/*! 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 62 An astronomieal telescope has an... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An astronomieal telescope has an angular magnification of magnitude 5 for distant object. The separation between the objective and the eyepiece is \(36 \mathrm{~cm}\) and the final image is formed at \(\infty\), The foeal length \(f_{0}\) of the objective and the focal length \(f_{e}\) of the eyepiece are (a) \(f_{0}=30 \mathrm{~cm}\) and \(f_{e}=6 \mathrm{~cm}\) (b) \(f_{0}=15 \mathrm{~cm}\) and \(f_{c}=12 \mathrm{~cm}\) (c) \(f_{0}=8.5 \mathrm{~cm}\) and \(f_{e}=12.9 \mathrm{~cm}\) (d) \(f_{0}=40 \mathrm{~cm}\) and \(f_{e}=11 \mathrm{~cm}\)

Short Answer

Expert verified
The correct focal lengths are: \( f_0 = 30 \text{ cm} \) and \( f_e = 6 \text{ cm} \), which corresponds to option (a).

Step by step solution

01

Understand the Relationship

We know that for an astronomical telescope, the angular magnification \( M \) is given by \( M = \frac{f_0}{f_e} \), where \( f_0 \) is the focal length of the objective, and \( f_e \) is the focal length of the eyepiece. Given that the magnification is 5, we have \( 5 = \frac{f_0}{f_e} \).
02

Use the Separation Formula

The separation between the objective and the eyepiece \( L \) is the sum of their focal lengths, i.e., \( L = f_0 + f_e \). We are given \( L = 36 \text{ cm} \). Therefore, \( f_0 + f_e = 36 \).
03

Set Up Equations

We have two equations: \( f_0 + f_e = 36 \) and \( \frac{f_0}{f_e} = 5 \). From the second equation, we can express \( f_0 = 5f_e \).
04

Solve the Equations

Substitute \( f_0 = 5f_e \) into the equation \( f_0 + f_e = 36 \):\[ 5f_e + f_e = 36 \]\[ 6f_e = 36 \]\[ f_e = 6 \text{ cm} \].Thus, \( f_0 = 5 \times 6 = 30 \text{ cm} \).
05

Verify the Solution

The computed values are \( f_0 = 30 \text{ cm} \) and \( f_e = 6 \text{ cm} \). Checking with the options given, option (a) matches our solution.

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.

Angular Magnification
Angular magnification is a key concept in understanding how telescopes work. In simple terms, it tells us how much larger or smaller an object appears when viewed through a telescope compared to the naked eye.
This is extremely important for astronomers who wish to see distant celestial objects like stars and planets more clearly.
In the case of astronomical telescopes, the angular magnification \( M \) can be calculated using the formula:
  • \( M = \frac{f_0}{f_e} \)
Where \( f_0 \) is the focal length of the telescope's objective lens, and \( f_e \) is the focal length of the eyepiece lens.
This formula shows us that the magnification is directly proportional to the objective's focal length and inversely proportional to the eyepiece's focal length. So, if you want to increase the magnification of a telescope, you would choose a telescope with a longer objective focal length or a shorter eyepiece focal length. Understanding this relationship helps in designing telescopes for specific observational needs.
Lens Separation
Lens separation in a telescope is the physical distance between the objective lens and the eyepiece lens.
This distance, denoted as \( L \), is crucial for setting up the telescope to form a clear image of distant objects. In an astronomical telescope set to infinity focus, this separation equals the sum of the focal lengths of the two lenses:
  • \( L = f_0 + f_e \)
Knowing the lens separation is essential, especially for amateur astronomers who might be assembling their telescopes.
It ensures that the telescope is properly configured to give a sharp image of the stars or planets they are interested in observing. By understanding the relationship between lens separation and focal lengths, one can readily calculate how to position the lenses or adjust the telescope's tubes.
Focal Length Calculation
The focal length of a lens is a measure of how strongly it converges or diverges light. In telescopes, calculating the focal lengths of the objective and eyepiece lenses is vital for determining the overall performance of the device.
From the exercise, we have two important equations:
  • \( f_0 + f_e = 36 \text{ cm} \)
  • \( \frac{f_0}{f_e} = 5 \)
By solving these equations step-by-step, we determine that:
  • \( f_e = 6 \text{ cm} \)
  • \( f_0 = 30 \text{ cm} \)
Using these calculations, astronomers and hobbyists can adjust their telescopes to achieve desired magnifications and ensure that the lenses are separated correctly.
This helps in capturing clear and precise images of astronomical phenomena.
Telescope Optics
Telescope optics is all about understanding how lenses work together to magnify and focus light from distant objects.
A fundamental part of telescope optics is how the objective lens collects light and how the eyepiece lens magnifies the image.
A well-designed telescope utilizes these optics to produce clear and detailed images.

A typical astronomical telescope has:
  • An objective lens: large and captures a lot of light.
  • An eyepiece lens: smaller, used to view the enlarged image formed by the objective lens.
When setting up telescope optics, it is crucial to ensure that these lenses are aligned perfectly and distanced correctly.
Otherwise, the images might be blurry or not focused properly. Proper understanding and application of telescope optics ensure that amateur astronomers can enjoy seamless stargazing experiences.

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

A small candle, \(2.5 \mathrm{~cm}\) in size is placed at \(27 \mathrm{~cm}\) in front of a concave mirror of radius of curvature \(36 \mathrm{~cm}\). At what distance from the mirror should a sereen be placed in order to obtain a sharp image? Deseribe the nature and size of the image. If the candle is moved closer to the mirror, how would the sereen have to be moved? (a) \(54 \mathrm{~cm}\) (b) \(27 \mathrm{~cm}\) (c) \(28 \mathrm{~cm}\) (d) \(475 \mathrm{~cm}\)

A convex mirror forms an image one-fourth the size of the object. If object is at a distance of \(0.5 \mathrm{~m}\) from the mirror, the focal length of mirror is (a) \(0.17 \mathrm{~m}\) (b) \(-1.5 \mathrm{~m}\) (c) \(0.4 \mathrm{~m}\) (d) \(-0.4 \mathrm{~m}\)

In order of obtain a real image of magnification 2 , using a converging lens of focal length \(20 \mathrm{~cm}\), where should an object be placed (a) \(50 \mathrm{~cm}\) (b) \(30 \mathrm{~cm}\) (c) \(-50 \mathrm{~cm}\) (d) \(-30 \mathrm{~cm}\)

An object is placed \(30 \mathrm{em}\) to the left of a diverging lens whose focal length is of magnitude \(20 \mathrm{~cm}\). Which one of the following correctly states the nature and position of the virtual image formed? \(\begin{array}{ll}\text { Nature of image } & \text { Distance from lens }\end{array}\) (a) inverted enlarged \(60 \mathrm{~cm}\) to the right (b) erect, diminished \(12 \mathrm{~cm}\) to the left (c) irverted, enlarged \(60 \mathrm{~cm}\) to the left \(\begin{array}{ll}\text { (d) erect, diminished } & 12 \mathrm{~cm} \text { to the right }\end{array}\) (c) imverted, enlarged \(12 \mathrm{~cm}\) to the left

A small objeet is placed \(10 \mathrm{~cm}\) in front of a plane mirror. If you stand behind the object, \(30 \mathrm{~cm}\) from the mirror and look at its image, for what distance must you focus your eyes? (a) \(20 \mathrm{~cm}\) (b) \(60 \mathrm{~cm}\) (c) \(80 \mathrm{~cm}\) (d) \(40 \mathrm{~cm}\)

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.