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\(\bullet\) Galileo's telescopes, I. While Galileo did not invent the telescope, he was the first known person to use it astronomically, beginning around \(1609 .\) Five of his original lenses have survived (although he did work with others). Two of these have focal lengths of 1710 \(\mathrm{mm}\) and 980 \(\mathrm{mm}\) . (a) For greatest magnification, which of these two lenses should be the eye- piece and which the objective? How long would this telescope be between the two lenses? (b) What is the greatest angular magnification that Galileo could have obtained with these lenses? (Note: Galileo actually obtained magnifications up to about \(30 \times\) but by using a diverging lens as the eye- piece.) (c) The Moon subtends an angle of \(\frac{10}{2}\) when viewed with the naked eye. What angle would it subtend when viewed through this telescope (assuming that all of it could be seen)?

Short Answer

Expert verified
Objective: 1710 mm, Eyepiece: 980 mm; Length: 2690 mm; Max magnification: 1.75; Moon angle: 2.86.

Step by step solution

01

Identifying Telescope Components

In a simple refracting telescope, the lens with the greater focal length should be used as the objective lens, whereas the lens with the shorter focal length should be the eyepiece. This setup maximizes magnification.
02

Assigning Focal Lengths

Given focal lengths are 1710 mm and 980 mm. Therefore, the lens with a focal length of 1710 mm should be the objective lens, and the lens with 980 mm focal length should be the eyepiece.
03

Calculating Telescope Length

The length of the telescope is the sum of the focal lengths of the objective lens and the eyepiece lens. Thus, the length of the telescope is 1710 mm + 980 mm = 2690 mm.
04

Calculating Angular Magnification

The angular magnification (M) of a telescope is given by the ratio of the focal length of the objective lens (\( f_o \)) to the focal length of the eyepiece lens (\( f_e \)). Thus, \( M = \frac{f_o}{f_e} = \frac{1710}{980} \approx 1.75.\) This is the greatest magnification using the given lenses.
05

Calculating Subtended Angle of the Moon

The angle subtended by an object through a telescope is the angle subtended by the object with the naked eye divided by the magnification.The Moon's angle with the naked eye is \( \frac{10}{2} = 5.\)Therefore, the angle subtended through the telescope is \( \frac{5}{1.75} \approx 2.86.\)

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Key Concepts

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

Understanding the Refracting Telescope
A refracting telescope is a device used to observe distant objects. It uses lenses to bend and focus light. The word 'refracting' comes from refraction, which is the bending of light as it passes through different materials. There are two main lenses in a refracting telescope:
  • The objective lens, which gathers light from a distant object and focuses it into an image.
  • The eyepiece lens, which magnifies the image formed by the objective lens so that it can be viewed clearly by the human eye.
These telescopes have been used for centuries to explore the heavens, famously by Galileo. They can provide clear images of celestial bodies like the moon and planets, making them important tools in astronomy.
The basic principle behind their operation is straightforward, and it involves optical physics.
Exploring Focal Length
The focal length of a lens is the distance from the center of the lens to its focus point, where light converges to form a clear image. In a refracting telescope, each lens has a specific focal length.
Focal length is crucial for determining both the magnification and the size of the telescope:
  • A longer focal length in the objective lens allows more light to be captured, creating a brighter and more detailed image.
  • A shorter focal length in the eyepiece magnifies this image, enhancing the details further for observation.
This combination helps in optimizing the telescope's power and clarity. For instance, in the exercise about Galileo’s experiments, choosing the right focal lengths impacts both the telescope’s length and the magnification it provides.
Diving into Angular Magnification
Angular magnification is how much larger an object appears when viewed through a telescope compared to the naked eye. It is a critical measurement for telescope performance:
  • The formula used is \( M = \frac{f_o}{f_e} \), where \( f_o \) is the focal length of the objective lens and \( f_e \) is that of the eyepiece lens.
  • A higher magnification indicates that the object appears larger through the telescope.
In Galileo's telescope, the choice of lenses with specific focal lengths allowed his telescope to have a notable angular magnification, even though technological limitations were present at that time. Understanding how to calculate this magnification helps astronomers to choose the right lenses for their specific needs.
The Role of the Objective Lens
The objective lens is perhaps the most pivotal part of the refracting telescope. Its main job is to collect light and focus it to create a clear image.
Here’s how it works:
  • It is the larger of the two lenses, ensuring it can gather more light from the object being observed.
  • With a longer focal length, it brings distant images into sharper relief, which is ideal for astronomical observations.
In historical telescopes like Galileo's, the objective lens was carefully chosen to optimize performance, highlighting its significance in the overall design and efficiency in viewing celestial bodies.
Eyepiece Lens Explained
The eyepiece lens is an essential component that works in concert with the objective lens to allow for effective observation. Its primary duty is magnifying the focused image provided by the objective lens.
  • The eyepiece lens is smaller and positioned close to the viewer’s eye, making it convenient for detailed viewing.
  • A shorter focal length in the eyepiece equates to higher magnification, enabling the viewer to see finer details of the object being observed.
The eyepiece lens was vital in Galileo's telescopes as it allowed him to study celestial phenomena with greater detail and precision. Choosing the right eyepiece is key in achieving the desired level of magnification in any optical telescope.

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Most popular questions from this chapter

. A camera with a 90 -mm-focal-length lens is focused on an object 1.30 m from the lens. To refocus on an object 6.50 \(\mathrm{m}\) from the lens, by how much must the distance between the lens and the film be changed? To refocus on the more distant object, is the lens moved toward or away from the film?

\(\bullet\) During a lunar eclipse, a picture of the moon (which has a diameter of \(3.48 \times 10^{6} \mathrm{m}\) and is \(3.86 \times 10^{8} \mathrm{m}\) from the earth) is taken with a camera whose lens has a focal length of 300 \(\mathrm{mm}\) . (a) What is the diameter of the image on the film? (b) What per- cent is this of the width of a 24 \(\mathrm{mm} \times 36 \mathrm{mm}\) color slide?

A slide projector uses a lens of focal length 115 \(\mathrm{mm}\) to focus a 35 \(\mathrm{mm}\) slide (having dimensions 24 \(\mathrm{mm} \times 36 \mathrm{mm} )\) on a screen. The slide is placed 12.0 \(\mathrm{cm}\) in front of the lens. (a) Where should you place the screen to view the image of this slide? (b) What are the dimensions of the slide's image on the screen?

\(\bullet\) You are examining a flea with a converging lens that has a focal length of 4.00 \(\mathrm{cm}\) . If the image of the flea is 6.50 times the size of the flea, how far is the flea from the lens? Where, relative to the lens, is the image?

. A 135 mm telephoto lens for a 35 mm camera has \(f\) -stops that range from \(f / 2.8\) to \(f / 22\) . (a) What are the smallest and largest aperture diameters for this lens? What is the diameter at \(f / 11 ?\) (b) If a 50 \(\mathrm{mm}\) lens had the same \(f-\) stops as the telephoto lens, what would be the smallest and largest aperture diameters for that lens? (c) At a given shutter speed, what is the ratio of the greatest to the smallest light intensity of the film image? (d) If the shutter speed for correct exposure at \(f / 22\) is 1\(/ 30\) s, what shutter speed is needed at \(f / 2.8 ?\)

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