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Chapter 39: Problem 11

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. $$\text { Magnifying power }=-\frac{\text { Focal length of objective }}{\text { Focal length of eyepiece }}=-\frac{60 \mathrm{~cm}}{3.0 \mathrm{~cm}}=-20$$ The image is inverted.

Short Answer

Expert verified
The magnifying power is -20, indicating 20x magnification and an inverted image.

Step by step solution

01

Understand the Formula for Magnifying Power

The magnifying power of a telescope for which the lens is adjusted to focus parallel rays is given by the formula \( -\frac{f_o}{f_e} \), where \( f_o \) is the focal length of the objective lens, and \( f_e \) is the focal length of the eyepiece lens.
02

Identify Given Values

From the problem, the focal length of the objective lens \( f_o \) is \( +60 \text{ cm} \), and the focal length of the eyepiece lens \( f_e \) is \( +3.0 \text{ cm} \).
03

Substitute Values into Formula

Substitute the given focal lengths into the formula for magnifying power: \(-\frac{60}{3.0}\).
04

Perform the Division

Calculate the division: \(-\frac{60}{3.0} = -20\).
05

Interpret the Result

The magnifying power of \(-20\) means the telescope magnifies the image 20 times larger, and because it is negative, the image is inverted.

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Key Concepts

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

Focal Length
The focal length is a fundamental concept in understanding how telescopes work. It represents the distance between the center of a lens, or a mirror, and its focus point鈥攚here parallel rays of light converge. In telescopes, both the objective and eyepiece lenses have specific focal lengths that are crucial in determining the device鈥檚 magnifying power.
- **Positive focal length** indicates a converging lens.
- Longer focal lengths in objective lenses general offer higher resolution and magnification.
Understanding focal length helps in comprehending the capacity of a telescope to magnify distant objects, forming part of the key formula for calculating magnifying power.
Objective Lens
The objective lens is the primary lens in a telescope. It has two main functions: collecting light and creating an initial image of the observed object. This lens is typically larger to capture more light, which enhances image brightness and clarity.
In our exercise, the objective lens has a focal length of 60 cm, which affects the entire setup of the telescope. The focal length is instrumental in forming a sharp and clear image, which is then focused to form an inverted and magnified image once it passes through the eyepiece lens.
Eyepiece Lens
The eyepiece lens is the second lens you look through in a telescope. It magnifies the image created by the objective lens and makes the viewable image larger and sharper.
In the problem, we have an eyepiece lens with a focal length of 3.0 cm. A shorter focal length in the eyepiece results in higher magnification since the formula involves dividing the focal length of the objective by that of the eyepiece.鈥攁 shorter focal length increases the result of this division.
The eyepiece not only increases magnification but also contributes crucially to the overall user experience by assembling the viewable image in its final form.
Inverted Image
When using a telescope, you may notice the image depends on the magnifying power's negative sign, indicating the image is inverted. This is a common characteristic due to the way the lenses bend and focus the light. - **Inversion** occurs naturally because of the optical path light follows through both the objective and eyepiece lenses.
- An inverted image is typical in astronomical telescopes.
While this might be unusual for an observer, the priority is usually magnification over orientation, especially when observing far-off celestial bodies. Understanding this inversion helps users anticipate how images will appear, ensuring they achieve the best possible focus and clarity.

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Most popular questions from this chapter

A farsighted woman cannot see objects clearly that are closer to her eye than \(60.0 \mathrm{~cm}\). Determine the focal length and power of the spectacle lenses that will enable her to read a book at a distance of \(25,0 \mathrm{~cm} .\)

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?

What is the magnifying power of a lens of focal length \(+2.0 \mathrm{~cm}\) when it used as a magnifying glass (or simple microscope)? The lens is held close to the eye, and the virtual image forms at the distance of distinct vision, \(25 \mathrm{~cm}\) from the eye.

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 or positive lenses are prescribed). Keep in mind that for virtual images formed by a convex lens \(\left|s_{i}\right|>s_{o}\). We have $$\frac{1}{f}=\frac{1}{25}-\frac{1}{75} \quad \text { or } \quad f=+37.5 \mathrm{~cm}$$ and $$\text { Power }=\frac{1}{0.375 \mathrm{~m}}=2.7 \text { diopters }$$

A nearsighted person named George cannot see distinctly objects beyond \(80 \mathrm{~cm}\) from the eye. What is the power in diopters of the spectacle lenses that will enable him to see distant objects clearly? The image, which must be right-side-up, must be on the same side of the lens as the distant object (hence, the image is virtual and \(\left.s_{i}=-80 \mathrm{~cm}\right)\), and nearer to the lens than the object (hence, diverging or negative lenses are indicated). Keep in mind that for virtual images formed by a concave lens \(s_{0}>\left|s_{i}\right| .\) As the object is at a great distance, \(s_{o}\) is very large and \(1 / s_{o}\) is practically zero. Then $$\frac{1}{s_{e}}+\frac{1}{s_{i}}=\frac{1}{f} \quad \text { or } \quad 0-\frac{1}{80}=\frac{1}{f} \quad \text { or } \quad f=-80 \mathrm{~cm} \text { (diverging) }$$ and \(\quad\) Power in diopters \(=\frac{1}{f \text { in meters }}=\frac{1}{-0.80 \mathrm{~m}}=-1.3\) diopters
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