/*! 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 89 Mars subtends an angle of \(8.0 ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 26: Problem 89

Mars subtends an angle of \(8.0 \times 10^{-5}\) rad at the unaided eye. An astronomical telescope has an eyepiece with a focal length of \(0.032 \mathrm{~m}\). When Mars is viewed using this telescope, it subtends an angle of \(2.8 \times 10^{-3}\) rad. Find the focal length of the telescope's objective lens.

Short Answer

Expert verified
The focal length of the objective lens is 1.12 m.

Step by step solution

01

Understand the magnification formula for telescopes

The magnification of a telescope can be calculated using the formula: \( M = \frac{\theta'}{\theta} \), where \( \theta' \) is the angle subtended by the image when viewed through the telescope, and \( \theta \) is the angle subtended by the object to the unaided eye.
02

Substitute the given angles into the magnification formula

Substitute \( \theta' = 2.8 \times 10^{-3} \) rad and \( \theta = 8.0 \times 10^{-5} \) rad into the formula to find the magnification: \[ M = \frac{2.8 \times 10^{-3}}{8.0 \times 10^{-5}}.\]
03

Solve for magnification

Calculate the magnification: \[ M = \frac{2.8 \times 10^{-3}}{8.0 \times 10^{-5}} = 35.\] Thus, the magnification of the telescope is 35.
04

Understand the relationship between magnification and focal lengths

The magnification \( M \) is also related to the focal lengths of the objective lens \( f_o \) and the eyepiece \( f_e \) by the formula: \[ M = \frac{f_o}{f_e}.\]
05

Substitute the known magnification and eyepiece focal length into the formula

The known focal length of the eyepiece is \( f_e = 0.032 \) m and we calculated the magnification is 35. Use these in the formula:\[ 35 = \frac{f_o}{0.032}.\]
06

Solve for the focal length of the objective lens

Rearrange the formula to solve for \( f_o \): \[ f_o = 35 \times 0.032 = 1.12 \text{ m}.\] Thus, the focal length of the objective lens is 1.12 meters.

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 91影视!

Key Concepts

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

Astronomical Telescopes
Astronomical telescopes are important tools for stargazers and astronomers. They allow us to see celestial objects in much greater detail than with the naked eye. The main components of a typical astronomical telescope are the objective lens and the eyepiece.
The objective lens is the large lens at the front of the telescope. Its main role is to gather light from a distant object and focus it into an image. The eyepiece is the smaller lens you look through. It magnifies the image produced by the objective lens.
Together, these two lenses work to enhance our view of the universe. With a strong enough telescope, you can observe planets, stars, and distant galaxies. Some telescopes use mirrors instead of lenses, but the basic function remains the same. The quality of the lenses or mirrors and their sizes significantly affect a telescope's performance. A high-quality and well-sized objective lens or mirror provides clearer and brighter images.
Focal Length
The focal length is a critical property of any lens or mirror in a telescope. It refers to the distance from the lens or mirror to the point where it brings incoming light to focus. For astronomical telescopes, both the objective lens and the eyepiece have their own focal lengths.
A longer focal length in the objective lens means the telescope can gather more light and produce a brighter image. This is why larger telescopes, with longer focal lengths, are capable of capturing faint objects that smaller telescopes cannot. On the other hand, a shorter focal length in the eyepiece means a higher magnification. It is important to balance the different focal lengths to achieve clear and detailed observations.
  • Objective lens focal length is linked to the telescope's light-gathering power.
  • Eyepiece focal length influences the magnification level.
Knowing the focal lengths helps in calculating the telescope's magnification and its effectiveness in viewing distant objects.
Angular Magnification
Angular magnification is a key concept describing how much larger a telescope can make objects appear. Simply put, it tells us how many times the object's angular size increases when viewed through the telescope. If an object subtends an angle of certain radians when viewed with the naked eye, a telescope can magnify this angle, making the object appear larger.
The formula for angular magnification is straightforward: \( M = \frac{f_o}{f_e} \), 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.
This concept is crucial when it comes to choosing the right telescope for specific observations. A higher angular magnification means you can see more details of planets or stars. However, it is not always better to have the highest magnification. Too much magnification can lead to blurry images, so finding the right balance is essential.
  • High magnification allows for more detailed views.
  • Excessive magnification can cause image distortion.
Understanding angular magnification helps in selecting the proper combination of lenses for a desired viewing experience.

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

The contacts wom by a farsighted person allow her to see objects clearly that are as close as \(25.0 \mathrm{~cm}\), even though her uncorrected near point is \(79.0 \mathrm{~cm}\) from her eyes. When she is looking at a poster, the contacts form an image of the poster at a distance of \(217 \mathrm{~cm}\) from her eyes. (a) How far away is the poster actually located? (b) If the poster is \(0.350 \mathrm{~m}\) tall, how tall is the image formed by the contacts?

A farsighted person can read printing as close as \(25.0 \mathrm{~cm}\) when she wears contacts that have a focal length of \(45.4 \mathrm{~cm}\). One day, however, she forgets her contacts and uses a magnifying glass, as in Figure \(26-40 b\). It has a maximum angular magnification of 7.50 for a young person with a normal near point of \(25.0 \mathrm{~cm}\). What is the maximum angular magnification that the magnifying glass can provide for her?

A ray of sunlight is passing from diamond into crown glass; the angle of incidence is \(35.00^{\circ} .\) The indices of refraction for the blue and red components of the ray are: blue \(\left(n_{\text {diamond }}=2.444, n_{\text {crown glass }}=1.531\right),\) and red \(\left(n_{\text {diamond }}=2.410, n_{\text {crown glass }}=1.520\right)\) Determine the angle between the refracted blue and red rays in the crown glass.

The moon's diameter is \(3.48 \times 10^{6} \mathrm{~m},\) and its mean distance from the earth is \(3.85 \times 10^{8} \mathrm{~m}\). The moon is being photographed by a camera whose lens has a focal length of \(50.0 \mathrm{~mm}\). (a) Find the diameter of the moon's image on the slide film. (b) When the slide is projected onto a screen that is \(15.0 \mathrm{~m}\) from the lens of the projector \((f=\) \(110.0 \mathrm{~mm}\) ), what is the diameter of the moon's image on the screen?

Mars subtends an angle of \(8.0 \times 10^{-5} \mathrm{rad}\) at the unaided eye. An astronomical telescope has an eyepiece with a focal length of \(0.032 \mathrm{~m}\). When Mars is viewed using this telescope, it subtends an angle of \(2.8 \times 10^{-3} \mathrm{rad}\). Find the focal length of the telescope's objective lens.
See all solutions

Recommended explanations on Physics Textbooks

Thermodynamics

Read Explanation

Atoms and Radioactivity

Read Explanation

Engineering Physics

Read Explanation

Wave Optics

Read Explanation

Fundamentals of Physics

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation
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.