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The large telescope at Mt. Palomar has a concave objective mirror diameter of \(5.0 \mathrm{~m}\) and radius of curvature \(46 \mathrm{~m}\). What is the magnifying power of the instrument when it is used with an eyepiece of focal length \(1.25 \mathrm{~cm}\) ?

Short Answer

Expert verified
The magnifying power of the telescope is 1840.

Step by step solution

01

Understand the Variables

We have a concave mirror with a diameter of \(5.0 \mathrm{~m}\) and a radius of curvature \(R = 46 \mathrm{~m}\). The eyepiece has a focal length \( f_e = 1.25 \mathrm{~cm} = 0.0125 \mathrm{~m}\). We are asked to find the magnifying power of the telescope.
02

Calculate the Focal Length of the Objective

For mirrors, the focal length \(f_o\) is half of the radius of curvature. Therefore, we have:\[ f_o = \frac{R}{2} = \frac{46 \mathrm{~m}}{2} = 23 \mathrm{~m} \]
03

Use the Telescope Magnifying Power Formula

The magnifying power \( M \) of a telescope is given by the formula:\[ M = \frac{f_o}{f_e} \]where \( f_o \) is the focal length of the objective and \( f_e \) is the focal length of the eyepiece.
04

Calculate the Magnifying Power

Substitute the focal lengths into the magnifying power formula:\[ M = \frac{23 \mathrm{~m}}{0.0125 \mathrm{~m}} \]Perform the division:\[ M = 1840 \]

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Key Concepts

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

Focal Length Calculation
In the realm of telescopes, understanding focal lengths is pivotal to grasping how these instruments magnify distant objects. The focal length is the distance from the mirror or lens to the point where parallel rays of light converge to a single point. This is known as the focal point. Knowing this, the focal length directly influences how magnified and clear the final image will appear.

In many telescopes, including the one at Mt. Palomar, we deal with a concave objective mirror. The special property of mirrors is that the focal length (denoted as \( f_o \)) is half the radius of curvature. The radius of curvature is the distance from the mirror's surface to its center of curvature. You can think of this as the mirror's "big picture curve."

For simple calculations, if the radius of curvature is given, the focal length calculation formula is succinct:
  • \( f_o = \frac{R}{2} \)
For the Mt. Palomar telescope, given a radius of curvature \( R = 46 \mathrm{~m} \), the calculated focal length is \( f_o = 23 \mathrm{~m} \). This straightforward calculation is the stepping stone to understanding telescope optics.
Objective Mirror
The objective mirror is the heart of reflecting telescopes. Large telescopes, like the one at Mt. Palomar, employ large concave mirrors as their primary collecting surface for light. This design is not just a choice of style; it serves pivotal optical objectives.

An objective mirror's primary role is to capture as much light as possible from distant stars or galaxies and focus it to a point. The diameter of the mirror dictates the amount of light it can collect. Larger diameters, such as the Mt. Palomar’s \(5.0 \mathrm{~m}\) mirror, mean better light-gathering power.
  • More light means clearer and brighter images.
  • Larger mirrors allow for better resolution, revealing finer details of celestial objects.
The focused light at the focal point is taken in by the eyepiece, turning a faint image into something visible and analytical for astronomers.
Eyepiece Focal Length
The eyepiece focal length is another critical element in determining a telescope's magnification capabilities. While the objective mirror handles light collection and initial focusing, the eyepiece is responsible for the viewing or magnification of the image.

The focal length of the eyepiece (\( f_e \)) works together with the objective's focal length to define how much the telescope magnifies an image. By changing the eyepiece, one can vary the magnification of the telescope. The relationship between the objective's focal length and the eyepiece's focal length is encapsulated by the magnifying power formula:
  • \( M = \frac{f_o}{f_e} \)
In simpler terms, a smaller eyepiece focal length relative to the objective's focal length means greater magnification. In the hand of astronomers, adaptable eyepieces give flexibility in viewing celestial wonders, whether one's aim is a wide-angle view or an enlarged close-up.
For Mt. Palomar, an eyepiece focal length of \(1.25 \mathrm{~cm} = 0.0125 \mathrm{~m}\) plays a vital role in achieving the telescope's impressive magnifying power of 1840.

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Most popular questions from this chapter

A farsighted person named Amy cannot see clearly objects closer to the eye than \(75 \mathrm{~cm}\). Determine the power of the spectacle lenses which will enable her to read type at a distance of \(25 \mathrm{~cm}\). The image, which must be right-side-up, must be on the same side of the lens as the type (hence, the image is virtual and \(s_{i}=75 \mathrm{~cm}\) ), and farther from the lens than the type (hence, converging on positive lenses are prescribed). Keep in mind that for virtual images formed by a convex lens \(\left|s_{i}\right|>s_{0}\). We have $$ \frac{1}{f}=\frac{1}{25}-\frac{1}{75} \quad \text { or } \quad f=+37.5 \mathrm{~cm} $$ and $$ \text { Power }=\frac{1}{0.375 \mathrm{~m}}=2.7 \text { diopters } $$

Compute the magnifying power of a telescope, having objective and eyepiece lenses of focal lengths \(+60\) and \(+3.0 \mathrm{~cm}\), respectively, when it is focused for parallel rays. The image is inverted.

An engraver who has normal eyesight uses a converging lens of focal length \(8.0 \mathrm{~cm}\), which he holds very close to his eye. At what distance from the work should the lens be placed, and what is the magnification of the lens? Method 1 When a converging lens is used as a magnifying glass, the object is between the lens and the focal point. The virtual erect, and enlarged image forms at the distance of distinct vision, \(25 \mathrm{~cm}\) from the eye. For a virtual image \(s_{i}<0\). Thus, Method 2 By the formula, $$ M_{A}=\frac{d_{n}}{f}+1=\frac{25}{8.0}+1=4.1 $$ Note that in this simple case \(M_{T}=M_{A}\).

An opera glass has an objective lens of focal length \(+3.60 \mathrm{~cm}\) and a negative eyepiece of focal length \(-1.20 \mathrm{~cm} .\) How far apart must the two lenses be for the viewer to see a distant object at \(25.0 \mathrm{~cm}\) from the eye?

A nearsighted man cannot see objects clearly that are beyond 50 \(\mathrm{cm}\) from his eye. Determine the focal length and power of the glasses that will enable him to see distant objects clearly.

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