens can be placed at either of two locations. Find the two object distances.

Concept Simulation 26.1 at illustrates the concepts that are pertinent to this problem. A ray of light is traveling in glass and strikes a glass-liquid interface. The angle of incidence is \(58.0^{\circ},\) and the index of refraction of glass is \(n=1.50 .\) (a) What must be the index of refraction of the liquid such that the direction of the light entering the liquid is not changed? (b) What is the largest index of refraction that the liquid can have, such that none of the light is transmitted into the liquid and all of it is reflected back into the glass?

An engraver uses a magnifying glass \((f=9.50 \mathrm{~cm})\) to examine some work, as in Figure \(26-40 b\). The image he sees is located \(25.0 \mathrm{~cm}\) from his eye, which is his near point. (a) What is the distance between the work and the magnifying glass? (b) What is the angular magnification of the magnifying glass?

A person working on the transmission of a car accidentally drops a bolt into a tray of oil. The oil is \(5.00 \mathrm{~cm}\) deep. The bolt appears to be \(3.40\mathrm{~cm}\) beneath the surface of the oil, when viewed from directly above. What is the index of refraction of the oil?
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