Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 33: Problem 70

Two refracting telescopes are used to look at craters on the Moon. The objective focal length of both telescopes is \(95.0 \mathrm{~cm}\) and the eyepiece focal length of both telescopes is \(3.80 \mathrm{~cm} .\) The telescopes are identical except for the diameter of the lenses. Telescope A has an objective diameter of \(10.0 \mathrm{~cm}\) while the lenses of telescope \(\mathrm{B}\) are scaled up by a factor of two, so that its objective diameter is \(20.0 \mathrm{~cm}\). a) What are the angular magnifications of telescopes \(A\) and \(B\) ? b) Do the images produced by the telescopes have the same brightness? If not, which is brighter and by how much?

Short Answer

Expert verified
How does the brightness of the images produced by the telescopes compare? Answer: Yes, telescopes A and B have the same angular magnification of 25. However, the images produced by telescope B are 4 times brighter, as telescope B collects 4 times more light than telescope A due to its larger objective lens diameter.

Step by step solution

01

Calculate angular magnification for telescopes A and B

Using the angular magnification formula, we can calculate the angular magnification for each telescope. Since both telescopes have the same focal lengths, we expect them to have the same magnification. \(f_{objective} = 95.0\,\text{cm}\) \(f_{eyepiece} = 3.80\,\text{cm}\) Angular Magnification = \(\frac{f_{objective}}{f_{eyepiece}} = \frac{95}{3.8}\) Calculating the ratio, we get: Angular Magnification = \(25\) So, the angular magnifications of telescopes A and B are \(25\). #b)
02

Calculate the light-gathering power for telescopes A and B

Now, we will calculate the light-gathering power for each telescope, which is proportional to the area of the objective lens. For telescope A: \(d_A = 10.0\,\text{cm}\) Area \(A = \pi(\frac{d_A}{2})^2\) For telescope B: \(d_B = 20.0\,\text{cm}\) Area \(B = \pi(\frac{d_B}{2})^2\)
03

Compare the light-gathering power and brightness

Now, we compare the area of telescopes A and B: \(\frac{\text{Area B}}{\text{Area A}} = \frac{\pi(\frac{d_B}{2})^2}{\pi(\frac{d_A}{2})^2} = \frac{(\frac{d_B}{2})^2}{(\frac{d_A}{2})^2} = (\frac{d_B}{d_A})^2\) Because \(d_B = 2d_A\), substituting into the equation gives: \(\frac{\text{Area B}}{\text{Area A}} = (2)^2 = 4\) Thus, telescope B collects 4 times more light than telescope A, and the images produced by telescope B are 4 times brighter.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

As a high-power laser engineer you need to focus a 1.06-mm diameter laser beam to a 10.0 - \(\mu\) m diameter spot \(20.0 \mathrm{~cm}\) behind the lens. What focal length lens would you use?

A person with a near-point distance of \(24.0 \mathrm{~cm}\) finds that a magnifying glass gives an angular magnification that is 1.25 times larger when the image of the magnifier is at the near point than when the image is at infinity. What is the focal length of the magnifying glass?

The object (upright arrow) in the following system has a height of \(2.5 \mathrm{~cm}\) and is placed \(5.0 \mathrm{~cm}\) away from a converging (convex) lens with a focal length of \(3.0 \mathrm{~cm}\). What is the magnification of the image? Is the image upright or inverted? Confirm your answers by ray tracing.

An object is \(6.0 \mathrm{~cm}\) from a converging thin lens along the axis of the lens. If the lens has a focal length of \(9.0 \mathrm{~cm}\), determine the image magnification.

A telescope is advertised as providing a magnification of magnitude 41 using an eyepiece of focal length \(4.0 \cdot 10^{1} \mathrm{~mm}\). What is the focal length of the objective?
See all solutions

Recommended explanations on Physics Textbooks

Modern Physics

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Linear Momentum

Read Explanation

Mechanics and Materials

Read Explanation

Fluids

Read Explanation

Medical Physics

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.

Sign-up for free