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The objective lens of an astronomical telescope forms an image of a distant object at the focal point of the eyepiece, which has a focal length of \(5.0 \mathrm{cm} .\) If the two lenses are \(45.0 \mathrm{cm}\) apart, what is the angular magnification?

Short Answer

Expert verified
The angular magnification is 8.0.

Step by step solution

01

Identify Given Values

We're given the focal length of the eyepiece as \(f_e = 5.0\, \text{cm}\) and the distance between the lenses \(L = 45.0\, \text{cm}\).
02

Understand Telescope Configuration

Since it's an astronomical telescope, the objective lens creates an image at the focal point of the eyepiece. Thus, the focal length of the objective lens \(f_o\) can be found by \(L = f_o + f_e\).
03

Calculate the Focal Length of the Objective Lens

Using the equation \(f_o = L - f_e\), we substitute the given values: \(f_o = 45.0\, \text{cm} - 5.0\, \text{cm} = 40.0\, \text{cm}\).
04

Calculate Angular Magnification

The angular magnification \(M\) is given by \(M = \frac{f_o}{f_e}\). Substituting the found focal lengths \(f_o = 40.0\, \text{cm}\) and \(f_e = 5.0\, \text{cm}\), we get \(M = \frac{40.0}{5.0} = 8.0\).

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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 crucial characteristic of telescopes, especially when studying distant objects. It refers to how much larger (or smaller) an object appears through the telescope compared to the naked eye. Imagine looking at a far-away star. Without a telescope, it appears tiny. With angular magnification, a telescope can make it seem much bigger.In the case of the astronomical telescope, the angular magnification is determined by the ratio of the focal lengths of the objective lens and the eyepiece. The formula used is:\[ M = \frac{f_o}{f_e} \]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
A greater angular magnification means that the image appears larger. However, practical limits exist, as over-magnifying can distort images. Understanding how to calculate this helps us make wise choices about eyepieces and objective lenses.
Objective Lens
The objective lens is an essential component of an astronomical telescope. Its primary role is to gather light from a distant object and create a focused image. This lens is usually larger than the eyepiece to collect as much light as possible, which is vital for viewing faint stars and galaxies.The power of an objective lens is defined by its focal length, which is the distance over which the lens converges light to form a sharp image. For an astronomical telescope, the objective lens's focal length is calculated using the distance between the lenses and the eyepiece's focal length.In the given exercise, the objective lens's focal length is determined by the equation:\[ f_o = L - f_e \]Using such formulas ensures the proper functioning of the telescope, as they directly influence the final image quality.
Eyepiece
The eyepiece of an astronomical telescope is a small lens or series of lenses located at the viewing end of the telescope. Its main function is to magnify the image formed by the objective lens so that it can be viewed comfortably. In simple terms, the eyepiece acts as a magnifying glass, allowing you to see the details of celestial objects. The choice of eyepiece influences the angular magnification, with different eyepieces providing different levels of magnification. Moreover, the focal length of the eyepiece is a significant parameter. Changing the eyepiece can either magnify the image even more or decrease its size based on its focal length. Astronomers often have a range of eyepieces to swap, adjusting the magnification as needed for different observations.
Focal Length
Focal length is a fundamental concept in understanding how lenses function in a telescope. It represents the distance from the lens to the point where it focuses a parallel beam of light (like starlight). In telescopes, both the objective lens and the eyepiece have their own focal lengths, which collectively determine the telescope's overall properties. For example, a longer focal length in the objective lens allows for higher resolution and detailed views of distant objects, whereas the eyepiece's focal length is pivotal in setting the magnification level. In the exercise given, you can see how crucial focal length is in calculating the angular magnification. Proper balance of these parameters allows for clear and precise images, unlocking the vast beauty of the universe for exploration.

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Most popular questions from this chapter

A converging lens with focal length \(3.00 \mathrm{cm}\) is placed $4.00 \mathrm{cm}$ to the right of an object. A diverging lens with focal length \(-5.00 \mathrm{cm}\) is placed \(17.0 \mathrm{cm}\) to the right of the converging lens. (a) At what location(s), if any, can you place a screen in order to display an image? (b) Repeat part (a) for the case where the lenses are separated by \(10.0 \mathrm{cm}.\)
Two lenses, separated by a distance of \(21.0 \mathrm{cm},\) are used in combination. The first lens has a focal length of \(+30.0 \mathrm{cm} ;\) the second has a focal length of \(-15.0 \mathrm{cm} .\) An object, $2.0 \mathrm{mm}\( long, is placed \)1.8 \mathrm{cm}$ before the first lens. (a) What are 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) What is the angular size of the Moon as viewed from Earth's surface? See the inside back cover for necessary information. (b) The objective and eyepiece of a refracting telescope have focal lengths \(80 \mathrm{cm}\) and \(2.0 \mathrm{cm}\) respectively. What is the angular size of the Moon as viewed through this telescope?
The eyepiece of a microscope has a focal length of \(1.25 \mathrm{cm}\) and the objective lens focal length is \(1.44 \mathrm{cm} .\) (a) If the tube length is \(18.0 \mathrm{cm},\) what is the angular magnification of the microscope? (b) What objective focal length would be required to double this magnification?
Cameras You would like to project an upright image at a position \(32.0 \mathrm{cm}\) to the right of an object. You have a converging lens with focal length \(3.70 \mathrm{cm}\) located \(6.00 \mathrm{cm}\) to the right of the object. By placing a second lens at \(24.65 \mathrm{cm}\) to the right of the object, you obtain an image in the proper location. (a) What is the focal length of the second lens? (b) Is this lens converging or diverging? (c) What is the total magnification? (d) If the object is \(12.0 \mathrm{cm}\) high, what is the image height?
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