/*! 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 38 The largest refracting telescope... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The largest refracting telescope in the world is at Yerkes Observatory in Wisconsin. The objective lens is \(1.02 \mathrm{~m}\) in diameter and has a focal length of \(19.4 \mathrm{~m}\). Suppose you want to magnify Jupiter, which is \(138,000 \mathrm{~km}\) in diameter, so that its image subtends an angle of \(\frac{1}{2}^{\circ}\) (about the same as the moon) when it is \(6.28 \times 10^{8} \mathrm{~km}\) from earth. What focal-length eyepiece do you need?

Short Answer

Expert verified
The focal length of the eyepiece needed is approximately \( 0.112 \, \text{m} \) or \( 112 \text{ mm} \).

Step by step solution

01

Understanding Magnification Formula

To determine the focal-length eyepiece needed, we first need to understand the formula for the magnification of a telescope: \[ M = \frac{f_o}{f_e} \]where \( M \) is the magnification, \( f_o \) is the focal length of the objective lens, and \( f_e \) is the focal length of the eyepiece. Our goal is to find \( f_e \).
02

Calculate Angular Size of Jupiter

To find the necessary magnification, we begin by calculating the angular size of Jupiter in the sky without any magnification. The formula to find the angular size \( \theta \) in radians is:\[ \theta = \frac{d}{D} \]where \( d = 138,000,000 \, \text{m} \) (the diameter of Jupiter) and \( D = 6.28 \times 10^8 \, \text{km} = 6.28 \times 10^{11} \, \text{m} \) (the distance from Earth to Jupiter).
03

Calculate Magnification Factor

The desired angular size of Jupiter's image is \( \frac{1}{2}^\circ \). We convert this into radians:\[ \frac{1}{2}^\circ = \frac{\pi}{360} \text{ rad} \approx 0.00873 \text{ rad} \]
04

Find Required Magnification

With the original angular size \( \theta \) found in Step 2, and the desired angular size \( 0.00873 \text{ rad} \), the magnification needed is:\[ M = \frac{0.00873}{\theta} \]
05

Calculate Focal Length of Eyepiece

Using the magnification needed from Step 4, and the formula \[ M = \frac{f_o}{f_e} \] we solve for \( f_e \), given \( f_o = 19.4 \, \text{m} \). Doing so, we find:\[ f_e = \frac{f_o}{M} \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Magnification Formula
Understanding how a telescope magnifies objects starts with the magnification formula. This is essential for calculating the details of what we see through a telescope. The formula used is:
  • \( M = \frac{f_o}{f_e} \)
Here, \( M \) is the magnification, \( f_o \) represents the focal length of the objective lens, and \( f_e \) stands for the focal length of the eyepiece.
This simple equation tells us how much larger an object will appear when viewed through a telescope.
By changing the eyepiece, one can adjust how strongly the telescope magnifies the object in view.
Magnification is crucial because it determines the size and clarity of the image produced.
Angular Size of Celestial Objects
The angular size of an object in the sky tells us how large it appears to be from our point of view on Earth.
To calculate this, we use:
  • \( \theta = \frac{d}{D} \)
In this formula, \( d \) is the actual size of the object (e.g., a planet's diameter), and \( D \) is the distance from Earth to the object.
In the given scenario, Jupiter's diameter and its distance from Earth allow us to find its angular size.
This measurement helps determine the initial size of what appears in the telescope without magnification.
Understanding the angular size is important because it sets the groundwork for how much magnification we need to apply to see a celestial object in detail.
Focal Length Calculations
Focal length calculations involve determining how much we need to focus light rays to form a clear image.
Each lens in a telescope has its focal length, impacting how the telescope operates.
In our exercise, the objective lens has a focal length of \( 19.4 \, \text{m} \).
This length is crucial as it helps define the base magnification of the telescope.
The eyepiece's focal length is what we need to calculate since it adjusts the telescope's total magnification.
By understanding how focal lengths work together, we can better manipulate telescopes to achieve the desired view.
Objective and Eyepiece Lenses
A telescope's ability to magnify and focus light starts with its lenses.
The objective lens is the main lens that gathers light from distant objects and focuses it to form an image.
The Yerkes Observatory's objective lens, for example, is quite large at \( 1.02 \, \text{m} \) in diameter.
This size allows it to collect a significant amount of light, crucial for viewing faint celestial bodies.
The eyepiece lens is what we look through; it takes the light focused by the objective and magnifies the image further.
The selection of the eyepiece, specifically its focal length, determines the telescope's overall magnifying power.
Careful coordination of these lenses allows for a wide range of observations, from scanning the night sky to examining particular celestial phenomena in detail.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

You want to take a full-length photo of your friend who is \(2.00 \mathrm{~m}\) tall, using a \(35 \mathrm{~mm}\) camera having a 50.0 -mm-focal-length lens. The image dimensions of \(35 \mathrm{~mm}\) film are \(24 \mathrm{~mm} \times 36 \mathrm{~mm},\) and you want to make this a vertical photo in which your friend's image completely fills the image area. (a) How far should your friend stand from the lens? (b) How far is the lens from the film?

A camera has a lens with an aperture diameter of \(8.00 \mathrm{~mm}\). It is used to photograph a pet dog. What aperture diameter would correspond to an increase in the intensity of the dog's image on the film by a factor of \(2 ?\)

The crystalline lens of the human eye is double convex and has a typical range of optical power from 115 diopters to 150 diopters. (a) What is the range of focal lengths the eye can achieve? (b) At minimum power, where does it focus the image of a very distant object? (c) At maximum power, where does it focus the image of an object at the near point of \(25 \mathrm{~cm} ?\)

The focal length of an \(f / 4\) camera lens is \(300 \mathrm{~mm}\). (a) What is the aperture diameter of the lens? (b) If the correct exposure of a certain scene is \(\frac{1}{250} \mathrm{~s}\) at \(f / 4,\) what is the correct exposure at \(f / 8 ?\)

A \(135 \mathrm{~mm}\) telephoto lens for a \(35 \mathrm{~mm}\) 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 \(1 / 30 \mathrm{~s},\) what shutter speed is needed at \(f / 2.8 ?\) Calculate \(m_{1}\) and \(M_{2}\) for the two lenses and do not make the approximation that leads to Equation 25.4

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.