/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 55 The Yerkes refracting telescope ... [FREE SOLUTION] | 91Ó°ÊÓ

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The Yerkes refracting telescope of the University of Chicago has an objective 1.02 \(\mathrm{m}\) in diameter with an \(f\) -number of \(19.0 .\) (This is the largest-diameter refracting telescope in the world.) What is its focal length?

Short Answer

Expert verified
The focal length is 19.398 m.

Step by step solution

01

Identify the Given Values

First, identify the information provided in the problem. We are given the diameter of the lens, which is \(1.02\, \mathrm{m}\), and the \(f\)-number, which is \(19.0\).
02

Understand the Formula for Focal Length

The \(f\)-number (or focal ratio) is defined as the ratio of the focal length \(f\) to the diameter \(D\) of the objective lens: \[f\text{-number} = \frac{f}{D}\]We can rearrange this formula to find the focal length \(f\): \[f = (f\text{-number}) \times D\]
03

Calculate the Focal Length

Use the formula to calculate the focal length:\[f = 19.0 \times 1.02\] Perform the multiplication: \[f = 19.398\, \mathrm{m}\]
04

Verify Units and Context

Check that the units are correct and that the answer makes sense in the context of the problem. The diameter was given in meters, and the focal length is also expressed in meters.

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Key Concepts

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

Refracting Telescope
A refracting telescope uses lenses to form an image. This type of telescope gathers light through a large lens called the objective lens. The Yerkes refracting telescope, mentioned in the exercise, is an example of this technology. It uses a 1.02-meter diameter lens, making it the largest refracting telescope in the world.

The functioning principle relies on two lenses. The objective lens captures the light and focuses it to create an image. An eyepiece lens then magnifies this image for viewer observation.
  • The lens curvature bends light, focusing parallel rays to converge at a focal point.
  • Light gathering and image resolution depend on the size of the objective lens.
  • Larger diameters allow for better image clarity.
Refracting telescopes are perfect for observing planets and large celestial objects like the moon. They offer excellent clarity and are often used in educational settings to demonstrate basic principles of optics.
Optics in Physics Education
Optics, a branch of physics, deals with the study of light behavior and its interactions with matter. Teaching optics enhances students’ understanding of fundamental physics principles. This includes knowledge of how light travels, refracts, and reflects.

Applying concepts of optics helps students understand everyday phenomena and diverse technologies, including glasses, cameras, and telescopes. Refracting telescopes such as the one at the University of Chicago serve as practical examples of optics in action.
  • Optics teach about the wave and particle nature of light.
  • Educational tools often include practical observations to reinforce learning.
  • Experimentation with lenses and light sources reveals insights into focal length and magnification.
Understanding optics is vital for fields like astronomy, medical imaging, and even art. Physics education fosters critical thinking by encouraging students to question and experiment with how light works.
F-number Definition
An F-number, or focal ratio, is a critical concept in optics. It defines the relationship between a lens's focal length and its aperture diameter. This indicates how much light the lens can gather and its depth of field in photography and optics.

Formally, the F-number (f-number) is expressed as: \[f ext{-number} = \frac{f}{D}\] where \(f\) is the focal length and \(D\) is the diameter of the aperture.
  • A lower f-number means a wider aperture, allowing more light.
  • Conversely, a higher f-number increases the depth of field but reduces light intake.
  • An f-number of 19 in the Yerkes telescope means it has a long focal length compared to its aperture diameter.
In telescopes, choosing an appropriate f-number impacts image brightness and sharpness. This is essential when observing distant celestial objects, ensuring the right balance between light capture and detail resolution.

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Most popular questions from this chapter

You have a camera with a 35.0 -mm focal length lens and 36.0 -mm-wide film. You wish to take a picture of a 120 -m-long sailboat but find that the image of the boat fills only \(\frac{1}{4}\) of the width of the film. (a) How far are you from the boat? (b) How much closer must the boat be to you for its image to fill the width of the film?

(a) Where is the near point of an eye for which a contact lens with a power of \(+2.75\) diopters is prescribed? (b) Where is the far point of an eye for which a contact lens with a power of \(-1.30\) diopters is prescribed for distant vision?

A microscope with an objective of focal length 8.00 \(\mathrm{mm}\) and an eyepiece of focal length 7.50 \(\mathrm{cm}\) is used to project an image on a screen 2.00 \(\mathrm{m}\) from the eyepiece. Let the image distance of the objective be 18.0 \(\mathrm{cm}\) . (a) What is the lateral magnification of the image? (b) What is the distance between the objective and the eyepiece?

An object is 16.0 \(\mathrm{cm}\) to the left of a lens. The lens forms an image 36.0 \(\mathrm{cm}\) to the right of the lens. (a) What is the focal length of the lens? Is the lens converging or diverging? (b) If the object is 8.00 \(\mathrm{mm}\) tall, how tall is the image? Is it erect or inverted? (c) Draw a principal-ray diagram.

The left end of a long glass rod 8.00 \(\mathrm{cm}\) in diameter, with an index of refraction 1.60 , is ground and polished to a convex hemispherical surface with a radius of 4.00 \(\mathrm{cm}\) . An object in the form of an arrow 1.50 \(\mathrm{mm}\) tall, at right angles to the axis of the rod, is located on the axis 24.0 \(\mathrm{cm}\) to the left of the vertex of the convex surface. Find the position and height of the image of the arrow formed by paraxial rays incident on the convex surface. Is the image erect or inverted?

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